PROPOSITION I. To three numbers given, to find a fourth in a Di­rect proportion.

To operate this proportion Multiply thē third term, by the second term, and their product divide by the first term, the Quotient shall be a fourth term required. Examp. 1. Admit the Circumference of a Circle whose Diameter is 14 parts be 44 parts, what is the Circumference of that Circle, whose Diameter is 21 parts? Now according to the Rule if you multiply the third term 21, by the second term 44, it produceth 924; which divided by the first Term 14, the Quotient is 66, and so the Circumference of the Circle, whose Diameter is 21, will be 66 parts, and so for any other in a direct proportion.

PROP. II. To three numbers given, to find a fourth in an Inversed proportion.

To operate this proportion, Multiply the first term, by the second term, and their pro­duct divide by the third term, the Quotient is the fourth term required: Examp. Admit that 100 Pioneers, be able in 12 hours to cast a More of a certain length, breadth, and depth; in what time shall 60 Pioneers do the same? Now if according to the Rule, you Multiply the first term 100, by the second term 12, their pro­duct is 1200; which divided by the third term 60, the Quotient is 20, so I say that in 20 hours, 60 Pioneers shall do the same, and so for any other in an Inversed proportion.

PROP. III. To three numbers given, to find out a fourth in a Duplicate proportion.

The nature of this proposition is to discover the proportion of Lines, to Superficies, and Superfi­cies, to Lines; for like Plains are in a duplicate Ratio; that is as the Quadret of their Homologal sides; therefore to Operate any Example in this proportion, Square the third term, and its square multiply by the second Term, their product di­vide by the square of the first Term, the Quotient is the 4th. term sought; Examp. Admit there be two Geometrical squares; now if the side of the grea­ter [Page 4] square be 50 feet, and require 3000 Tiles to pave it; what number shall the lesser square require, whose side is 30 feet? To operate this according to the Rule, I square the third Term 30, whose square is 900: then I multiply it by the second Term 3000, its product is 2700000, which divided by 2500, the square of the first Term 50, the Quotient is 1080, and so many Tiles will pave the lesser square, whose side is 30 feet.

PROP. IV. To three numbers given, to find a fourth in a Triplicate proportion.

THE nature of this proposition is to disco­ver the proportion of Lines to Solids, and So­lids to Lines; for like Solids, are in a Triplicate Ratio, that is to the Cubes, of their Homolo­gal sides: Therefore to operate any Question in this proportion, Cube the third Term, and his Cube multiply by the second Term, and their product divide by the Cube of the first Term; the Quotient is the fourth Term sought. Examp. Admit an Iron Bullet whose diameter is 4 Inches, weigh 9 pounds; what is the weight of that Bullet whose Diameter is 6 Inches? Now to operate this proportion; first according to the Rule I Cube the third Term 6 whose Cube is 216, then I multiply its Cube by the second Term 9, the product is, 1944, which divided by 64, the Cube of the first Term; the Quotient is 30 24/64 pounds which is equal unto 30l. 6 ounces: which is the weight [Page 5] of the propounded shot; and so for any other.

PROP. V. To two numbers given, to find out a third, fourth, fifth, sixth, &c. Numbers in a continual pro­portion.

To operate this proportion, you must multi­ply the second number by it self, and that pro­duct divide by the first Term, the Quotient is a third proportional: Again you must multiply the third Term by it self, and its Quadret di­vide by the second Term, the Quotient is a fourth proportional, and so after this manner a fifth, sixth; or as many more proportionals as you please may be found: Examp. Let it be re­quired to find six numbers in a continual pro­portion to one another; as 4 to 8. To operate this first according to the Rule, I multiply the second Term 8 by it self the product is 64, which divided by the first Term 4, the Quoti­ent is 16: so is 4, 8, and 16 in a continual pro­portion; And so observing the Rules prescribed, proceed in your operation untill you have found your six numbers in a continual propor­tion; which in this Example will be 4, 8, 16, [...]2, 64, and 128, and so will you have form'd six numbers in a continual proportion.

PROP. VI. Between two numbers given, to find out a mean Arithmetical proportional.

THIS proposition might be performed without the help of the rule of proportion: ne­vertheless because it conduceth to the Resolu­tion of the next ensuing proposition, I insert it in this place; To operate it this is the Rule: add half the difference of the given Terms, to the lesser Term, so that Agragate, is the Arith­metical mean required: Examp. Admit 20 and 50 to be the two numbers propounded: Now to operate this proposition, first according to the Rule, I find that the difference of the two given Terms 20, and 50, is 30, whose half is 15, which being added to the lesser Term 20, it makes 35, so is 35, a mean Arithmetical pro­portion betwixt 20, and 50, given.

PROP. VII. Between two numbers given, to find out a mean Musical Proportional.

BOETIUS hath this Rule for it, where­fore take his own words:In his second Book of his Arithmetick and the 38 Chapter; where he saith that this proportion hath, Magnam vim in Musici modulaminis tempera­mentis, & in Speculatione naturalium questionum: i. e. Great force in Mu­sical composition, (or in the Composure of Musick) and in the discovery of the Secrets of Nature. saith he, ‘Differen­tiam terminorum in mi­norem terminum multi­plica, & post junge ter­minos, & juxta cum qui inde confectus est; com­mitte illum numerum, [Page 7] qui ex differentiis & ter­mino minore productus est, cujus cum latitudi­nem inveneris, addas eam minori termino, & quod inde colligitur me­dium terminum pones.’ That is, Multiply the difference of the Terms, by the lesser term, and add likewise the same Terms together: this done if you divide the product, by the sum of the Terms, and to the Quotient thereof, add the lesser Term; the last Sum is the Musical mean desired: Examp. Admit the two numbers given be 6, and 12. I say that if the difference of the Terms which is 6, were Multiplied by the les­ser Term 6, it would produce 36; then if you add the two terms 6, and 12, together: their sum would be 18, now if you divide 36, by 18, the Quotient is 2; lastly if to the Quotient 2, you add the lesser Term 6, the sum thereof will be 8, which is a Mean Musical proportional required.

PROP. VIII. How to find the Square-Root of any whole num­ber, or Fraction.

Defin. To Extract the Root of any Square number propounded, is to find out another number, which being Multiplied by it self, pro­duceth the Number propounded. Now for the more easie and ready Extraction of the Square-Root of any number given, This Table [Page 8] here under annexed will be usefull; which at first sight giveth all single Square numbers, with their respective Roots.

The Explication of the Table.

In the uppermost rank of this Table, is pla­ced the respective root of every single Square­number, and in the other the single Square­numbers themselves; so that if the Root of 25 were demanded, the Answer would be 5, so the Square root of 49, is 7, of 81 is 9; and so for the Rest, and so contrarily the Square of the Root 5 is 25, of 7 is 49, of 9 is 81, &c.

Example: If the Square root of 20736, were required, first they being wrote down in or­der as you see, draw the Crooked-line,As you use to doe in Division to represent the Quotient. then to prepare this or any o­ther number for Ex­traction, make a point over the place of Unites; and so on every other figure towards the Left-hand; as you see in the Margent. [Page 9] Then find the Root of [...] the first Square 2, which is 1; place it in the Quo­tient, and also under 2; then draw a line, and substract 1 from 2, there remains 1; which place under the line, then to the last remainder 1, bring down the next Square 07; and then there will be this num­ber 107, which number I call a Resolvend: Then double the Root in the Quotient 1, whose double is 2, which 2 place under the place of tens in the Resolvend, un­der 0; so is this 2 called a Divisor; and 10 called a Dividend.

Then demand how often the Divisor 2, can be had in the Dividend 10, it permitteth but of 4, which place in the Quotient, and under 7 the place of Unites in the Resolvend, and there will appear this number 24; Then Mul­tiply this 24, by 4, (the last Square placed in the Quotient) it produceth 96, which place orderly under 24, as you see, and this 96 is called a Ablatitium; (but some calleth it a Gnomon:) then draw a line under it, and sub­stract 96, the Ablatitium, out of the Resolvend 107, there remains 11, which place orderly under the last drawn line, then thereunto bring down the next Square 36, so will there be a new Resolvend 1136; then double the whole Root 14 in the Quotient, whose double is 28; [Page 10] place it under the Resolvend 1136 as was a­fore directed; so shall 28 be a new Divisor, and 113 be a Dividend; then I find the Divisor 28 can be had in the Dividend 113, 4 times, which four place in the Quotient, and under the place of Unites in the Resolvend, so there appeareth this number 284, which number, multiplyed by 4, the last figure in the Quotient, produceth a new Ablatitium 1136; which place orderly under the Resolvend 1136, and then draw a line, then substract the Ablatitium 1136, from the Resolvend 1136; and the remainder is 00, or nothing: and thus the work of Ex­traction being finished, I find the Root of the Square number 20736, to be 144; and so must you have proceeded gradually step by step, if the number propounded, had consisted of some 4, 5, 6, or more Squares; still observing the aforegoing Rules and Directions.

PROP. IX. How to find the Cube-Root of any whole Num­ber, or Fraction.

Defin. To Extract the Cube-Root of any Num­ber propounded, is to find out another Num­ber, which being multiplied by it self, and that product by the number again, shall produce the number propounded; Now for the more easie and ready Extraction of the Cube-root of any number propounded, this Table hereafter annexed will be usefull, which at first sight giveth the Cube-root of any whole number under 1000; which are called single Cube-numbers.

[Page 13]

The Explication of the Table.

In the uppermost rank of the Table is pla­ced the respective Roots of every single Cube, and in the other the respective single Cube­Numbers; for if the Cube-root of 512 were desired, the Answer would be 8, of 64 is 4; and so of the rest: and if the Cube of the Root 7 were desired, it would be found 343; of 9 it would be 729, &c.

Examp. Admit the Cube root of the Num­ber 262144, were required, first they be­ing wrote down in order as you see, draw the Crooked-line.

Then place a point o­verAs you use to do in Division, to represent the Quotient. the place of Unites, and another over the place of Thousands; and so on still in­termitting two places between every adjacent point; and observe that as many points, as in that order are placed over any number pro­pounded, of so many figures doth the Root [Page 14] consist of: so that in this [...] Example, there being two points, therefore the Root consisteth of two places as you see in the Quotient; Now first find the Root of the first Cube 262; which permitteth but of 6, place 6 in the Quotient, and subscribe its Cube 216, under 262, and then draw a line un­der it, and substract 216, out of 262, and the re­mainder is 46, which place in order under the last drawn line as you see. Then to the Remainder 46, bring down the next Cube-number 144, so will there appear 46144, which I call a Resol­vend: then draw a Line under it, and square the Number in the Quotient 6, whose square is 36; Then Triple it and it will be 108, Then subscribe this Triple square 108, under the Re­solvend, so that the place of Unites in the Tri­ple Square 8, may stand under 1 the place of Hundreds in the Resolvend: Then Triple the Root in the Quotient 6, whose Triple is 18, Then subscribe the Triple 18, under the Re­solvend, so that the place of Unites 8 in the Triple, may stand under 4 the place of Tens in the Resolvend, and so draw a Line under neath it, and add the Triple Square 108, and the Triple 18 together in such order as they [Page 15] stand, their Sum is 1098, which may be cal­led a Divisor, and the whole Resolvend 46144, except 4 the place of Unites a Dividend; then draw another line.

Then seek how many times 1098 the Divi­sor, can be had in 4614 the Dividend, it per­mitteth but of 4, which subscribe in the Quo­tient; Now Multiply the Triple square 108, by 4, it produceth 432, which in order subscribe under the Triple square 108: Then square 4, the figure last placed in the Quotient, whose square is 16; and Multiply it by 18 the Triple, it produceth 288, which subscribe under the Triple orderly, then subscribe the Cube of 4 (last placed in the Quotient) which is 64, in Order under the Resolvend. Then draw a [...]ine underneath it, then add the three num­bers, viz. 432, 288, and 64, together in such order as they are placed, their sum is 46144: Then draw another line under the Work, subtracting the said total 46144, from the Resol­ [...]end 46144, there remains 00, or nothing, which remainder subscribe under the last drawn [...]ine, thus the work being finished I find the Cube root of 262144 the number propounded, to be 64: And thus you must have proceeded orderly step by step, if the number propounded [...]ad arisen to some 3, 4, 8, 10, or more places, observing the direction prescribed untill all had [...]bserved compleated.

PROP. I. Any Number given under 10000, or 100000, to find the Logarithm corresponding thereunto.

1. If the number propounded consist of one place whose Logarithm is required to be found, as suppose (5,) look for 5, in the top of the left hand Column under the LetterSignifies the Num­ber, or figure sought. N, and right against 5, and in the next Column under LOG.Signifies the Loga­rithm answering to the number Opposite. you will find this number or rank of figures, 0698970, which is the Logarithm of the number 5 required.

2. If the number consisteth of two places as if it were 57, look 57 under N, and opposite to it and under LOG. you will find this num­ber 1. 755875, which is the Logarithm of 57, the number propounded.

3. If the number propounded consist of three places as 972, look for 972, under N, and opposite to 972, and under [...]o) the Column, you shall find this number 2. 987666, which is the Logarithm of 972, the number which was propounded.

4. But if the number consist of four places as 685, look the three first figures 168, under the Column N, and opposite to that, and un­ [...]er 5 at the top of the page, you will find this number 3. 226599, which is the Logarithm of 1685, the number propounded.

5. But if the number☞ Note this Rule well, for this explains the use of the Table of proportional parts, printed at the end of this Book. given be above 10000, and under 100000, you may find its Logarithm by the Table of parts proportional, printed at the latter end of this Book. Thus if the Lo­garithm of 35786, be sought, first seek the Log. of 3578, which will be 553649, and the com­mon Difference under D is 121; with this dif­ference 121, Enter the Table of parts propor­tional, and finding 121 in the first Column un­der D, you may then lineally under 6, find the number 72, which add to the Log. of 3578, that is 553649, it produceth, 553712, which is the Log. of 35786 the number propounded: now be­cause the number propounded 35786, ariseth to the place of X. M. therefore there must be the figure 4 prefixed before its Logarithm, and then it will be thus 4, 553712, which 4, is cal­led the Index, as shall be hereafter shewed.

Now before we proceed to find numbers cor­responding to Logarithms, it will be necessary to explain the meaning of the first figure to the left hand of any Logarithm placed, Mr. Briggs calleth it a Cha­racteristickDefinition. The Rule to find the Characteristick or Index appertaining to any Lo­garithm. or Index, which doth represent the distance of any the first figure of any whole number from Unity, whose Index is 0, a Cypher; so the Index o [...] 10 is 1, and so to 100 whose Index is 2, and s [...] to 1000 whose Index is 3, and so to 10000 whose Index is 4, and so if you persist furthe [...] the Characteristick is always one less in dignity [Page 21] than the places or figures os the number pro­pounded.

PROP. II.

First as is before shewed if it be a Vulgar Fraction, find the Log. of the Numerator, and the Log: of the Denominator, then substract the Log: of the Numerator, from the Log: of the Denominator, the remainder [...] is the Log: of the Fraction propounded: Now if you would find the Logarithm of 5/7, do as is prescribed whose Log. I find to be 0. 146121, Now to find the Log. of a Mixt Number, reduce it into an Improper Fraction, and then do as before, so the Log of 15 ⅖, Improper 77/5;, is 1, 187, 52, and so do for any other Mixt number.

PROP III. A Logarithm propounded to find the whole, or Mixt number, corresponding thereunto.

For the more speedy finding the number, answering unto the Logarithm propounded, ob­serve that if the Index be 0, then the Number sought may be found between 1 and 10; If 1, [Page 22] between 10 and 100; if 2, between 100, and 1000; if 3 between 1000 and 10000, and so on still observing the Rules of the Characteris­tick, or Index, therefore loo, in the Table un­till you find the Logarithm proposed, and a­gainst it in the Margent according to the afore­going directions under N, you shall find the number belonging thereunto. This Rule holds in force in Mixt Numbers also.

PROP. I. To Multiply one number by another.

Admit 90, be to be multiplied by 42, what is the product? To find which first find the Log. of the Multiplicand 90, whose Log. is 1. 95424: Then find the Log. of [...] the Multiplier 42, whose Log: is 1 62324, then add these two Log: together, viz. the Log: of the Multiplicand, and Mul­tiplier, their sum is 3, 57748, which is the Log: of 3780, the product of 90; and 42, Multipli­ed together.

PROP. II. To Divide one number by another.

Admit the Dividend (or number to be divi­ded) be 648, and the Divisor 72, what is the number that the Quotient shall consist off? To find which, first write down the Logarithm of the Dividend 648, which is 2. 81157 and al­so [Page 24] write down the Logarithm of the Divisor 72, which is 1. [...] 85733. Now substract the Log: of the Divisor, out of the Log: of the Dividend, the remainder is 0. 95424, which is the Loga­rithm of 9, so I conclude that the Divisor 72, is contained in the Dividend 648, 9 times, and so do for any other.

PROP. III. To find the Square-Root of a Number.

Admit it be required to Extract the Square­Root of the Number 144, to perform which first write down the Log: of 144 which is 2. 15836. Then take the half thereof which is 1. 07918 which number 1. 07918, is the Log: of 12, the Root of 144 propounded, and so do for any other.

PROP. IV. To find the Cube Root of any Number.

Admit it be required to Extract the Cube Root of 1728, to perform which, First write down the Log of 1728 which is 3. 23754, then take the third part thereof which is 1. 07918, which is the Log. of 12; which is the Cube­root of the Number propounded 1728, and so for any other. Note on the contrary if you multiply the Log. of any Number propounded by 3, it produceth the Log. of the Cube thereof.

PROP. V. A Summ of Money being forborn for any number of years, to find how much it will amount unto, reckoning Interest on Interest, according to any Rate propounded.

Admit 300 pounds Sterling, be put out for 4 years, for Compound Interest at 6 l. per Cent. what will it amount to when the four years are expired? To find which substract the Log of [...]00 l. the principal, whose Log. is 2. 477121, out of the Log. of 318 l. Principal and Interest for a year whose Log. is 2. 502427, the re­mainder is 0. 025306, which being multiplyed by 4, the number of years of its continuance, produceth 0. 101224, which added to the Log. of the principal 300l. to wit, to 2. 477121, makes [...], 578345, which is the Log. of 378 l. 14 s. 10d. 2q. very near, and so much will 300 l. amount to.

PROP. VI. A Summ of Money being to be paid hereafter, to find what it is worth in [...]eady Money.

Admit 100 pounds Sterling, to be paid at 30 years end; I demand how much it is worth in ready Money? after the rate of Interest of 6 l. per Cent. To find which substract the Logarithm of 100 the principal, whose Log. is 2. 000000 from the Log. of 106 Principal and Interest, whose Log. is 2. 025306, the remainder is 0 025306, which Multiplyed by 30 the num­ber of years to succeed, produceth 0. 759180, which substracted out of 2. 000000, leaveth 1. 240820, which is the Log. of 17 411/1000, which sheweth the said 100 l. is worth but 17 l. 8s. 2d 3q. fere.

PROP. VII. A yearly rent, or Annuity to continue any number of years, to find what it is worth in ready Money, at any Rate of Interest propounded.

What is 100 pound per annum to continue 30 years, worth in ready money at 6 l. per Cent. To find which first substract the Log of 100 l. the principal, which is 2. 000000 from the Log. of 106 l principal and interest for a year, whose Log. is 2. 025306 the remainder is 0. 025306: Then Multiply 0. 025306, by 30 the number of years of its continuance, it produceth the number 0. 759180; Then Divide 100 l. by 6 [Page 27] the rate of interest and the Quotient is 16 6667/10000, &c. which: 16 6667/10000, is the proportional parts of 100 l. the principal, then add the Log. there­of which is 1. 221829 to the former Log. 0. 759180 it produceth 1. 981009, which is the Log. of 95 7215/10000 parts the Arrearages with the said some for that Time, then from those Arrea­rages 95 7215/10000, substract the parts proportional of 100, to wit 16 6667/10000, the remainder is 79 [...]48/10000, which is the bare Arrearages for that proportio­nal part; Then take the Log. of 79 0548/10000, which is 1. 897929, out of the which take the Log. found by Multiplication of years, to wit 0. 759180, there remains 1. 138749, which is the Log. of the value of the Arrearages in ready money, Then to the Log. 1. 138749, add the Log. of 100 l. principal, 2. 000000, it produ­ceth this number 3. 138749; the Log. of 137 6 48/100, reduced is 1376 l. 9. sh. 7d. 80/100 or ⅘ fere: and so much is the said Annuity worth in ready money.

PROP. I. To find the right Sine, or Tangent of any Arch or Angle of a Triangle containing any number of Degrees and Minutes.

IF the Angle orAnd here'tis ne­cessary to understand, that every Circle is supposed to be divided into 360 E­qual parts which are cal­led Degrees, and every of those Degrees into 60 Minutes, and every Mi­nute into 60 Seconds, and every Second into 60 Thirds, &c. so that a Semi-Circle contains 180 Degrees, and a Qua­drant 90 Degrees; Now an Arch or Angle of a Triangle, is the Inter­section of its two sides, and the measure thereof, is an Arch of a Circle, which cutteth each of the two sides equidistant from the Angular point, (which is the Center.) Now the Logarithm Sine, or Tang­of any such Arch of a Triangle, containing a­ny Number of Degrees or Minutes of the Quadrant, may be found in the Tables, printed at the End of this Book, where they are plainly expressed, and are found as directed in the precedent Rules. Arch of the Tri­angle propounded beless than 45 Deg. the Sine, [Page 29] or Tangent belonging thereunto, is found in the Column under the Title SINE, or TANGENT, at the top of the Table; and if there be any Mi­nutes annexēd unto the Degrees, you must find them out in the first Co­lumn under M. signify­ing Minutes, and oppo­site to those Minutes, and under the title aforesaid, you shall have the Loga­rithm of the Sine or Tan­gent, of the Arch or An­gle required.

But if the Arch or Angle of a Triangle exceed 45 Degrees, you must then look for the Sine or Tangent belonging thereunto, in the bottom of the said Table, and if thereunto are Minutes annexed, you must look for them in the first Column to the Right hand under M. and so opposite to those Minutes in the Column above the Title, Sine, or Tang; there have you the Log. of the Sine, or Tangent, of the Arch or Angle, of the Triangle propoun­ded.

Examp. Suppose it were required to find the Log-Sine or Log-Tangent; of an Angle of 25 D. 37 M. whose Log Sine, whereof according to the former directions I find to be 9. 635833. and Tangent thereof to be 9. 680768. and so for any other under 45 degrees.

Again, suppose it were required to find the Log-Sine or Log-Tangent, of an Angle of 64D. 23M the Sine whereof, I find to be this num­ber'9, 955065, and the Tangent thereof, 10. 319231, and so for any other Arch, or Angle of a Triangle, above 45 degrees.

PROP. II. To find the Co-Sine or Co-Tangent of any Arch, or Angle propounded.

Defin. The Co-sine or Co tangent, of an Angle or Arch, is the remaining part of the Angle propounded, to a Quadrent or 90 De­grees; and is by some called the Complement of an Angle, thus the Arch or Angle of 64D. 23M. taken out of 90D. leaves 25D. 27M. for its Complement, on the contrary if 25D. 37M. were taken out of 90 Degrees, there would remain 64D. 23M. for its Complement. So you see that these two Angles, are the Complements of each other, because they two are equal to a Quadrent or 90 Degrees.

Now the Logarithm of the Complement, may be exactly found with ease, for the Sines and Tangents of every degree, and Minute of the Quadrent in one Column is joyned with his Complement in the next Column, so that [Page 31] without substracting the Angle from 90D. you may readily find the Complement thereof ei­ther the Arch in Degrees and Minutes, or the Log. Sine, or Tangent thereof, as you have oc­casion: Thus the Log. of the Sines Comple­ment before mentioned, to wit, 64D. 23M. Comp. is 25D. 37M. is 9. 635833, Tang. is 9. 680768; so 64D. 23M. is the others Compl. whose Sine is 9. 955065, and his Tang. is 10, 319231; so for any other.

PROP. III. To find the Secant of any Arch or Angle propounded.

In this little Book I have not room to set down the Tables of Artificial Secants at large, as I have done with the Sines and Tangents: Nevertheless I will not here omit to shew how they may be easily found out, by the Tables of Sines. The method is thus, substract the Lo­garithm Sine, of the Sines compl of an Angle, from the double Radius of the Tables, and the remainder shall be the Secant required: As if I desire the Secant of 25D. 37M. I find the Loga­rithm-sine of his complement to be 9. 955065, which substracted from the double Radius, that is 20. 000000: there remains 10, 044935 which is the Secant of it, and so the Secant of 64D. 23M. is 9. 955065; which is the Complement of the former, because they both are Equal to 20. 000000, the double Radius; and so may a­ny other be found out.

PROP. I. To erect a perpendicular on any part of a line assigned.

LET the Line be A, B, and on the point D,Fig. 1, 'tis required to raise a perpendicular to A, B, To operate which first open your Compasses to any convenient distance, and placing one foot thereof in D, with the other make the two marks C, and E, equidistant from D; then open the Compasses to some other convenient distance, and set one foot in E, and describe the Arch FF; then likewise in C, describe the Arch GG, then through the Intersections of these two Arches, and to the point D, draw H D, per­pendicular to A B; as was required.

PROP. II. To Erect a Perpendicular, on the End of a Line.

Let the given line be A B, and on the EndFig. 2, thereof at B, 'tis required to raise a Perpen­dicular [Page 34] line: To perform which open your Compasses to the distance B D, then on B as a Center, describe the Arch D, E, F, then from D, to E, place BD; then placing one foot in E, describe the Arch CF, then remove yourFig. 2. Compasses to F, and draw the Arch CE; Lastly through their Intersection draw C B, which is a Perpendicular to AB, on the end B; as required.

PROP. III. From a Point above to let fall a Perpendicular on a Line.

Let the line given be B A, and 'tis requiredFig. 3. from the point above at C; to let fall a Per­pendicular to the said Line: To perform which place one foot of your Compasses in C, and o­pen them beyond the given line A B, and de­scribe the Arch EF; divide EF, into two parts in D; Lastly draw CD, which shall be per­pendicular unto AB, falling from the point above at C, as was so required.

PROP. IV. To draw a right line Parallel to a right line, at a­ny distance assigned.

Let the distance assigned be O E, and theFig. 4. Lime given be A B, and 'tis required to draw C D, Parallel to A B; at the distance O E: To perform which, take in your Com­passes the distance O E, and on A, describe [Page 35] the Arch H, and on B, the Arch K; then draw C D, so as it may justly touch the two Arches, but cut them not, so shall C D; be pa­rallel to A B, at the assigned distance O E, as was required.

PROP. V. To Protract an Angle of any Quantity of Degrees propounded.

Let it be required to Protract, or lay down an Angle, of 40 degrees: To perform which first draw a right line as A B, then open y [...]r Compasses to 60 degrees, in your line of Chords: and with that Distance on A, de­scribeFig. 5. the Arch E F, then take 40 degrees in your Compasses out of your line of Chords, and place it on the Arch, from F, to H; Lastly through the point H, and from A draw A C; so shall the Angle CAB contain 40 degrees as required.

PROP. VI. To measure an Angle already protracted.

Let the Angle given be C A B, and 'tis [...]equired to find the Quantity thereof: To [...]erform which take in your Compasses 60 de [...]rees from your line of Chords; and on A,Fig. [...]. [...]escribe the Arch EF; then take in your Com­ [...]asses the Distance FH, and apply it to your [...]ne of Chords; and you will find the Angle, [...] AB to contain 40 degrees.

PROP. VII. To divide an Angle into two Equal parts.

Let the Angle given be BAC, and 'tis required to divide it into two equal parts: To perform which do thus: first take in your Com­passes any convenient distance, and placing one foot in A, describe the Arch FKHE,Fig. 6. then on H, describe the Arch KK, and on K, the Arch HH; lastly through the Intersections of these two Arches, draw the line AD, to the Angular point A; so shall the Angle BAC, be divided into two equal parts, viz. BA [...], and DAC; as required.

PROP. VIII. To divide a right line into any Number of Equal or Unequal parts; or like to any divided line propounded.

Let the line A B, be given to be divided into 5 equal parts; as the line CD. To per­form which do thus: first on the point C, draw out a line making an Angle with CD at plea­sure: then make CF, equal to AB; and joyn their Extremities FD, then draw Parallel linesFig. 7. to FD, through all the 5 points of CD, (by the 4 prop. aforegoing) which shall divide AB, into 5 equal parts; as required: This way is to be observed, when the line given to be divided, is greater than the divided line propounded.

PROP. IX. How to Protract or lay down any of the Regular Figures, called Polygons.

To perform which divide 360 degrees, (the number of degrees in a Circle) by the number of the Poligon his sides: as if it be a Pentagon by 5, if a Hexagon by 6, &c. the Quotient is the Angle of the Center; its Complement to 180D. (or a Semi circle) is the Angle at the Figure, half whereof is the Angle of the Tri­angle at the Figure: Now I will shew how to delineate any Poligon three ways, viz. 1 by the Angle at the Center, 2. by the Angle at the Figure, 3. by the Angle of the Triangle at the Figure: I have hereunto annexed a Table, which gives at the first sight, (without the trouble of Division) 1. the quantity of the Angle at the Center; 2. the quantity of the Angle at the [Page 38] Figure; and 3 the Quantity of the Angle at the Triangle of the Figure, from a Triangle to a Decigon.

PROP. X. To divide a line according to any assigned pro­portion.

Admit the right line given to be AB, and 'tis required to divide the same into two parts, bearing proportion the one to the other as the lines E, and F doth: To perform which, first draw the line CD, equal to the given line AB: Then draw the line HC, from C, to containFig. 12. an Angle at pleasure. Then from C to G, place the line F, and from G, to H, place the line E: Then draw the line HD. And lastly, draw GK parallel to HD, (by the 4 prop. precedent) so is the line DC, equal to AB, and divided into two parts, bearing such proporti­on to each other, as the two given lines E, and F, as was required.

PROP. XI. To two lines given, to find a third proportional to each of them.

Admit the two given lines be A and B, andFig. 13. 'tis required to find a third proportional to A, as A, to B: First make an Angle at pleasure; as HIK. Then place the line B, from I, unto P; and the line A, from I, unto L; and draw PL. then also place the line A, from I unto M, and draw QM, parallel unto LP, (by 4 prop.) so shall the line IQ, be a third proportional unto the two given lines A, and B, as was required. For as B, is to A, so is A, unto the proportional found IQ.

PROP. XII. To three lines given to find out a fourth proportio­nal unto them.

Admit the three given lines to be A, B, and C; and 'tis required to find a third proportional to them, which shall have such proportion unto A, as B, hath unto C. To perform which, first make an Angle at pleasure as DKG, now see­ing the line C, hath such proportion to B, as the line A, unto the line sought: Therefore placeFig. 14. the line C, from K, unto H and B, from K, to F, and draw FH. Again, place the line A, from K, to I, and draw IE, parallel unto FH, (by 4 prop.) until it cutteth DK, in E; so have you the line KE, a fourth proportional, as was required. For as C, is unto B, so is A, unto the found line KE.

PROP. XIII. To find a mean proportional Line between any two right lines given.

Let the two given lines be A, and B, be­tween which it is required to find a mean pro­portional line. To perform which, first joyn the two lines A, and B together, so as they make the right line CED: Then describe thereon aFig. 15. Semicircle CFD. Then on the point E, erect the perpendicular EF, (by 1 prop.) to cut the limb of the Semi-circle in F, so shall EF, be a mean proportional line, between the two given lines A, and B, as required.

PROP. XIV. To find two mean proportional Lines between any two right Lines given.

Let the two given lines be A, and B; be­tweenFig. 16. which 'tis required to find two mean proportionals. To per­form which, first make anAnd this of all other the Inventions of Plato, Apollonius, Sporus, Archi­tas, Diocles, Nicomedus; & many other famous Geo­metricians and Philoso­phers, I like best for the ready performance of this Conclusion, whose several Methods I could here de­scribe, but for brevity sake do omit them. Angle containing 90 deg. making the sides CD, and CE of a convenient length: then from C, place the line B, unto F, and the line A, from C, unto G; and draw FG, which divide equally in H, and describe the Se­mi-circle F K G. Then take the line B in your Compasses, and place­ing one soot in G, with the other make a mark [Page 43] in the limb of the Semi-circle in K, then draw ST, in such sort that it may justly touch the Semi-circle in K, and may cut through the two sides of the Angle, equidistant from the Cen­ter of the Semi-circle H; so shall SF, and TG, be two mean proportionals, betwixt the two gi­ven lines A, and B, as required.

PROP. XV. To make a Geometrical square equal to divers Geo­metrical squares.

Let there be given the 5 sides of five Geome­trical Squares, viz. A, B, C, D, E; and 'tis requi­red to make one Geometrical Square, equal to the said five Sqares: To perform which first make a Right Angle as ABC, making its contain­edFig. 17. sides of a convenient length. Then from B, place A, to D, and from B, place B, to E, and draw Ed. Then place Ed, from B, to F, and C, from B, to G; and draw GF. Then place GF, from B, to H, and D, from B, to I; and draw [...]H. Lastly from B, unto K, place IH, and from B, unto L, place the line E; and draw LK. So shall LK, be the side of a Square, equal to the five Squares propounded.

PROP. XVI. To make a Circle equal to divers Circles propounded.

Let the two Circles propounded be A,Fig. 18. and B, and 'tis required to make a third Circle, [...]e [...]ual to the said Circles propounded. To per­form [Page 44] which, first take the Diameter, of the les­ser Circle A, and place it as a Tangent, on the Diameter of the greater Circle B, at right An­gles;Fig. 18. as ECD. Then draw the Diagonal ED, which divide equally in F, on which as a Cen­ter describe the Circle K, making E D, the Diameter of which Circle K shall be equal unto the two given Circles A, and B, as requiredSo after the same manner, may divers Cir­cles be added into one by the help of the former proposition well understood.

PROP. I. To find the superficial Content of a Geometrical square.

Let the side of the Square AA be 4 Perch,Fig. 19. what is the Area, or superficial content thereof? To find which multiply its side 4, by its self, it produceth 16, which is the content of that Square AAAA, propounded.

PROP. II. To find the superficial content of a Parallelogram, or long Square.

Multiply the length in parts, by the breadth in parts; the product is the content thereof. So in the Parallelogram, or long Square ABCD, the length of the side AB, or CD is 20 Paces,Fig. 20. and the breadth AC, or BD is 10 paces, and his superficial content is required. I say there­fore if according unto the Rule, you multi­ply the length 20, by the breadth 10, it produ­ceth 200 Paces; which is the content of the Pa­rallelogram or long Square ABCD.

PROP. III. To find the superficial Content of any Right-lined Triangle.

Although right-lined Triangles are of seve­ral kinds, and forms; as first in respect unto their Angles, they are either Right-angled, or [Page 46] Oblique-angled, i. e. Acute-angled, or Obtuse­angled. Secondly in respect of their sides, they are either an Equilateral, Isosceles, or Scaleni­um Triangle: But now seeing they are all measur'd by one and the same manner, I shall therefore add but one Example for all; which take for a general Rule: which is,

Multiply the length of the Base, by theThe Rule. length of the Perpendicular, half their product is the Area or superficial content thereof. So if the content of the Triangle ABC, be required. To find which first from the Angle B, let fall the Perpendicular DB, on the Base AC, (by prop. 3. §. 1.) let therefore the length of theFig. 21. Perpendicular BD be 24, and the Base AC 44 parts. Now if the Base AC 44, were multiply­ed by BD 24, the product is 1056, half where­of is 528, the Content of the Triangle ABC, pro­pounded.

PROP IV. To find the superficial Content of a Rhombus.

First let fall a Perpendicular from one of the Obtuse-angles, unto its opposite side, (by prop. 3. §. 1.) and then Multiply the length of the side thereof, by the length of the Perpendicu­lar, their product is the Content thereof.

So in the Rhombus ABCD, the side AC, or BD is 16 Inches, and the Perpendicular KC isFig. 22. 14 Inches, which multiplyed into the side 16, produceth 224 Inches; which is the Area, or superficial Content, of the Rhombus ABCD, pro­pounded.

PROP. V. To find the superficial content of a Rhomboides.

Frst let fall a Perpendicular, as in the former proposition, then the length thereof multiply by the length of the Perpendicular; the pro­duct is the Area, or superficial content thereof. For in the Rhomboides EDAH, whose length AH, or ED is 32 Feet, and the length of theFig. 23. Perpendicular HK is 16 Feet, which multiply­ed together produceth 512 Feet, which is the Area or superficial content of the Rhomboides AHED, propounded.

PROP. VI. Te find the superficial Content of any Poligon, or many equal sided Superficies.

First from the Center unto the middle of either of the sides of the Poligon, let fall a Per­pendicular, (by 3. prop §. 1.) Then multiply the length of half the Perifery, by the Perpen­dicular, the product shall be the Superficial Con­tent of the Poligon.

Admit the Poligon to be an Hexagon AAAAFig. 24. AA, whose side AA is 22 Feet, and the Per­ [...]end [...]cular BE 19 Feet; now, if 66 half the Pe­rifery, be multiplyed by 19 it produceth 1254 Feet; which is the Content of the Poligon AA, &c. as required.

PROP. VII. To find the superficial Content of a Circle.

Multiply half the Circumference, by one half of the Diameter, their product is the superficial Content thereof.

Admit the Circumference of a Circle ACBD, be 44 Inches, what is the Area or Content thereof. (by the 9. prop. §. 2.) I find the Dia­meterFig. 25. to be 14 Inches, therefore I say if 22, half the Circumference, be multiplied by 7, half the Diameter, it shall produce 154 Inches; which is the superficial Content of the Circle ACDB, as required.

PROP. VIII. By the Diameter of a Circle given, to find the Cir­cumference.

Suppose the Diameter be 14, what is the Circumference? The Analogy or Proportion holds thus, as 7, to 22, so is 14, unto 44, the Circumference required.

PROP. IX. By the Circumference of a Circle given, to find the Diameter.

Suppose the Circumference of a Circle be 44 what is the Diameter? the Analogy or Proportion is, as 22, to 7, so is 44, unto 14, the Diameter re­quired.

Now the proportion of the Diameter, unto the Circumference is as 7, unto-22; or as 113, to 355; or as 1, unto 3, 1415926, &c. so is the Diameter to the Circumference.

PROP. X. By the Content of a Circle given, to find the Cir­cumference.

Suppose the Content of a Circle be 154, what is the Circumference, the Analogy or Proportion?

As 7, unto 4 times 22, which is 88, so is 154 the Content of the given Circle; to the square of the Circumference 1936, whose root being Extracted, as is taught (in prop. 8. §. 1. chap. 1.) gives the Circumference 44, as required.

PROP. XI. By the Content of a Circle given, to find the Dia­meter.

Suppose the Superficial Content of a Circle be 154 parts, what is the Diameter thereof? to find which this is the Analogy or Proportion.

As 22,

To 4 times 7, which is 28,

So is 154, the given Content,

To the Square of the Diameter 196, whose Root being Extracted (by 8 prop chap. 1. §. 1.) [...]iveth the Diameter 14, as required.

PROP. XII. By the Diameter of a Circle given to find the side of a square equal thereto.

To find which this is the Analogy or Propor­tion.

As 1, 000000,

To 0, 886227.

So is the Diameter of the Circle propounded.

To the side of a Square, whose superficial Content, is equal unto the superficial Content, of the Circle propounded.

PROP. XIII. By the Circumference of a Circle given, to find the side of a square equal to it.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 282093.

So is the Circumference of the Circle pro­pounded, to the side of a Square equal to the Circle.

PROP. XIV. By the Content of a Circle given to find the side a square equal to it.

To do which, Extract the Square-Root o [...] the Content propounded, (by prop. 8 chap. [...] §. 1.) so is the Root, the side of a Geometrica [...] Square, equal thereunto.

PROP. XV. By the Diameter of a Circle given, to find the side of an Inscribed square.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 707107,

So is the Diameter of the Circle propounded, To the side of the inscribed Square.

PROP. XVI. By the Circumference of a Circle given, to find the side of an Inscribed Square.

This is the Analogy, or Proportion.

As 1. 000000,

To 0. 225079.

So is the Circumference of the Circle pro­pounded, To the side of the inscribed Square.

PROP. XVII. [...]o find the Superficial Content of an Oval, or El­leipsis.

Let the Oval given be ABCD, and 'tis re­ [...]uired to find the Area or Superficial Content [...]ereof? To do which multiply the length A [...] 40 Inches, by the breadth CD 30 Inches, theFig. 26. [...] is 1200. Which divide by 1. 27324; [...]e Quotient is 942 48/100 parts. Which is the [...]ea or Superficial Content of the Oval ABCD [...]opounded

PROP. XVIII. To find the Superficial Content of any Section, or Portion of a Circle.

Multiply half the Circute of the Section, by the Semidiameter of the whole Circle, and the product thence arising is the Area or superfi­cial Content thereof.

Suppose there be a Circle whose Diameter isFig. 27. 14 parts, and the Circute of the Quadrent ABC is 11 parts, and the Content of the said Qua­drent is desired? To find which multiply 5 ½ or, 5., 5 half the Circute of the Quadrent, by 7 the Semidiameter, the product is 38 5/10, which is the Content of the Quadrent ABC pro­pounded.

PROP. I. To find the solid Content of a Cube.

Multiply the side into its self, and that pro­duct by its side again; their product is the so­lid Content thereof.

Suppose there be a Cube A, whose side isFig. 28. 2 Feet; and his solid Content is required? I say if his side 2, be multiplyed by its self, it pro­duceth 4, which again multiplyed by 2, it pro­duceth 8 Feet, which is the solid Content of the Cube propounded.

PROP. II. To find the solid Content of a Parallelepipedon.

First get the Superficial Content of the End, (by prop. 1, or 2, §. 2.) which multiply into the length, the product is the solid Content.

Suppose there be a Parallelepipedon B, whoseFig. 29. sides of the Base is 40, and 30 Inches, and length 120 Inches, and his Solid Content is demanded? I say if you multiply 30, by 40, the product is 1, 200, which is the superficial Content at the Base. Which multiplyed by the length 120 In­ches produceth 144000 Inches, which is the so­lid Content of the Parallelepipedon B, propounded.

PROP. III. To find the solid Content of a Cylinder.

First get the superficial Content of the Circle at the Base, (by prop. 7. §. 2.) and by it multi­ply its length, their product is the solid Content thereof.

Suppose there be a Cylinder as D, whoseFig. 30. Diameter of the Circle at the Base is 7 parts, and the length of the Cylinder is 14 parts, and 'tis required to find the solid Content thereof? First I find the superficial Content of the Base to be 38. 5, which multiplied into 14 the length, giveth 539 parts, which is the solid Content of the Cylinder propounded.

PROP. IV. To find the solid Content of a Pyramid.

First get the superficial Content of the Base of the Pyramid, (by some of the aforegoing pro­positions in Planometria) and then multiply that into ⅓ of his Altitude, the product is the solid Content thereof.

Suppose there be a Pyramid H, whose side of the Base is 4 ½ parts, or 4 5/10, and his Altitude 12 parts, and his solid Content is required? First I find, (by prop. 1. §. 2.) the superficial ContentFig. 31. of the Base to be 20 25/100 or 20 ¼, which multiply­ed by 4, (which is ⅓ of the Altitude 12) produ­ceth 81 parts, for the solid Content of the Py­ramid propounded.

PROP. V. To find the solid content of a Cone.

First find the superficial Content of the Circle at the Base, (by prop. 7. §. 2.) then multiply it by ⅓ of its Altitude or Heighth, the product is the solid Content thereof.

Suppose there be a Cone as B, whose Dia­meterFig. 32. of the Base is 7, and his Altitude or Heighth is 15 parts, and his solid Content is re­quired? First I find the superficial Content of the Base to be 38½ or 38. 5; which multiply­ed into 5, ⅓ of its Altitude or Heighth) produ­ceth 192. 5, or ½, which is the solid Content of the Cone propounded.

PROP. VI. By the Diameter of a Globe to find his solid Content.

This is the Analogy or Proportion.

As 6 times 7, which is 42.

Is to 22,

So is the Cube of the Diameter of the Sphere, or Globe propounded.

To the solid Content thereof.

Suppose there be a Sphere or Globe, whose Diameter is 12 Inches; what is the solid Con­tent thereof? say, (see the Globe R.)

As 42,

Is to 22,Fig. 33.

So is 1728, the Cube of the Diameter,

To the solid Content 905 6/42 or 1/7 of the Globe, or Sphere propounded: This and all other such☞ Note that every Sphere is equal unto two Cones, whose Height and Diameter of the Base is the same with the Axis of the Sphere. And a Sphere is two thirds of a Cylinder, whose Height and Diameter of the Base is the same with the Axis of the Sphere; according unto the 9th. Manifestation of the first Book of Archimedes of the Sphere and Cylinder like Propositions, are performed by the help of the first Proposition, of the first Chapter of this Book.

PROP. VII. By the Circumference of a Sphere, or Globe, to find his solid Content.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 016887,

So is the Cube of the Circumference of the Globe or Sphere propounded

To the solid Content thereof.

PROP. VIII. By the Axis of a Globe, to make a Cube equal there­unto.

This is the Analogy or Proportion.

As 1. 00000,

To 0. 80604,

So is the Axis of the Sphere propounded,

To the [...]u [...]-Root, which shall be equal to it.

PROP. IX. By the Circumference of a Globe, to make a Cube equal thereunto.

This is the Analogy or Proportion.

As 1. 000000,

To 0 256556.

So is the Circumference of the Globe propoun­ded,

To the Cube-Root, which shall be equal to the Sphere, or Globe, propounded.

PROP. X. By the solid Content of a Sphere or Globe, to make a Cube equal thereunto.

Extract the Cube-root of the solid Content of the Sphere or Globe, (by prop. 9. § 1. chap. 1.) so shall the Root, so found, be the side of a Cube, equal unto the Globe or Sphere propounded.

PROP. XI. A Segment of a Sphere being given to find the solid Content thereof.

To find which first say, As the Altitude of the other Segment, is to the Altitude of the Seg­ment [Page 58] given: so is that Altitude of the other Segment increased by half the Axis, unto a fourth: Then say, As 1, to 1, 0472, so is the product of the Quadrant of half the Chord of the Circumfe­rence of that Segment, multiplyed by that fourth, To the solid Content of the Segment propounded.

PROP. I. Two Angles and the Base of a Rectangled Trian­gle given, to find the other parts.

ADmit the Triangle given be ABC: NowFig. 34. the Angle at B, is an Angle of 90°, or a Right-angle; And the Angle at C is 57° 35', and the Base AB; is 736 parts.

Now first I find the Angle at A, to be 32° 25': it being the Complement of the Angle at C, unto 90°: Secondly, to find the Cathe­tus, or Perpendicular, this is the analogy or proportion. [...]

Add the Log. of the third and second Terms together, and from their Sum, deduct the Log. ofObserve this for a ge­neral Rule in Trigonome­try. the first number, so is the Remainder, the Log. of [Page 63] the fourth Term, or Number sought, as you see in the aforegoing Example.

Thirdly to find the Hypothenuse AC, the Analogy or Proportion hold thus.

As S. V, C 57° 35',

To Log. Base AB 736 [...]arts.Fig. 34

So Radius or S. 90°,

To Log. Hypothenuse AC 871 8/10 parts re­quired: Thus are the three required parts, of the given Triangle ABC found, viz. the Angle A to be 32° 25', the Cathetus BC to be 467 4/10 parts, and the Hypothenuse AC to be 871 8/10 parts, as was so required to be found.

PROP. II. The Hypothenuse, Base, and one of the Angles Of a Rectangled Triangle given, to find the o­ther parts thereof.

In the Triangle ABC, the Hypothenuse AC is 871 8/10 parts, the Base AB is 736 parts, and the Angle at B, is known to be a Right-angle; or 90°: First to find the Angle at the Cathetus C, the analogy or proportion holds thus.

As Log. Hypothen: AC 871 8/10 partsFig. 34.

To Radius or S. 90°.

So Log. Base AB 736 parts,

To the S. V. at Cathetus C 57° 35'.

Secondly, now having found the Angle at the Cathetus C, to be 57° 35'; I say the Angle of the Base A is 32° 25', being the Compl. of the Angle C, unto 90°.

Thirdly to find the Cathetus BC, this is the [...]nalogy, or proportion.

As Radius or S. 90°,

To Log. Hypothen. AC 871 8/10 parts,

So S. V. at Base A 32° 25'.Fig. 34.

To Log. Cathetus BC, 467 4/10 parts required. It may also be found, as in the former Proposi­tion.

PROP. III. In a Rectangled Triangle, the Base, and Cathetus given to find the other parts thereof.

In the Triangle ABC, the Base AB is 736 parts, and the Cathetus BC is 467 4/10 parts, and the Angle B, between them is a right angle o [...] 90°: And here you may make either side of the Triangle, Radius, but I shall make BC the Cathetus Radius, and then to find the Angle at the Cathetus C, this is the Analogy or [...] ­portion.

As Log. Cathet, BC 467 4/10 parts,Fig. 35.

To Radius or S 90°.

So Log. Base AB 736 parts,

To T. V. Cathe C 57° 35', as required.

Secondly, I find the other Angle, at A to be [...] 32° 25', it being the Complement, to C 57 35', unto 90°.

Thirdly, To find out the Hypothenuse AC this is the analogy or proportion.

As S. V. Cathe C. 57° 35',

To Log. Base AB 736 parts,

So Radius or S. 90°,

To Log. Hypothenuse AC 871 8/10 parts, [...] ­quired. But making the Base AB Radius, yo [...] may find the Hypothenuse AC, by this anal [...] or proportion.

[Page] [Page]

As Radius or S. 90°,

To Log. Base AB 736 parts.

So Sc. V. Base A 32° 25',

To Log. Hypothenuse AE 871 8/10 parts required,Fig. 35. and thus you have all the parts of the Trian­gle propounded.

PROP. IV. The Base, and Hypothenuse, with the Angle be­tween them given, to find the other parts of a Rect-angled Triangle.

In the Triangle ABC, the Base AB is 736 parts, and the Hypothenuse AC is 871 8/10 parts, and the Angle A included between them is 32° 25'. First to find the Angles, and first remem­ber that the Angle B is a right Angle; or 90°. Secondly, that the Angle at C, is the Comple­ment to the Angle at A 32° 25' unto 90°:Fig. 35. and therefore is 57° 35': Now these being known, you may find the Cathetus, by this analogy or proportion.

As S. V. Cathe. C. 57° 35',

To Log. Base AB 736 parts.

So S. V. Base A 32° 25',

To Log. Cathe. BC 467 4/10 parts required. Thus I have sufficiently explained all the Cases of Plain Rect-angle Triangles, for to these rules they may be all reduced.

PROP. I. Two Angles, and a side opposite, in an Oblique-Ang­led Triangle given, to find the other parts there­of.

IN the Triangle ABC, the Angleat A is 50°, and at C is 37°, and the side AB is 30 parts, and opposite to the Angle C

First, to find the Angle B, remember that (as 'tis said, in the third Maxim aforegoing) 'tis the Complement, to the Angles A 50°, and C 37°, to 180°, and therefore is the Angle at B 93°.

Secondly, having thus found the Angles, theFig. 36. two unknown sides, may be found by the pro­portion they bear to their opposite Angles, for that proportion holds also in these; thus to find the side BC, this is the analogy or proportion.

As S. V. C 37° 00',

To Log. side AB 30 parts.

So S. V. A. 50° 00',

To Log. side BC 38 19/100 parts required to be found.

But it may be more readily found, and per­formed in such case as this, where you have a Sine, or Tangent, in the first place, by the A­rithmetical Complement thereof, and so save the Substraction.

Now the readiest way to find the Arithmetical☞ The Rule to find the Complement Arithmeti­cal, of any Logarithm Number. Complement is that of Mr. Norwood, in his Doctrine of Triangles; which is thus: begin with the first Figure towards the left hand of any Number and write down the Complement, or the re­mainder thereof, unto 9:9. 962398. And so do with all the0. 037602. rest of the Figures, as you see here done. Saying 9, wants of 9, 0: and again 9, wants 0: 6, wants 3; 2, wants 7: 3, wants 6; 9, wants 0: only when you come to the last Figure to the right hand, take it out of 10, so 8, wants 2; of 10: Thus you may readily find the Co-Ar. of any Sine, al­most as soon as the Sine it self.

But if you want the Complement Arithmetical of any Tangent, you may take the Co-tang. which is exactly the Co-Arith. of the double Radius, so that the Tangent, and Co-tangent, of an Arch makes exactly 20. 000000.

Now if the Radius be in the first place, then there is no need of taking the Co-Arith. of the first Number, only you must cut off, the first I, to the left hand thus X, and you will have the Logarithm of the Number desired.

Thirdly, now to find the side AC, by the opposite Angle B; which is 93° 00': And see­ [...]ng the Angle B, exceeds 90°, you must work [...]y the Complement to 180°) as in the seventh [...]ork in page 61 is taught.

Thus having found all the parts of the Tri­angle [ [...]] Fig. 36. propounded, Viz. The Angle B, to be 93° 00', the side AC to be 49 78/100 parts, and the side BC to be 38 19/100 parts, as was required to be found.

PROP. II. Two sides, and an Angle opposite to one of them in an Oblique-angled Triangle given, to find the other parts thereof.

In the Triangle ABC, the side AB is 30 parts, and the side AC, is 49 78/100 parts, and the opposite Angle C, is 37° 00'.

First, To find the Angle at B, this is the A­nalogyFig. 36. or Proportion.

As Log. cr. AB 30 parts,

To S. V. at C 37° 00'.

So Log. cr. AC 49 78/100 parts,

To Sc. V. B 93° 00', as was required to be found.

Now seeing that the Angle C, is 37° 00', and the Angle B, is 93° 00', which makes 120° 00', therefore must the Angle A be 50° 00'; the [Page 69] Complement to 180°: so having found all the three Angles, you may find the other side CB,Fig. 36. 38 19/100 parts, as afore in the first proposition, by his opposite Angle.

PROP. III. Two Sides of an Oblique-angled Triangle, with the Angle included between them given, to find the other parts thereof.

In the Triangle ABC, the side AC is 49 78/100 parts, the side DB is 30 parts, and the Angle A between them is 50° 00'; and 'tis required to find the other parts of the Triangle propoun­ded. To resolve this Conclusion, let fall a Perpendicular DB, from the Angle B, on the side AC; (by prop. 3. §. 1. chap. 4) and then pro­ceed thus.

First, Seeing the Oblique-angled Triangle,Fig. 37. ABC is divided into two Rectangled Triangles, Viz. ADB, and BDC: Now I will begin with the Triangle ADB, in which is given the An­gle A 50° 00', and the Angle D is a right An­gle, or 90°, and the side AB 30 parts, and the sides AD, and DB, and the Angle at B, are required.

First to find the Angle at B, remember that it is the Complement unto the Angle A 50° 00', unto 90° 00', and therefore must the Angle B be 40° 00'; Now for to find the Cathetus BD, (as in prop. 1. and 2 §. 2. chap. 5.) by the Rule of opposition, the Analogy or Proportion holds thus.

As Radius or S. 90°,

To Log. Hypoth. AB 30 parts.

So S. V. at A 50° 00',

To Log. Cath. BD 22 98/100 parts sought.

And AGAIN, say.

As Radius or S. 90°,

To Log. Hypoth. AB 30 parts.

So S. V. at B 40° 00',

To Log. Base AD 19 28/100 parts sought.

Thus in the Triangle ADB, you have found the Angle B, to be 40° 00', the Cathetus BD, to be 22 98/100 parts; and the Base AD to be 19 28/100 parts, as was so required.

Now for the other Triangle which is BDC, in which there is given the side BD, 22 98/100 parts, and the Angle at D, is a Right-angle, or 90°, and the sides DC, and CB, and the Angles B, and C, are required.

First to find the side DC, substract AD, 19 28/100Fig. 37. parts, out of AC, 49 78/100 parts; there remains the Base DC; 30 50/100 parts: Thus have you the two sides of the Triangle, to wit the Base DC, 30 50/100 parts, and the Cathetus BD, 22 98/100 parts, and the Angle D between them is a Right-angle or 90°. Now you may find the Angle at B, by the Tangent (as in prop. 3. §. 2. chap. 5.) thus.

As Log. Cath. BD, 22 98/100 parts,

To Radius or S. 90°.

So Log. Base CD 30 50/100 parts.

To T. V. B. 53° 00'.

Secondly, For the Angle C, remember 'tis the Complement of the Angle B, 53°, to 90°; and therefore is the Angle C, 37° 00', required.

Thirdly, To find the Hypoth. BC, this is the Analogy or Proportion.

As S. V. B. 53° 00',

To Log. Base DC 30 50/100 parts.

So Radius or S. 90°,

To Log. Hypoth. BC 38 19/100 parts: Thus have you found all the required parts of the Tri­angle ABC propounded, viz. the Angle C to be 37° 00', the Angle B, to be 93° 00', That is equal unto the two Angles B 40° and B. 53° as afore in the former Proposition. and the Side BC, 38 19/10 [...] parts, as required to be found.

Another way to perform the same.

Take the Sum of the[ [...]] two sides, and the diffe­rence of the two sides; and work as followeth.

Now to find the twoFig. 37. Angles B, and C, this is the Manner, and by this Analogy or Proportion, they are found out and known.

As Log. Z. cr^s. AB, and CA, 79 78/100 parts,

To Log. X. cr^s. AB, and CA; 19 78/100 parts,

So T. of ½ VV unknown, 65° 00',

To T. ½X. of VV, 28° 00'.

This difference of Angles 28° 00', add unto 65° 00', (half the difference of the unknown Angles) and it shall produce 93° 00', which is the greater Angle, and substracted from it, leaves 37° 00', which is the lesser Angle C: so have you the required Angles.

PROP. IV. The three sides of an Oblique-angled Triangle gi­ven, to find the Angles.

In the Triangle ABC, the side AC, is 49 78/100 parts, the side AB, is 30 parts, and the side BC, is 38 19/100 parts; and the three Angles of the Tri­angle are required.

The resolution of [...] this Conclusion is thus. Take the Summ and Dif­fer.Fig. 37. of the two sides AB, and BC; And then work as follows: To find a Segment of the Base AC, to wit CE; say:

As Log. Base AC, 49 78/100 parts,

To Z. cr^s. AB, and BC; 68 19/100 parts,

So X. cr^s. AB, and BC; 8 19/100 parts,

To Log of a Segment of the Base AC, to wit C E 11 22/100 parts.

This Segment of the Base CE, 11 22/100 parts, being substracted from the whole Base AC, 49 78/100 parts, the remainder is EA 38 56/100 parts, in the middle of which as at D, the Perpendicu­lar DB, will fall from the Angle B; and so di­videFig. 37. it into two Rectangled Triangles, to wit, ADB, and CDB, whose Base DA is 19 28/100 parts, which taken from AC 49 78/100 parts, leaves the Base of the greater Triangle CD 30 50/100 parts.

Now having the two Bases of these two Tri­angles, and their Hypothenuses; to wit CD 30 50/100 parts, DA 19 28/100 parts, CB 38 19/100 parts, and BA 30 [Page 73] parts; you may find all their Angles, by the Rule of Opposite sides, to their Angles as afore.

PROP. I. Case 1. A Side and an Angle adjacent thereunto being gi­ven, to find the other Side.

In the Triangle ABC, there is given the Side AB 27° 54'; and the Angle A 23° 30', and the Side BC is required, to find which this is the Analogy or Proportion.Fig. 38. [...]

PROP. II. Case 2. A Side and an Angle adjacent thereunto being gi­ven, to find the other Oblique-angle.

In the Triangle ABC, there is given the Side AB 27° 54', and the Angle A 23° 30',Fig. 38. and the Angle at C is required, to find which say by this Analogy or Proportion.

As the Radius or S 90° 00',

To Sc. of cr. AB 27, 54.

So is S. V. at A 23, 30,

To Sc. V. at c 69, 22 required.

PROP. III. Case 3. A Side and an Angle adjacent thereunto being given, to find the Hypothenuse.

In the Triangle ABC, there is given the SideFig. 38. AB 27° 54', and the Angle at A 23° 30', and the Hypothenuse AC, is required; which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. of V. at A 23, 30.

So is Tc cr. AB, 27, 54.

To Tc. Hypothenuse AC, 30, 00 required.

PROP. IV. Case 4. A Side and an Angle opposite thereunto being given, to find the other Oblique-angle.

In the Triangle ABC, there is given the SideFig. 38. BC 11° 30', and the Angle A 23° 30', and the Angle C is required, to find which, say by this Analogy or Proportion.

As Sc. cr. BC, 11° 30',

To Radius or S. 90, 00.

So is Sc. V. at A, 23, 30,

To S. V. at C. 69, 22, as required.

PROP. V. Case 5. A Side and the opposite Angle given, to find the Hypothenuse.

In the Triangle ABC, there is given theFig. 38. side BC 11° 30', and the Angle at A 23° 30', and the Hypothenuse AC, is required, which may be found by this Analogy or Proportion.

As S. V. at A 23° 30',

To Radius or S. 90, 00.

So is Ser. BC 11. 30,

To S. Hypothenuse AC 30, 00. as required.

PROP. VI. Case 6. A side and the opposite Angle given, to find the other side.

In the Triangle ABC, there is given the side BC 11° 30', and the Angle at A 23° 30', and the side AB is required, to find which this is the Analogy or Proportion.

As Radius or S 90° 00',Fig. 38.

To Tc. of V. at A. 23. 30,

So is T. cr. BC 11, 30,

To S. of cr. AB 27. 54 as was required.

PROP. VII. Case 7. The Hypothenuse, and an Oblique Angle given, to find the side adjacent thereunto.

In the Triangle ABC, there is given the [Page 79] Hypothenuse AC, 30° 00', and the Angle A 23° 30', and the side AB, is required, which is found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. V. at A, 23, 30.Fig. 38.

So is T. Hypoth. AC, 30, 00,

To T. cr. AB, 27, 54, as was required.

PROP. VIII. Case 8. The Hypothenuse, and an Oblique-angle given, to find the opposite Side.

In the Triangle ABC, there is given the Hypothenuse AC, 30° 00', and the Angle at A 23° 30', and the Side BC, is required, which is found by this Analogy or Proportion.Fig. 38.

As the Radius or S. 90° 00',

To S. Hypoth. AC, 30, 00.

So is S. V. at A, 23, 30,

To the S. cr. BC, 11, 30. which was required.

PROP. IX. Case 9. The Hypothenuse, and an Oblique-angle given, to find the other Oblique-angle.

In the Triangle ABC, there is given the Hy­pothenuse AC 30° 00', and the Angle A, 23° 30', now the Angle at C, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. Hypoth. AC, 30, 00.

So is T. of V. at A, 23, 30,Fig. 38.

To Tc. of V. at C. 69, 22, as was required.

PROP. X. Case 10. The sides given, to find the Hypothenuse.

In the Triangle ABC, there is given the side AB 27° 54', and the side BC 11° 30', and the Hypothenuse AC is required, to find which say by this Analogy or Proportion.

As the Radius or S. 90° 00',Fig. 38.

To Sc. cr. BC. 11, 30.

So is Sc. cr. AB 27, 54,

To Sc. Hypothenuse AC 30, 00. required.

PROP. XI. Case 11. The sides given, to find an Angle.

In the Triangle ABC, there is given, the side AB 27° 54', and the side BC 11° 30', and the Angle at A, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',Fig. 38:

To S. cr. AB. 27, 54.

So is Tc. cr. BC. 11, 30,

To Tc. of V. at A. 23. 30. as required.

PROP. XII. Case 12. The Hypothenuse, and a side given, to find the o­ther side.

In the Triangle ABC, there is given, the Hy­pothenuseFig. 38. AC 30° 00', and the side AB 27° 54' and the side BC is required, which may be [Page 81] found by this Analogy or Proportion.

As Sc. cr. AB. 27° 54',

To Radius or S. 90 00.

So is Sc. Hypothenuse AC. 30° 00',

To Sc. cr. BC. 11° 30' as required.

PROP. XIII. Case 13. The Hypothenuse, and a Side given, to find the contained Angle.

In the Triangle ABC, there is given the Hy­pothenuseFig. 38. AC 30° 00', and the side AB 27° 54', and the Angle at A is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To T. cr. AB. 27° 54'

So is Tc. Hypoth. AC 30° 00',

To Sc. of V. at A, 23° 30', as required.

PROP. XIV. Case 14. The Hypothenuse, and a Side given, to find the oppo­site Angle.

In the Triangle ABC, there is given the Hy­pothenuse AC 30° 00', and the side AB 27° 54', [...]ow the Angle C, is required, which may beFig. 38. [...]ound by this Analogy or Proportion.

As the S. Hypoth. C, 30° 00',

To Radius or S. 90° 00'.

So is S. of cr. AB, 27° 54',

To S of V. at C. 69 22, as required.

PROP. XV. Case 15. The Oblique Angles given, to find either Side.

In the Triangle ABC, there is given the An­gle A 23° 30', and the Angle at C 69° 22', and the side BC, is required, which may be found by this Analogy or Proportion.

As the S. of V. at C, 69° 22',Fig. 38.

To the Radius or S. 90° 00'.

So is the Sc. of V. at A, 23° 30',

To the Sc. of cr. BC, 11° 30', as required.

PROP. XVI. Case 16. The Oblique-angles given, to find the Hypothenuse.

In the Triangle ABC, there is given the An­gle A 23° 30', the Angle C, 69° 22', and the Hypothenuse AC, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',Fig. 38.

To Tc. of V. at C. 69°, 22',

So is Tc. of V. at A, 23 30,

To Sc. Hypoth. AC, 30 00, as required.

PROP. I. Case 1. Two Sides, and an Angle opposite to one of them given, to find the other opposite Angle.

IN the Triangle ADE, there is given the SideFig. 39. AE, 70° 00', the Side ED, 38° 30', and the Angle A, 30° 28', now the Angle at D, is re­quired, to find which this is the Analogy or Proportion.

As S. cr. DE, 38° 30',

To S. V. at A, 30 28.

So is S. cr. AE, 70 00,

To S. V. at D, 130 03,Note that in thes [...] Operations, for the more facility of the learner, I omit Seconds, which doth belong unto the Angles, &c. required.

PROP. II. Case 2. Two Angles and a Side opposite to one of them gi­ven, to find the Side opposite to the other.

In the Triangle ADE, there is given the An­gleFig. 39. at D, 130° 03', the Angle E, 31° 34', and the Side AE, 70° 00', now the Side AD, is re­quired, which may be found by this Analogy or Proportion.

As S. V. at D, 130° 03',

To S. cr. AE, 70 00.

So is S. V. at E, 31 34,

To S. cr. AD. 40 00, required.

PROP. III. Case 3. Two Sides and an Angle included between them being known, to find the other Angles.

In the Triangle ADE, there is given the SideFig. 39. AE, 70° 00', the Side AD, 40° 00', and the An­gle A 30° 28', Now the Angles D, and E, are re­quired, which is thus found: take the Sum and Difference of the two Sides, and work as follow­eth, saying. [...]

As S. ½ Z. cr^s. AE and AD, 55° co',

To S. ½ X. cr^s. AE and AD, 15 00.

So is Tc. ½ V. at A, 15 14,☞ And here ob­serve that if the Sum of the two contained Sides exceed a Semicircle, then substract each side severally from 180°, and proceed with those Complements, as with the sides given, the Operation produceth the Complements of the Angles sought, unto a Semicircle or 180 Degrees.

To T. ½ X. VV. D and E. 49 1430".

AGAIN.

As Sc. ½ Z. cr^s. AE and AD, 55° 00',

To Sc. ½ X. cr^s. AE and AD, 15° 00'.

So is Tc. ½ V. at A, 15° 14',

To T. ½ Z. VV. D and E, 80 48 30".

This difference of the Angles unknown D and E, 49° 14' 30", being added unto the half Sum of the Angles 80° 48' 30", (unknown) produceth the Greater Angle D 130° 03', and substracted from it, leaves the Lesser Angle E, to wit 31° 34'.

PROP. IV. Case 4. Two Angles, and their Interjacent side being known, to find the other sides.

In the Triangle ADE, there is given the Angl [...] Fig. 39. at A 30° 28', and the Angle at D 130° 03', and their Interjacent-side AD 40° 00', and the Sides DE and EA, are required: Which is thus found.

Take the Sum and Dif­fference of the two An­gles,☞ And here observe that if the Sum of the two given Angles excede a Semicircle or 180°, substract them from a Se­micircle, and proceed with the Residues, the Operation will produce each side's Complement to a Semicircle, or 180 Degrees. and work as fol­loweth, saying.

As S. ½ Z. of VV. A and D, 80° 15' 30", Fig. 39. [...]

To S. ½ X. of VV. A and D, 49 47 30.

So is T. ½ cr. AD, 20 00 00,

To T. ½ X. cr^s. DE and EA, 15 45 00.

AGAIN Say.

As Sc. ½ Z. of VV. A and D, 80° 15' 30",

To Sc. ½ X. of VV. A and D, 49 47 30.

So is T. ½ cr. AD, 20 00 00,

To T. ½ cr^s. Z. DE and AE. 54 15 00.

Add the half Difference of the Sides DE and AE, 15° 45', unto half the Sum of the Sides DE and AE, 54° 15'. It produceth the greater Side, the Side AE 70° 00', but if deducted from it, leaves the lesser Side ED, which is 38° 30', as was required.

PROP. V. Case 5. Two Sides and an Angle opposite to one of them gi­ven, to find the third side.

In the Triangle ADE, there is given the SideFig. 39. AE 70° 00', the Side DE 38° 30', and the Angle A 30° 28', the Side AD is required.

First by Case 1. Prop. 1. I find the Angle at D to be 130° 03', and then proceed thus

First take the Sum and Difference of the two Angles; then also find the Difference of the two Sides given, and then work as followeth. [...] [...]

Now say,

As S. ½ X. VV. D and A, 49° 47' 30",

To S. ½ Z. VV. D and [...], 80 15 30.

So is T. ½ X. cr^s. AE and ED, 15 45 00,

To T. ½ cr. AD. 20° 00' 00": which doubled giveth the Side AD, 40 00 00, as was required.

PROP. VI. Case 6. Two Angles and a Side opposite to one of them gi­ven, to find the third Angle.

In the Triangle ADE, there is given the An­gleFig. 39. A 30° 28', the Angle D 130° 03', and his opposite Side AE 70° 00', and 'tis required to find the Angle at E.

First by Prop. 2. Case 2. I find the Side DE, opposed to the Angle A; to be 38° 30', then proceed thus.

Fi [...]st find the Sum and Difference of the Sides. Then find the Difference of the Angles. [...] [...]

Now say,

As S. ½ X. cr^s. DE and AE 15° 45',

To S. ½ Z. cr^s. EA and DE 54 15.

So is T. ½ X. VV. D and A 49 47 30",

To Tc. ½ V. at E 15° 47' 00". which doubled giveth the Angle at E 31 34, as required.

PROP. VII. Case 7. Two Sides and an Angle opposite to one of them gi­ven, to find the Included Angle.

In the Triangle ADE, there is given the SideFig. 39. AE 70° 00', the Side ED 38° 30', and the An­gle opposite thereunto at A 30° 28', and the Angle E is required.

First by Prop. 1. Case 1. I find the Angle D, opposite to AE, to be 130° 03', then proceed thus.

First find the Difference of the Angles, then find the Sum and Difference of the Sides. [...] [...]

Now say,

As S. ½ X. cr^s. AE and ED 15° 45',

To S ½ Z. cr^s. AE and ED 54 15.

So is T. ½ X. of VV. D and A 49 47 30",Fig. 39.

To Tc. ½ V. at E 15° 47'. Which doubled is the Angle at E 31° 34', as was required.

PROP. VIII. Case 8. Two Angles and a Side opposite to one of them be­ing known, to find the Interjacent Side.

In the Triangle ADE, there is given the An­gle E 31° 34', the Angle D 130° 03', and his opposite Side AE 70° 00', Now the Side ED is required.

First by Prop. 2. Case 2. I find AD opposed to E, to be 40° 00', and then work thus.

Take the Sum and Difference of the Angles, then also find the Difference of the two Sides:Fig. 39. [...] [...]

Now say,

As S. ½ X. VV D and E 49° 14' 30",

To S. ½ Z VV D and E 80 48 30.

So is T ½ X cr^s. AD and AE 15 00 00",

To T. ½ cr^s. ED, 19° 15' 00", which being doubled is the Side ED 38° 30', as required.

PROP. IX. Case 9. Two Sides and their Included Angle being known, to find the third Side.

In the Triangle APZ, there is given the SideFig. 40. ZP 38° 30', the Side PA 70°, and the Angle P, let be 31° 34, and the Side AZ is required.

The Resolution of this Case depends on the Catholike proposition of the Lord of Marchiston, by supposing the Oblique-Triangle to be divided (by a supposed Perpendicular falling either within or without the Triangle) into two Rect­angulars.

Now in the Triangle AZP, let fall the Per­perpendicular ZR; so is the Triangle AZP divi­ded into two Rectangulars ARZ and ZRP. Now the Side AZ may be found at two Opera­tions thus: say,

As the Radius or S. of 90° 00'

To Sc. of the included V, P. 31 34.

So is T. of the lesser Side PZ. 38 30,

To T. of a fourth Arch. 34 08.

If the contained Angle be less than 90°, take this fourth Arch from the greater Side; but if it be greater than 90°, from its Complement unto 180°, the Remainder is the Residual Arch: Now again say,

As Sc. of the fourth Arch. 34° 08'

To Sc. Residual Arch. 35 52

So Sc. of the lesser Side PZ. 38 30

To Sc. AZ the Side required. 40 00

Note that if the Angles at the Base be both of one kind, that then the Perpendicular falls within the Triangle: if of diverse kinds, without the Triangle. But note that many times the Perpendicu­lar will fall without the Triangle, as it dothFig. 41. now within; in such case the Sides of the Tri­angle must be continued, so will there be two Rectangulars, the one included within the o­ther: as in the Triangle HIK, the Perpendicular let fall is KM, falling on the Side HE, and so the two Rectangulars found thereby will be IM K, and KMH, and so by the directions in the former proposition find out the Side IK, if re­quired to be found.

PROP. X. Case 10. Two Angles and their Interjacent Side known, to find the third Angle.

In the Triangle AZP, there is given the Side ZP 38° 30', the Angle P 31° 34', and the An­gle Z 130° 03', and the Angle at A is requi­red.

First the Oblique-Triangle AZP, being redu­ced into two Rectangulars ARZ, and ZRP, byFig. 40. Case 9 aforegoing, I find the Angle RZP, to be 64° 19', (in the Triangle ZRP.) which ta­ken out of Angle AZP 130° 03', leaves the Angle AZR 65° 44': Now the Angle A is found by this Analogy or Proportion.

As S. V. PZR, 64° 19',

To S. V. AZR 65 44,

So is Sc. V. at P 31 34,

To Sc. V. at A 30 28: which was required to be found out and known.

PROP. XI. Case 11. Three Sides given, to find an Angle.

In the Triangle APZ, the Side AZ is 40° 00', the Side ZP is 38° 30', the Side AP is 70° 00', and the Angle Z is required. To find which do thus.

Add the three Sides: together, and from halfFig. 40. their Sum, deduct the Side opposite, to the re­quired Angle: and then proceed as you see in the Operation following. [...] Fig. 40.

½Sum is 65° 07' 30", the Sc. ½ V. at Z which doubled is 130° 03' 12"; the Angle at Z required.

PROP. XII. Case 12. Three Angles given, to find a Side.

In the Triangle AZP, the Angle A is 30° 28'Fig. 40. 11", the Angle Z 130° 03' 12", the Angle P is 31° 34' 26", and the Side AZ, opposite to P, is required.

This Case is likewise performed as the former Case or Proposition, the Angles being conver­ted into Sides, and the Sides into Angles, by taking the Complement of the greatest Angle unto 180°: see the work. [...] Fig. 40. which being doubled, gives the Side AZ 40° 00 required to be found out and known

☞ But if the greater Side AP were required the Operation would produce the Complem [...] thereof unto a Semicircle or 180°; therfo [...] [Page 95] substract it from 180°, it leaves the remaining required Side sought.

Thus I have laid down all the Cases of Tri­angles, both Right-lined and Spherical; either Right, or Oblique-angled; I might hereunto have annexed many Varieties unto each Case, and some fundamental Axioms, which somewhat more would have Illustrated and Demonstrated those Cases, and Proportions; but because of the smallness of this Treatise, which is intended more for Practice than Theory, I have for brevi­ty sake omitted them, and refer you for those things to larger Authors, who have largely discoursed thereon to good purpose.

PROP. I. The Distance of the Sun from the next Equinoctial point (either Aries or Libra) being known, to find his Declination.

THE Analogy or Proportion.

As Radius or S. 90°,

To S. of the Sun's distance from the next Equi­noctial point,

So it S. of the Sun's greatest Declination,

To the S. of the Sun's present Declination sought.

PROP. II. The Sun's place given, to find his Right-Ascen­sion.

This is the Analogy or Proportion.

As Radius or S. 90°,

To T. of the Sun's Longitude from the next Equi­noctial point,

So is the Sc. of his greatest Declination,

To T. of his Right-Ascension from the next Equi­noctial point.

PROP. III. To find the Sun's place or longitude from Aries, his Declination being given.

This is the Analogy or Proportion.

As S. of the Suns greatest Declination,

To Radius or S. 90° 00',

So is S. of his present Declination,

To S. of the Suns Place or Longitude from AriesIf the Sun's Declina­tion be North, and in­creasing, this Proportion finds the Sun's distance from ♈; but if decreasing from ♎ in the northern Sines. But if the Sun's Declinati­on be South, and increasing from ♎; if decreasing from ♈, among the Southern Sines.

PROP. IV. By knowing the Suns Declination, to find his Right Ascension.

This is the Analogy or Proportion.

As Radius or S. 90°,

To Tc. of the Suns greatest Declination,

So is T. of the Declination given,

To S. of the Suns right Ascension requiredFrom the next Equi­noctial point either ♈, or ♎..

PROP. V. By knowing the Latitude of a Place, and the Suns Declination, to find the Ascensional Difference.

This is the Analogy or Proportion.

As Radius or S. 90°,

To Tc. of the Latitude given,

So is T. of the Suns Declination given,

To the S. of the Ascensional difference required.

☞ Note that if you reduce the degrees, &c. of the Ascensional difference into hours, it will shew you how much the Sun riseth, or setteth before, or after six a Clock, in that Latitude.

PROP. VI. To find the Suns Oblique Ascension or Descension.

First find the Ascensional Difference by the 5th Proposition, and his Right-ascension by the fourth: Now if the Suns Declination be Northerly, de­duct the Ascentional Difference out of his Right Ascension, from the beginning of ♈, (for the six Northern Signs ♈ ♉ ♊ ♋ ♌ ♍) it leaves the Oblique Ascension; and added unto the Right­ascension, giveth the Oblique-descension.

But if the Suns Declination be Southerly, the Ascentional Difference, added to the Right-ascensi­on, (for the six Southern Signs ♎ ♏ ♐ ♑ ♒ ♓) giveth the Right-ascension, and substracted there from leaves the Oblique-descension.

[Page 104] [Page] [Page]

PROP. VII. By knowing the Suns Declination, and the Latitude of a Place, to find the Suns Amplitude.

This is the Analogy or Proportion.

As Sc. of the Latitude,

To the Radius or S. 90°.

So is the S. of the Suns Declination,

To the S. of the Amplitude from the East or West Points of the Horizon.

PROP. VIII. By knowing the Suns Declination and Amplitude, from the North part of the Horizon, to find the Latitude.

This is the Analogy or Proportion.

As Sc. of the Amplitude from the North,

To Radius or S. 90° 00'

So is S. of his Declination given,

To Sc. of the required Latitude.

PROP. IX. By knowing the Latitude of a place, and the Sun's Declination, to find at what time the Sun will be on the true East or West Points.

The Analogy or Proportion is.

As T. of the given Latitude,

To T. of the Sun's Declination propounded,

So is Radius or S. 90° 00',

To, Sc. of the Hour from Noon.

PROP. X. By knowing the Sun's Declination, and Latitude of a place, to find his Altitude at six a Clock.

This is the Analogy or Proportion.

As Radius or S. 90° 00',

To S. of the Sun's Declination,

So is S. of the Latitude of the place,

To S. of the Sun's Altitude at six a Clock.

PROP. XI. By knowing the Latitude of a place, and the Sun's Declination, to find the Azimuth at six.

This is the Analogy or Proportion.

As Radius or S. 90° 00',

To the T. of the Sun's Declination,

So is Sc. of the Latitude of the place,

To the T. of the Azimuth sought.

PROP. XII. By knowing the Latitude of a place, and the Sun's Declination, to find the Sun's Altitude when he i [...] on the true East or West points.

This is the Analogy or Proportion.

As S. of the Latitude,

To the Radius or S. 90° 00',

So is the S. of the Declination,

To the S. of the Sun's Altitude being due Ea [...] or West.

PROP. XIII. To find the Sun's Altitude at any time of the day.

The Analogy or Proportion is.

As Radius or S. 90° 00',

To Tc. of the Poles height,

So is S. of the Sun's Distance,

From the Hour of Six,

To the T. of an Arch: which being substract­ed from the Sun's Distance from the Pole; say,

As Sc. of the Arch found,

To Sc. of the remaining Arch of the Sun's Di­stance from the Pole,

So is S. of the Poles height,

To the S. of the Sun's Altitude at the Hour required.

PROP. XIV. By knowing the Latitude of a Place, with the Sun's Declination, and Altitude, to find the Hour of the Day.

To solve this Conclusion do thus: get the Sum of the Complements of the Latitude, Declina­tion and Altitude givenAs in Case 11 of Oblique Spherical Tri­angles., Then find the Difference betwixt their half Sum, and the Complement of the Altitude; then say,

As Radius or S. 90° 00',

To Sc. of the Sun's Altitude,

So is Sc. of the Latitude of the Place,

To a fourth Sine: then again say,

As the fourth S.

To the S. of ½ Z. of the Lat. Declin. and Alt.

So is the S. of X. of the Altitude to the ½ Z,

To a fifth S. unto which Sine, if you add the Radius or 90° 00', half that Sum shall be the Sine of an Arch, whose double Complement is the Hour from the Meridian.

PROP. XV. To find the Time of the Sun's Rising or Setting, and consequently the Length of the Day or Night.

To resolve this Conclusion, first by prop. the 5. find the Ascensional Difference, which redu­ced into Hours, and Minutes of Time, by allow­ing for every 15 Deg. one Hour, and for every Deg. less than 15°, 4', of Time, and for every 15 Min. one Minute of Time.

Secondly, If the Sun's Declination be Norther­ly, the Ascentional Difference added unto 6 Hours, gives the Time of Sun-setting, and sub­stracted therefrom, leaves the Time of Sun­Rising: On the contrary, if the Sun's Declinati­on be Southerly, the Ascentional Difference added unto 6 Hours, gives the Time of Sun-Rising, and deducted therefrom, the Time of Sun­setting.

Thirdly, If you double the Time of Sun­Rising, it gives you the length of the Night; and the Time of Sun-setting, the length of the Day.

PROP. XVI. The Sun's Declination, Altitude and Azimuth known, to find the Hour of the Day.

The Analogy or Proportion is.

As the Sc. of the Sun's Declination,

To the S. of the Azimuth,

So is the Sc. of the Altitude,

To the S. of the Hour from Noon: which con­verted into Time, will shew the Hour of the Day.

PROP. XVII. By knowing the Sun's Declination, Altitude, and Hour from Noon, to find the Azimuth.

The Analogy or Proportion is.

As Sc. of the Sun's Altitude,

To S. of the Hour from Noon,

So is Sc. of the Sun's Declination,

To the S. of the Azimuth, required.

PROP. XVIII. By knowing the Latitude of a place, the Altitude of the Sun, and the Hour from Noon, to find the Angle of the Sun's Position.

This is the Analogy or Proportion.

As Sc. of the Sun's Altitude,

To S. of the Hour from Noon,

So is Sc. of the Latitude,

To S. of the V. of the Sun's Position, at the time of the Question.

PROP. XIX. By knowing the Sun's Altitude, Declination, and Azimuth; to find the Latitude.

The Analogy or Proportion is.

As S. of the Sun's Azimuth,

To S. of his Distance from the North-pole,

So is S. of V of the Sun's Position,

To Sc. of the Latitude required.

PROP. XX. To find the length of the Crepusculum, or Twilight

The Crepusculum or Twilight, is nothing else but the Refraction of the Sun's Beams in the Density of the Air. Which the Learned Pet. Nonnius found the length of the Crepusculum (by his many strict obser­vations Watched the Time after Sun-setting when the Twilight in the West was shut in, so that no more Twilight than in any other part of the Skie near the Horizon appea­red there: then by one of the known fixed Stars, having found the true Hour of the Night, he found the length of the Twilight, to be as in the rule is mentioned.) to continue from the time of the S [...] passing below the Hori­zon of a place, untill the Sun had run below the said Horizon 18° 00', and then followed the shut­ting in of the Twilight, and untill the Sun was [Page 111] departed so low the Twi­light continued. — To find which observe this Ana­logy or Proportion.

As Radius or S. 90°,

To Sc. of the Sun's Declination,

So is Sc. of the Poles-height,

To a fourth Sine: which keep.

Then out of the Sun's Distance from the South-Pole, subduct the Complement of the Pole; and of that remains and the degrees 62, being added to it, their Sum and Difference found, say again.

As the fourth Sine found,

To S. ½ Z of the remainder and 62° 00',

So is S. ½ X. of the remainder and 62° 00',

To a Number, which being multiplyed by the Radius is equal unto the Quadrat of the Sine of the ½ Angle of the Sun's Distance at the Ending of the Twilight, from Noon next ensuing.

Then from the Sun of the whole Angle con­verted into Hours, substract the Hour of the Sun's setting Or ½ Diurnal Arch., it gives you the length of the Crepusculum, or Twilight.

But the Sun being in the Winter Tropick, makes the Twilight longest of a­nyTo find the length of the least Crepusculum or Twilight. Twilight, the whole Winter half year: Now in a certain Parallel, be­twixt that Tropick, and the Equinoctial is the shortest Crepusculum: the Declination of which Parallel, is thus found.

As the Tc. of the Pole,

To the S of the Pole,

So is the T. of 99° 00',

To S. of the Declination of the Parallel, in which the Sun maketh the shortest Crepusculum of the Year.

But before the Crepusculum come to be short­est, there is another Parallel, in which the Crepusculum is equal to that of the Equinoctial: the Declination of which is found thus.

As the Radius or S. 90° 00',

To S. of the Poles Elevation or Altitude,

So i [...] S. of 18° 00',

To S. of the Declination of the Parallel, in which the Sun maketh the Crepusculum equal to that in the Equinoctial.

PROP. XXI. To find the Quantity of the Angles, which the Cir­cles of the 12 Houses make with the Meri­dian.

This is the Analogy or Proportion.

As the Radius or S. 90°,

To T. of 60°: for the 11th, 9th, 5th and 3d House, Or to the T. 30° for the 12th, 8th, 6th, and 2d House,

So is the Sc. of the Pole,

To the Tc. of any House with the Meridian.

PROP. XXII. To find the Right Ascension of the Point in the Equinoctial: and also the Point in the Ecliptick; called Medium Coeli or Cor Coeli.

First, To find the Right Ascension of the Point of the Equinoctial; called Medium Coeli, vel Cor Coeli, find out the Sun's Right Ascension, by prop. 2. Then reduce the whole Time from Noon last past into degrees, which add unto the right Ascension of the Sun, so shall their A­gragat, be the right Ascension of the point, which in the Equinoctial, is called Medium Coeli, vel Cor Caeli, required to be found.

Secondly, By the 2 propositions aforegoing, you may find the right Ascension of the point in the Ecliptick Culminant in the Meridian, cal­led Medium Coeli vel Cor Coeli, which is the Cuspis of the tenth House: and his Declination by prop. the first.

PROP. XXIII. To find the Angle of the Ecliptick with the Meri­dian.

The Analogy or Proportion is.

As the Radius or S. 90°,

To S. of the Sun's Greatest Declination,

So is Sc. of the Sun's right Ascension, from the next Equinoctial point,

To Sc. of the V. of the Ecliptick, with the Me­ridian.

PROP. XXIV. To find the Angle of the Ecliptick with the Ho­rizon.

The Analogy or Proportion is.

As Radius or S. 90°,

To Sc. of the Altitude of Cor Coeli,

So is S. of the V. Ecliptick with the Meridian,

To Sc. of the V. of the Ecliptick and Horizon sought.

PROP. XXV. To find the Amplitude Ortive of the Ascendent, or Horoscopus.

This is the Analogy or Proportion.

As Radius or S. 90°,

To S. of Altitude of Med. Coeli,

So is T. of V. Ecliptick with the Meridian,

To Tc. of the Amplitude Ortive of the Ascen­dent, or the distance of the Azimuth from the Meridian.

PROP. XXVI. To find the Ascendent degree of the Ecliptick, or the Cuspis of the first House.

The Amplitude Ortive of the Ascendent, is equal to the Distance of the Azimuth of 90°, from the Meridian, wherefore the Cuspis of the [Page 115] first House, or Ascendent Degree of the Ecliptick, may be found thus.

As Radius or S. 90°,

To Sc. of the V. Ecliptick with the Meridian,

So is Tc. of the Altitude of Med. Coeli,

To T. of the Distance of Med. Coeli, from the Ascendent Degrees.

PROP. XXVII. To find the Distance of the Cuspis of any House, from Med. Coeli.

This is the Analogy or Proportion.

As Sc. of the remaining part of V. of the E­cliptick with the Meridian, (found by prop. 28.)

To Sc of the former part of the V,

So is T. of the Altitude of Med. Coeli,

To T. of the Distance of the Cuspis of that House sought, from Med. Coeli.

PROP. XXVIII. To find the parts of the Angle of the Ecliptick with the Meridian, cut with an Arch perpendicular to the Circle of any of the Houses.

The Analogy or Proportion is:

As Radius or S. 90°,

To Sc. Altitude of Med. Coeli,

So is T. of the Circle of any House with the Meridian,

To Tc. of that part of that Angle which is next the Meridian:

Then substract that part found, out of the whole Angle, for the remaining or latter part

PROP. XXIX. To find the Pole's Altitude, above any of the Circles of the Houses.

The Analogy or Proportion is.

As the Radius or S 90°,

To S. of V. of the Circle of the House with the Meridian: (found by the 21 prop.)

So is the S of the Poles Elevation, above the Horizon of the Place,

To S. of the Altitude of the Pole, above the Circle of Position.

PROP. XXX. By knowing the Latitude and Longitude of any fixed Star, to find his Right Ascension and De­clination.

The Analogy or Proportion is.

1. As Radius or S. 90°,

To S. of the Stars Longitude from the next Equinoctial point,

So is Tc. of the Stars Latitude,

To T. of a fourth Arch.

Which compared with the Arch of Distance betwixt the Poles of the World and the Ecliptick 23°, 30'; And if the Latitude and Longitude of the Star be both of one Dignity, i e. when the Star hath North Latitude in the six Northern Sines, ♈, ♉, ♊, ♋, ♌, ♍, or South Latitude in the six Southern Sines, ♎, ♏, ♐, ♑, ♒, ♓: Then shall the difference between this found Arch, and the

[Page] [Page 117] Distance of the Poles be your fifth Arch: But if the Latitude and Longitude of the Star be of contrary qualities, i. e the one Northern, and the other Southern, then add this fourth Arch to the Distance of the Poles 23° 30', and the Sum thereof shall be your fifth Arch; with which,

AGAIN, say.

2. As S. of the fourth Arch,

To S. of the fifth Arch,

So is T. of the Stars Longitude,

To T. of the Stars Right-ascension from the next Equinoctial point.

3. As Sc of the fourth Arch,

To Sc. of the fifth Arch,

So is S of the Stars Latitude,

To S. of the Stars Declination.

I might also shew how by having the La­titude and Longitude of any two fixed Stars, to find their Distance: but because 'tis the very same with finding the Distance of any two Places on Earth, I refer you to the Directions of Prop 1, 2 and 3. of Chap. 7, ensuing, where you will see the plain Demonstration thereof.

PROP. XXXI. By knowing the Pole's Altitude, to find when any fixed Star shall be due East or West.

This is the Analogy or Proportion.

As Radius or S. 90°,

To T. of the Stars Declination,

So is Tc. of the Pole,

To Sc. of the Stars Horary Distance from the Meridian.

PROP. XXXII. By knowing the Poles Altitude, to find the Eleva­tion of any fixed Star above the Horizon, being due East or West.

This is the Analogy or Proportion.

As S. of the Poles Altitude,

To Radius or S. 90°,

So is S. of the Stars Declination,

To S. of the Stars Elevation, above the Hori­zon, at due East or West.

PROP. XXXIII. To find out the Horizontal Parallax of the Moon.

The Analogy or Proportion.

As the Moons Distance from the Center of the Earth,

To the Earth's Semidiameter,

So is Radius or S. 90°,

So S. of the Moon's Horizontal Parallax in that Distance.

PROP. XXXIV. The Horizontal Parallax of the Moon being known, to find her Parallax in any apparent Latitude.

This is the Analogy or Proportion.

As Radius or S. 90°,

To S. of the Moon's Altitude,

So is S. of the Moon's Horizontal Parallax,

To S. of the Parallax in that Altitude.

PROP. XXXV. By knowing the Moon's Place in the Ecliptick, (having little or no Latitude) and her Paral­lax of Altitude, to find the Parallaxes of her Longitude and Latitude.

First, If the Moon be in the 90° of the Eclip­tick, she hath then no Parallax of Longitude, and the Parallax of the Latitude, is the very Parallax in that Altitude.

Secondly, But if the Moon be not in the 90th. Degree of the Ecliptick, to find the Parallaxes of the Latitude and Longitude, the Analogy or Proportion is,

1. As Radius or S. 90°,

To T. of the V. of the Ecliptick and Horizon,

So is Sc. of the Moon's Distance from the As­cendent, or Descendent deg. of the Ecliptick,

To Tc. of the Ecliptick's V, with the Azimuth of the Moon.

AGAIN say,

2. As the Radius or S. 90°,

To S. of that V. found,

So is the Parallax of the Moon's Altitude,

To the Parallax of her Latitude sought.

LASTLY say,

3. As the Radius or S. 90°, 00'

To Sc. of the former V. found,

So is the Parallax of the Moon's Altitude,

To the Parallax of his Longitude sought, which being added to the true Motion of the Moon, if she be on the East part of the 90° of the Eclip­tick. Or from it to be deducted if she be on the West part of the 90° of the Ecliptick.

PROP. XXXVI. How by knowing the Refraction of a Star, to find his true Altitude.

For the speedy performance of which I have annexed this Table of Refractions of the Stars observed by Tycho Brabe a Nobleman of Den­mark, and a most famous Astronomer.

The USE of which Table is thus.

PROP. I. How to find the Distance of any two Cities or Places, which differ onely in Latitude.

IN this Proposition there are two Varieties which are these.

• 1. If both the Places lie under one and the same Meridian, and on one and the same side the Equinoctial, either on the North or South side thereof; then substract the lesser Latitude from the greater, and convert their difference into Miles (by allowing 60 Miles for a Degree) so have you the distance of the two Places propounded.
• 2. But if the two Places lie under one and the same Meridian, but the one on the South side of the Equinoctial, and the other on the North side, then add both their Latitudes toge­ther their Sum is their Distance.

PROP. II. To find the Distance of any two Places which differ only in Longitude.

There are also in this Proposition two Va­rieties.

• 1. The two Places may both lie under the Eqinoctial, and so have no Latitude: and if so, [Page 183] substract the lesser Longitude out of the greater, and convert the remainder into Miles; so have you the distance of any two Places so posuited.
• 2. But if the two Places differ only in Longi­tude, and lieth not under the Equinoctial, but under some other Intermediate Parallel of Lati­tude, between the Equinoctial, and one of the Poles: Then to find their distance, this is the Analogy or Proportion.

  As Radius or S. 90°,

  To Sc. of the Latitude:

  So is S. of ½ X. of Longitude,

  To S. of ½ their distance, which being doubled, and converted into Miles, giveth the required distance.

PROP. III To find the Distance of any two Places, which dif­fer both in Latitude, and in Longitude.

In this Proposition three Varieties do present themselves to our View.

• 1. One of the Places may lie under the Equi­noctial and have no Latitude, and the other under some Parallel of Latitude between the Equinoctial and one of the Poles. In such case observe this Analogy or Proportion.

  As Radius or S. of 90°,

  To Sc. of their X. of Longitude:

  So is Sc. of their Latitude,

  To Sc. of their Distance required.

• [Page 184]2. But if both the Places proposed shall be without the Equinoctial, but on the one side either both towards the North, or both towards the South, then their Distance may be found, by this Analogy or Proportion.

  As Radius or S. 90° 00',

  To Sc. of their X. of Longitude:

  So is To. of the greater Latitude,

  To the T. of a fourth Arch, which substracted from the Complement of the lesser Latitude, the remainder must be the fifth Arch; Then say,

  As Sc. of the fourth Arch,

  To Sc. of the fifth Arch:

  So is S. of the greater Latitudes,

  To Sc. of the Distance of the two proposed Places.

• 3. The two Places propounded may be so situated, that one of them may lie on the North, and the other on the South side the Equinoctial: the Distance of Places so situated may be obtai­ned, by this Analogy or Proportion.

  As Radius or S. 90°,

  To Sc. of X. of Longitude:

  So is Tc. of the greater Latitude,

  To T. of a fourth Arch, which being substrac­ted out of the Summ of the other Latitude, and the Radius or 90° Deg. the remainder is a fifth Arch; Then say,

  As Sc. of the fourth Arch,

  To Sc. of the fifth Arch:

  So is S. of the Latitude first taken,

  To Sc. of the Distance required.

These are all the Varieties of the Positions of Places on the Terrestrial Globe: For if the Di­stance of any two Places be required, they must fall under one or other of these Varieties, and may be obtained by one or other of the Proportions, mentioned in the three afore­going Propositions.

Also if you know the Latitude and Longitude of any two fixed Stars, or their Right Ascension and Declination, then by these Rules their Distance may be found, which is of good use to Astronomy. It may also be applyed to Circular Sailing; of all other ways the most perfect: which is treated of in its due Place.

PROPI. The Rumb, and Distance sailed thereon being gi­ven,The Do­ctrine of Rightli­ned Trian­gles, both Right, and Oblique­Angled, applied to Propositi­ons of Plain Sai­ling. to find the Difference of Latitude, and the Departure from the Meridian.

Admit a Ship sails N. W. by N. 372' or Miles, or 124 Leagues, I demand her Difference of Latitude and departure from the Meridian?

In the Triangle ABC, the Hypothenuse AC representeth the distance sailed, or Rumb-line,Fig. 44. BC the departure from the Meridian, and AB the difference of Latitude.

1. To find which say,

As Radius or S. 90°,

To Log. distance sailed 372'.

So is Sc. V. A of the Course 56° 15',

To Log. cr. AB 309 3/10 Minutes, which being divided by 60' giveth 5°. 9'. 18" for the Diffe­rence of Latitude.

2. To find the Departure from the MeridianFig. 44. say,

As Radius or S 90°,

To Log. Rumb-line AC 372'.

So is S. of V. of the Course A 33° 45',

To Log. cr. BC 206 6/10 Minutes, the departure from the Meridian, which divided by 60 giveth 3°. 26'. 36" for the difference of Longitude.

Note that by this Proposition you may keep an Account how much you have sailed either East or West, North or South.

PROP. II. By the Rumb and Difference of Latitude given, To find the Distance, and the Departure from the Meridian.

Admit a Ship sail N. W. by W. untill herFig. 44. difference of Latitude be 309 3/10 Minutes, I de­mand her distance sailed, and her departure from the Meridian?

1. To find the distance, say,

As Sc. of V. of the Course A 56° 15',

To Log. cr. AB the X. of Lat. 309 3/10 Minutes.

So is the Radius or S. 90°,

To Log. AC 372 the distance sailed.

2. For the Departure, say,Fig. 44.

As Sc. of A V. of the Course 56° 15",

To Log. cr. AB. X of Lat. 309 3/10 Minutes.

So is S. of V. of the Course A 33° 45',

To Log. cr. AB 206 6/10 Minutes, the Departure required.

By the help of this Proposition, when your Latitude by Observation doth not agree with your dead reckoning, (kept by the former Proposition) Then according to this Rule, you may make your way saild agree with your Observed Latitude, and so correct your Ac­count or dead Reckoning.

PROP. III. By knowing the Distance of the Meridians of two Places, and their Difference of Latitude, to find the Rumb, and Distance.

Admit A, to represent the Lizard, AB the Parallel thereof, C. St. Mary's Islands, being one of the Azores, and CB the Meridian thereof.

In the Triangle ABC, there is given the side AB 816 Minutes, the Distance of the Lizard, from the Meridian of St. Marys, and the side CB their difference of Latitude 768 Minutes, IFig. 45. demand the Rumb: i.e. the Angle at C, and the Distance of the Lizard from St. Marys?

1. For the Rumb or Angle at C, say,

As Log. cr. CB 768',

To radius [...] S 90°.

So is Log. cr. AB 816',

To T. of the B [...], or Angle at C 46° 44', and is from the Lizard unto St. Marys to the fourth Rumb of the Meridian, and 1° 44' more, Viz. S. W. and [...] 44'. Westerly, or from St. Marys, to the Lizard, N. E. and 1° 44' Easterly: and thus it shall be by the Plain Chart.

2. For their Distance AC, say,

As S. Rumb. or V. at C. 46° 44',

To Log. cr. AB 816' Minutes.

So is Radius or S. 90°,

To Log. Hypoth. AC 1120' 1/10 which is the Di­stanceFig. 45. of the Lizard, unto St. Marys Istand, and such should be the distance by the Plain Chart.

PROP. IV. Admit two Ships to set sail from one Port, one Ship sails W. S. W. 40', the other W. by N. so far untill she finds the first Ship to bear from her S. E. by E. I demand the second Ships distance from the Port, and their Distance asunder?

In the Triangle ADE, let A represent the Port, AD the W. S. W. course, and AE theFig. 46. Course W. by North.

1. To find the second Ships distance from the Port, say,

As S. of V. at E. 22° 30',

To Log. cr. AD 40' Minutes.

So is S. of V. at D 123° 45',

To Log. cr. AE 86 98/100 Minutes, which is theFig. 46. distance required.

2. To find the two Ships their distance Asun­der, say,

As S. of V. at E 22° 30',

To Log. cr. AD 40 Minutes.

So is S. of V. at A 33° 45',

To Log. cr. DE 58 12/100 Minutes, which is the Distance required.

PROP. V. Two Ships sets sail from two Ports, which lie N. and South of each other, the one sails from the Northermost Port 72 29'/100, and then meets she other Ship, which came from the Southermost Port, on a N. W. Course, and had sailed from thence 56 80'/100 I demand the Rumb on which the first ship made her way, and also the Distance be­tween the two Ports?

In the Triangle ADE, let A be the Souther­mostFig. 47. Port, AD the Course and way of the second Ship N. W. 56 80'/100, let E be the Norther­most Port, ED the Course and Way of the other Ship 72 29'/100, and D the Place where they both meet.

1. To find the Rumb on which the first Ship sailed, say,

As Log. cr. DE 72 29/100 Minutes,

To S. of V. at A 45° 00'.

So is Log. cr. DA 56 80/100 Minutes,

To S. of V. at E 33° 45', which sheweth the Course of the first Ship to be S. W. by South.

2. To find the Distance between the twoFig. 47. Ports A and E, say,

As S. of V. at A 45° 00',

To Log. cr. DE 72 29/100 Minutes,

So is S. of V. at D 101° 15',

To Log. cr. EA 100', which is the required Distance.

PROP. VI. Admit a Ship coming off the Main Ocean and I had sight of a Promontory or Cape, by which it is my desire to sail, I find it to bear from me S. S. E. and distant by Estimation 33', or Miles: But keeping still on my Course S. untill the Evening, having sailed 36' or Miles, I would then know how the Cape bears, and its distance from the Ship?

In the Triangle ADE, admit that at A, I do observe the Cape D, to bear from me S. S. E. 33', and having sail'd from A, to E 36' South, I desire to know its Distance, and bearing. In the Triangle, there is therefore given, AD 33',Fig. 48. AE. 36', and the Angle at A 22° 30'.

1. To find the Angle at E, say,

As Z. cr^s. AE, and AD 69',

To X. cr^s. AD, and AE 03'.

So is T. ½ VV unknown D and E 78° 45',

To the T. of 12° 20', which taken from 78° 45', leaves the Angle at E 66° 25', so that the [Page 193] Cape D then bears from me E. N. E. and 01° 05' Northerly.

2. To find the Distance of the Cape ED from the Ship, say,

As S. V. at E 66° 25',

To Log. cr. AD 33'.

So is S. V. at A 22° 30',

To Log. cr. ED 13 78/100 Miles distant, so that the Cape is then distant from the Ship 13 78/100 Miles.

PROP. VII. Two Ports both lying in one Latitude, distant 64' or Miles, the Westermost of those Ports lieth op­posite to an Island, more Northerly distant there­from 47' or Miles, which Island is also distant from the Eastermost Port, 34' or Miles, I de­mand the Course from the Westermost Port to that Island?

In the Triangle ADE, let A be the Westermost Port, and E, the Eastermost Port, distant Asun­der 64'; and let D be the Island, distant from A 47', and from E 34': Then is the Angle atFig. 49'. A required, which is the Course or Rumb, from the Westermost Port, unto the Island: To find which, say,

As Log. cr. AE 64',

To Log. Z. cr^s. AD, and ED 81'.

So is Log. X. cr^s. AD, and ED 13',

To Log. os a certain line AO 16 454/ [...].

Which added to AE 64, is 80 454/1000,

The ½ whereof is AB, 40 227/1000

Then again say,

As Log. cr. AD 47', [Page 194] To Radius or S. 90°.

So is Log. AB 40 227/1000,

To Sc. V. at A, 58° 51', that is N. E. by E. 2° 36' Easterly, which is the Course from the Westermost Port A, unto the Island D.

PROP. I. To find by the Table, what Meridional parts are contained in any Difference of Latitude.

In this Proposition three Varieties present themselves unto our View.

• 1. When one Place is under the Equinoctial, the other having North, or South Latitude, his Meridional parts corresponding, is to be esteem­ed for the Meridional Difference of Latitude.
• 2. When both Places are towards one of the Poles, then the Meridional parts of the lesser, taken from the Meridional parts of the greater Latitude, the remainder is the Meridional dif­ference required.
• 3. When one Place hath North, and the o­ther South Latitude, their corresponding Me­ridional parts added together gives the Meridio­nal difference of Latitude sought: thus having sound them out they may thus be applyed.

PROP. II By knowing the Latitudes, and the difference of Longitude of any two Places, to find the Rumb, and Distance.

Admit there be a Port in the Latitude of 50° 00' North, and another in the Latitude of 13° 12' North, and their Difference of Longitude is 52° 57' West, I demand the Rumb and Di­stance?

In the Triangle A b c, let A b represent the proper difference of Latitude, bc the Departure, Ac the distance sailed, A the Angle of theFig. 50. Course, c the Complement of the Course.

In the Triangle ABC, AB is the Meridional difference of Latitude, BC the Difference of Lon­gitude, A the Angle of the Rumb, C the Compl. of the Angle of the Rumb: These things being understood the work evidently appears to be the same as in Rightangled Plain Triangles.

There is then required first the Difference of Latitude, and this falls under the second Va­riety. [...]

1. To find the Rumb or Course say,

As Merid. X. Lat. 2676',

To Radius or S. 90°.

So is X. of Longitude 3177',

To T. of the Rumb 49° 53', the Course there fore is S. W. ½ W, &c.

2. To find the Distance,Fig. 50. [...]

As Sc. Course 40° 07',

To proper X. of Lat. 2208'.

So is Radius or S. 90°,

To the Distance 3426 Minutes as required.

PROP. III. By knowing the Latitudes, and distance of two Places, to find the Rumb, and Difference of Lon­gitude.

1. To find the Rumb or Course say,

As the Distance sailed,

To Radius or S. 90°.

So is the X. of Latitude,

To Sc. of the Rumb required.

2. To find the Difference of Longitude say,

As Radius or S. 90°,

To the X. of Latitude in Merid. Parts.

So is T. of the Rumb,

To the X. of Longitude required.

PROP. IV. By knowing the Latitudes, and Rumb of two Pla­ces, to find their Distance, and Difference of Longitude.

1. To find the Distance say,

As Sc of the Rumb,

To the X of Latitude.

So is Radius or S. 90°,

To the Distance required.

2. To find the Difference of Longitude say,

As Radius or S. 90°,

To the X. of Latitude in M. Parts.

So is T. of the Rumb,

To the X. of Longitude required.

PROP. V. By knowing the Rumb, Difference of Longitude, and one Latitude, to find the other Latitude, and the Distance.

1. To find the other Latitude say,

As T. of the Rumb,

To the X of Longitude in parts.

So is Radius or S. 90°,

To the Merid. X. of Latitude required.

2. To find the Distance say,

As Sc. of the Rumb,

To the X. of Latitude.

So is Radius or S. 90°,

To the required Distance,

PROP. VI. By knowing the Distance, one Latitude, and Rumb, to find the other Latitude, and Difference of Longitude.

1. To find the Difference of Latitude say,

As Radius or S. 90°,

To the Distance.

So is Sc. of the Rumb,

To the X. of Latitude required.

2. To find the Difference of Longitude say,

As Radius or S. 90°,

To the Merid. X. of Latitude.

So is T. of the Rumb,

To the X. of the Longitude required.

PROP. I. Two Places, the one under the Equinoctial, the other in any Latitude given; also their difference of Longitude given, to find.

• 1. Their Distance in the Arch of a great Circle.
• 2. The direct Position of the first place from the second.
• 3. And of the second Place from the first.

Here we call the Angles which the Arch makes with the Meridians of the places pro­pounded, the Angles of the Direct Positions of those places one from the other: because the Arch of a great Circle. drawn between two places is the nearest distance; and the most di­rect way of the one, to the other Place. Now I shall not here demonstrate it by Schemes, as I have done in the other two Sections, but shall only lay down the proportions, whereby the re­quired parts may be found; and so leave the in­genious Seamen to practice it with Schemes at his leasure: and,

1. To find the nearest distance from Place to place, in the Arch of a Great Circle: Say ac­cording to the 10 Case of Rectangled Spherical Triangles.

As the Radius,

To Sc. of X. of Longitude.

So is Sc. of X. of Latitude;

To Sc. of the Distance in the Arch required.

2. For the Direct Position, say by the 11 Case thus,

As the Radius,

To S. of X. of Latitude.

So is Tc. of X. of Longitude,

To Tc. of V. of Position required.

3. For the Direct Position of the second Place from the first, say by the 11 Case thus,

As the Radius,

To S. of the X of Longitude.

So is Tc. of X. of Latitude,

To Tc. of V. of Position required.

PROP. II. Two Places proposed, the one lying under the Equi­noctial, the other in any Latitude given; with their distance in a great Circle of the same Places being also known, to find.

• 1. Their Difference of Longitude.
• 2. The direct Position from the first to the second Place.
• 3. And from the second to the first Place.

1. For their Difference of Longitude, say by Case 12,

As Sc. of the Latitude,

To Radius.

So is Sc. of their Distance in the Arch,

To Sc. of their Difference of Longitude requi­red.

2. Now to find the Direct Position from the first place to the second, say by the 13 Case; and thirdly, for the Direct Position from the second place to the first, say by the 14 Case of Rectan­gulars.

PROP. III. Two Places lying in one Latitude given, their dif­ference of Longitude being also known, to find.

• 1. The nearest distance of those two Places.
• 2. The direct Position of one Place from the other.

The Resolution of this Proposition depends on the 9 Case of Oblique Spherical Triangles: by sup­posing the Oblique Triangle, to be transfigured or converted into two Rectangulars, by a supposed Perpendicular: and then,

1. To find the nearest distance in the Arch, say by the 8 Case of Rectangulars.

As the Radius,

To Sc. of the Latitude.

So is S. of half X. of Longitude,

To S. of half the required distance, which doubled giveth the distance of the two places in the Arches, as sought.

2. For the Direct Position, say by the 9 Case.

As the Radius,

To S. of the Latitude.

So is T. of half X. of Longitude,

To Tc. of V. of Position required.

PROP. IV. Two Places lying both in one Latitude given, and the nearest distance being also known, to find.

• 1. Their Difference of Longitude.
• 2. The direct Position of the one Place from the other.

The Resolution of this Proposition falls un­der the 11 Case of Oblique Spherical Triangles: for here you have the three sides of the Triangle given, viz. the Arch of Distance, and the other two sides (are both equal) being the Comple­ments of the places Latitude: and here seeing the two sides are equal, therefore the two Angles of Position are also equal: now there is required the three Angles,

1. To find their Difference of Longitude, add the double of the Complement of Latitude to the Arch of Distance; then from half this Sum, de­duct the Arch of Distance, and then proceed in all points as you see in Case the 11th. So shall their Difference of Longitude be obtained.

2. To find their direct Position:

First, to the double Complement of Latitude, add the Arch of Distance, then from half that agra­gate, deduct the Complement of Latitude, and then work as before, so shall the direct Position be attained.

PROP. V. Two Places proposed lying in one Latitude, and the distance of those Places in their Parallel given; to find.

• 1. Their Difference of Longitude,
• 2. Their distance in the Arch of a great Circle,
• 3. The direct Position of the one from the other.

Now you must understand, that as the Semi­diamiter of a Parallel, is in proportion to the Semidiamiter of the Equinoctial: so is any num­ber of Miles in that Parallel, to the Minutes of Longitude answering to those Miles: fo that if we suppose the Semidiameter of the Equinoctial to be Radius, then the Semidiameter of any Pa­rallel is the Sine of that Parallel's distance from the Pole, that is the Sc. of the Latitude of that Parallel: Therefore,

1. To find the Diff. of Longitude say,

As Sc. of the Latitude,

To the Radius,

So is the Distance in that Parallel,

To the Diff. of Longitude required.

2. Now the Difference of Latitude being ob­tained, the nearest distance may be found, as in the third proposition aforegoing: 3. so likewise may the Angles of Position also.

PROP. VI. By knowing the nearest Distance of two Places, their Difference of Longitude, and one of their Latitudes; to find the Direct Position thereof from the other.

This Proposition falls under the first Case of Oblique Spherical Triangles, and is thus resolved: therefore,

As S. of the Distance of the two Places,

To S. of their X. of Longitude.

So is Sc. of the Latitude of the one Place gi­ven,

To S. of the Direct Position from the other as was so required.

PROP. VII. By knowing the Latitudes of two places, and like­wise their Difference of Longitude; to find,

• 1. The distance in the Arch.
• 2. The direct Position from the first to the second place.
• 3. The direct Position from the second to the first place.
• 4. The Latitudes and Longitudes by which the Arch passeth.
• 5. The Course and Distance from Place to Place through those Latitudes and Longitudes according to Mercator.

I shall here make use of M. Norwood's example of a Voyage from the Summer-Islands, unto the [Page 210] Lizard: now because the work is various I have therefore illustrated it with a Scheme, and shall be as brief and facile as possible. Therefore,

In the Triangle ADE, let A be the Summer­Islands, whose Latitude is 32° 25', AD the Com­plementFig. 51. thereof 57° 35', let E represent the Lizard whose Latitude is 50° 00', and ED the Complement thereof 40° 00', and let their Dif­ference of Longitude, namely the Angle ADE be 70° 00', now Drepresenteth the North-Pole, and AE an Arch of a great Circle passing by these two Places: now see the operation.

1. By having the Complements of the Latitudes of the two Places, viz. AD 57° 35', and ED 40° 00', and their Difference of Longi­tude, namely the Angle EDA 70° 00': you may find the nearest distance EA to be 53° 24'; by Case the 9. § 5. chap. 5.

2. Then having found the nearest distance inFig. 51. the Arch EA to be 53° 24', (or 3204 Miles) the Angle of Position from the Summer Islands to the Lizard, namely the Angle DAE, may be found by Case the 1. § 5. chap. 5. to be 48° 48', that is N. E. and 03° 48' Easterly.

3. And also by the same Case, may the Direct Position from the Lizard, to the Summer-Islands, namely the Angle AED befound to be 81° 10', that is W. by N. and 2° 25' Westerly.

4. In order to the finding the Latitudes and Longitudes by which the Arch passeth, first let fall the Perpendicular DB, so is the ObliqueFig. 51. Triangle ADE converted into two Rectangulars, viz. ABD, and DBE: secondly, by Case the 8. § 4. chap. 5. you may find the length of the [Page 211] Perpendicular DB to be 39° 26', whose Comple­ment is 50° 34', which is the greatest Latitude by which the Arch ABE passeth, so the greatest Ob­liquityFig. 51.

of the Equinoctial from that Circle is 50° 34'. — Thirdly, by Case the 9. § 4 chap. 5. you must find the vertical Angles, viz. ADB, and BDE, which will appear, the Angle ADB to be 58° 31', and EDB to be 11° 29': now these things being obtained, the Lati­tudes by which the Arch passeth at every tenth degree of Longitude from A, may be found by resolving the several Right-Angled Triangles, viz. BDc, BDf, &c. substracting 10° from ADB 58° 31', there remains BDc 48° 31', and so for the rest as in the Table. Now by knowing these Angles last found, and the Perpendicular BD before found to be 39° 26', you may by Case the 3. §. 4. chap. 5. find the Latitudes of the several points A. c. f. g. h. i. B. and E. to be as in the subsequent Table.

5. Thus havingFig. 51.

found the Latitudes and Longitudes of the Arch, and the other required parts afore­mentioned, we now come to shew how the Course, and the Distance from place to place according to Mercator may be found. So to find, first the Course and Distance [Page 212] Ac. now there is given the Latitude of A 32° 25', and of c 38° 51', and their Difference of Longitude is 10° 00', now the Proper DifferenceFig. 51. of Latitude is 6° 26', or 386', and Meridional Difference of Latitude is 475'. Now knowing these things by proposition 2. § 2. chap. 8. yon may find the Course from A to c, to be N. E. 51° 38'; and the Distance Ac to be 622', and so those Rules prosecuted will shew the course and distance from c to f; from f to g; from g to h, &c. So of the rest, which for brevity sake I shall omit, and leave the Ingenious Seaman to Calculate at his Pleasure.

I might hereunto annex many more proposi­tions of Circular Sailing, but because of the smallness of this Treatise, and that those Pro­positions already handled, being by the Inge­nious Seaman well understood, will be sufficient to enable him to perform any other Conclusion in Circular Sailing whatsoever, I therefore here omit, and hasten forwards unto the other parts of this Mathematical Treasury.

PROP. I. By the Protractor, to Protract an Angle of any quantity of degrees propounded.

AN Angle may be laid down easily accord­ing to the directions of Prop. 5. §. 1. Ch. 4.Fig. 52. but because this is more usefull in Surveying, Know that if it be required to protract an Angle of 50 deg. having drawn the line A B at plea­sure, place the Centre of your Protractor on C, and moving it by your Protracting Pinn, untill the Meridional line thereof be directly on the line A B, then make a Mark by the division of 50° on the limb of the Protractor as at D, and [...] [Page 214] [...] [Page 215] [Page 216] draw the line CD, so shall the Angle DCB, be an Angle of 50 degrees.

PROP. II. By the Protractor given, to measure an Angle given.

This is performed by the line of Chords also, according to prop. 6. §. 1. chap 4. and by the Protractor is found thus: Suppose DCB were an Angle whose Quantity were desired, to find which, first the Center of the Protractor applyedFig. 52. unto the Angular point C, and its Meridional line lying justly with CB; you shall perceive the Point D, to touch the limb of the Circle at 50 deg. Therefore I conclude the Measure of the Angle DCB, to be 50 degrees.

PROP. I. How to take the Plot of a Field, by the Semicircle at one Station taken in any part thereof, from whence all the Angles may be seen, and measu­ring from the Station unto every Angle thereof.

SUppose ABCDEF were a Field, and 'tis re­quired to take the Plot thereof: Having placed marks at all the Angles thereof, and made choice of your Station, which let be K; at which, place your Instrument, and turn­ingFig. 53. it about untill the Needle hang over the Meridian Line of the Chart, there screw it fast: Then directing your sight to A, you'l find the Degree out by the Index to be 40° 15': Then measuring KA with your Chain it appears to be 5 Chains and 20 Links, which note down in your Field-book: and so do by all the rest untill you have found all the Angles and Distances from your Station K, to each respective Angle, which finished your work will stand thus.

PROP. II. How to delineate on Paper any Observation taken according to the Doctrine of the last Proposition.

Upon your Paper draw a Line to represent the Meridian line as M, H, then Placing the Center of your Protractor on the point K, lay­ing the Meridian line of the Protractor on the Meridian line M, H, then seeing the Angle at A was 40° 15', make a Mark against 40° 15' of the Protractor, as at A, and so do with all theFig. 53. other Angles, as you find them in your Table: Then remove your Protractor, and draw the Lines KA, KB, &c. This done lay down on each line his respective Measure, as it appear­eth in the Table. Lastly draw the Lines AB, BC, &c. So have you on the Paper the exact Figure of the Field.

PROP. III. How by the Semicircle to take the Plot of a Field at one Station in any Angle thereof, from whence you may view all the other Angles, by measuring from the Stationary-Angle, unto all the other Angles.

Admit A, B, C, D, E, F, G, to be a Field, whose Plot is required: Place your Semicircle at G, and turning it about untill the Needle hang over the Meridian line of the Chart, and there screw it fast: Then direct your sights to the several Angles, viz. B, C, D, &c. in order one after the other, and so shall eace respectiveFig. 54. Angle be found, as in the subsequent Table: Then with your Chain, measure from your Sta­tionary-Angle G, to all the other respective Angles, which done you have finished, and the work standeth thus.

PROP. IV. How to delineate any Observation taken according to the Doctrine of the last Proposition.

Upon your Paper draw a streight line as M, N, then take a point therein as G, to represent the Stationary-Angle, to which point apply the Center of your Protractor, (in all respects as isFig. 54. before taught) then according to the Notes in the Table, prick off all the Angles, viz. B, C, &c. according to their due quantity, then draw all the lines, viz. GB, GC, GD, &c. and on them place their respective measure (as appeareth in your Notes) lastly draw the lines AB, BC, CD, &c. So is there on the Paper the exact Figure of the Field, as was required.

PROP. V. How by the Semicircle to take the Plot of a Field at two Stations, by measuring from each Station to the visible Angles: the Field being so Irregu­lar that from no one Place thereof, all the Angles can be seen.

Admit A, B, C, D, E, F, G, H, I, K, to be the Figure of a Field, whose Plot is required: having made choice of your two Stations, viz. Q, and P, and placed Marks in all the Angles: Then place your Semicircle at Q, and thereFig. 55. six it with the Needle hanging over the Meri­dian of the Chart, represented by R, Q, X, and direct your sights unto all the visible Angles, viz. [Page 221] A, B, C, D, E, and F, and note down the Quantity of each Angle in your Field-book: Then measure with your Chain from your Station Q, to the Angles A, B, C, D, E, and F, and their length so found, note down in your Field-book also.Fig. 55.

This done direct your sight unto your second Station P, and note down in your Field-book the degree of Declination, of your second-station P, from the Meridian. Then measure the Stationary Distance PQ with your Chain, and note it down in your Field-book also.

Then remove the Instrument unto P, your second-station, and there fix it with the Needle hanging over the Meridian line of the Chart re­presented by TPB, then direct your sights to the several visible Angles at this second Station, viz. F, G, H, I, and K, in order one after ano­ther, and note down the Quantity of each An­gle in your Field-book: Then with your Chain measure from your Station P, to these several Angles G, H, I, and K, (in all respects as at the first station Q.) and their length so found note down in your Field-book likewise: So have you finished your Observation, and your work stan­deth thus.Fig. 55.

The Observation taken at the first Station Q.

The Declination of the Station P, from the Meridian R Q X, is 30° 00', and the Stationary distance Q P is 9 Chains.

The Observation taken at the second Station P.

☞ Note that the manner of taking the Plot of a large Champain Field, at many Stations, is almost the same with this Proposition; for he that can do the one, can also perform the o­ther: therefore for brevity sake I here omit it as superfluous.

PROP. VI. How to delineate any Observation taken according to the Doctrine of the last Proposition.

Upon your Paper draw the Meridian-line R Q X, then place the Center of your Protrac­tor on Q, (representing your first Station) and its Meridional-line lay equal to R Q X, then prick off the Angles visible at your first Station Q, viz. A, B, C, D, E, and F, Of their due quantity, then draw Q A, Q B, &c. laying on them their corresponding measure, noted in your Field-book. Now because your second Station P, doth decline 30° 00', from the Me­ridian RQX, prick off 30° 00', and draw PQ,Fig. 55. making it 9 Chains as in your Field-book ap­peareth, so doth P represent your second Stati­on. Then in all respects as before, place your Protractor at P your second Station, and draw the Meridian T P B parallel to R Q X, then prick off the several Angles, viz. F, G, H, I, and K, Of their due quantity, and then draw PF, PH, PI, &c. of their due length. Lastly draw the lines AB, BC, CD, &c. and so shall you have on your Paper the exact Figure of the Field as required.

PROP. VII. How by the Semicircle, to take the Plot of a Field at t [...] Stations, which lieth remote from you, when either by opposition of Enemies you may not, or by some other Impediment you cannot come into the same.

Admit the Figure A, B, C, D, E, F, to be a Field into which by no means you can possibly enter, and yet of necessity the Plot thereof must be had, for the obtaining of which chuse any two Stations, it mattereth not whether near at hand or far off, so that all the Angles may be seen. Let your two Stations be H and L, (the full length of the Field if possible) then place your Instrument at H, and fixing it as isFig. 56. afore shewed, direct your sights to the several Angles of the Field, viz. A, B, C, &c. orderly one after another, observing their degrees as is afore taught, noting it down in your Field-book: then take up your Instrument, leaving a mark in its room at H, And measure with your Chain from Hunto L, your second Station, which note down in your Field-book; Then placing your In­strument at L, your second Station, and as is be­fore taught, fixing it there, make the like Obser­vation to the several Angles, viz. A, B, C, D, &c. as at the first Station H, and note it down in your Field-book also, And having so done you have finished, and your Work standeth thus.

Observations at the first Sta­tion H, are

The Angle from H the first Station, unto L the se­cond Station, is 180° 00', theFig. 56. Stationary distance HL, is 60 Chains.

Observations at the second Sta­tion L, are

PROP. VIII. How to delineate any Observation taken according to the Doctrine of the last Proposition.

Upon your Paper draw a Line as HL, which make equal to 60 Chains, then placing the Center of your Protractor on H, your first Sta­tion, prick off all the Angles A, B, C, &c. asFig. 56. you find them in your Field-book, and draw HA, HB, HC, &c. at pleasure: then remove your Protractor unto your second Station L, pla­cing it as before, and prick off all the Angles [Page 226] A, B, C, D, &c. as you find them in your Field notes; and draw the lines LA, LB, LC, &c. at length untill they intersect the former lines, HA, HB, &c. in the Points A, B, C, &c.Fig. 56. which Points of Intersection are the Angles of the Field. Lastly draw AB, BC, CD, &c. So shall you have on your Paper the Figure of your Field, required.

PROP. IX. How by the Semicircle, to take the Plot of a great Champain-Plain, Wood, or other overgrown Ground, by measuring round about the same, and making Observation at every Angle thereof.

Admit A, B, C, D, be the figure of a Large overgrown Champain-Field; whose Plot is requi­red. First Place your Instrument at A, laying the Index on the Diameter; and turn it about, untill you espy the Angle at D, and there fix it fast: and direct your sights to B, and note the Degree cut by your Index, in your Field-book, (as afore is taught) then remove your Instru­ment to B, and there make the like observation, and so to C, and D, noting it down in your Field-book, as asore. Then with your Chain, measure the Sides AB, BC, CD, and DA, whoseFig. 57. length note down in your Field book, and so you have finished and your work standeth thus.

PROP. X. How to delineate any Observation taken according unto the Doctrine of the last Proposition.

Upon your Paper draw the line AB, at Plea­sure,Fig. 57. and placing the Center of your Protractor on the Point A, prick off an Angle of 100°, and draw AD, setting on it, and also on AB, their corresponding measure, in your notes: Then on B, protract an Angle of 117° 15', draw BC of its due length: Then draw the line CD, so have you the exact figure of the Field, on your Paper.

PROP. XI. How to take the Plot of any Field, by the help of the Chain only.

Admit the Figure A, B, C, D, E, to represent a Field whose Plot is required. To obtain theFig. 58. which, first measure the sides CD, CB, and BD, and note their due length down in your Field­book, [Page 228] and then measure the Sides CA, and BA, and then note [...] down their Length in your Field­book. Then measure the sides BE, and ED, for the sides BC, andFig. 58. BD, were be­fore known) which note down in your Field­book. So is your Field A, B, C, D, E, reduced into three Triangles, viz. CBD, CAB, and BED, the length of whose sides are all known, thus you have finished, and the works stands as you see.

PROP. XII. How to delineate any Observation, taken according to the Doctrine of the last Proposition.

Upon your Paper, draw a streight line, as CD, make it 5 Chains, 97/100, take CB in your Compasses, and strike an Obscure Arch; then take BD, and with that extent in D, cross the for­merFig. 58. Arch in B, and draw BC, and BD. Then take in your Compasses BE, and on B, strike an Obscure Arch, then take DE, and also cross [Page 229] the former Arch in E, and draw BE, and ED. Lastly take the line CA, and on C strike an Obscure Arch, then take AB, and on B, inter­sectFig. 58. the former Arch in A, then draw CA, and AB, so have you on your Paper the exact figure of the Field A, B, C, D, E, as was re­quired.

PROP. I. How to find the Area or superficial Content of a Trapezium.

Trapeziums are Quadrangles of sundry forms: yet take this as a general Rule, whereby their Content may be found. Admit it be required to find the Area or superficial Content of the Tra­pezium ABCD, to find which, first by drawing [Page 231] the Diagonal AD, you reduceth it into two Tri­angles, ABD, and ADC. Then by prop. 3. §. 1. of Chap. 4 let fall the two Perpendiculars on AD, from B, and C, Then by prop 3. §. 2 Ch.Fig. 59. 4. find the superficial Content of the two Trian­angles ABD, and ADC, which added together, is the Content os the Trapezium; by which Rule the Content of the Trapezium, A, B, C, D, is found to be 630 Perches.

PROP. II. To find the Area or superficial Content of a many­sided Irregular Figure.

Admit A, B, C, D, E, F, G, to be an Irregu­lar many-sided Figure, representing a Field whose Content is required: now in regard the Field is Irregular, therefore reduce it into Triangles, viz. ABC, ACG, EDG, DEG, and DFG, andFig. 60. then find the Content of all the said Triangles, by prop. 3. §. 2. Chap. 4 and add their Con­tents together; so shall that Sum be the Con­tent of the said Figure; and so do for any other.

PROP. III. How to reduce any Number of Perches into Acres, and on the contrary, Acres into Perches.

To find how many Acres are contained in a­ny Number of Perches given, you must consi­der that 160 Perches do make a Statute Acre, therefore if you divide the Number of Perches [Page 232] propounded, by 160, the Quotient is the number of Acres contained therein; and if there be a remainder which exceed 40, then divide it by 40, the Quotient shall be Roods, and the remainder Perches.

But on the contrary, if it were required to find how many Perches are contained in a cer­tain Number of Acres propounded. You must multiply the Number of Acres, by 160: the product shall be the Perches contained there­in.

It may be here expected, that I should shew how to reduce customary Measure to statute Mea­sure; and also that I should treat of the Division and Separation of Land. But because Mr. Rath­borne, and of late Mr. Holwell, hath sufficiently explained the same, by many varieties, I shall for brevity sake omit it, and leave you to consult those Authors.

PROP. I. How by the Semicircle to take an Accessible Alti­tude.

ADmit AB, be the Height of a Tower, which is required to be known. First placing your Semicircle at D, (with the Arch downwardsFig. 61. and the two sights fixed) place it HorizontalWhich to do is no more than thus; with a Thread and Plummet fastened at the Center of the Semicircle, so that it hath liberty to play, move the Semicircle until the Thread playeth against 90 deg. then screw it fast, and it is Horizontal. and screw it fast; Then move your Index, till through the sights thereof, you espy the top of the Tower at B, and observe what degree the lower part of the Index cutteth and that will be equal unto the Angle at D 50 deg Then measure the di­stance DA, which let be 299 Feet. Now the heighth of the Tower AB, is found, according to prop. 1. §. 2. Chap. 5. thus,

As Sc. V. at A 50° 00',

To Log. cr. DA 299 Feet.

So is S. V. at A 50 00,

To Log. AB 356 3/10 Feet the height of the Tower AB required.

PROP. II. How by the Semicircle to take an Inaccessible Al­titude, at two Stations.

Let AB be a Tower whose height is required; having placed your Instrument at E, as before direct your sights unto the Top of the Tower at B, and finding the Degree cut by the Index,Fig. 61. to be 23° 43', I say it is the Quantity of the Angle at E: Now by reason of Water, or such like Impediment, you can approach no nearer the Base of the Tower, than D, Therefore measure ED, which is found to be 512 Feet, then at D, make the like Observation, and the Angle at D, appeareth to be 50° 00', whose Complement is the Angle DBA, 40° 00', and the Complement of the Angle E 23° 43', is the An­gle EBA 66° 17': Now if the lesser Angle at B, be taken out of the greater, the remainder is 26° 17', the Angle EBD: Now first to find the side BD, of the Trangle EBD, say according to prop. 1. §. 3. chap. 5. thus.

As S. of V. EBD, 26° 17',

To Log. cr. ED 512 Feet.

So is S. of V. at E 23° 43',

To Log. cr. BD 465 2/10 Feet required.

Now to find the Height of the Tower AB, say according to prop. 2. §. 2. chap. 5. thus.

As Radius or S. 90°,

To Log. cr. DB 465 2/10 Feet found.

So is S. of V. BDA 50° 00',

To Log. cr. BA 356 3/10 Feet, which is the heightFig. 61. of the Tower required.

☞ Note that in taking any manner of Alti­tude the height of your Instrument must be added unto the height found, and that will give you the True Altitude required.

PROP. III. How by the Semicircle to take an Inaccessible Dis­tance at two Stations.

Admit A, and B, be the two Stations, from either of which it is required to find the distance unto the Church at C; placing your Instrument at B, the Index lying on the Diameter, and di­rect your sights unto the Church at C, fasten your Instrument, and turn your sights about un­till you see through your sights, your second Station at A, so will you find your Index to cutFig. 62. 30° 00', which is the Quantity of the Angle ABC. Then measure the distance AB, which is found to be 250 Yards, then with your In­strument at A, make the like Observation as before, and you will find the Angle BAC to contain 50° 00'. Now by the third Maxim of Plain Triangles §. 1. Chap. 5 you find also the Angle ACB, to be 100° 00': now to find the distance AC, and BC, you may by their oppo­site proportion according to prop. 1. §. 3. chap. 5. find the distance of AC, thus.

As S. of V. at C 100° 00',

To Log. cr. AB 250 yards.

So is S. of V. B 30° 00',

To Log. cr. AC 127 yards. Which is the di­stance of the Church from A.

Now to find the distance BC, say,

As S. of V. at A 100° 00',

To Log. cr. AB 250 yards.

So S. is of V. at A 50° 00',

To Log. cr. BC 194 4/10 yards, which is the di­stance of the Station B, from the Church at C.

PROP. IV. How to find the Horizontal line of any Hill or Mountain, by the Semicircle.

Let Figure 63 be a Mountain, whose Hori­zontal-line AB is required to be found: to find which, place your Instrument at A, and ha­ving caused a Mark to be placed on the Top of the Mountain at C; (of the just height of your Instrument) then move your Index, untill through the sights thereof you espy the Mark at C, so will you find the Quantity of the An­gle CAD, to be 50° 00', and by consequenceFig. 63 the Angle ACD to be 40° 00', then measure up the Hill AC, which is 346 yards. Now ha­ving obtained these several things, 'tis required to find the length of AD part of AB; to find which say,

As Radius or S. 90°,

To Log. cr. AC 346 Feet.

So is Sc. of V. at A 50° 00',

To Log. cr. AD 222 4/10 Feet.

Now seeing the Hill or Mountain descendeth on the other side, you must place your Instru­ment at C, and direct your sights unto the Bottom at B, and the Angle DCB will be found 50° 00', and the Angle CBD 40° 00'. Then measuring down the Mountain as CB, it appea­reth

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4. 63. [...]. [Page] [Page 237] to be 415 Feet; then have you the An­gles DCB, and CBD.

To find DB, part of AB, say,

As Radius or S. 90°,

To Log. cr. CB 415 Feet.

So is S. of V. BCD 50° 00',

To Log. cr. DB, 318 Feet: Now AD 222 4/10Fig. 63. Feet added thereunto produceth AB 540 4/10 Feet, which is the Horizontal line required of the Mountain ACBD.

☞ Note that when you come to delineate a Field wherein are Hills, you must protract the line AB, instead of the Hypothenusal Lines AC, and CB, and 'twill be necessary to distinguish those kind of Fields, by shadowing them off with Hills and Dales.

PROP. I. To Measure a Piece of Board, Plank, Glass, &c.

In Measuring of Board, Glass, &c. Carpen­ters and other Mechanicks measure by the Foot, 12 Inches unto the Foot; so that a Foot of Board, or Glass, contains 144 Square Inches.

Now if a Piece of Board, Plank, or Glass, be required to be measured, let it be either a Parallelogram, or Tapering Piece: first by the Rules aforegoing find the Content thereof in Inches, and that Product divide by 144, the Quotient is the Content of that Superficies in Feet.

PROP. II. To measure Tiling, Flooring, Roofing, and Parti­tioning-works.

In Tiling, Flooring, Roofing, and Partitioning­work, Carpenters, and other Workmen, reckon by the Square, which is 10 Feet every way; so that a Square containeth 100 Feet: Exam­ple.

There is a Roof 14 Feet broad, what length thereof shall make a Square? Divide 100 by 14, it yields 7 1/7 Feet.

Now if you have any Number of Feet gi­ven, and the Number of Squares therein contai­ned are required, divide that Number by 100, the product is Squares.

PROP. III. To measure Paving, Plaistering, Wainscotting, and Painting-work.

In Paving, Plaistering, Wainscotting, and Painting-work, Mechanicks reckon by the Yard Square, so each Yard is equal unto 9 Square Feet.

By the Rules aforegoing find the Superficial Content of the Court, Alley, &c. in Feet: which divide by 9, the Quotient is the Number of Yards in that work contained.

PROP. I. How to Measure any kind of Timber, or Stone, whether Three-square, Four-square, Many-square, Round, or of any other fashion, provided it be streight and equal all along.

To perform which first by the Rules afore­going in Chap. 4. §. 2. get the Superficial Content at the End, and then say,

As 144, the Inches of the Superficial Content of the End of a Cubick Foot,

To a Cubick Foot containing 1000 parts;

So is the Superficial Content of the End of any piece of Timber,

To the Solid Content of one Foot length of the said piece of Timber.

According to which Mr. Phillips calculated the ensuing Table, which I have thought fit hereunto to annex.

Case 2 Or the solid Content in Feet, &c. may be found otherwise thus.

By the Rules aforegoing find the Content of the End of the piece of Timber in Inches, which Content multiply by the length of the said piece of Timber, or Stone in Inches, and that Product divide by 1728, it produceth the Solid [Page 246] Content of that Piece of Timber, or Stone, in Feet, and parts of a Foot.

PROP. II. To measure Round Timber which is Hollow: or a­ny other Hollow Body.

If Hollow Timber be to be measured, first measure the Stick as though it were not Hollow, then find the Solidity of the Concavity, as though it were Massie Timber, then substract this last found Content, out of the whole Content be­fore found, the remainder is the Content of that Hollow Body.

PROP. III. To Measure Tapering Timber, or Stone.

Those Tapering Bodies are either Segments of Cones, or Pyramids: now the way to measure such bodies, is demonstrated in Prop. the 4. and 5. §. 3. Chap. 4: But now to find the Content of these Segments do thus: measure the Solidity of the whole Cone, or Pyramid, and then find the Content of the Top part thereof cut off, (as if it were a Cone, or Pyramid of it self) and the Content thereof, deduct from the Content of the whole Cone, or Pyramid: so shall the re­mainder be the Content of the Segment requi­red: which reduced into Feet gives the Solid Content of that Piece of Timber in Feet. Now to find the length of the Top part cut off, from the Cone, or Pyramid, say,

PROP. IV. How to find the Solid Content of any Solid Body, in any strange form, such as Geometry can given: no Rule for the measuring thereof.

These strange forms are either Branches in Metal, Crowns, Cups, Bowles, Pots, Screws, or Twisted Ballisters,Whose Surface is bounded by a Line called by Proclus a Helicoides, but it may also be called a Helix, a Twist or Wreath, &c. or any other Irregular-Solid, that keep not in thick­ness one Quantity, but are thicker in one place, than in another, so that no man by Geometry, is possible to measure their Solidity.

Now for the finding the Content of any such like Irregular Body in Inches or Feet, do thus: Cause to be made a Hollow Cube, or Parallelepipe­don, so that you may measure it with an Inch­Rule without Difficulty, and so to know the true Content of the whole, or any part thereof at pleasure within the Concavity: Then take some other convenient Vessel, and put pure Spring­water therein; then having filled the Vessel to a known Measure, make a Mark precisely round the very edge of the Water, then take the solid body and put it therein, then take out as much of the Water (as by means of the body put therein) is arisen above the Mark, untill the Water do justly touch at the Mark again: then put the Water taken forth into the [Page 249] Hollow Cube, and find the solid Content thereof (being transformed into a Cubick Body) in Feet, Inches, and parts of an Inch: Which Con­tent is the just solidity of the Body put into the Water. (Archimedes by this Proposition found the deceit of the Crown of Gold which Gelo the Son of Hiero had vowed unto his Gods: now the Workmen had mixed Silver with the Gold, which Theft was disco­vered by the great skill of Archimedes)See Procl. lib. 2. cap. 3. & Viturvius lib. 9. cap 3. And herein you must be very curious not to spill any of the Water, or take out of the Vessel, or put into the Hollow Cube, any more than the just quantity arisen above the Mark, for if you do it will produce infinite Errours, and thus may the Solidity of any Irregular Body be found.

PROP. I. To find the Solid Content in Inches of any Cask.

I Shall follow Mr. Oughthred's method, which is, Take the Diameter of the Cask both at Head and Bung, by which find the Area's of their Circles, which done, then take two thirds of the Area of the Bung, and one third of the Area at the Head, which added together, shall be the Mean Area of the Cask; which multi­plyed into the length of the Vessel, it will shew how many solid Inches are contained therein.

Example: Suppose the Diameter at the Head of a Vessel be 18, and at the Bung 32, and length is 40 Inches.

Now I find the Aggregate of the two Circles to be 620, and 989, Cubick Inches: which mul­tiplyed by 40, the length, produceth 24839, 56/100 Cubick Inches, for the whole Content of that Cask in Cubick Inches.

PROP. II. To find the Content of a Vessel in Wine, or Ale Gallons.

The Wine Gallon is established by the Con­sent of Artists, in these and other Nations, to contain 231 Cubick InchesSee Mr. Oughthred in his Book of the Cir­cles of Proportion, page the 57. and Mr. Edm. Gunter in his Book of the Cross-staff, part 21. chap. the 4.. Yet Dr. Wybard affirms it to be somewhat less, to wit 225, at most: The Ale Gallon contains 282 Cubick Inches, accor­ding to the Establish­ment of Excise. Here­in Artists differ somewhat in their Experi­ments.

Now having already shewed how to find the Content in Inches of any Cask, I now come to shew how to find the Content in Gallons, of any Beer, Ale, or Wine Cask, which is thus: Divide the Number of Inches given by 231, for Wine Measure, and 282, for Ale Measure. In the former Example I find the said Cask to con­tain 107, 53 Wine Gallons, and 88, 8, &c. Gal­lons in Ale Measure.

PROP. III. How to Gauge or Measure Brewers Tuns, &c.

Those Tuns are most commonly Segments of Cones or Pyramid, whose Basis is either a Square Parallelogram, Circle, or Oval; to measure which, [Page 253] let their form be what it will you must do thus. By the former Rules of Measuring such Seg­ments or Bodies, you must find their Solid Con­tent in Cubick Inches, (as in prop. 3. §. 2. chap. 10.) which Content divide by 282 Inches, (the Inches in one Gallon) it sheweth the Content in Gallons, and dividing the Gallons by 36, (the number of Gallons in a Barrel) it shews the Content in Barrels.

PROP. I. To Gauge a Ship, thereby to find how many Tuns her burthen is.

IN the Gauging or Measuring of Ships, Naupegers, or Ship-Wrights, observe these three Particular Rules: First, that if you measure the Ship within, you shall find the Content, or the Burthen the Ship will hold or take in. Secondly, if the Ship be measured on the outside, to her light mark as she swims be­ing unladen, you shall have the Content of the Empty Ship. Thirdly, but if you measure from the light mark, to her full draught of Water be­ing laden, you shall have the true Burthen of the Ship.

Now to find the Content of the King's Royal Ships: Measure the length of the Keel, the breadth of the Mid ship Beam, and the depth of the Hold: which three multiply into one ano­ther, and divide their Product by 100; so shall you find how many Tuns her Burthen is.

But for Merchant's Ships, which give no al­lowance for Ordnance, Masts, Sails, Cables, Anchors, &c. which are all a Burthen, but no Tonnage, you must divide the product by 95, so shall their true Burthen be found.

PROP. II.

First you shall multiply the Keel Cubically; and in like manner every Beam; the Mid ship Beams multiply them Cubically; and also the Reaking of the Ship, both at Stem, and Stem­Post, multiply them Cubically; likewise the principal Timbers, that doth mould the Ship, multiply them Cubically; and the depth of the Hold, multiply it Cubically; and so conse­quently every Place, or Places, which doth lead any work, multiply them Cubically; then if it be required to have a Ship as big again, or thrice as big; double, or triple each respec­tive Cubical number; then by prop. 9. §. 1. chap. 1: Or by prop. 4. §. 2. chap. 2. find the Cube­roots thereunto belonging; then according unto these respective Numbers, make your Keel, your Timbers, Beams, &c. which being done, you shall make a Ship of the Mould and Pro­portion desired.

PROP. I. How to draw the Hour-lines on an Equinoctial Plain.

AN Equinoctial Plane, is such which lieth Parallel unto the Equinoctial, and is an Horizontal Plane, under the Pole. This is the first and plainest kind of Dials, and is made after this manner: First describe the Circle AE, W, E, R, for your Planes, then Cross it with the two Diameters EW, and AER. ThenFig. 64. divide the Semicircle E, W, R, into 12 equal parts in the points ☉, ☉, ☉, &c. Then from the Center Q, and through the said points [Page 257] draw streight lines, which shall be the true Hour-lines belonging unto this Equinoctial Plane. Now because these Planes are capable of re­ceiving all the Hour-lines from Sun-rising unto the Sun-setting, in Summer; therefore the Hour­lines of 4, and 5, in the Morning; and 7, and 8, in the Evening; must be delineated as you see done in the Figure: These Hours may be sub-di­vided into half Hours, and Quarters: TheFig. 64. Stile of this Dial, must be a streight Pin, or Wyre set Perpendicular, to the Plain, on the Cen­ter Q. and of any convenient length. This Dial may be made for any Latitude, and is of good use for Seamen, and others.

PROP. II. How to draw the Hour-lines on a Polar Plane.

A Polar Plane is one that lies Parallel unto the Pole, and under the Equinoctial is an Hori­zontal Dial: the way to make this Dial is thus. First draw the line AB, for the Horizontal line of the Plane; and cross it at the Middle at right angles, with the line 12, Q, 12, which is theFig. 65. Meridian or Hour line of 12; Then upon the line 12, Q 12, either above or below the point Q, assume any point as S, then setting one foot of your Compasses in S, describe the Semicircle CED, which divide into 12 Equal parts, in the points ☉, ☉, ☉, &c. Then lay a Ruler un­to S, and unto the several points ☉, ☉, ☉, &c. and it will cross the line AB, in the points x, x, x, &c. Then through those points draw (by prop. 4. §. 1. chap. 4.) right lines all Parallel [Page 258] unto 12 Q 12, and so is your Dial finished. Then according unto the breadth of the Plane, you may proportionFig. 65. your Stile,Which may be ei­ther a Pin of the length of Q S, placed on Q, and Perpendicular unto the Plane, or it may be a piece of brass or elsewhat of the breadth of 12, to 3, or 9. Whose height must be equal to the di­stance between the two Hour-lines 12, and 9, or 12, and 3, and then will the shadow of the upper edge thereof shew the Hour of the day: The height of the Stile, is also found thus.

As the Tangent of the Hour-line 4 or 5,

To the Distance hereof from the Meridian.

So is the Radius,

To the Height of the Stile.

Then for the other Hour-line, say,

As the Radius,

To the Height of the Stile.

So is the Tangent of any other Hour-line,

To the Distance thereof from the Meridian line.

PROP. III. How to draw the Hour-lines on a Meridian Plane, which is an East, or West Dial.

A Meridian Plane stands upright directly in the Meridian, and hath two Faces, one to­wards the East, and the other towards the West.

Now admit it be required to make a direct East Dial, in the Latitude of 51° 32': let A, B,Fig. 66. C, D, be a Dial-plane, on which you would de­scribe a Direct East Dial, on the point D, de­scribe [Page 259] an obscure Arch HG, with the Radius of [...]our line of Chords, then take 38° 28', the Complement of your Latitude, place it from G to L; then draw DL quite through the Plane; Then to proportion your Stile unto your Plane, so that all the Hours may be placed thereon, from Sun-rising to 11 a Clock. Assume twoFig. 66. points in the line LD, as K, for 11; and I for the 6 a Clock Hour lines; then draw 6, 16, and 11, K 11, Perpendiculur to LD. This done, with the Radius of your line of Chords on L, strike the Arch OP, and from P, to O, place 15° 00'; and draw OK, to cut 6 I 6, in M, so shall IM be the height of the Stile proportioned unto this Plane; which may be a Plate of Brass, whose breadth must be equal to the distance between the Hour-lines of 6, and 9, which must be placed Perpendicular to the Plane, on the line 6, I 6, whose shadow of the upper edge, shall shew the Hour of the day. Now to draw the Hour-lines, with the Radius of your line of Chords, on M strike the Arch QN, which divide into 5 equal parts in the points •, •, •, &c. Then lay a Ruler from M un­to each of those points, and it will cut the line JK in the points *, *, *, &c. through which points (by prop. 4. § 1. chap. 4.) draw Parallels to 6 I 6, as the lines 77, 88, &c. which shall be the true Hour-lines of an East Plane, from 6Fig. 66. in the Morning, till 11 before Noon. Then for the Hour-lines of 4, and 5, you must prick off 5 as far from 6, as 6 is from 7; and 4, as far as 6 is from 8; and draw the Hour-lines 55, and 44, as before. Thus is your Dial compleat­ed, and in the forming of which, you have [...] [Page 258] [...] [Page 259] [Page 260] made both an East, and a West Dial; which is the same in all respects, only whereas the Arch H G, through which the Equinoctial passed in the East Dial, was described on the right hand of the Plane, in the West it must be drawn on the left hand, and the Hour-lines 4, 5, 6, 7, 8, 9, 10, and 11, in the Forenoon in the East Dial, must be 8, 7, 6, 5, 4, 3, 2, and 1, in the West inFig. 67. the Afternoon; as in the Figure plainly appear­eth: Now you may find the distance of the Hour-lines from the Substile, by this Analogy or Proportion.

As the Radius,

To the Height of the Stile.

So is the Tangent of any Hours distance from 6,

To the distance thereof from the Substile.

PROP. IV. How to draw the Hour-lines on a direct South, and North Plane,

This Plane or Dial must stand upright, ha­ving his face or Plane, if it be a South Dial, di­rectly opposite unto the South; but if a North Plane, directly opposite unto the North; now admit it be required to make a Direct South Dial,Fig. 68. for the Latitude of 51° 32': To make which first describe the Circle ABCD, to represent an E [...]ect direct South Plane, cross it with the Dia­meters CB, and AD, then out of your Line of Chords take 38° 28', the Complement of the La­titude, and set it from A, unto a, and from B, unto b, Then lay a Ruler from C unto a, and it will cut the Meridian ARD, in P, the Poles of

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Fig. Fig. 68. Fig. 69. [Page 262] other hath the North Pole of the World eleva­ted above it, and beholdeth the North part of the Meridian. The Hour-lines of 9, 10, 11, or 1, 2, and 3, is not expressed on this Plane, because 12, representeth 12, at Midnight; neither are the other said Hours expressed, because the Sun is never above the Horizon, at those Hours; Therefore the North Dial is capable only toFig. 69. receive these Hours, namely 4, 5, 6, 7, and 8, in the Morning; and 4, 5, 6, 7, and 8, at Night; as doth plainly appear in the Figure: Now the distance of the Hour-lines from the Meridian, may be found by this Analogy, or Propor­tion.

As Radius or S. 90°,

To Sc. of the Latitude.

So is T. of the Hour from Noon,

To T. of the Hour-line from the Meridian.

PROP. V. How to draw the Hour-lines on an Horizontal Plane.

This Horizontal Plane, or Dial, is one of the best and most usefull Dials in our Oblique He­misphere: Admit it be required to make an Horizontal Dial, for the Latitude of 51° 32': To make which, first describe the Circle ABFig. 70. CD, which representeth your Horizontal Plane, Then cross it with the two Diameters ARC, and BRD, Then take 51° 32' out of your Line of Chords, and set it from B, to a, and from C, to b, Then lay a Ruler from A, unto a, and it will cut the Meridian BD, in P, the Pole of the [Page 263] World, Then lay a Ruler from A, unto b, and it will cut ABD the Meridian, in the point AE, where the Equinoctial cutteth the Meridian, then through the three points A, AE, and C, drawFig. 70. the Equinoctial Circle, whose Center is at H; (and found as in the former proposition) Then divide the Semicircle ADC into 12 equal parts, in the points •, •, •, &c. Then lay a Ruler to R the Center of the Plane, and on those points, so shall the Equinoctial Circle AAeC, be by it divided into 12 unequal parts in the points *, *, *, *, &c. Then a Ruler laid unto P the Pole of the World, and those Points, shall cut the Semicircle CDA in those Points I, I, I, &c. Lastly, from the Center R, and through those Points, let there be drawn right lines, which shall be the true Hour-lines of such an Horizontal Plane, from 6 in the Morning, untill 6 at Night; but for the Hours of 4 and 5 inFig. 70. the Morning; and 7 and 8 in the Evening; they are delineated by producing 4 and 5 in the Evening, through the Center R, and 7 and 8 in the Morning; extending them out, unto the other side of the Plane, so shall you have those Hour-lines also on your Plane delineated as you see in the Figure. The Stile of this Plane may be a thin Plate of Brass, cut exactly unto the Quantity of an Angle of 51° 32', and set Per­pendicular on the Meridian line, for the forming of this Stile take out of your Line of Chords 51° 32', and set it from D, unto e, and draw Re, which shall be the Axis of the Stile, you may also prefix the Halves, and Quarters of Hours, in the very same manner as the Hours themselves were drawn.

Now to find out the distance of the Hour-lines from the Meridian, say,

As the Radius or S. 90°,

To the S. of the Latitude.

So is the T. of the Hour from Noon,

To the T. of the Hour-line, from the Meridian Line.

These kinds of Dials being so frequently usedFig. 70. with us, in this Oblique Sphere, for the help of the speedy delineating of them, I have an­nexed hereunto the Table of Longomontanus, wherein the Hour-lines, for many Latitudes, are calculated.

PROP. VI. How to draw the Hour-lines, on an Erect declining Plane.

These Planes are made to set on the sides of Houses, wherein the Meridian is always a Per­pendicular,Fig. 71. drawn on the Plane, in whose top is the Center, where the Substile, and the Hour­lines all meet.

Now before we can delineate the Hour-lines on any such Planes, two things must be given: As the Latitude of the Place, and the Planes De­clination; by having which we must find these three things: viz. The Poles height above the Plane. The distance of the substile from the Meri­dian. And the Plane's difference of Longitude.

For the finding of which Requisites, by Geo­metrical Projection, we describe on the Dial Plane, these Circles of the Sphere, viz. The Horizon, Meridian, and Equinoctial, which be­ing described in their true Position, on the Plane, we proceed thus.Fig. 71.

Admit it be required to make a Direct South Dial, on an Erect, Direct South Plane, Declining Westward 24° 20', in the Latitude of 51° 32'.

Now in order to find the requisites before mentioned, describe the Circle ZHNO, and cross it with the two Diameters ZQN, and H QO: now Z is the Zenith, N the Nadir, ZQN the Hour-line of 12, HQO the Horizon. Now seeing the Plane declines S. W. 34° 20': make Na, and Ob, each equal to 34 20: Then a Ruler layed from Z, to a, will cut the Horizon in S, the [Page 267] South point of the Horizon, through which draw the Meridian ZSN, whose Center is at Y, found as in the fourth Proposition aforegoing: Then a Ruler laid from Z to b, will cut the Hori­zon in W, the West point thereof. Now the Horizon and the Meridian being projected on the Plane, take out of your line of Chords 51° 32', which place from H, unto c, and from N, unto d; then lay a Ruler from W, unto c, and it cutteth the Meridian in P, the Pole of the World. Then through P and Q,Fig. 71. draw the line PQD, which representeth the Axis of the World, and the Substilar line of the Dial, then lay a Ruler from W, to d, it cutteth the Meridian in AE, so is W AE two points through which the Equinoctial must pass, whose Center is found as afore to be at M, (being always in the Axis of the World) so have you on your Plane the Horizon HQO, the Meridian ZPSAe N, and the Equinoctial LAeKWG, described on the Plane as required.

Now first to find the Poles height above the Plane, which in this Scheme is represented by BP, Lay a Ruler from G, unto P, and it shall cut the Plane in V, then measure the di­stance BV, on your line of Chords, and youFig. 71. will find it to contain 34° 33', which is the Poles height above the Plane.

Secondly, To find the distance of the Substile from the Meridian represented in the Scheme by the Arch ZB, or ND, which measured as afore will appear to be 18° 08', the distance of the Substile from the Meridian.

Thirdly, To find the Plane's Difference of Lon­gitude, which in the Scheme is represented by [Page 268] the Angle AEPK, lay a Ruler from P, unto AE, and it cutteth the Plane in X, then measure the Arch DX, as afore, and so will you find the Planes Difference of Longitude, to be 30° 00': Thus by Geometrical Projection have we foundFig. 71. all the three Requisites: Now to find them by Arithmetical Calculation observe these Analogies or Proportions.

1. For the Poles height above the Plane, say,

As Radius or S. 90°,

To Sc. of the Latitude 38° 28'.

So is Sc. of the Declination 65° 40',

To S. of the Poles height above the Plane 34° 33'.

2. For the Distance of the Substile, from the Meridian, say,

As the Radius or S. 90° 00',

To the S. of the Plane's Declination 24° 20'.

So is Tc. of the Latitude 38° 28',

To the T. of the Substilar Distance from the Meridian 18° 10'.

3. For the Plane's Difference of Longitude, say,

As the Sc. of the Latitude 38° 28',Fig. 71.

To the Radius or S. 90° 00'.

So is S. of the Substilar Distance 18° 10',

To the S. of the Difference of Longitude 30 Deg.

Or, it may be found thus.

As the S. of the Latitude,

To the Radius.

So is the T. of the Declination,

To the T. of the Difference of Longitude requi­red.

These things found, we come now to shew how the Hour-lines may be projected. To pro­ject which observe, First, to lay a Ruler from P the Pole of the World, to AE the Intersection ofFig. 71. the Equinoctial with the Meridian, and it will cut the Plane in x, where begin to divide the Semicircle L x G, into 12 Equal parts in the Points •, •, •, •, &c. Then lay a Ruler from Q, to every of those parts, and it shall cut the Equinoctial; and divide it into 12 unequal parts, in the points *, *, *, *, &c. Then a Ruler laid from P the Pole of the World unto each of these points, it will divide the Plane into 12 unequal parts in the Points I, I, I, I, &c. Then by a Ruler laid from the Center Q, to those points, draw right lines, which shall be the true Hour-lines proper unto such a Declining Plane, as you see plainly demonstrated by the Scheme.

Now the Substilar line falleth in this Dial, just on the Hour-line of 2, in the Afternoon, be­cause the Plane declineth Westerly. The Angle of the Stile is DQR 34° 33'. which may be either a Plate or Wyre, brought into such an Angle, which must be placed Perpendicular to the Plane, and directly over the Substilar line QD 2.

Now the distance of the Hour-lines, from the Substilar line, may also be found by this Analogy or Proportion.

As the Radius,Fig. 71.

To the S. height of the Pole above the Plane.

So is the T. of the Hour-line from the Meridian of the Plane,

To the T. of the Hour-line from the Substile.

Thus have you compleated your Dial; as you see in the Scheme, and here you may take notice that having finished a West Decliner, you have also made an East Decliner; if you only convert the Hour-lines of the West Decliner, in such manner as you see in Fig. 72. on the EastFig. 72. Decliner, and compleat all as you see in that Scheme.

Thus I have explained the making and de­lineating of the best and most usefull Dials both by Geometrical Projection, and also by A­rithmetical Calculations, in as brief and compen­dious a manner as possible. There are sundry other kind of Dials, as Incliners, Decliners, and Recliners, which being not so usefull, for bre­vity sake, they are here omitted: As for Instru­mental Dials, as Quadrants, Rings, Cylinders, &c. Which depend on the Sun's height, I refer you to Mr. Edm. Gunter's Book, wherein they are largely described.

As for the Beautifying and Adorning of those Dials, &c. by describing on them the Equi­noctial, Tropicks, Parallels of Declination, Paral­lels of the Sun's Place, Length of Days, the Sun's Rising and Setting, Jewish, Italian, and Babylo­nish Hours, Almicanthars, Azimuths, Circles of Position, the Signs Right Ascending, Descending, Culminating, &c. I do advise you to consult Mr. Gunter, Mr. Foster, Mr. Wells, and Mr. Holwel's Works, all which Authors have very learnedly shewed the describing of them, by several large Schemes, and Figures, for the plainer Illustration thereof.

Now seeing the Latitude of a Place must be first known, before a Dial can be made to it,

[Page] [Page] [Page]

Fig. 72. [Page] [Page 271] I have therefore hereunto annexed a Table of the Latitudes of all the principal Cities, Towns, and Islands, in and about Great Britain and Ire­land; so that if you are to make a Dial, for any of those parts, you may have recourse to this Table, and make your Dial to the Latitude of that place, which you find to be the nearest to the Place, for which you are to make your Dial.

PROP. I. To fortifie a Hexagon according to this Author's Proportion.

First describe the Hexagon PPP, &c, thenFig. 74. divide the Interior Polygon PP, into 1000 equal parts, take 125 for the Gorges, and set it from P to C. Then on C raise a Perpendicular, make it equal to 138 parts, for your Flanks CF, then draw the Face AF, falling on the third part of the Curtain CC, at D, and so do on every Ba­stion, untill the work is compleated.

PROP. II. To fortifie a Hexagon according to the Proportion of De la Mont.

First describe your Hexagon P, P, P, &c. Now supposing your Interior Polygon PP, 1000 parts, the Capital 333, the Gorge 200, and the Flank 150 parts, take out of your Triangular Scale Fig.Fig. 74. 75, (which is made for the more speedy de­lineation according to this proportion of De la Mont) PA for the Capital, and prick it off from PA, on all the Bastions. Then take PC, and prick off all the Gorges from P to C. Then take FC and prick it off at Right Angles, from C to F. Lastly draw all the Faces AF, AF, &c. so is your Hexagon compleat, as required.

PROP. III. To fortifie a Hexagon according to Manesson Mallet's Proportion.

Now our Authour Monsieur Manesson Mallet, in his Works intituled Travaux de Mars, de­viates from our former Authour, only in this: that as De la Mont did place his Flanks at Right Angles, he places them at 98 Degrees with the Curtains, and leaves no second Flank in all his Fortifications.

Therefore having described the Polygon PP,Fig. 74. &c. divide PP into 1000 parts, prick off the Capitals PA 333, and the Gorges PC 200, then lay off the Flanks CF, 150 parts, at an Angle [Page 282] of 98 deg. with the Curtain CC (by prop 5. §. 1. chap. 4.) and draw all the Faces, AF, AF, &c. Falling on C the point of the Flank and Curtain, so shall your Hexagon be fortified as was required.

PROP. IV. To fortifie a Hexagon according to the Emperour's Proportion.

First describe the Polygon PPP, &c. divide P P, &c. into 22 parts, take 8 for the Capitals [Page 283] PA, which prick off all round from P to A, take 5 for the Gorges; which prick off all round from C to P, then take 4 for the Flank CF,Fig. 74. which prick off all round at Right-angles from C to F, lastly draw the Faces AF, AF, AF, &c. So is the Hexagon compleated as was required.

PROP. V. How to fortifie a Hexagon according to Count Pagan's Proportion.

To delineate this Work draw a line, about the middle whereof as at M, set off MA, the half of the Exterior Polygon 500 parts, which makes the Exterior Polygon 1000, then on M (by prop. 1. §. 1. chap. 4.) raise the Perpendi­cularFig. 74. Mm, which make Mt, MT Equal to 150, then draw ATC, and ATC, then take 185, and place it from T to C, and to C, and draw CC for the Curtain, then on the points C raise Perpendiculars CF, to the line of defence CA, for the Flanks, so have you also the Faces FA. Then on the Points A set off half the Angle of the Figure, to wit 60° (as you see in the Ta­ble in page 38) and draw the lines OA and O A, so shall O be the Center of the Figure, and PC the Gorge, and AP the Capitals: then finishFig. 74. each Bastion at your own discretion, and the Work is finished as required.

PROP. VI. To fortifie a Hexagon according to the way prescri­bed by His Majesty Carolus II.

His late Majesty C. II. hath much facilitated this Work, as will appear in this following Ex­ample, by making the line of Defence, stand at Right-angles with the Flank of the Interior Polygon, by this Table, which supposes the Inte­riorFig. 74. Polygon to be 1000. Then

Now describe the Hexagon PP, &c. Then divide the Interior Poligon PP, into 1000 parts, take 367 and prick off all the Capitals PA; Then take 203 and prick off all the Gorges from P to C. Now draw the lines of defence AC andFig. 74. AC, &c. Then at C, set the Flanks at Right Angles with the line of Defence AC, so shall FC be the Flank, and FA the Faces, then finish every Bastion, and your Hexagon is for­tified as was required.

☞ Thus have I set down the several Ways and Rules, for laying the fundamental Ground­line, from the most con­siderable Inginiers ofObserve this for a ge­neral Rule in Regular Fortification. this last Age, out of all which it's most agreeable to those Authors, and to practice, to take ⅓ of the Interior Polygon for the Capital, ⅕ for the Gorge, and Flank, which leaves 6/10 for the Cur­tain, and let this be taken for a general Rule, where the Flank, and Curtain, stand at Right Angles.

PROP. VII. By the Semicircle to lay down on the Ground, any of the former Fortifications.

Having drawn the Plot of your Fort on Im­perialFig. 76. paper, or Vellom, and if it be a Regular Fort you need not describe it but two half Ba­stions from the Center, for that will be sufficient. Having such a Plate whose length is set down on each respective line, and all proper Angles expressed, will not only be usefull for laying down the Work, but for finding the Solidity of the Ramparts, Parapets, and the other Earth Works See Fig 76.

If it be in such a Place, that from the CenterCase 1. of the Fort, all the Angles may be seen, place your Semicircle at Z, and lay off all the An­gles of the Center, which here is 60°; then mark out the Diametrical lines, and making them their due length, as by your Plate they appear to be, set Piquets, on all the P, P'^s upright with the Plane, Then take up your Instrument and place a Piquet at Z. Then lock-spit out all the Polygons PP. Then mark out the Gorges CP, then set out the Flanks CF, either at RightFig. 76. Angles, or as otherwise required. Then lock­spit out the Flanks CF, and the Faces AF, ha­ving first set off the Capital PA, so is the Fort lined out for the Ground-line.

But if there be Houses and Obstacles in theCase 2. way, that from the Center all may not be seen, then must you mark out any one side and mea­sure it, and at each End set off the Angles of [Page 287] the Polygon, (which here is 120°) and draw side after side, untill all be finished: Then fi­nish the Bastions as before, and here great care must be had, or else you will run into infinite Errours.

☞ But you have liberty Experimentally to alter any of the former proportions, as you have occasion, and as will best serve the Place; as you see by the fortifying a streight lined Figure: Fig. 77. wherein CountFig. 77. Pagan's or in Manesson's way it may not be al­lowed without some alterations.

PROP. VIII. How to lay down the Profile of the Work, according to this Table.

Now to lay down this Profile draw a line of a convenient length as RSOACGD for the level or Ground-line, then by your Scale of 20, o [...] 30, at most in an Inch, representing Feet. Take out of it 70 for RA, 16 for R S, 8 [Page 290] for OA, 3 for AE, 112 for EG, 18 for GC, and 60 for CD, and mark them off on your Paper (as in Fig. 78.) at S, O, I, K, C, raise orFig. 78. let fall Perpendiculars (by prop. 1, 2, or 3. §. 1. chap. 4.) then take 16 for ST, and OQ, 12 for IF and KH, and 6 for Cf, and draw RT, TQ, QA, EF, HG, cf, fd: Then from Q set off QL 12, LV 3, QM 2, and LD 1, and raise the Perpendiculars MP 4, DZ6, and VX 1½ Then draw VX, XY, YZ, ZP, and PQ, and make the little Ditch by its measure, so is th [...] Profile perfected: as for the Faus-Bray, they ar [...] now out of use, therefore I omit them.

The Solid Content of those Earth-Works may easily be attained by the former Rules which Content being got in Feet, divide that product by 324, the Quotient shall be the Soli [...] Flores contained therein, a Flore being 18 foo [...] square and 1 Foot deep.

PROP. IX. To fortifie a Triangle, with Demi-Bastions.

This Triangle may consist and be compre­hended of three equal or unequal sides in this Example: let it be an Equilaterial Triangle PPPFig. 84. Now divide PP into three parts, then take 1, and prick off the Capitals PA, &c. and the Gorges make equal thereunto, as PC, PC, &c. then make the Flanks FC to stand at Right Angles, and to be ½ of PC or PA, then draw the Faces AF, AF, &c. and the Work is finished as required.

PROP. X. To fortifie a Square with Demi-Bastions.

The sides of the Square may be from 100 to 200 Feet, let PP be 180 Feet, which divide in­to 3 parts, take one for the Gorges PC, and forFig. 85. the Capitals PA, and prick them off all round as you see, then take ⅙ of PP, and at Right Angles prick off the Flanks CF, then draw the Faces AF, AF, &c. and the Figure is compleated.

PROP. XI. To fortifie a Parallelogram with Demi-Bastions, and Tong.

First describe the Parallelogram, or Long­Square, PPPP, then divide PP into 6 parts (the side on which the Tong, or Tenaile, is placed) and make MC equal unto ⅙ thereof, and also MG, and MH. Draw CG, GC, and CH, HC,Fig. 86. then finish the Demi-Bastions as before, so shall the Work be compleated as was required. A Long Square may also be fortified as Fig. the 77.

PROP. XII. To fortifie a Star Redoubt of 4, 5, or 6 Points

• 1. A Star Redoubt of four points may have his side from 40 to 60 Feet: First describe the Square PPPP, then divide PP into two parts at M, take ¼ of PM, (and by prop. 1. §. 1. ch. 4.)
  Fig. 87.
  [Page 300] raise Perpendiculars round at M, make MA equal to ¼ of PM, and draw all as in the Fi­gure.
• 2. A Star Redoubt of five points is thus fortified. Describe the Pentagon PP, &c. then divide PP into halves at M, raise the Perpendiculars MA, make MA equal to ⅓ of PM, and draw the Fort
  Fig. 88.
  in all respects as the Figure representeth.
• 3. A Star Redoubt of six points is thus fortified. Describe the Hexagon PPP, &c. divide PP into two equal parts at M, then raise Perpendiculars at the M'^s, then make MA equal to ½ of PM,
  Fig. 89.
  or ¼ of PP, and draw every respective line as you see in the Figure.

PROP. XIII. To Delineate a Plain Redoubt

Plain Redoubts are called Grand Redoubts, which are used as Batteries in Approaches, whose side may be from 60 to 80 Feet, or Petit Redoubt,Fig. 90. which are used for a Court of Guards in the Trenches, and may be from 20 to 50 Feet, and are framed and delineated in all respects as you see in Fig 90.

The Profile's to be set on these several Works, and the Motes, are alterable and uncertain, for they being sometimes used in Approaches; then they do require the Breast-work at the Bottom to be 7 or 8 Foot wide, and the Interior Height 6, and the Exterior 5 Feet, and the Mote to be either 8, 10 or 12 Feet, sometimes 14 or 20 Feet wide at the bottom, and the height of 7, 8 or 9 Feet, to have two, or three [Page 301] ascents to rise to the Parapet. There are many other things belonging to this Art, which the limitation I am bound to, will not permit here to be treated of.

PROP. I. To Order any number of Souldiers into a Square Battail of Men.

Admit it were required to Martial into a Square Battail 16129 Men: To doe which ex­tract the Square Root of 16129 (by prop. 8. §. 1. chap. 1.) which is 127, therefore you are to place 127 Men in Rank, and also in File.

PROP. II. To Order any number of Souldiers into a Double Battail.

Admit 16928 Men were to be Martialled into a Double Battail, extract the Root of half the number of Men; i. e. of 8464, whose Root is 92, therefore I say that 92 Men must be pla­ced in File, and 184 in Rank, to order that number of Men propounded into a Double Bat­tail.

[Page 302] [Page] [Page]

PROP. III. To Order any number of Souldiers into a Battail of the Grand Front.

Admit 16900 Souldiers were to be Martialled into a Battail of the Grand Front, that is Qua­druple. Extract the Square Root of 4225 (that is ¼ of the Men) the Root is 65; therefore I say 65 must be placed in File, and 260 in Rank, to form a Battail of the Grand Front.

PROP. IV. Any number of Men, together with their distance in Rank and File, being propounded, to Order them into a Square Battail of Ground.

Admit 2500 Souldiers were to be Martialled into a Square Battail of Ground, in such sort that their distance in File should be 7 feet, and in Rank 3 feet, and 'tis required to know how many Men must be placed in Rank and in File to draw up 2500 Men into Square Battail of the Ground. According to prop. 1. §. 1. ch. 1. say, As — 7 to 3, So is 2500 to 1071, &c. whose Square Root is 32, &c. Therefore I say 32 Men are to be placed in File. Now to find how many Men are to be placed in Rank, di­vide 2500 by 32, the Quotient is 78, which are the number of Men to be placed in Rank, and 4 Men to be disposed elsewhere.

PROP. V. Any number of Souldiers propounded, to Order them in Rank and File, according to the reason of any two Numbers given.

Admit 6400 Souldiers are to be Martialled into Array, in such Order that the number of Men placed in File, shall bear such proportion to the number in Rank as 7 to 13; (according to prop. 1. §. 1. chap. 1.) say as 7 to 13, so is 6400 to 11885, &c. whose Square Root is 109, &c. the number of Men to be placed in Rank, by which divide 6400, it produces, 58, &c. the number of Men to be placed in File, and 78 Men to be employed elsewhere.