Thus much of the description of the In­strument and its parts: Now of the use thereof.

THe use of this Instrument is very great and various, but it consisteth [Page 5]especially in the resolution of all sorts of Rightlined Triangles, unto which all Geometricall conclusions are reducible, and in Arithmeticall proportions or the resolution of the Golden Rule.

In the resolution of Triangles, 4 things are considerable. 1 The quantity of the Angles. 2 The proportions of the sides or subtendents and perpendicular betwixt themselves. 3 The contents of the area in parts correspondent to the parts of the sides. 4 The reduction of those parts to a perpendicular and basis of another de­nomination: all which are performed by the Instrument in this manner.

To finde the quantity of any angle.

IF it be in Planimetry, having three sights placed on the Instrument, the first at the center of the left hand label, the second at the center of right hand label, & the third at the end of the right hand label, work thus. Lay the Instrument upon a Geo­metricall [Page 6]tripos or staffe, as the Semicircle or Circumferentor is usually laid, by those that measure Land, horizontally, then turne it till you can behold one of the markes through the sights at the center of the labels, afterward turne the right hand label till you see the other marke through the sight at its center, and that which is at the end thereof, and then looking on the Tangent line, you shall see the angle among the Tangents, where the intersection is of the label and the Tangent.

If the angle be above 90 gr. then you must have foure sights, and behold the first marke through the sights at the cen­ter, and the end of the left hand labell, and leaning the Instrument at that posi­tion, turne the right hand label till you can also behold the second mark through its sights, and then the angle which the two labels make where they crosse each other shall be the angle sought, which that you may know how great it is, adde together the angle found in the Tangent cut by the right hand label, and that on the Quadrant cut by the left hand, label and substract [Page 7]them both out of 180 gr. and you have the angle required.

In Altimetry work thus, set the Instru­ment on one edge upright, so as that the lower lamina, containing the distance betwixt the centers of the two labels, may stand horizontall, and the right and left hand laminae Perpendicular; then lift up the right hand label, till, through the sights thereof, you can behold the object, and then in the Tangent line shall you see the angle. This is of use to find the altitude of Sun, Moon [...]or Stars, as also the altitude of any Tower, Steeple, Hill, &c.

Two angles being known, to represent any Rightlined Triangle.

TRiangles of all sorts are distinguished two waies.

First, By their sides, and so they are said to be. 1 Equilaterall, having all sides a like. 2 Equicrurall having only [Page 8]two sides equall; and the third unequall. 3 Scalene, viz. Triangles neither of whose sides are equall.

Secondly, By their angles, and they are, 1 Rectangular containing one Right-angle. 2 Obtuseangular wherein one of the angles is obtuse or greater then a Rightangle, that is above 90 gr. 3 A­cuteangular, wherein all the angles are lesse then 90 gr.

Note that the three angles of every Triangle are equall to two Right angles, that is 1 80 gr.

To represent any Rectangular Triangle.

TUrne back the left hand label to the lamina of the Tangent, for then it maketh a right angle with the Radius, then turne the right hand label to the degree of the other known angle in the Tangent line, and so the two labels and the Radius shall give the Triangle, toge­ther [Page 9]with the proportions of the sides betwixt themselves.

As suppose, the two known angles be 90. and 45. the distance betwixt the cen­ters (1) the Radius will be 100 parts, the right hand label shall cut the left hand label also in 141 parts.

The one of these sides is the Radius, the second a Tangent of 45. equall al­ways to the Radius, the third, the Secant of 45. being a line drawn from the center through the limbe of the circle till it meet the tangent; if one of the known angles be above 45: and the other 90 gr. then place the labell at the tangent of its com­plement to 90. and then shall the parts on the labell on the left hand represent the Radius, and the Radius of the Instru­ment shall represent the tangent, and the labell on the right hand shall repre­sent the secant.

Thus supposing the known angles to be 90. and 63 gr. 30′. because if I put the right hand labell to 63 gr. 30′ it will not crosse the left hand labell, I therefore [Page 10]take its complement (1) 26 gr. 30′. and applying the right hand labell unto this degree in the Tangent, the Triangle comprehended, as in the former example, shall be the Triangle required; and the parts discovered in every intersection or angle shall shew the proportion of the Ra­dius, Tangent, and Secant, as here the Radius that is represented by the label on the left hand, shall be 500, the Tan­gent represented by the Radius of the In­strument 1000, and the Secant represent­ed by the right hand label 1120 parts, which if reduced to a Radius of 1000, the proportion holds thus, the Radius 1000, the Tangent 2000, the Secant 2240, which agreeth to the tables of Tangents and Secants, unto a Unit, which is as neer as can be expected by Instrument.

If one of the known angles be lesse then 45, and the other 90 gr. then you need no more but apply the right hand label to the degree of the Tangent line, and the Radius of the Instrument shall be the Radius or bases of the Triangle, and [Page 12]the proportions of the other sides will be shewed as abovesaid.

To represent any obtuse angular Triangle.

AS suppose the two known angles be 60 & 100 gr. this triangle must needs be obtuse anglar, because 100 gr. is more then 90 gr. & because neither of the labels can make an angle more then 90 gr. therefore the obtuse angle must be found in the intersection of the two labels; there­fore to find out this triangle work thus, add the two known angles together, and they will make 160: substract this out of 180, and there remaineth 20, which is the third angle; therefore to represent this, put one labell in 20 gr. and the other in 60 gr. and where they crosse each other, the intersection shall be an angle of 100 gr. because all three angles must make ex­actly 180 gr. and the labels thus applied give you the Triangle required, and the [Page 12]divisions of the labells in their intersecti­on together, with the Radius shew you the proportions of the three sides thereof, as in the former examples.

To represent any Acuteangular Triangle.

THis is to be done only by applying the labels unto the degrees of the angles known, both in the tangent line and quadrant.

But if any of the angles be above 60 gr. then it must be supplied, by putting one of the labels to the complement of both added together, unto 180 gr. as suppose two known acute angles be 60 and 80: the labels being applied to those degrees in the Tangent and Quadrant, will not meet to crosse each other; there­fore put the one label in 60, and the other in 40 gr. which is the complement of the sum of both 140 to 180: and so you shall have the Triangle required, the pro­portion [Page 13]of whose sides is to be found out as aforesaid.

To find the Content of any of these Triangles.

TO effect this, the perpendicular must be known, which being multi­plied into half the basis, the product shall give the Area.

I have shewed before how to find the proportions of every side, and so by con­sequence the length of the basis; with as great facility is the perpendicular also to be found: for when you have applied the labels unto each other, so as that the Triangle is by them represented, doe but cast your eye directly into that intersecti­on, and under the labels amongst the pa­rallel lines; you shall find the distance from the basis, or the length of the per­pendicular: which multiplyed as afore­said, shall give the Area.

Or to work by the Instrument, I say thus.

As 1 to the Multiplicator,

So the Multiplicand, to the Product.

Supposing then the perpendicular to be 5, and the basis 10: whose half is 5, that is to be multiplyed by the perpendi­cular.

I apply 10 in the left hand labell, (which also may be reckoned for 1 or 100 or 1000: as occasion serveth) unto 5, a­mongst the parallels, then looking for 5, upon the said labell, I finde it directly meeting with 25 amongst the parallels, which is the number sought.

And thus may any number be multi­plied, only remembring to observe that the same Figures and divisions stand sometimes for units, sometimes for de­cads, 100. 1000, &c. as occasion serveth. So also working by the contrary way, may any be divided.

For, as the divisor is to 1;

So the dividend, to the quotient.

Therefore apply the number belong­ing to the divisor on the labell to 1, a­mongst the parallels, and then over a­gainst [Page 15]the number of the dividend in the labell, shall be the quotient amongst the parallels. As if I would divide 9 by 3; here 3. is the divisor, and 9 the dividend.

I therefore bring down the 3: upon the labell, till it come exactly upon the pa­rallel of 1: and then over against 9: which was the dividend, I find the pa­rallel marked with the figure 3, which is the quotient

For 3. is contained in 9. three times.

But here is to be noted, that it some­times falleth out more convenient in working proportions on this Instrument to find the first number amongst the pa­rallels, and the second upon the label, and then the third upon the parallels, will exactly answer to the 4^th or quotient upon the labell.

To reduce the Area of a Trian­gle found of one denominati­on, to a perpendicular and ba­sis of another denomination.

THe reason of the inserting of this proposition is, because that in measu­ring of Triangles, whose Area we would finde either in Acres, Perches, Yards, Feet, or Inches; it seldome falleth out that the perpendicular or basis measured, giveth just the same number as it doth upon the Instrument: yet notwithstan­ding the Triangles and their sides hold proportion unto each other; and therefore there is a necessity of working by pro­portion, to reduce the one into the o­ther.

As first, supposing we have a rectangle equicrurall Triangle containing 90.45, 45 gr. in its angles on the Instrument, the basis will be 100, and the perpendi­cular [Page 17]100. which perpendicular multipl [...]y­ed into half the basis, giveth 5000 for the Area. Now suppose again, that by measure we find the basis to be but 30, in a Trian­gle consisting of such angles, I say then.

So 100 the perpendicular on the In­strument, to 30 the perpendicular inquired

Which being multiplyed into 15. half the basis, giveth the Area 450.

Or 2. supposing the Triangle be sca­lene, having all its sides unequall, whose angles shall be 30 gr. 40 gr. and 110 gr. upon the Instrument the basis will be found to be 100, and the perpendicular about 35.

The Question is what the perpendi­cular shall be if the bases of such a Tri­angle being measured shall be 80 perches.

Bring down the left hand label there­fore [Page 18]till the figure of 10 come just to the parallel of 8, and then just over against 35 on the label shall be found amongst the parallels, the perpendicular required, (viz.) 28; by which multiply halfe the bases, and the Area will be found 1120.

For, As 1. Is 28.

So is 40.61 halfe the bases, to 1120.

Bring down therefore the figure of 10 upon the label, (which here standeth but from 1) to 28 amongst the parallels, and over against 4. or 40. on the label shall be found amongst the parallels, the num­ber aforesaid 1120; which in a large In­strument may easily be discerned.

And in this manner may most Arith­meticall Proportions be found out, and Questions of the Golden Rule resolved; as also Lines and Numbers may be in­creased in continuall proportion. And Mr. Gunter's Canons for Land measure, to find the contents of all oblong Super­ficies in Perches, Chaines, or Acres, may be made use of and resolved by the In­strument.

Thus hath been shewed its use in the re­solution of Triangles, and in working proportions.

BEsides it is appliable to many other Mathematicall conclusions, as to find the length in parts of any Line of Signes, Tangents, Secants, and Cords, and so by consequence is of great use in the Pro­jection of Spheares, describing of Dials, whether Sciotericall, or Instrumentall.

To find the length of the Tan­gent line of any degree, to a Radius of 10000.

TUrne backe the left hand label to the left hand lamina, so as that it may make a Rightangle with the Radius, then turne the right hand label to the degree required in the Tangent line, and marking the place of its intersection with the other label, you shall there see amongst the di­visions of the left hand label the number of parts that such a Tangent must be of.

As suppose the degree be 30, whose Tangent you would find; the labels being applyed unto each other, as aforesaid, the parts intercepted in the left hand label, betwixt the right hand label, and the Ra­dius, shall be 5773, which must be the length of that Tangent.

But note, that if the Tangent be above 45 gr. then the right hand label must be applyed to the complement thereof, and then the parts intercepted shall resemble the Radius, and the Radius of the Instru­ment shall represent the Tangent, and the proportion that the intercepted parts beare unto the Radius of the Instrument 10000, the same proportion shall the Radius of the Instrument 10000 beare unto the Tangent required.

As suppose you would find the Tan­gent of 60, apply the right hand label un­to its complement, that is the Tangent of 30 gr. and you shall find the parts inter­cepted, that represent the Radius of 60 gr. 5773. and the Radius of the Instru­ment that doth represent the Tangent 10000. I say then.

As 5773 the Radius found, to 10000, the Radius of the Instrument.

So 10000 the Radius of the Instru­ment, to 17320 the Tangent of 60 gr.

And thus by proportion, may be found the length of the Tangent line for any degree even unto 90 gr.

NOw then having known the length of the Tangent line in parts, it is ea­sie (as is commonly known) to make Scales and prick down a Tangent line to a Cir­cle of any Semidiameter, which when it is done, the use thereof is so commonly known, that I need speak nothing of it, but refer to Mr. Gunter and others that have written largely of its use.

To find the length of a Secant for any degree.

THis is to be done just in the same manner as in the former proposition the Tangent was to be found, only with this difference, that whereas the parts in­tercepted betwixt the intersection of the [Page 22]labels and the Radius of the Instrument on the left hand label were to be reckoned for the Tangent, or else for a Radius un­to the Tangent represented by the Ra­dius of the Instrument: here the Secant is always to be found by observing the number of parts intercepted betwixt the center of the right hand label, and its in­tersection with the other label. This in the first of the two former examples, the Radius being found 10000, the Tangent 5773, the Secant shall be .11547, which is a Secant of 30 gr. but if the degree be above 45: then must we work by propor­tion as we did in the last, & the proportion will hold thus. As the parts intercepted on the left hand labell, to the parts inter­cepted on the right hand label. So 10000 the Radius of the Instrument, to the parts of the Secant required. Thus in the forementioned example, the parts on the left hand label are 5773, on the right hand label 11547. I say then, As 5773, to 11547. So is 10000 to 20000.

Bring down therefore 5773 on the [Page 23]left hand label, to 11547 amongst the pa­rallels in the quadrant; and then 10000 upon the left hand labell will meet exact­ly with 20000 the secant of 60 gr. whose complement by which the worke was wrought is 30 gr.

The use of this line in Trigonometry and Horography, I leave to be searched out in their Books, that have written largely of the use thereof.

To find the length of a line of sines for any degree.

THere is no more required to find out this, but to look the degree in the quadrant; then observing amongst the pa­rallels, just meeeting with the degree: you shall find there the parts required. As if I would know the length of the Sine of 30 gr. casting my eye upon the arch of the circle, where it is cut with the degree of 30: I find meeting exactly in that intersection the parallel of 5: or [Page 24]5000. I conclude then, if the Radius be 10000. the sine of 30 gr. must needs be 5000.

The use of this line I also referre to the writings of others, because it is so com­monly treated of, and known of all.

To find the length of a line of Chords for any degree.

THis line so often made use for the di­viding circles, or finding the quan­tity of any arch or angle, may be found with as great facility and delight as the three former; indeed it may be done with­out any more trouble, by doubling the sine of halfe the arke, as if we would know the Chord of 60 gr. halfe its arke is 30 gr. the sine whereof by the former proposition, as 5000. which doubled, giveth 10000, the Chord required.

But it may be found with great delight and pleasure by the instrument another way: as followeth.

Apply the edge of the right-hand la­bel, [Page 25]exactly to the degree in the limbe of the quadrant; whose Chord you desire, and the parts intercepted betwixt that degree and the center of the label, are the Chord of that degree: thus in the former exam­ple, the labell being applied to the 60 gr. the parts intercepted shal be exactly 10000, that is the radius, if applyed to the 30 gr. the parts intercepted will be 5176: which is the Chord of 30 gr. directly dou­ble to the sine of 15 gr. 2588. which is half the arke.

To find a line or number in con­tinuall proportion, having two numbers given

SUppose the numbers be 4: and 8: I say then.

As 4. to 8. So 8. to 16. So 16. to 32. &c.

Bring down 8. on the left hand labell till it meet with 4 amongst the parallels, then against 8 amongst the parallels, shal be found 16, on the label, & against 16 a­mongst the parallels shal be found 32 on the labell, and so forward.

Having the Square and its Root to find the Cube there­of.

FOr as the Root is to the Square,

So is the Square unto the Cube.

As suppose, the Root be 3: and the Square 9: the Cube will be found to be 27.

work this.

Apply 9, on the left hand labell to 3: amongst the parallels, and then 9, a­mongst the parallels shall meet exactly with 27: the Cube required.

Having the Cube and its Square whereof it was made to find the Root of both.

THis is to be wrought by working di­rectly contrary to the former pro­position.

For.

As the Cube is to the square, so is the [Page 27]square unto the root, apply therefore the number, standing for the cube and square together, finding the one on the label the other amongst the parallels, and then just over against the number for the square shall stand the root. Always remembring that if the first number be found upon the label, the third number must then be searched there also: but if it be amongst the parallels then the third must be a­mongst them to, for the third number in these proportions is always to be found on that line or Scale where the first was; and the fourth on that line where the se­cond was: whether it be on the label or amongst the parallels.

Many other things might be wrought by this Instrument; I have only given hints to those that are more curious and have greater leasure to follow these stu­dies, who I doubt not, but by their in­dustry will be able from what I have said to draw new conclusions, and from many pleasing propositions conducing much to the satisfaction of those that are ingeni­ous.