PROBLEME. 1. Two numbers being given, to find a third Geometrically propertional unto them, and to three a fourth, and to four a fifth, &c.

GEometrical proportion is when divers numbers being compar'd together, differ among themselves, increasing or decreasing, after the rate or reason of these numbers, 2.4 8.16.32. for as 2 is half 4. so is 8 half 16. and as this is continued, so it may be also discontinued, as 3.6.14.28; for though 3 is half 6 and 14 half 28. yet 6 is not half 14, not in proportion to it as 3 is to 6: there is also Arithme­tical, and Musical proportion; but of that in other more large discourses, being not material to our present purpose (though I may hint it after­ward.)

To find this by the numbers, ex­tend the Compasses upon the line of numbers, from one number to ano­ther, this done, if you apply that ex­tent (upwards or downwards, as you would either increase or diminish) from either of the numbers propoun­ded, the moveable point will stay on the 3d proportional number required. Also the same extent applyed the same way from the third, will give you a fourth, and from the fourth a fifth, &c.

PROB. 2. One number being given to be multi­plyed by another number given, to find the product.

Extend the Compasses from 1 to the Multiplicator, and the same ex­tent applyed the same way from the Multiplicand, will cause the movea­ble point to fall on the product.

PROBLEM 3. Of Division.
One number being given to be divided by another, to find the Quotient.

Extend the Compasses from the [Page 31]Divisor to 1, and the same extent will reach from the Dividend, to the Quo­tient, or extend the compasses from the Divisor to the Dividend the same extent shall reach the same way from 1 to the quotient; as for example;

Let 750 be a number given to be divided by 25 (the Divisor) I extend the Compasses downward, from 25 to 1, then applying of that extent the same way from 750, the other point of the Compasses will fall up­on 30, the Quotient sought; or you may say, as 25 is to 750, so is 1 to 30.

2. Let 1728 be given to be divi­ded by 12, say as 12 is to 1, so is 1728 to 144. Extend the Compasses from 12 to 1, and the same extent shall reach the same way from 1728 to 144. or as before, as 12 to 1728. so 1 to 144.

3. If the number be a decimal fraction, then you work as if it were an absolute whole number; but if it be a whole number, joyned to a de­cimal fraction, it is wrought here as [Page 32]properly as a whole number, example, [...] would divide 111.4 by 1.728. ex­tend the Compasses from 1. 728. to 1, the same extent applyed from 111 4. shall reach to 64. 5. so again 56. 4. being to be divided by 8. 75. the Quotient will be found to be 6. 45.

Now to know of how many Fi­gures any Quotient ought to cosist, it is necessary to write the Dividend down, and the Divisor under it, and see how often it may be written un­der it; for so many figures must there be in the Quotient, as in Dividing this number 12231, by 27. according to the rules of Division. 27. may be written 3 times under the Dividend: therefore there must be 3 Figures in the Quotient; for if you extend the Compasses from 27 to 1, it will reach from 12231, to 453. the Quotient sought for.

Note, That in this, and also in all other questions, it is best to order it so, as that the Compasses may be at [Page 33]the closest extent; for you may take a close extent more easily and exactly, than you can a large extent, as by ex­perience you will find.

PROB. 4. To three numbers given to find a fourth in a direct proportion, (or the rule of 3. direct.)

Extend the Compasses from the first number to the second; that done, the same extent applyed the same way from the third, will reach to the fourth proportional number required.

PROBL. 5. Three numbers being given to find a fourth in an inversed proportion, (or the back Rule of 3.)

Extend the Compasses from the first of the numbers given, to the second of the same denomination, if that distance be applyed from the third number backwards, it shall reach to the fourth number sought.

If 60 pence be 5 shillings, how much is 30 pence? facit 2.5. two shillings, five tenths of a shilling, that is, being reduced, 2 shillings 6 pence.

Again, If 60 men can raise a Brest­work of a certain length and breadth in 48 hours, how long will it be ere [Page 35]40 men can raise such another? Ex­tend the Compasses from 60. to 40. numbers of like denomination, viz. of Men; this done, that extent apply­ed the contrary way from 48. will reach to 72, the fourth number you look for. Therefore I conclude that 40 men will perform as much in 72 hours as 60 men will do in 48 hours.

Note, That this back Rule of 3, may for the most part be wrought by the direct Rule of 3.

If you do but duely consider the order of the Question, for you must needs grant that fewer men must have longer time, and the contrary there­fore the answer must be in proportion to the question, which might have been wrought thus as well. The Ex­tent from 40 to 60 will reach the same way from 48 to 72 in direct proportion: Or contrarily, as 60 to 40, so is 72 to 48. which you see is but by Turning the question to its di­rect opperation, according to the true reason of the question.

Thus you have the way for the di­rect and reverse rule of 3: for the double rule of 3, and Compound rule of 3, this is the rule for it.

Alwayes in the double rule of 3, 5 terms are propounded, and a sixth is required, 3 of which terms are of supposition, and two of demand; now the difficulty is in placing them, which is best done thus; as in this Example.

If 5 Scholars spend 20 l. in 3 months, How many pounds will serve 9 Scholars for 6 months?

Note here the terms of suppositi­on are the first three, viz. 5, 20, and 3, and the terms of demand are 9 and 6.

Then next for the right placing them observe which of the terms of Supposition is of the same denomina­tion with the term required, as here the 20 l. is of the same denomina­tion with the how many pounds re­quired. Set that always in the second place, and the two terms of suppo­sition one above another in the first [Page 37]place, and the terms of demand one above another in the last place thus,

5-20-9

3—pounds—6

Then the work is performed by two single rules of 3 direct, thus:

Extend the Compasses from 5 to 20. the same extent applyed the same way from 9 shall reach to 369 a 4th. this is the first operation: Then as 3 to 36: the 4th. so is 6 to 72 the number of pounds required. By the line of numbers the double rule is wrought as soon as the compound: therefore I shall wave it now:

Four Questions and their Answers, to shew the various forms of working on the line of numbers.

1. If 12 men taise a frame in ten days, in how many days might 8. men raise the like frame? Reason tells me, that fewer men must needs have longer time; therefore the work is thus, as 12 is to 8, so is 10, to 15, by the last Rule, or 8 to 10, so is 12 to 15.

[Page 38] 2. If 60 yards of stuff, at 3 quar­ters of a yard broad, would hang about a certain room; how many yards of stuff of half a yard broad, will serve to hang about the same room? the work is thus, as 510^th to 710^th ½ so is 60, to 90, or as 75 to 5, so is 60 to 90, wrought backwards.

3. If to make a foot superficial, 12 inches in breadth, do require 12 in­ches in length, the breadth being 16 inches, how many inches in length must I have to make a foot superfi­cial? the work is thus, as 16 is to 12, so is 12 to 9, the number of inches to make a foot.

4. If to make a foot solid, a base of 144 inches require 12 inches in height, a base being given of 216 in­ches, how much in height makes a foot solid? the work then is, as 216 is to 144, so is 12 to 8. or otherwise thus, as 12 is to 216, so is 144 to 8. the height sought.

PROB. 6. To three numbers given to find a fourth in a doubled proportion.

This Problem concerns questions of proportions between lines and su­perficies: now if the denomination of the first and second terms be of lines, then extend the Compasses from the first term to the second, (of the same kind or denomination,) this done, that extent applyed twice, the same way from the third term, the move­able point will stay upon the fourth term required.

PROBL. 7. To three numbers given to find a fourth in a triplicated proportion.

The use of this Problem, consist­eth in questions of proportion, be­tween lines and solids, wherein if the first and second terms have denomi­nation of lines, then extend the Com­passes from the first to the second, that extent applyed three times from the third, will cause the moveable point to stay on the fourth proportional re­quired.

PROB. 8. Betwixt two numbers given to find a mean arithmetically proportional.

This may be done without the help of the line of numbers: nevertheless, because it serves to find the next fol­lowing, I shall here insert it, though I thought to pass both this and the next over in silence, yet to set forth the excellency of number, I have set them down; and the Rule is this, Add half the difference of the given terms to the lesser of them, and that aggre­gate (or sum) is the Arithmetical mean required.

PROB. 9. Betwixt two numbers given to find a mean musically proportional.

Multiply the difference to the [Page 44]terms by the lesser term, and add likewise the sa [...]e terms together; this done, if you divide that Pro­duct by the sum of the terms, and to the Quotient add the lesser term, that last sum is the Music [...] mean required: or shorter thus,

Multiply the terms one by another, and divide the Product by their sum, and the Quotient doubled is the Mu­sical mean required.

PROB. 10. Betwixt two numbers given to find a mean Geometrically proportional.

Divide the space on the line of numbers, between the two extreme numbers, into two equal parts; and the point will stay at the mean pro­portional required. So the extreme numbers being 8 and 32, the middle point between them will be found to be 16.

PROB. 11. Betwixt two numbers given to find two means Geometrically proportional.

Divide the space between the two extreme numbers into 3 equal parts, and the two middle points, dividing the space shall shew the two mean proportionals. As for example, let 8 and 27, be two extremes, the two means will be found to be 12, and 18, which are the two means sought for.

PROB. 12. To find the Square root of any number under 1000000.

The Square root of every number [Page 46]is always the mean proportional be­tween 1, and that number for which you would find a square root; but yet with this general caution, if the fi­gures of the number be even, that is 2, 4, 6, 8, 10, &c. then you must look for the unit (or one) at the beginning of the line, and the number in the se­cond part, and the root in the first part, or rather reckon 10 at the end to be the unit, and then both root and square will fall backwards to­ward the middle in the second length or part of the line: but if they be odd, then the middle one will be most convenient to be counted the unity, and both root and square will be found from thence forwards to­ward 10. so that according to this rule, the square of 9, will be found to be 3, the square of 64, will be found to be 8, the square of 144, to be 12. the square of 1444, to be 38. the square of 57600; to be 240. the square of 972196, will be found to be 986. and so for any other number. [Page 47]Now to know of how many figures any root ought to consist, put a prick under the first figure, the third, the fifth and the seventh, if there be so many; and look how many pricks, so many Figures there must be in the Root.

PROB. 13. To find the Cubique Root of a Number under. [...]

The Cubique root is always by the first of two mean proportionals be­tween 1 and the Number given, and therefore to be found by dividing the space between them into three equal Parts: So by this means the root of 1728 will be found to be 12, the root of 17280 is neer 26, the root of 172800 is almost 56, although the point on the Rule representing all the square numbers is in one place, yet by altering the unit it produceth various points and numbers, for their respe­ctive proper roots. The Rule of find [Page 48]which is in this manner: You must set (or suppose pricks to be set) pricks under the first figure to the left hand, the fourth figure, the seventh and the tenth; now if by this means the last prick to the left hand shall fall on the last figure, as it doth in 1728, then the unit will be best placed at 1 in the middle of the Line, and the Root, the Square and Cube will all fall forward toward the end of the Line.

But if it fall on the last but 1, as it doth in 17280 then the unit may be placed at 1 in the beginning of the Line, and the Cube in the second length, or else the unit may be pla­ced at 10 in the end of the Line, and the Cube in the first part of the Line, you may help your self, as in the first Problem of the 2 Chapter.) But if the last prick fall under the last but two, as in 172800, it doth then place the unit always at 10 in the end of the Line, then the Root, the Square, and Cube, will all fall backward, and be found in the second part, between [Page 49]the middle 1 and the end of the Line. By these Rules it doth appear that the Cube root of 8 is 2, of 27 is 3, of 64 is 4 of 125 is 5, of 216 is 6, of 345 is 7, of 512 is 8, of 729 is 9, of 1000 is 10. As you may see by this following Table of Square and Cu­bique roots.

Thus you have the chief use of the line of numbers in general, and they that have skill in the rule of three and a little knowledge in plain triangles, may very aptly apply it to their par­ticular purposes, Yet for their sakes for whom it is intended, I shall in­large; to some more particular appli­cations in measuring all sorts of Su­perficies, and Solids; wherein I do judge it will be most serviceable do them that be unskilful in Arithme­tick, as before said.

PROB. 1. The breadth of an Oblong superficies given in Foot-measute, to find how much in length makes a Foot.

Extend the Compasses from the breadth to 1, the same extent applyed [Page 53]the same way from 1, will reach to the length required.

So the breadth being 8 tenths, or 0, 80, the length to make a Foot Su­perficial will be found to be 1. 25. Or shorter thus, as 8 tenths (or 80 of a 100) is to 1, so is 1 to 1, 25 of an hundred.

PROB. 2. Having the length and breadth of any Superficies given in Foot-measure, to find the content of that Superficies in Foot-measure.

Extend the Compasses from 1 to the breadth, the same extent applyed the same way from the length, will reach to the Content.

PROB. 3. Having the breadth and length of an Oblong Superficies given in Inches, to find the content in Inches.

As 1 Inch to the breadth in Inches, so the length in Inches to the Con­tent in square Superficial Inches.

So the breadth 30 Inches, and the length 183, the Content will be found to be 5490.

Or else, as 1 to 183, so is 30 to 5490 Inches.

PROB. 4. Having the breadth and length of an Oblong superficies given in Inches, to find the Content in feet.

As 144 the number of Inches in one Foot, is to the breadth in Inches; so is the length in Inches unto the Con­tent in Feet. So as 144 to 30, so is 183 to 38, 250, that is, to 38 Foot and a quarter.

PROB. 5. Having the breadth of an Oblong su­perficies given in Inches, and the length in Feet and parts, to find the Content in Feet, and such like parts as the length was.

As 12 to the breadth in Inches, so is the length in Feet to the Content in Feet.

As 12 unto 30, so is 15 to 37, 50.

PROB. 6. Having the breadth in Inches, to find how much makes a Foot in Inch-measure, (that is, how many Inches makes a Foot.)

As the breadth in Inches to 144, so is 1 to the length in Inches. As 30 to 144 so is 1 to 48 Inches, or as 12 to the breadth in Inches, so is 12 the con­trary way to the number of Inches; for if you extend the Compasses from 12 to 6, that extent applyed the con­trary way from 12 shall reach to 24 the Inches required.

PROB. 7. Having the length and breadth of an Oblong superficies, to find the side of a square equal to it.

Divide the space between the length and the breadth into two equal parts, and the middle point shall shew the side of the Square that shall be e­qual in area, or quantity, to that Ob­long; so that a Square made of 11, 32, is equal to an Oblong of 16 one way, and 8 the other way.

PROB. 8. Of a Circle. Having the Diameter of a Circle, to find the side of a square equal to that circle.

As 10000 to 8862, so is the Di­ameter 15 to the side of the Square 13, 29, that is equal to the Circle.

PROB. 9. Having the Circumference of a Cir­cle, to find the side of a square equal to the same Circle.

As 10000 to 2821, so is the Cir­cumference 47, 13 to the side of the Square 13, 29 equal to the Circle.

PROB. 10. Having the Diameter to find the Cir­cumference.

As 1 is to the Diameter, so 3142 to the Circumference. Or, as 7 to 22, so is the Diameter to the Circumfe­rence: or as 113 to 355.

So the Diameter being 15, the Cir­cumference wil be about 47 13 parts

PROB. 11. Having the Circumference to find the Diameter.

As 3142 is to 1, so is the Circum­ference to the Diameter: Or, as 22 is to 7, so is the Circumference to the Diameter: Or as 355 to 113.

So the Circumference being 47, 13. the Diameter is 15.

PROB. 12. To find the side of a square that may be inscribed within a Circle, by having the Diameter.

The Extent from 1 to 7071 will reach from the circumference to the side of the Square required. So the Diameter being 15, the side of the square inscribed will be 10, 60.

PROB. 13. By having the Circumference, to find the side of the inscribed square.

The extent from 1 to the Circumfe­rence, will reach from 2251 to the side of the Square required.

So the Circumference being 47 13 the side of the inscribed square will be 10.60 as before.

PROB. 14. Having the Diameter to find the Su­perficial content of a Circle.

The extent from 1 to the Diameter, being twice repeated (the same way) from 7854, will reach to the content required.

PROB. 15. Having the Circumference to find the Superficial content of a Circle.

The extent from 1 to the Circum­ference, being twice repeated from 07958, will reach to the content.

PROB. 16. Having the content to find the Dia­meter.

The extent from 1. to 1273, will reach from the content to another [Page 59]number, whose square root is the Di­ameter required.

PROB. 17. Having the content to find the Cir­cumference.

The extent from 1 to 12, 75, will reach from the content to another number, whose square root is the Cir­cumference required.

PROB. 18. Having the content of a Circle to find the side of a square equal to it.

Extract the square root of the con­tent by the 12 Probleme of the last Chapter, and that is the side required.

PROB. 19. To find the content of a Circle two ways.

Multiply the Diameter by it self, and multiply that product by a 11, and divide this last product by 14, and the quotient shall be the content re­quired.

PROB. 20. How to measure a Circle, a Semicircle, or a quarter of a Circle, or any part that goeth to the Centre of the sup­posed Circle.

PROBL. 21. How to measure a Triangle.

Take half the base, and the whole perpendicular; and work with them two, as if it were an Oblong square figure, or you may take the whole base and half the perpendicular.

PROBL. 22. How to measure a Rhombus, or a Rhomboides.

A Rhombus is a Diamond-like fi­gure, as a quarrie of glass is, contain­ing 4 equal sides and two equal oppo­sit angles; but a Rhomboides is a figure made of two equal opposite sides and two equal opposit angles: and to measure them you must take any one side, and the nearest distance from that side to his opposite side for the other side, and then reckon it as an Oblong square.

PROB. 23. How to measure a Trapezium.

A Trapezium is a figure compre­hended of 4 unequal sides, and of 4 unequal angles: and before you can measure it, you must reduce it into two triangles, by drawing a line from any two opposite corners, then deal with it as two triangles; or you may save some work thus, the line you draw from corner to corner, will be the common base to both triangles: [Page 62]then say as 1 is to half the perpendi­culars of both the triangles put toge­ther, so the whole base to the con­tent.

PROB. 24. How to measure a many-sided irregu­lar figure or Polygon.

You must reduce it into triangles, or to trapeziums, by drawing of lines from convenient opposite corners: and then the Work is all one with that of the last Problem.

PROB. 25. How to measure a many sided regular figure, commonly called a regular Po­lygon.

Measure all the sides, and take the half of the sum of them for one side of a Square, and the neerest distance from the center of the Polygon to the middle of one of the sides, for the other side of a Square; and with them two numbers work as if it were a Square Oblong figure, and it will give the Content of the Polygon desired.

PROB. 26. How to reduce Feet into Yards, Ells, or other parts.

First for yards, if 9 foot make one yard, how many shall 36 foot make.

The extent from 9 to 2, will reach from 36 to 4, for so many yards is 36 foot. But if you were to measure any quantity by the yard, as the Plai­stering or Painting of a House, then I would advise you to have a yard to be divided into a hundred parts, (which is as neer as commonly Workmen go to, or else, into a 1000, if you do re­quire more exactness) and measure all your lengths and breadths with that, and set them down thus, 2. 25 (or by 1000 thus 2. 250) and the length thus 10.60, and multiply them toge­ther, and the product is the true con­tent of that long square, the like holds for ells, or poles, furlongs, or any o­ther kind of measure. Again, for a yard in length, if 3 foot make one yard, then what shall 30 make? it maketh 10, or the contrary if 10 yards make [Page 64]30 foot, what shall 12 make? the ex­tent from 10 to 12, will reach from 30 to 36 foot, but if it be given in feet inches, then say as 9 to the breadth so is the length in feet and inches (or decimal parts) to the content in yards required.

PROB. 1. Having the breadth and length of an oblong Superficies given in perches, to find the content in perches.

As 1 perch to the breadth in per­ches, so the length in perches to the content in perches.

PROB. 2. Having the length and breadth in per­ches, to find the content in square acres.

As 160 to the breadth in perches, so is the length in perches to the con­tent in acres.

As 160 unto 30, so is 183 to 34, 31 (in acres and perches.)

PROB. 3. Having the length and breadth of an Oblong superficies given in Chains, to find the Content in Acres.

It being troublesome to divide the Content in Perches by 160, we may measure the length and breadth by Chains, each Chain bein 4 perches in length, and divided into 100 links, then will the Work be more easie in Arithmetick, or by the Rule; for as 10 to the breadth in Chains, so the length in Chains to the content in Acres.

PROB. 4. Having the Base and Perpendicular of a Triangle given in Perches, to find the Content in Acres.

If the Perpendicular go for the length, and the whole base for the breadth, then you must take half of the oblong for the content of the tri­angle, by the second probleme, as 160 to 30, so is 183 to 34.31, or else with­out halfing, say as 320 to the perpen­dicular, so is the base to the content in acres; as 320 unto 30, so is 183 to 17, 15.

PROB. 5. Having the perpendicular and base given in chains, to find the content in acres.

As 20 to the perpendicular, so is the base to the content in acres.

As 20 to 7, 50, so is 45, 75, to 17, 15 parts.

PROB. 6. Having the concent of a Superficies after one kind of perch, to find the con­tent of the same Superficies according to another kind of perch.

As the length of the second perch, is to the first, so is the content in acres to a fourth number, and that 4^th num­ber to the content in acres required. Suppose a Superficies be measured with a chain of 66 feet or with a perch of 16 1/2, and it contain 34. 31, and it be demanded how many acres it would contain, if it were measured with a perch of 18 foot? these kind of proportions, are to be wrought by the backward rule of Three, after a duplicated proportion: wherefore I extend the Compasses from 165. unto 18, 0, and the same extent doth reach backward, first from 34.31 to 31.45, and then from 31. 45 to 28. 84, the content, in those larger acres of 18 foot to a perch.

PROB. 7. Having the Plot of a field with the content in acres, to find the scale by which it was Plotted.

Suppose a plain contained 34 acres 31 Centesmes, if I should measure it with a scale of 10 in an inch, the length should be 38 Chains and 12 Centesmes, and the breadth 6 Chains and 25 Centesmes, and the content according to that dimension, would by the 3 Probleme of this Chapter be found to be 23, 82. whereas it should be 34.31, therefore to gain the truth, I divide the distance between 23. 82 and 34, 31 into two equal parts, then setting one foot of the Compasses upon 10, the supposed true scale, I find the other to extend to 12, which is the length of the scale required.

PROB. 8. Having the length of the Oblong, to find the breadth of the acre.

As the length in perches to 160, so is one acre to the breadth in perches. As 40 to 160, so is 1 to 4. Again, as [Page 73]50 to 160, so is 1 to 3.20, so is 2 to 6 40; or again if you measure by chains, As the length in Chains to 10, so is 1 acre to his breadth in Chains; as 12 50 unto 10, so 1 to 0. 80, or if the length be measured by foot measure, then as the length in feet unto 43560 so is 1 acre to his breadth in foot measure.

So the length of the oblong being 792 feet, the breadth of one acre will be found to be 55 foot, the breadth of 2 acres 110 feet.

The use of this Table is to shew you how many Inches, Centesmes of a Chain, feet, yards, paces, perches, chains, acres, there is in a mile, either long or square, or consequently any of them all, in any of the other that is less; as for example, I would know how many Inches there is in a long berch, I look on the uppermost row for perches, and in the next row under, I find 198 for the quantity of inches in a long perch. But if I would know now many inches there is in a square [Page 74]perch, then look for perch on the left hand, and in the inch colume you have 39204, for if you multiply 196 by 196, it will produce 39204.

The like is for any other number in the whole Table, an [...] is of very good use to reduce one number into another, or one sort of measure into another: as inches into feet, and feet into Yards, and Yards into Perches, and Perches to Chains, and Chains into Acres, and Acres into Miles, or the contrary either long-wise or square- wise: as is well known to them that have occasion for these measures. Thus much shall suffice for Superficial measure, the practice of which will make it plain to any ordinary capaci­ty.

PROB. 1. By Foot-measure. A peice of [...]imber being to be mea­sured and not just square, how to make it square.

Divide the Space between the breadth, and the thickness, into two equal parts, and the Compasses shall stay at the side of the Square, equal to the oblong made of that breadth and thickness; which is the mean pro­portional between them. The breadth being 18, and thickness 6, the side of the Square will be found to be 10, 38.

PROB. 2. Having the side of a square, equal to the base of any Solid given in Foot-measure, to find how much makes a Foot Solid in Foot-measure.

As the side of the square in Foot-measure unto 1, so is 1 to a 4^th num­ber, and that 4^th to the length. As 2 120 unto 1.000, so 1.000 unto 0, 471, and that to 0. 222. or thus, the extent from 2. 120 to 1, will reach from 1, twice repeated to 0. 222, [...]nd so much is the length to make a Foot Solid, (at that squareness.)

PROB. 3. To find how much in length makes a Foot, any breadth and depth with­out squaring.

As 1 to the breadth in Foot-mea­sure, so is the depth to a fourth num­ber, as that 4^th number to 1 so is 1 to the lenth in Foot-measure.

PROB. 4. Having the side of a square, equal to the Base of a Solid given, and the length thereof in Foot- measure, to find the content in Feet.

As 1 to the side of the square in Foot-measure, so the length in Feet to a fourth number, and that fourth to the content in Foot-measure. The extent from 1 to 2. 12, twice repea­ted from 15.25, shall reach unto 68.62.

PROB. 5. Having the length breadth and depth of a Square Solid given in Foot-measure, to find the content in Feet.

As 1 to the breadth in Foot-mea­sure, so is the depth to the Base in Feet; as 1 to that Base, so the length in Feet to the content in Feet.

As 1 to 2, 50, so 1. 80 to 4. 50, then as one 1 to 4. 50, so is 15.25, unto 68. 625. The content required.

PROB. 6. By Inches, (only) and Feet and Inches. Having the side of a Square, equal to the base of any Solid given in Inches, to find how many Inches in length will make one Foot.

The side of the Square is found as in the first Problem of this Chapter, or by the 7^th of Board measure. Then as the side of the square in Inches to 41, 57, so is one Foot to a 4^th number, and that 4^th to the length in Inches, and tenth parts of an Inch.

The extent from 25, 45 unto 41, 57 twice repeated from 1 will reach to [Page 80]2, 67, or more easie if it be squared, as [...]he side of the square is to 12, so is 12 to a 4^th and that fourth to the length required. The extent from 25, 45 to 12 being twice repeated from 12, will stay at 2, 667, or more short 267.

PROB. 7. Having the breadth and depth of a squared Solid given in inches, to find the length of a Foot in Feet and Inches.

As 1 to the breadth in Inches, so the depth to a fourth number, which is the content of the base in Inches, then as this 4 number is to 1728, so is 1 to the length of a Foot Solid in Inch measure. As 1 to 21, 6, so is 30 to 648, then as 648 to 1728, so is 1 to 2, 667.

PROBL. 8. Having the side of the square and the length thereof given in Inch-mea­measure, to find the content in Feet.

As 41.57, to the side of the square in Inches, so is the length to a fourth Number, and that fourth to the con­tent in Foot-measure. As 41. 57, to 25. 45, so 183, twice repeated unto 68, 62.

PROB. 9. Having the side of a Square equal to the base of any solid given in Inch­measure and the length in Foot-mea­sure, to find the content in Feet.

As 12 to the side of the square in Inches, so the length in Feet to a fourth Number, and that fourth to the content in Foot-measure. As 12 to 25, 45, so 15. 25 to 32. 55, and 32. 55 to 68. 62. Or the extent from 12 to 25, shall reach to 68. 62, the con­tent sought.

PROB. 10. Having the length, breadth and depth, of a Squared Solid given in Inches, to find the content in Inches.

As 1 to the breadth in Inches, so the depth to the base, then as 1 to the base, so the length to the content in Inches. As 1 to 21. 6, so 30 to 648. as 1 to 648, so that 183 to 118584.

PROBL. 11. Having the length, breadth, and depth given in Inches, to find the content in Feet.

As 1 to the breadth in Inches, so the depth in Inches to the base in In­ches; Then as 1728 to the base, so is the length in Inches to the content in Feet. As 1 to 21, 6, so 30 to 648, as 1728 to 648, so 183 to 68. 62.

Or you may say,

As 12 to 21.6, so 30 to 54, as 144 to 54, so 18, 3 to 68.62.

PROB. 12. Having the breadth and depth of a squared solid given in Inches, and the length in Feet, to find the con­tent.

As 1 to the breadth in Inches, so the depth in Inches to a fourth number. Then as 144 to that fourth, so is the length in Feet to the content in Feet.

As 1 to 216, so is 30 to 648; then as 144 to 15.25, so is 648 unto 68.62, Or as 144 to 21.6, so 30 to 4. 50: as 1 to 4. 50, so 15.25 to 68.62. Or again,

As 12 to 21.6, so 30 to 54: then as 12 to 54, so 15.25 to 68.62, the content required.

PROB. 1. Having the Diameter of a Cylinder given in Foot-measure, to find the length of a Foot-solid in Foot mea­sure.

As the Diameter in Feet to 1.128, so is 1 to a fourth, and that fourth to the length in Foot-measure.

The Extent from (the Diameter) 1. 25; to 1. 128, being twice repeated from 1, will reach to 8. 148, the length sought.

To find how much is in a foot long at any Diameter given.

Say, As 1. 128 to the feet and parts in Diameter, So is 1 to a 4th. [Page 85]and that 4th. to the feet and parts in one foot long.

PROB. 2. Having the circumference given in Foot-measure to find the length of a Foot solid in Foot- measure.

As the Circumference in Foot-measure is 3. 545; so is 1 to a fourth, and that fourth to the length sought.

As 3 f 927 p unto 3. 545, so is that distance twice repeated from 1 to 0, 815 the length of a Foot- solid.

The Circumference given to find how much is in a foot long at that compass, as 3, 545 to the feet about, the same extent applyed twice the same way from 1 shall reach to the quantity in one foot long. At 5 foot about one foot in length, is two foot neer in the content, which multiply­ed by the feet long gives the whole content.

PROB. 3. Having the Diameter and length of a Cylinder given in Foot-measure, to find the content in Foot- measure.

As 1. 128 to the Diameter in Foot measure, so is the length in Foot-measure to a fourth, and that fourth to the content in Foot-measure.

The Extent from 1. 128 to 1. 25. being twice repeated from 8. 75, will reach to 10. 737, the content sought for.

PROB. 4. Having the Circumference and length of a Cylinder given in Foot-measure, to find the content in Foot-measure.

As 3. 545 to the Circumference in Feet, so is the length in Feet to a fourth, and the fourth to the content in Foot-measure.

The Extent from 3.545 to 3.927, being twice repeated from 8. 75, will reach to 1074, the content in Foot- measure.

PROB. 5. By Inch-measure. Having the Diameter of a Cylinder given in Inches, to find how many Inches makes a Foot-solid.

As the Diameter to 46.90, so is 1 to a fourth number, and that fourth number to the length in Inches.

The Extent from 15 to 46.90, will reach from 1 (being twice repeated) to 9.778, the length sought.

PROB. 6. Having the Circumference given in Inches, to find the length of a Foot-solid.

As the Circumference to 147.36 so is 1 to a fourth number, and so that fourth to the length in Inches.

The Extent from 47. 13, to 147. 36, being twice repeated from 1, will reach to 9. 78, the length sought.

PROB. 7. Having the Diameter and length given in Inches to find the content in Inches.

As 1 128 to the Diameter in In­ches, so is the length to a fourth, and [Page 88]that fourth to the content in Inches. The extent from 1 128 to 15, being twice repeated from 105, will reach to 18555, 34 the content in Inches.

PROB. 8. Having the Circumference and length given in Inches, to find the content in Inches.

As 3, 545 to the Circumference in Inches, so is the length in Inches to a fourth number, and that fourth to the content in Inches. The extent from 3, 545 to 47, 13, being twice repeat­ed from 105 will reach to 18555 the content in Inches.

PROB. 9. By Feet and Inches. Having the Diameter given in Inches and the length in Feet to find the content in Feet.

As 13, 54 to the Diameter, so the length to a fourth number, and that fourth to the content in Feet. The ex­tent from 13, 54 to 15, being twice repeated from 8, 75 will reach to 10, 74, the content sought for.

PROB. 10. Having the Circumference given in Inches, and the length in Feet-mea­sure to find the content in Feet.

As 42, 54 to the Circumference in Inches, so is the length in Feet to a fourth, and that fourth to the content. The extent from 42, 54 to 47, 13; being twice repeated from 8, 75, will reach to 10, 74 the content sought.

PROB. 11.

It being an ordinary way in mea­suring of round Timber, such as Oake, Elme, Beech, Pear-tree, and the like, (which is sometimes very rugged, and uneven, and knotty) to take a line and girt about the middle of it, and then to take the fourth part of that, for the side of a Square equal to that Circumference: But this mea­sure is not exact, but more then it should be. But either because of al­lowance for the faults abovesaid, or for Ignorance, the custome is sill used, and men commonly think them­selves wronged if they have not such [Page 90]measure. Therefore I have fitted you with a proportion for it both for Dia­meter and Circumference.

And first for the Diameter.

The Diameter given in Inches and the length in Feet, to find the content.

As 1, 536 to the Diameter, so is the length to a fourth, and that fourth to the content in Feet: according to the rate abovesaid.

The extent from 1. 526 to 9. 53, being twice repeated from 8, shall reach to 3. 12, the content.

PROB. 12. Having the Circumference in Inches, to find the content in the abovesaid measure.

As 48 to the Inches about, so is the length to a fourth number, and that fourth to the content. The extent from 48 to 30, being twice repeated from 8, shall fall upon 3. 12, the content required.

PROB. 13. How to measure Taper Timber, that is bigger at one end than at the other.

The usual way for doing of this, is to take the Circumference of the middle or mean bigness, but a more exact way, is to find the content of the base of both ends and add them together; and then to take the half for the mean, which multiplyed by the length, shall give you the true con­tent.

PROB. 14. To mensure a Cone, such as is a Spire of a Steeple, or the like by having the height and Diameter of Base.

Example; let a Cone be to be mea­sured, whose base is 10 Foot, and the height thereof 12 Foot, the content of the base will be found, by the 14 Problem of Superficial measure, to be 78, 54, Then this 78, 54 multiply­ed by 4, a third part of 12, the per­pendicular or height of the Cone will give 314, 4, for the content of the Cone required: By the numbers work thus, the extent from 1 to 4, will reach from 78, 54 to 314, 4. But be­cause there may be some trouble in getting the true perpendicular of a Cone, which is its height, take this rule; First, take half the Diameter, and multiply it in it self, which here is 25, then measure the side of the Cone 13, and multiply that by it self which here is 169, from which take the Square of half the Base, which is 25, your first number found, and the [Page 93]remain is 144, the Square root of which is the height of the Cone, or length of the perpendicular.

PROB. 15. To measure a Globe or Sphere arith­metically.

Cube the Diameter, then multiply that by 11, and divide by 21, gives you the true solid content; let a Sphere be to be measured whose Axis or Diameter is 14, that multiplyed by it self gives 196, and 196 again by 14 gives 2744, this multiplyed by 11 gives 30184, and this last divided by 21 gives 1437.67, for the content of the Sphere whole Diameter is 14. But more briefly, by the numbers thus, The extent from 1 to the Axis, being twice repeated from 3. 142, will reach to the Superficial content, that is, the Superficies round about. But if the same extent from 1 to the Axis be thrice repeared f [...]om 5238, it will reach to the solid content, as 1 to 14, so 3. 142 to 617 being twice re­peated, as 1 to 14, so 5278 to 1437. [Page 94]being thrice repeated. As for many sided figures if they have length, you have sufficient for them in the Chap­ter of Superficial measure, to find the base, and then the base multiplyed by the length giveth the content. But as for figures of roundish form, they coming very seldom in use, I shall not in this place trouble you with them, for they may be reduced to Spheres or Cones, or Triangles, or Cubes, and then measured by those Problems accordingly. And so much for the mensuration of Solids.

PROB. 1. The true content of a [...]olid measure being known, to find the Gage point of the same measure.

The Gage-point of a Solid mea­sure is the Diameter of a Circle, whose Superficial content is equal to the Solid content of the same measure so the Solid content of a Wine Gallon, (according to Winchester measure be­ing 231 Cube Inches, if you conceive a Circle to contain so many Inches, you shall by the 16 Problem of board measure find the Diameter thereof to be 17.15, for as 1 is to 1.273, so is 231 to 294. 1, whose square root by [Page 109]the 12 of the second, is 17. 15, the Gage-point for Wine measure.

Thus likewise you may discover the Gage point for Ale-measure, an Ale-Gallon (as hath been of late dis­covered) containing 282 Cubique Inches. For as 1 is to 1.273, so is 282 to 356.3, whose square root (by the 12 Problem of the second Chap­ter) is 19.95, the Gage-point of Ale measure, because of soil and wast ex­ceeding that of wine above 2 Inches.

Another way by having the Diame­ter, length, and true content of any Vessel.

Extend the Compasses on the line of numbers to half the distance be­tween the content and length of the Vessel, the same extent will reach from the Diameter to the Gage-point.

PROB. 2. To find the equated Dia­meter two ways first by Arithmetick, as in Problem 8 of the second Chapter.

Add the two Diameters at head [Page 111]and bung together, (being measured in Inches) and the half sum keep. Secondly, Substract the Diameter of the head out of the Diameter at the bung, and note the remainder, which is to be divided by 4, or 4 ½ (as by the number 5 is easie) to find out a fourth (or somewhat less) part of the difference. Thirdly, that fourth part (or somewhat less) is to be added to the half sum kept, to make up the mean Diameter sought. Examp.

A Vessel hath in Diameter at the head 18 in. 3 p. and at the bung 21.5 Inches, I would know the mean Dia­meter 18.3, and 21.5 added, is 39.8, the half is 19.9, the less taken from the greater, the remainder is 3 in. 2 p. which brought or reduced into 100 is 320. and in stead of dividing by 4 or 4 ½ say 45, and the product is 7 tenths and 3 over; which 7 tenths add to 19. 9, and it maketh 20. 6 tenths for the mean Diameter required to be found.

Another way by Geometry and somewhat more exact, Of the Diame­ter [Page 112]at the head, and bung, take the dif­ference, then say as 1 is to 7, so is the difference to a fourth sum, which fourth sum is to be added, to the least of both the Diameters, viz. that at head, as in the former example 18.30 and 21.50 added is 39 80, whose half is 19. 90. and the difference is 3 in. 2 p or 320, then say, as 1 is to 7, so is 320 to 2. 24, which 2.24 if you take with your Compasses, out of your Scale of Inches and add it to 18.30, you shall see it reach to 20 Inches 54 parts, and this is the true mean Dia­meter, to make it a perfect Cylinder; o [...] if you add 2 Inches 24 to 18. 30, it makes it 20. 54 as before.

Note, If the difference of the two Diameters be much, then you must add the 100 parts as in the Ta­ble annexed.

This Table is thus used when the difference between the head and bung in close Cask is above 4 inches; thus,

As 10 to 7. so is the difference be­tween the head and bung augmented by the 100 parts of an Inch, as in the Table, to the fourth number, which you must add to the Diameter at the head, to get a mean Diameter.

PROB. 3. Having the mean Diameter, and the longth of a Vessel, to find the content.

Extend the Compasses from the Gage-point to the mean Diameter, the same extent being twice repeated from the length, shall give the con­tent in galons and 100 parts.

PROB. 4. Having the Diameter and Content, to find the length.

Extend the Compasses from the Di­ameter to the Gage-point, the same extent twice repeated from the con­tent, shall give the length.

As 38. to 17. 15, so is 324 to about 66 twice repeated.

PROB. 5. Having the length and content, to find the Diameter of a Vessel.

Extend the Compasses to half the distance, between the length and content, the same extent shall reach from the Gage-point to the Diame­ter.

Divide the space between 66, and 324, in two equal parts, the same ex­tent shall reach from 17. 15 to 38, the Diameter abovesaid.

PROB. 6. Having measured a Vessel according to Wine measure to know what it holds in Ale measure, without knowing the Gage-point.

As 282 is to 231, so is the content in wine Gallons, to the content in Ale Gallons. Or the contrary, as 231 is to 282, so is the content in Ale Gallons, to the content in wine Gal­lons.

PROB. 7. To measure any vessel a more easie way.

There is yet a more easie way, and for exactness no way infer or to any extant, and that by taking the length in Inches and 10 parts, but the Dia­meters [Page 117]at head and Bung, with a line called Cughtred's Gage line, (and to be had at John Brown's house in the Minories neer Aldgate, Mathemati­cal-In [...]ument-maker) the use of which is thus, take the Diameter at the bung, with those divisions on the line aforesaid, from that en [...] where he di­visions begin to be numbred. And set that down twice: and the Diame­ter of the inside of the head (for so we understand all along,) and set that down once. In this manner. [...] and then add them together as here you see, the length in In­ches suppose to be (30. 82.) Thirty and eighty two of a hundred: then say, As 1 is to 1. 77, so is 30.82 to 54.55 hundred parts of a Gallon, being a little more then half a Gallon, which is 54 gallons ½ the content of a high Country Hogshed, whose measures were as be­fore.

Depth — [...]

Diam. at top — [...]

Diam. at bottom — [...]

Sum of the Diameters [...]

The Summe of the Diameters Squared — [...]

Product of Diam. [...]

The Residue [...]

[...]

The residue the fraction cut of

[...] [...]

To make the Number 1077 which is fixed for an Ale-Gallon, or 882 for a Wine Gallon, or the fixed Number for a Beer-Barrel which is 38796, for an Ale Barrel is 34485.

Multiply the Cube Inches in a Gallon or Barrel by 14, divide the Product by 11, and multiply the quo­tient [Page 120]by 3, produceth these fixed Numbers.

PROB. 1. Any Number of Souldiers being pro­pounded, to order them into a Square Battle of Men.

Find by the 12 Problem of the second Chapter, the Square root of the number given; for so much as that root shall be, so many Souldiers [Page 138]ought you to place in Ranck, and so many likewise in File, to make a Square Battle of men.

PROB. 2. Any Number of Souldiers being pro­pounded, to order them into a double Battle of men: that is, which may have twice as many in ranck as file.

Find out the square root ot half the number given, for that root is the number of men to be placed in file, & twice as many to be placed in ranck, to make up a double Battle of men.

PROB. 3. Any number of Souldiers being pro­pounded, to order them into a quadru­ple batail of men; that is, four times so many in Ranck as File.

Here the Square root of the fourth part of the number propounded, will shew the number to be placed in File, and four times so many are to be pla­ced in Ranck.

So 2048 being divided by 4, the quotient is 512, whose root is 22 (6) and so many are to be placed in File and 88 in Ranck, being four times 22, &c.

PROB. 4. Any number of Souldiers being given, together with their distances in Rank and File, to order them into a Square [...]attail of ground.

Extend the Compasses from the di­stance in File to the distance in ranck; this done, that extent applyed the same way from the number of Soul­diers propounded, will cause the moveable point to fall upon a fourth number, whose Square root is the number of men to be placed in File; by which, if you divide the whole number of Soldiers, the quotient will shew the number of men to be placed in Ranck.

PROB. 5. Any number of Souldiers being pro­pounded to order them in rank and file, according to the reason of any two numbers given.

This Problem is like the former, for as the proportional number given for the file, is to that given for the ranck, so is the number of Souldiers to a fourth number, whose square root is the number of men to be pla­ced in ranck, by which if you divide the whole, you may have the number to be placed in file.

PROB. 1. A sum of money put out to Use, and the Interest forborn for a certain time, to know what it comes to at the end of that times, counting Interest upon In­terest at any rate propounded.

Take the distance with your Com­passes between 100, and the Increase of 100 l. for one Year, (which you [Page 143]must do very exactly) and repeat it so many times from the principal as it is forborn years, and the point of the Compasses will stay on the Prin­cipal with the Interest, and increase according to the rate propounded.

PROB. 2. A sum of money being due at any time to come, to find what it is worth in ready money.

This question is only the inverse of the other; for if you take the space between 106 and 100, and turn it back from the sum proposed, as many times as there are years in the questi­on, it shall fall on the sum required.

PROB. 3. A yearly Rent, Pension, or Annuity being forborn for a certain term of years, to find what the Arrears come to at any rate propounded.

First you must find the principal that shall answer to that Annuity, then [Page 145]find to what sum the Principal would be augmented at the rate and term of years propounded; then if you sub­stract the principal out of that sum the remainder is the Arrears required.

PROB. 4. A yearly Rent or Annuity being pro­pounded, to find the worth in ready money.

First, find by the last what the ar­rears come to at the term propoun­ded, and then what those arrears are worth in ready money, and that shall be the value of it in ready money.

PROB. 5. A sum of money being propounded, to find what Annuity to continue any number of years, at any rate pro­pounded, that sum of money will pur­chase.

Take any known annuity, and find the value of it in ready money; this being done, the proportion will be thus: As the value found out is to the [Page 148]annuity taken, so is the summe pro­pounded to the annuity required.

PROB. 1.

Possibly that this little Book may meet with some that are well skilled in Arithmetick, and being much used to that way, are loth to be weaned from that way, being so artificial and exact, yet though they can multiply & divide very well, yet perhaps they know not this way, to save their di­vision and yet to take in all the fra­ctions together as if of one denomina­tion: I shall begin first with Foot-measure being the more easie, and I suppose my Two-foot-rule to be di­vided into 200 parts, and figured with 10. 20. 30. 40. 50.60.70. 80. 90. 100. And then so again to 200. as in the 3 Chap. and then the work is on­ly thus: set down the measure of one side of the square, or oblong thus, as for example, 7. 25, and 9. 88, and multiply them as if they were whole numbers, and from the product cut off 4 figures, and you have the content in Feet, and 1000 parts of a Foot, or Yard, Ell, Perch, or whatsoever else it be. Note the examples following.

[...] [...]

For any kind of flat Superficies, this is sufficient instruction to him that hath read the first part; but if it be Timber, or Stone, you must thus find the Base, and then another work will give you the other side, as in Chapter 5 Problem 2. or, Multiply the length by the Product of the breadth and thickness, and that Pro­duct shall be the content required.

PROB. 2. To Multiply Feet, Inches, and 8 parts of an Inch together without Reducti­on, and so to measure Superficial (and Solid) measure,

First, Multiply all the whole Feet, then all the Feet and Inches, across, and right on, then the parts by the Feet, and also the Inches, and parts, across and right on; then add them to­gether, and you shall have the answer in feet, long Inches, (that is, in pieces of a Foot long, and an Inch broad) square Inches, and 8 parts of a Square Inch: as for example.

Let a peice of Board be given to be measured that is 3. 3. 5. i. e. three Foot, three Inches, and 5 eights, one way, and 2. 3. 4, the other way. I set the numbers down in this Man­ner, [...] & then right on, first as the line in the Scheme from 2 to 3 leads.

I say thus, 3 times 2 is 6, set 6 right under 2. and 3 as in the example, in the left page: for 6 Foot, as is clear, if you consider the Scheme over the example, viz. the squares noted with f. then for the next I say cross-wise, 2 times 3 is 6, viz. long Inches, as you may perceive, by the 2 long squares marked with 9 L. and 6 L. [Page 167]which 6 I put in the next place to the right hand, as in the example; then for the next, viz. 3 times 3. is 9, (croswise, as the stroke from 3 to 3 shews) which 9 is also 9 long Inches, as the Scheme sheweth, and must be put under 6, in the second place toward the right hand, in the Scheme it is express'd by the 3 long Squares, marked with L 9. Then last­ly for the Inches, 3 times 3 is 9, go­ing right up, as the stroke from the 2 threes lead you: but note, this 9 must be set in the next place to the right hand, because they are but 9 Square Inches, but had the Product been a­bove 12, you must have Substracted the 12 s. out, and set them in the long Inches place, and the remainder, where this 9 now standeth, and this 9 is express'd in the Scheme, by the little Square in the corner markt with (□ 9.)

Then now for the Fractions, or 8 parts of an Inch, first say, croswise as the longest prick line doth lead you [Page 168]to; 3 times 4 is 12, for which 12, you must set down 1. 6, that is 1 long Inch, and 6 Square Inches, the reason is, a piece 8 half quarters of an Inch broad, and 12 Inches long, is a long Inch, or the twelfth part of a Foot superficial, and if 8 be 12 Square Inches, then 4 must needs be 6 Square Inches: therefore, in stead of 12, I set down 1. 6, as you may see in the example, and in the least long Square of the Diagram, or Scheme. Then do likewise for the other long Square, which is also multiplyed across; as, two times 5 is 10. that is, as I said before, 1. 3, as the Example and Scheme make manifest, considering what I last said, and it is marked by the 2. 00. But if this or the other had come to a greater number, you must have Substracted 8 s. as oft as you could, and set down the remainder in the place of Square Inches, and the number of 8 s. in the place of Long Inches, as here you see.

Then for the two shorter Long [Page 169]Squares next the corner, say croswise again, Three times 5 is 15, that is 1, 7, because eight Half-quarters an Inch long do make one square Inch, as well as eight Half-quarters a Foot long made one Long Inch: Therefore I set 1 in the place of square Inches, and 7 in the next place to the right hand, and it is expressed in the Dia­gram by the small long square, and marked with * 1. 7.

Then again for the other little long square, say croswise, as the shorter prick line leads you, Three times 4 is 12, that is 1. 4; and do by this as the last: It is noted in the Scheme by 1-4.

Then lastly for 5 times 4, as the short prick line sheweth you, is 20: out of which 20 take the 8 s. and set them down in the last place, and the 4 remaining you may either neglect, (or set it down a place further) for you cannot see it on the Rule; there­fore, I thus advise, if it be under 4. neg­lect it quite, but if above, increase the [Page 170]next a figure more if 4 then it is a half, and so may be added; for note, 64 of these parts make but one square Inch; of which parts, the little square in the right hand lower corner of the Scheme is 20, for which I set down 2. 4, that is two Half-quarters, and 4 of 64. which is the last work, as you may see by the Scheme and Example.

Now to add them together say thus, 4 is 4, which I put furthest to the right hand, as it were useless, because not to be exprest; then 472 are 13, from which take 8, and for it carry 1 on to the next place, or as many times 1 as you find 8, and set down the re­mainder, which here is 5, then 1 I carried, and 13619, is 21, from which I take 12, and set down 9, be­cause 12 square Inches, is one long Inch: then 1 I carried, (or more, had there been more 12 s.) and 1169 is 18, from which take 12, as before, there remains 6, that is, 6 long Inches, and so had there been more 12 s. so many you must carry to the next [Page 171]place, because 12 long Inches is one Foot, lastly 1 I carried, and 6 is 7 Foot, so that the work stands thus, [...] and so for any other measure Superfi­cial or Solid.

1. Multiplication by the Sliding-rule.

Set the 1 on any side, (which being found, I call the first side) to the Multiplicator on the other (or se­cond) side; then seek the Multipli­cand on the first side, where 1 was: and right against it on the second, is the Product required.

2. Division by the Rule.

Set the Divisor found always on the first side, to 1 on the second side, then right against the Dividend found [Page 215]out on the first side, on the second is the Quotient required.

3. The Rule of 3. direct.

Set the first term of the question sought out on the first side, to the se­cond term of the question on the se­cond (or other) side: then right against the third term, found out on the first side, on the second side is the fourth proportional term required.

4. To work the Rule of 3. reverse.

Set the first term sought out on the first side, to the second being of the same denomination on the second line or side, then seek the third term on the second side, and on the first you shall have the answer required.

5. To work the double Rule of 3 direct.

To perform this, you must have two working: As thus, for an

PROB. 1. The breadth given to find how much makes a Foot.

If the breadth be given in Inches, then set 12 on the first side, to the Inches on the second: then right against 12 on the second, on the first is the number of Inches required.

PROB. 2. The length and breadth given to find the content.

First the breadth given in Inches, and the length in Feet and Inches: set 12 on the first side, to the breadth on the second; then right against the length on the first, on the second is the content required.

PROB. 3. The breadth given in Feet and Inches, and the length also in the same parts, to find the Content.

Set 1 on the first side to the Feet and Inches brought to a Decimal Fra­ction, or as near as you can guess, (for 6 Inches is half, 3 Inches is one quar­ter, 9 Inches is the quarters, 4 Inches is one third, 8 Inches is two thirds, and 1 Inch is somewhat less than one tenth on Rule,) on the other or se­cond side; then right against the length found on the first, on the se­cond is the Content required.

PROB. 1.

To measure Timber by this Rule, is nothing else but to work the Double-Rule of Three.

PROB. 2. To measure a piece that is not square.

Set 12 if you use the Inches, (or 1 if you use Foot-measure) on the first side, to the Inches thick on the se­cond; then right against the Inches broad on the first side, on the second is a fourth proportional: then in the second operation, set 12 on the first side, to the fourth proportional on the second, then right against the length on the first side, on the second is the Content required.

PROB. 3. The Square given, to find how much makes a Foot.

Set the Inches square on the first side, to 12 on the second; then right against 12 on the first, on the second is a fourth proportional number: then in the second work, as the In­ches square to the fourth proportio­nal, so is 12 to the number of Inches required, to make a Foot of Timber.

PROB. 1. A piece of Timber being not square, to make it square.

Set the breadth on the left side, to the breadth on the right, then right against the Inch and quarters thick found on the left side, on the right is the Inches square required.

PROB. 2. The side of the square given, to find how much makes a foot.

For all pieces between 3 or 4 In­ches, and 42 Inches square, which are the most usual: this is the best way, set the Inches or Feet, and In­ches square, found out on the right-side, to one Foot on the left, then [Page 232]right against one Foot on the right, on the left is the Inches, or Feet, and Inches required, to make a Foot of Timber.

PROB. 3. The side of the square and length given to find the Content.

For all pieces between one Inch or 8/12 part of a Foot, and 100 Foot, [Page 233]this is the easiest way. Set the word square or 1 Foot to the length on the left, then right against the Inches, or Feet and Inches square on the right, on the left you have the Content.

PROB. 4. The square of a small piece of Tim­ber given, to find how much makes a Foot.

For all pieces from 12 Inches to 1 Inch square, do thus: Set the In­ches and (12s. or) quarters square, counting one Foot on the right side for one Inch, and 2 Foot for 2 In­ches, &c. found out on the right-side to 100 on the left; then right a­gainst the upper or small 12 on the [Page 235]right, on the left is the length requi­red, to make a Foot of Timber.

PROB. 5. Under 1 Inch square to find the length of a Foot.

Set 1 Foot 9 Inches, 6 Inches, or 3 Inches, found on the right-side, for one Inch ¾, ½, or ¼ of an Inch, a­gainst 10 on the left-side, counted for 100: then right against the small 12, you have the Feet in length re­quired.

PROB. 6. A great piece above 3 Foot ½ square, to find the length of a Foot.

Set the Feet and Inches on the right, to 100 on the left: then right against small 12, is the Inches and 12 s, or 12 s of a 12th. that goes to make a Foot.

PROB. 1. The number of Inches that a piece of timber is about, being given, to find how much makes a Foot.

First, for all ordinary pieces, set one foot on the left, to the inches or feet, and inches above on the right, then right against TR for true mea­sure, or round for the usual measure, is the feet, or feet and inches requi­red, to make a foot of timber, at that circumference about.

PROB. 2. For very small Wood to find a Foot in length.

But for very small pieces of un­der 4 inches about, set 1 foot, 2 foot, &c. on the right, (counted for 1 inch, 2 inches, 3 inches, or 4 inches) to one foot on the left, then right against TR or round, you have a number, which Multiplyed by 12 is the number of feet required.

PROB. 3. The inches, or feet and inches about, and length given to find the con­tent.

Set the word round, or TR for the usual, or true measure, to the length on the left: then right against the inches about on the right, on the left is the content required.

PROB. 4. To find the content of a very small piece.

Set the word round, or TR to the length on the left, as in the 3. Pro­positiou of this Chapter: then right against the inches about on the right, (calling one foot one inch, and 6 inches ½ an inch,) on the left is the 12 s of one inch,) or 12 s. of a 12th. required.

PROB. 1. The diameter given in inches, to find the length of a foot.

Set one foot on the left, to the in­ches diameter on the right: then right against TD for true diameter, or the word diameter for the usual allowance, (of a string girt about and doubled 4 times for the side of the square) you have the feet and inches required.

PROB. 2. The diameter and length given to find the true content.

For all ordinary pieces, set the word Diameter for the usual measure, or TD for true measure, always to the length on the left: then right against the inches, or feet and inches Diameter on the right, on the left is the content required.