# Chap. III.The use of the Line of Numbers in solid measure, such as Stone, Timber, and the like.

## SECT. I.The use of the Line of Numbers in solid measure, such as Stone, Timber, and the like.

### 1. Having the side of a Square equal to the Base of any Solid given in inch measure, to find the length of a foot Solid in inch measure.

The side of a Square equal to the Base of a Solid, may be found by dividing the space between the length and breadth into two equal parts, as in the seventh Proposition of broad measure.

• As the side of the Square in inches, to 41,57:
• So is 1 foot, to a fourth number; and that fourth to the length in inches.

So in the Solid AH, the side of the square equal to the Base EC, being about 25 inches 45 parts, the length of a foot Solid will be found about two inches 67 parts, and the length of two foot Solid 5 inches 34 parts.

• As 25,45, unto 41,57: so 1,00, unto 1,6:
• and so are 1,63, unto 2,67.

### 2. Having the side of a Square equal to the Base of any Solid given in foot measure, to find the length of a foot Solid in foot measure.

• As the side of the Square in feet, unto 1:
• So is 1, unto a fourth number:
• And that fourth, to the length in foot measure.

So in the Solid AH, the side of the Square equal to the Base EC, being about 2 foot 120 parts, the length of a foot Solid will be found about 222 parts of a foot.

• As 2,120, unto 1,000: so 1,000, unto 0,471.
• and so are 471, unto 222.

### 3. Having the breadth and depth of a squared Solid given in foot measure, to find the length of a foot Solid in foot measure.

• As 1, unto the breadth in foot measure:
• So the depth in feet to a fourth number:
• which is the content of the Base in foot measure,

Then,

• As this fourth number, unto 1:
• So 1, unto the length in foot measure.

So in the Solid AH, the breadth being 2 foot 50 parts, the depth 1 foot 80 parts, the content of the Base EC will be found 4 foot 50 parts, and the length of one foot Solid about 222 parts, the length of a two foot Solid about 444 parts of 1000.

• As 1,00 unto 2,50: so are 1,80 unto 4,50.
• As 4,50 unto 1,00: so 1,000 unto 0,222.

### 4. Having the breadth and depth of a squared Solid given in inches, to find the length of a foot Solid in inch measure.

• As 1 hath to the breadth in inches:
• So the depth in inches to a fourth number;
• Which is the content of the Base in inches.

Then,

• As this fourth number unto 1728:
• So 1 unto the length of a foot in inch measure.

As in the Solid AH, the breadth AC being 30 inches, and the depth AE 21 inches 60 parts, the content of the Base EC will be found to be 648 inches, and the length of a foot Solid about 2 inches 67 parts, the length of a two foot Solid 5 inches 34 parts.

• As 1 unto 21,6: so 30 unto 648.
• As 648 unto 1728: so 1 unto 267.
• Or as 12 to the breadth in inches:
• So the depth in inches to a fourth number.
• As this fourth number unto 144:
• So 1 unto the length of a foot Solid in inch measure.

So in the Solid AH, the breadth being 30 inches, the depth 21 inches 6 parts, the fourth number will be found to be 54, and the depth of a foot Solid 2 inches 67 parts.

• As 12 unto 21,6: so 30 unto 54.
• As 54 unto 144: so 1 unto 2,67.

### 5. Having the side of a Square equal to the Base of nay Solid, and the length thereof given in inch measure, to find the content thereof in feet.

• As 41.57 to the side of the Square in inches:
• So the length in inches to a fourth number;
• and that fourth number to the content in foot measure.

So in the Solid AH, the length AB being 183 inches, and the side of the Square equal to the Base EC about 25 inches 45 parts, the fourth number will be found about 112, and the whole Solid content about 68 feet 62 parts.

• As 41.57 unto 25.45: so 183 unto 112:
• and so are 112 unto 68.62.

### 6. Having the side of a Square equal to the Base of any Solid, and the length thereof given in foot measure, to find the content thereof 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 number to the content in foot measure.

So in the former Solid AH, the side of the square equal to the Base AE, being about 2 foot 12 parts, and the length AB 15 foot 25 parts, the content will be found to be about 68 foot 62 parts.

• As 1 unto 2.12: so 15.25 unto 32.35:
• and so are 32.35 unto 68.62.

### 7. Having the side of a Square equal to the Base of any Solid given in inch measure, and the length of the Solid given in foot measure, to find the content thereof in feet.

• As 12 to the side of the Square given in inches:
• So the length in feet to a fourth number;
• and that fourth number to the content in foot measure.

So in the former Solid AH, the side of the Square being 25 inches 45 parts, the content will be found to be about 68 feet 62 parts.

• As 12 unto 25.45: so 15.25 unto 32.35.
• and so are 32.35 unto 68.62.

### 8. Having the length, breadth and depth of a squared Solid given in inches, to find the content in inches.

• As 1 unto the breadth in inches:
• So the depth in inches unto the Base in inches.

Then,

• As 1 unto the Base:
• So the length in inches unto the Solid content in inches.

So in the Solid AH, whose breadth AC is 30 inches, depth AE 21 inches, and 6 parts of 10, and length AB 183, the content of the Base EC will be found 648 inches, and the whole Solid content about 118584 inches.

• As 1 unto 21.6: so are 30 unto 648:
• As 1 unto 648: so are 183 to 118,584.

### 9. Having the length, breadth, and depth of a squared Solid 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.
• As 1728 to that Base:
• So the length in inches to the content in feet.

So in the Solid AH, the content will be found to be about 68 feet 62 parts.

• As 1 unto 21.6: so 30 unto 648:
• As 1728 unto 648: so 183 to 68,62.
• Or as 12 to the breadth in inches:
• So the depth in inches to a fourth number.
• As 144 to that fourth number:
• So the length in inches to the content in feet.

And so also in the same Solid AH, the content will be found to be about 68 feet 62 parts.

• As 12 unto 216: so 30 unto 68.62.
• As 144 unto 54: so 183 unto 68.62.

### 10. Having the length, breadth, and depth of a squared Solid given in foot measure, to find the content in feet.

• As 1 unto the breadth in foot measure:
• So the depth in feet to the Base in feet.
• As 1 unto the Base:
• So the length in feet to the content in feet.

And thus in the former Solid AH, the breadth AC will be two foot 50 parts, the depth AE, 1 foot 80 parts, and the length AB 15 foot 25 parts; then working as before, the content of the Base AF will be found 4 feet 50 parts, and the whole Solid content about 68 foot 62 parts, which of all others may very easily be tried by Arithmetick.

• As 1 unto 2.50: so 1,80 unto 4.50,
• As 1 unto 4.50: so 15.25 unto 68.62.

### 11. Having the breadth and depth of a squared Solid given in inches, and the length in foot measure, to find the content thereof in feet.

• As 1 unto the breadth in inches:
• So the depth in inches unto a fourth number,
• which is the content of the Base in inches.
• As 144 hath unto the fourth number:
• So the length in feet to the content in feet.

And so in the same Solid AH, the content will be found to be about 68 feet 62 parts.

• As 1 unto 21.6: so 30 unto 648.
• As 144 unto 15.25, so 648 unto 68.62.
• Or as 144 unto the breadth in inches:
• So the depth in inches unto a fourth number:
• which is the content of the Base in feet.
• As 1 hath unto that fourth number:
• So the length in feet to the content in feet.

And so in the same Solid AH, the content will be found to be about 68 feet 62 parts.

• As 12 unto 21.6: so 30 unto 54.
• As 12 unto 54: so 15.25 unto 68.62.

All these varieties (and such like not here mentioned) do follow upon the making of the Base of the Solid to be EC; there would be as many more if any shall begin with the Base EH, and so likewise if they make the Base to be FD.

## SECT. II.Of the mensuration of Cylinders.

### 1. Having the Diameter of a Cylinder given in inch measure, to find the length of a foot Solid in inches.

• As the Diameter in inches unto 46.90:
• So is 1 unto a fourth number:
• And that fourth number to the length in inches.

So the Diameter of a Cylinder being 15 inches, the fourth number, will be about 3,12, and the length of a foot Solid 9 inches 78 parts.

• As 15 unto 46.90: so 1 unto 3.127.
• And so are 3.127 unto 9.78.

### 2. 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 unto 1.128:
• So is 1 unto a fourth number;
• and that fourth number to the length in foot measure.

So the Diameter being 1 foot 25 parts; the length of a foot Solid will be found about 8.14 parts of 1000.

### 3. Having the Circumference of a Cylinder given in inches, to find the length of a foot Solid in inch measure.

• As the Circumference in inches to 147.36:
• So is 1 to a fourth number;
• and that fourth to the length in inches.

So the Circumference being 47 inches 13 parts, the length of a foot Solid will be found about 9 inches 74 parts.

• As 47.13 unto 147.36: so 1.00 to 3.13.
• and so are 3.13 unto 9,78.

### 4. Having the Circumference of a Cylinder given in foot measure, to find the length of a foot Solid in foot measure.

• As the circumference in feet to 3.545:
• So is 1 to a fourth number;
• and that fourth to the length in foot measure.

So the Circumference being 3 foot 927 parts, the length of a foot Solid will be found to be about 815 parts.

• As 3.927 unto 3,545: so 1.000 unto 0.903.
• and so are 903 unto 815.

### 5. Having the side of a Square equal to the Base of a Cylinder, to find the length of a foot Solid.

The side of a square equal to the Circle, may be found by the eighth Proposition of broad measure, and then this Proposition may be wrought by the first and second Proposition of Solid measure.

### 6. Having the Diameter of a Cylinder, and the length given in inches, to find the content in inches.

• As 1.128 unto the Diameter in inches:
• So the length in inches to a fourth number;
• and that fourth number to the content in inches.

So the Diameter being 15 inches, and the length 105, the content of the Cylinder will be found to be about 18555 inches.

• As 1.128 unto 15: so are 105 unto 1395.87:
• and so are 1396.87 unto 18555.34.

### 7. Having the Diameter and length of a Cylinder in foot measure, to find the content in feet.

• As 1.128 to the Diameter in feet:
• So the length in feet to a fourth number;
• and that fourth number to the content in feet.

So the Diameter being 1 foot 25 parts, and the length 8 foot and 75 parts, the content of the Cylinder will be found about 10 foot 75 parts.

• As 1.128 unto 1.25: so 8.75 unto 9.69;
• and so are 9,69 unto 10.74.

### 8. Having the Diameter of a Cylinder, and the length given in inches, to find the content in feet.

• As 46.90 to the Diameter in inches:
• So the length in inches to a fourth number;
• and that fourth to the content in feet.

So the Diameter being 15 inches, and the length 105, the content will be found about 10 foot 74 parts.

• As 46.906 unto 15: so 105 unto 33.58;
• and so are 33.58 unto 10.74.

### 9. Having the Diameter of a Cylinder, given in inches, and the length in feet, to find the content in feet.

• As 3.54 to the Diameter in inches:
• So the length in feet, to a fourth number;
• and that fourth to the content in feet.

So the Diameter being 15 inches, and the length 8 foot 75 parts, the content will be found about 10 foot 74 parts.

• As 13.54 unto 15: so 8.75 unto 9.69:
• and so are 9.69 unto 10.74.

### 10. Having the Circumference and length of a Cylinder given in inches, to find the content in inches.

• As 3.545 to the Circumference in inches:
• So the length in inches to a fourth number;
• and that fourth to the content in inches.

So the Circumference being 47 inches 13 parts, and the length 105 inches, the content will be found about 185.55 inches.

• As 3.545 unto 47.13: so 105 unto 1396.
• and so are 1396 unto 18555.

### 11. Having the Circumference and length of a Cylinder given in inches, to find the content in feet.

• As 147.36 to the Circumference in inches:
• So the length in inches to a fourth number;
• and that fourth to the content in feet.

So the Circumference being 47 inches 13 parts, and the length 105 inches, the content will be found about 10 foot 74 parts.

• As 147.36 unto 47.13: so 105 unto 33.58.
• and so are 33.58 unto 1074.

### 12. Having the Circumference and length of a Cylinder given in foot measure, to find the content in feet.

• As the 3.545 to the Circumference in feet:
• So the length in feet to a fourth number;
• and that fourth to the content in feet.

So the Circumference being 3 foot 927 parts, and the length 8 foot 75 parts, the content will be found to be 10 foot 74 parts.

• As 3.545 unto 3,927: so 8.75 unto 9.69.
• and so are 9,69 unto 10,74.

### 13. Having the Circumference of a Cylinder given in inches and the length in foot measure, to find the content in feet.

• As 142,54 to the Circumference in inches:
• So the length in feet to a fourth number;
• and that fourth to the content in feet.

So the Circumference being 47 inches 13 parts, and the length 8 foot 75 parts, the content will be found as before 50 foot 74 parts.

• As 142,54 unto 47,13: so 8,75 unto 9,69:
• and so are 9,69 unto 10.74.

## SECT. III.Of the Mensuration of Cones.

### 1. The Diameter of the base and the length of the side of a Cone being given, to find the superficial content thereof.

• As 7 is to 22; Or 113 to 355.
• So is ½ the Diameter 6 multiplied in 18 the side,
• To the Superficial Content 339.29.

So the Diameter of the Base of a right Cone being 12 inches, and the side thereof 18 inches, the Area will be found to be 339,29. For,

If you extend the Compasses from 7 to 22, or from 113 to 355, the same extent will reach from 108, (which is half the Diameter multiplied in the side) to 339,29 the Area, or Superficial content.

### 2. The Diameter and Axis of a right Cone being given, to find the Solid Content.

• As 28, Is to 22:
• So is the Square of the Diameter 144, multiplied by 1/3 of the Axis viz. 86.8.
• To the Solid Content of the Cone 678.85.

So the Axis of a Cone being 18 18 inches, and the Diameter 12 inches, the Solid content will be found to be 678.85.

Extend the Compasses from 28 to 22. The same extent will reach from 864 (1/3 of the Axis multiplied in the Square of the Diameter) to 678,85 the solid content.

## SECT. IV.Of the Mensuration of Spheres.

### 1. The Diameter of a Sphere being given, to find the Superficial content.

• As 7 is to 22, or 113 to 355.
• So is the Square of the Diameter 144
• To the superficial content.

Thus a Sphere whose Diameter is 12 inches, the superficial content thereof will be found to be 452.57.

Extend the Compasses from 7 to 22, the same extent will reach from 144 (the square of the Diameter) to 452.57 the superficial content.

### 2. The Superficies of a Sphere being given, to find the Axis.

• As 22, is to 7:
• So is the Superficies
• To the square of the Diameter.

So a Sphere whose Superficies is 452,57 inches, the Diameter thereof will be found to be 12 inches.

Extend the Compasses from 22 to 7, the same extent will reach from 452.57 (the Superficies) to 144, the Square of the Diameter, the ½ distance between 144 and 1 is 12 the Diameter.

### 3. The Axis of a Sphere being given, to find the Solid content.

• As 42, Is to 22:
• So is the Cube of the Diameter
• To the Solidity.

So if the Axis of a Sphere be 12 inches, the Solid content thereof will be found to be 590.62.

Extend the Compasses from 42 to 22, the same extent will reach from 1728 (the Cube of the Diameter) to 905.14, the Solid content.

### 4. The Solidity of a Sphere being given, to find the Axis.

• As 22, Is to 42:
• So is the Solidity
• To the Cube of the Diameter, or Axis.

So a Sphere whose Solid content is 905.14. the length of the Axis will be found to be 12 inches.

Extend the Compasses from 22 to 42, the same extent will reach from 905.14, the Solidity, to 1728 the Cube of the Axis.

## SECT. V.Of the Mensuration of Prismes.

A Prisme is a Solid figure contained under Planes; whereof the two opposite are equal, like, and Parallel; but the other are Parallelograms. Euclid. Defin. 13. Lib 11.

### 1. To find the Solid content of a Triangular Prisme.

Suppose a piece of Timber or Stone to be an Equilateral Triangle at the ends, each side thereof being 2.25 foot, and the length of the piece 17.75 foot, this is called a Triangular Prisme.

1. Find the content of the Triangle at the end of the piece (by the tenth aforegoing) which will be found to be 2.19. Then say,

• As 1, Is to the Area of the Base:
• So is the Length of the Piece
• To the content of the Piece in foot measure.

Extend the Compasses from 1 to 2.19 (the Content of the Area of the base in feet) the same extent will reach from 17.75 (the length of the piece in feet) to 38.87, the content of the Piece in feet.

### 2. To find the Solid Content of a Regular Solid, whose sides at the end thereof are equal, and more than 3. As 4, 5, 6, 7, 8, or 10, &c.

Suppose a Regular Solid, as of Timber or Stone, the Plane at the Base or end thereof being a Pentagon, or Figure of 5 equal sides and angles, each side being 12 inches, or one foot, and the length of the Solid 14 foot.

1. Find the Content of the Base (or Pentagon) at the end, by the 1. of the second Section beforegoing, which will be found to be 1.725 foot, the Perpendicular of the Pentagon being 0.69 parts of a foot.

Note, The Perpendiculars in these Regular Polygons may be found exact enough for these kinds of Mensurations by taking the least distance from the Center to one of the sides of the Polygon.

Then say,

• As 1, Is to the Content of the Base in feet 1.725:
• So the length of the piece 14 foot,
• To the Content of the Piece in feet 24.15.

Extend the Compasses from 1to 1.725, (the Content of the Base) the same extent will reach from 14 foot, the length of the Piece, to 24.15, the Content of the Piece in feet.

And in the same manner, if the side of an Octagon were 12 inches or 1 foot, the perpendicular would be found to be 1.64, and the length 21.5 feet, the Solidity would be found to be 103.20.

## SECT. VI.Of the Mensuration of Pyramides.

A Piramide is a Solid figure compregended under divers Planes, set upon one Plane, (which is the Base of the Pyramide) and gathered together to one Point. Euclid. Lib. II. Defin. 12.

The Base of Pyramids may be either Triangle, Squares, Pentagons, Hexagons, &c. as the Prismes were; Wherefore to measure any Pyramis, you must first find the Area, or Content of the Base, and then say,

• As 1, Is to the Area or content of the Base 2.25:
• So is one third part of the height 15 feet,
• To the Content of the Piece in feet 24.15.

Suppose a Pyramis, whose Base is a Square, each side being 18 inches, or 1.5 feet, and the height of the same Pyramis were 45 feet, and it were required to find the Solidity. The Area of the Base by the second of the fifth Section beforegoing, will be found to be 2.25 feet.

Extend the Compasses from 1, to 2.25, (the Content of the Base) the same extent will reach from 15, (one third part of the height) to 33.75, the Solid content of the Pyramid in feet.

And the like of any other.

## SECT. VII.Of the Mensuration of Frustrums or Segments of Pyramids or Cones.

The Solidity of every Cone or Pyramid is found by multiplying the Area of the Base (of what form soever) into one third part of the Altitude; Therefore in a Cone whose Base is Circular, and the Diameter of that Circle is in Foot measure 2.50, its Area will be found by what is delivered in the foregoing Sections to be 4.91, and its Altitude 56.25 foot; I say,

• As 1, To the Area of the Base:
• So is one third of the Altitude
• To the Solid Content.

So the Area of the Base of a Cone or Pyramis being 4.98, and the Altitude 56.25, the Solid Content thereof will be found to be 92.06 foot.

Extend the Compasses from 1to 4.91 the Area of the base, the same extent shall reach from 18.75, the third part of the Altitude, to 92.06, the Solid Content of the Cone or Pyramis.

But if this Cone or Pyramis were cut off at 18 foot from the Greater end, and then the lesser Bases Area shoul be found to be in foot measure 2.27, what shall the Solidity of the Frustrum be? And in this nature do most Timber Trees grow, and so being cut off ought to be measured, being either Squared or Growing; And no greater Error is here committed in the Measuring of Timber, it being in this form, than by the vulgar way of measuring such Timber, which is, by finding out the Square in the Middle of the Piece, and taking of that for the true Square, but this always makes the Content of the Piece less than it is; The Genuine and true way is this.

Multiply the Area of the two Bases together, and from the Product extract the Square Root, then add this Root, and the two Areas together, which sum multiplied by one third part of the Length of the Frustrum or part shall give the Solid Content of that piece.

So a Piece of Stone or Timber whose Area at one end is 4.91 (as in the former Piece) and at the Smaller end 2.27, and its length 18 foot; the Solidity by the former Rule will be found to be 63.48 foot. For,

• As 1, Is to the Greater base:
• So is the Lesser Base
• To a fourth number,

Whose Square Root being Extracted, and added to the two former Areas, will produce another number. Then say,

• As 1, Is to this number last found:
• So is one third of the length of the Piece,
• To the Solid Content of the Piece.

Therefore extend the Compasses from 1 to 4.91, the greater Base, the same extent shall reach from 2.27, the lesser Base, to 11.15. A mean Proportional between 1 and 11.15 will be found to be 3.34, which added to the other two Areas 4.91, and 2.27, (as is done in the margin,) will produce 10.52, which is your other number sought for: Then,

 Greater Base 4 91 Lesser Base 2 27 Square Root of 11.15 3 34 10 52

Extend the Compasses from 1 to 10.52, the same will reach from 6, (the third parts of 18 the Length,) to 63.12, the Solid Content of the Piece which is 63 foot, and half a quarter of a foot.

And now for Proof of this Work to be true, let us find the Solidity of the upper or lesser part of the whole Cone which was 56.25 foot long.

The Lesser Base, 18 foot being cut off of the whole Length is found to be 2.27, and 18 being taken from 56.25 the whole length, there will remain 38.25, and the third part whereof is 12.75, which multiplied by 2.27, the Base produceth 28.94 for the Solidity of the Lesser Cone or Pyramis, and this being added to 63.12, the Content of the Frustrum produceth 92.05, the which is equal to the whole Cone or Pyramis, both the parts equal to the whole, which proveth the Work to be true.

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