The First Book of the Cross-Staff

CAPT. VI. The Use of the Line of Numbers.

The Line of Numbers here noted 1, 2, 3, 4, unto 10, is compleat in those divisions which are between 1 and 10: the other like dicisions at the beginning of the Line do serve rather to answer to the first degrees of the other two lines of Sines and Tangents, than for any necessity, which is the cause, why some of them are omitted. And here, as in the use of other Scales, the figures 1, 2, 3, 4, are set down upon the Line do sometimes signifie themselves alone, sometimes 10, 20, 30, 40, sometimes 100, 200, 300, 400, and so foreward, as the matter shall require. The first figure of every Number is always that which is here set down, the rest must be supplied according to the nature of the question.

1. Having two numbers given, to find a third in continual proportion, a fourth, a fifth, and so foreward.

Extend the Compasses from the first Number unto the second; then may you turn them from the second to the third, and from the third to the fourth, and so forward.
Let the two numbers given be 2 and 4, extend the Compasses from 2 to 4, then you may turn them from 4 to 8, nad from 8 to 16, and from 16 to 32, and from 32 to 64, and from 64 to 128.
Or, if one foot of the Compasses being set to 64, the other fall out of the Line, you may set it to another 64 nearer the beginning of the Line, and there the other foot will reach to 128, and from 128 you may turn them to 256, and so forward. Or if the two first Numbers given were 9 and 10: extend the Compasses from 10 at the end of the Line, back unto 9, then you may turn them from 9 unto 8,1, and fro 8,1 unto 7,29. And so, if the two first numbers given were 1 and 9, the third would be found to be 81, the fourth 729, with the same extent of the Compasses.
In the same manner, if the first two numbers were 10 and 12, you may find the third proportional to be 14,4, the fourth 17,28. And with the same extent of the Compasses, if the first two Numbers were 1 and 12, the third would be found to be 144, and the fourth to be 1728.

2. Having two extreme Numbers given, to find a mean proportional between them.

Divide the space between the extreme Numbers into two equal parts, and the foot of the Compasses will stay at the mean proportional. So the extreme Numbers given being 8 and 32, the mean between them will be found to be 16, which may be proved by the former Proposition, where it was shewed, that as 8 to 16, so are 16 to 32.

3. To find the square Root of any Number given.

The square Root is always the mean proportional between 1 and the Number given, and therefore to be found by dividing the space between them into two equal parts; So the Root of 9 is 3, and the Root of 81 is 9, and the Root of 144 is 12, and the Roor of 1440 is almost 38.

If you suppose Pricks under the Number given, (As in Arithmetical extraction), and the last Prick to the left hand shall fall under the last figure, which will be as oft as there be odd figures, the unity will be best placed at 1 in the middle of the Line: so the Root and the Square will both fall forward toward the end of the Line. But if the last Prick shall fall under the last figure but one, which will be as oft as there be even figures, then the unity may be placed at 1 in the beginning of the Line, and the Square in the second length, or rather the unity may be placed at 10 in the end of the Line of the Root, and the Square will both fall backward toward the middle of the Line, in the second length.

4. Having two extreme Numbers given, to find two mean Proportionals between them.

Divide the space between the two extreme numbers given into three equal parts. As if the extreme Numbers given were 8 and 27, divide the space between them into three equal parts, the feet of the Compasses will stand in 12 and 18.

5. To find the Cubic Root of a Number given.

The Cubic Root is always 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 the Root of 1728 will be found to be12. The Root of 17280 is almost 26: and the Root of 172800 is almost 56. If you suppose a Prick under the Number given after the manner of Arithmetical extraction, and the last Prick to the left hand shall fall under the last figure, as doth in 1728, the unite will be best placed at the middle of the Line, and the Root, the Square, and the Cube, will all fall forward toward the end of the Line.

If the last Prick shall fall under the last figure but one, as in 17280, the untite may be placed at 1 in the beginning of the Line, and the Cube in the second length, or the unite may be placed at the 10 in the end of the Line: and the Cube in the first length, or if the Cube falls out of the Line, you may help your self, as in the first Proposition. But if the last Prick shall fall under the last figure but two, as in 172800, then place the unite always at 10 in the end of the Line: so the Root, the Square, and the Cube, will fall all backward, and be found in the second Length between the middle and the end of the Line.

6. To multiply one number by another.

Extend the Compasses from 1 to the Multiplicator; the same extent applied the same way, shall reach from the Multiplicand to the Product.

As if the Numbers to be multiplied were 25 and 30: either extend the Compasses from 1 to 25, and the same extent will give the distance from 30 to 750; or extend them from 1 to 30, and the same extent shall reach from 25 to 750.

7. To divide one Number by another.

Extend the Compasses from the Divisor to 1, the same extent shall reach from the Dividend to the Quotient. So if 750 were to be divided by 25, the Quotient would be found to be 30.

8. Three Numbers being given, to find a fourth Proportional.

This golden Rule, the most useful of all others, is performed with like ease. For extend the Compasses from the first Number to the second, the same extent shall give the distance from the thrid to the fourth.

As for example, the proportion between the Diameter and the Circumference, is said to be such as 7 to 22: if the Diameter be 14: how much is the Circumference? Extend the Compasses from 7 to 22, the same extent shall give the distance from 14 to 44: or extend them from 7 to 14, and the same extent shall reach from 22 to 44.

Either of these ways may be tried on several places of the Line; but that place is best, where the feet of the Compasses may stand nearest together.

9. Three Numbers being given, to find a fourth in a duplicate proportion.

If any have daily use of this Proportion, he may cause another Line of Numbers be made.

This Proposition concerns questions of proportion between Lines and Superficies; where if the denominations be of Lines, extend the Compasses from the first to the second Number of the same denomination: so the same extent being doubled, shall give the distance from the third Number unto the fourth.

The Diameter being 14, the content of the Circle is 154; the Diameter being 28, what may the content be? Extend the Compasses from 14 unto 28, the same extent doubled will rech from 154 to 616.

For first, it reaches from 154 unto 308; and turning the Compasses once more: it reaches from 308 unto 616; and this is the content required.

But if the first denomination be of the superficial content, extend the Compasses unto the half of the distance, between the first Number and the second of the same denomination: so the same extent shall give the distance from the third to the fourth.

The content of a Circle being 154, the Diameter is 14: the content being 616, what may the Diameter be? Divide the distance between 154 and 616 into two equal parts, then set one foot in 14, the other will reach to 28, the Diameter required.

10. Three Numbers being given, to find a fourth in a triplicate proportion.

This Proposition concerneth questions of proportion between Lines and Solids; where the first denomination be of Lines, extend the Compasses from the first Number to the second of the same denomination: so the extend being tripled, shall give the distance from the third Number unto the fourth.

Suppose the diameter of an Iron Bullet being 4 inches, the weight of it was 9 l. the Diameter being 8 inches, what may the weight be? Extend the Compasses from 4 to 8, the same extent being tripled, will reach from 9 unto 72. r first, it reacheth from 9 unto 18; then from 18 unto 36; thirdly, from 36 unto 72. And this is the weight required.

But if the first Denomination shall be of Solid content, or of the weght, extend the Compasses to a third part of the distance between the first number and the second Denomination; so the same extent shall give the distance from the third Number unto the fourth.

The weight of a Cube being 72 l. the side of it was 8 inches: the weight being 9 l. what may the side be? Divide the distance between 72 and 9, into 3 equal parts; then set one foot to 8, the other will reach to 4, the side required.