Cocker's decimal arithmetick wherein is shewed the nature and use of decimal fractions ... together with tables of interest and rebate ... : whereunto is added, his Artificial arithmetick, shewing the genesis ... of the logarithmes ... : also, his Algebraical arithmetick, containing the doctrine of composing and resolving an equation, with all other rules requisite for the understanding of that mysterious art according to the method used by Mr. John Kerley in his incomparable treatise of algebra / composed by Edward Cocker ... ; perused, corrected, and published by John Hawkins ...

CHAP. XI. The Use of the Square and Cube Roots in solving some Questions Arithmetical and Geometrical. (Book 11)

PROP. I. To find a mean proportional be∣tween two given Numbers.

MUltiply the given Numbers the one by the other, and extract the square Root of the product, so shall that square Root be the mean proportional sought.

Example.

Let the given Numbers be 12 and 48, and let it be required to find a mean proportional be∣tween them; first multiply the given numbers 12 and 48 the one into the other, and their pro∣duct is 576, the Square Root of which is 24, so

that I conclude 24 to be a mean proportional between 12 and 48, for, 〈 math 〉〈 math 〉 the square of the mean being equal to the pro∣duct of the extreams.

This proposition is useful in finding the side of a square that shall be equal to any given paralel∣logram; for, (according to the first Proposition of the Eighth Chapter of this Book,) if you mul∣tiply the contiguous sides of a Rectangular para∣lelogram the one by the other, that product will be its content, and if you extract the square root of that content, it will give you the side of a square, (in the same measure your paralelogram was) which will be equal to the given Paralelo∣gram.

PROP. II. To find the side of a Square that shall be equal to the Content of any given superficies.

FInd out the Content of the given superficies by the Rules laid down in the Eighth Chap∣ter, and then extract the Square Root of the

Content, so will that Root be the side of a square equal to the given superficies.

Example.

There is a Rectangled Triangle whose base and perpendicular are 16 and 18, I demand the side of a Square that will be equal to the given Triangle.

According to the second Proposition of the Eighth Chapter, I find the Content of this Triangle tobe 144, the square root of which is 12, and is the side of a square equal to the said Triangle.

In like manner, if you extract the square root of the Content of a Circle, Pentagon, Hexa∣gon, &c. or of any other Figure regular, or ir∣regular, it will give the side of a square equal to that superficies.

PROP. III. Having any two of the sides of a Right-angled plain Triangle, given to find the third side.

THis most excellent and useful proposition is generally called Pythagoras his Theoreme, and in the 47 Prop. of E•…•…clides Elements of

Geom. it is demonstrated, and proved that the Square made of the Hypothenuse, or slant side of a Right angled plain Triangle is equal to the sum of the squares made of the base and perpendicular.

As for Example.

In the Triangle ABC, the Base AB is 48, and the perpendicular BC is 36, now I demand the length of the Hypothenuse AC.

To find out

[illustration]

an answer to this, first, I square the base AB, (48) which is 2304, then square the Perpendicular (36) and its square is 1296, the sum of which two squares is 3600, which is equal to the Square of the Hypothenuse AC, therefore the square Root of 3600 will give the length of AC, which is 60.

PROP. IV.

THere is a Tower about which there is a Moat that is 48 foot wide, and a scaleing Ladder that is 60 Foot long, will reach from the outside of the Moat, to the top of a Wall, that is within the said Moat, now I demand the height of the said Wall above the Water?

Let the Base AB in the foregoing Triangle be the breadth of the Moat, and let the Hypo∣thenuse AC be the scaling. Ladder, then is the perpendicular BC the height of the Wall above the Water. Now it is plain that (be∣cause the Square of AC is equal to the sum of the squares of AB and BC) if from the square of AC which is 3600 you subtract the square of AB which is 2304, there will remain 1296, which is the square of CB, therefore I extract the square Root of 1296, and find it to be 36, which is the height of the said Wall above the Wa∣ter as was required.

By the help of this Proposition may be found the true perpendicular height of a Cone, or of a Pyramid; for, in a Cone, if you square the slant height, (which is the length of a line drawn from its vertical point, to the Circumfe∣rence of its base) and from the square of that, subtract the square of the semidiameter of its base, there will remain the square of the per∣pendicular height of that Cone.

Also, In a Pyramid, if from the square of the slant height of it, you subtract the square of that line which being drawn from the Centre of its base, shall touch the end of the said slant line, (whether they meet at an Angle or not) the remainder will be the square of the perpen∣dicular height of that Pyramid, and its square Root will give the height it self.

PROP. V. By the Content of a Circle to find its Diameter.

The proportion is

AS 22.

Is to 28.

So is the given Content

to the square of the Diameter.

Example.

There is a Circle whose superficial Content is 153.9385, I demand its Diameter?

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The Square Root of which is 13.99 (very near 14) for the Diameter required.

PROP. VI. By the Content of a Circle to find its Circumference.

The proportion is

AS 7

Is to 88

So is the given Content

to the square of the Circumference.

The Square Root of which is the Circumfe∣rence required.

Example.

There is a Circle whose superficial content is 153.9385, I demand the Circumference of that Circle? 〈 math 〉〈 math 〉 The square Root of which is 44 fere which is the Circumference required.

II. The Cube Root is that by help of which we resolve all Questions Mathematical that con∣cern solidity, and by which we increase solid bo∣dies according to any given proportion. By it we discover the solidity of a body that is capable of length, breadth, and depth, (or thickness,)

and by having the solidity given, we discover the side or Diameter of such a body.

Some Questions pertinent thereto may be such as follow.

PROP. VII.

THere is a Cube whose side is 4, I demand what shall be the side of a Cube whose soli∣dity is double to the solidity of that Cube?

To answer this proposition, find out the Cube of 4 (the side of the given Cube) which is 64, and double it, which is 128, then extract the Cube Root of 128, and it makes 5.0397 fere, and that is the side of the Cube which is double to the Cube whose side is 4.

PROP. VIII.

THere is a Cube whose solidity is 128 foot, I demand the side of a Cube whose solidity is half as much?

Take ½ of 128 = 64 the Cube Root of which (viz, 4.) answers the Question.

PROP. IX.

HAving the solid Content of a Globe to find the side of a Cube whose solidity shall be equal to the given Globe?

Extract the Cube Root of the given solid Con∣tent of the Globe, and it will give you the side of the Cube Required.

Example.

There is a Globe whose solid Content is 1728 Inches, I demand the side of a Cube equal there∣to?

Having extracted the Cube Root of 1728, I find it to be 12, which is the side of the Cube re∣quired.

PROP. X.

HAving the Diameter and Weight of a Bul∣let, to sind the Weight of another Bullet, whose Diameter is given.

As the Cube of the given Bullets Diameter,

Is to its weight, or solidity.

So is the Cube of the Diameter of any other Bullet,

To its weight, or solidity.

Examp'e.

There is a Bullet whose Diameter is 4 Inches, and its weight is 9 Pound, I demand the weight of another Bullet, whose Diameter is 6•…•… or 6.25 Inches?

The Cube of 4 is 64.

The Cube of 6.25 is 244.140625

Then I say

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So that the weight required is 34.33227 pounds, and if you Reduce the Decimal to the known parts of Averdupois weight, you will find the answer to be 34 lb. −05 oz.−05 dr.

This kind of proportion is by Artists Termed triplicate proportion.

In like manner, the Diameters of two Bullets, or Globes being given, and the solidity of one of them to find out the solidity of the other, it may be done by the same proportion, only chan∣ging the middlemost Term.

PROP. XI.

TO find the side of a Cube equal to a given paralelepipedon.

Find out the solidity of the given Paralelepipe∣don by the Eighth Prop. of the Eighth Chapter, then is the Cube Root thereof, the required side.

Example.

There is a paralelepipedon having the sides of its base 10 Foot 4 Inches, and 5 Foot 2 Inches, and its length is 20 Foot 8 Inches, I desire to know what is the side of a Cube whose content shall be equal to the given paralelepipedon?

The superficial Content of the base is 7688 in∣ches, which drawn into 248 the length in inches, the product is 1906624 inches for its solid Con∣tent, the Cube Root of which is 124 inches for the side of a Cube equal to the given paralele∣pipedon.

In like manner if you would find at any time the side of a Cube equal to any solid Body whe∣ther Regular, or irregular: First, Find the so∣lid

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Content of that Body, and then extracting the Cube Root of its solid Content you have your desire.

PROP. XII.

BEtween two given Numbers to find two mean proportionals.

Divide the greater extream by the lesser, and extract the Cube Root of the Quotient, and by the said Cube Root multiply the lesser extream, then will the product give you the lesser mean proportional, then multiply the said lesser mean by the said Cubique Root, and that product will give you the greater mean proportional.

Example.

Let the two given extreams be 6 and 48 be∣tween which it is required to find 2 mean pro∣portionals.

First, I divide 48 (the Greater Extream) by 6 (the Lesser Extream) and the Quotient is 8, the Cube Root of which is 2, then by (the Cube Root) 2 do I multiply 6 (the lesser extream, and the product is 12 for the lesser mean pro∣portional, and 12 being multiplyed by 2 (the Cube Root) the product is 24, for the greater mean proportional sought. Thus have I found 12 and 24 to be two mean proportionals between 6 and 48, for

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In like manner between 3 and 81 will be found 9 and 27, for two mean proportionals.

PROP. XIII.

THE, Concave Diameter of two Guns being known, and the quantity of Gun-powder that will charge one of them, to find out how much will be sufficient to charge the other.

The Capacities are one to another, as are the Cubes of their Diameters, and also the propor∣tion is direct.

Example.

If .25 pound of Gun-powder be sufficient to charge a Gun, whose Concave Diameter is 1½ Inches, or 1.5 Inch. how much powder will be sufficient to charge a Gun, whose Concave Dia∣meter is 7 inches? Answer, 25.47.

The Cube of 1.5 is 3.375 and the Cube of 7 is 343. wherefore the proportion is as followeth.

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Or thus,

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PROP. XIV.

THE Concave Diameters of two Guns being given, and the quantity of a weaker sort of Gun-powder sufficient to charge one of them,

to find out how much Gun-powder of a stron∣ger sort (the proportion of the strength and weakness of the Gun-powder being also given) will be sufficient to charge the other Gun.

This is solved by two operations in the Rule of proportion, first to find out how much of the stronger sort of Gun-powder will be of equi∣valent strength with the given quantity of the weaker sort, and this proportion is Reciprocal; The second is the same with that in the fore-going Proposition.

Example.

There is a Gun whose Concave Diameter is 1½ inches, and it requireth 25 pound of powder to Charge it, now there is another sort of Gun-powder which is much stronger than the for∣mer, and the proportion between their strength is as 5 to 2, now I demand how much of the strongest powder is sufficient to charge a Gun, whose concave Diameter is 7 inches?

To answer this, First, I find out how much of the strongest powder will charge that Gun▪ which is 1½ inch in its Concave Diameter, which is done by the following proportion, viz.

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Thus have I found that 1/10 of a pound of the strongest powder will charge a Gun whose Con∣cave diameter is 〈◊〉〈◊〉 inch. And according to the last proposition, I find by a direct Proportion that 10.16 pounds of the same will be sufficient to charge a Gun whose concave diameter is 7 inches, viz.

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