Of the Construction and Use of Coggeshall’s Sliding-Rule for Measuring.
This Rule is framed three Ways; for some have the two Rulers composing them sliding by one another, like Glaziers Rules; and sometimes there is a Groove made in one Side of a Two-Foot Joint-Rule, in which a thin sliding Piece being put, the Lines put upon this Rule, are placed upon the said Side. And lastly, one Part sliding in a Groove made along the Middle of the other, the Length of each of which is a Foot: the Form of this last being represented by Fig. 12.
Fig. 12
Upon the sliding Side of the Rule are four Lines of Numbers; three are double Lines, or Lines of Numbers to two Radius’s, and one a Angle broken Line of Numbers, marked by the Letters A, B, C, and D.
The three double Lines of Numbers A, B, C, are figured 1, 2, 3, 4, 5, 6, 7, 8, 9; and then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; they being the same as the Line A, and the two Lines on the Sliding-Piece C, upon Everard’s Sliding-Rule; and their Construction, Use, and Manner of using, are also the same.
The single Line of Numbers D, whose Radius is exactly equal to the two Radius’s of either of the Lines of Numbers A, B, C, is broke, for easier measuring of Timber, and figured thus, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40; this Line is called the Girt Line: from 4 to 5 it is divided into 10 Parts, and each Tenth into two Parts, and so on from 5 to 10; then from 10 to 20, it is divided-into 10 Parts, and each Tenth into 4 Parts, and so on from 20 to 40, at the End, which is right against 10, at the End of either of the double Lines of Numbers.
The Lines on the Back-side of this Rule, are these; a Line of Inch-Measure from 1 to 12, each Inch being subdivided into Halfs, Quarters, and Half-quarters: another Line of Inch-Measure from 1 to 12, and each Inch subdivided into 10 equal Parts: a Line of Foot-Measure, being one Foot divided into 100 equal Parts, and figured, 10, 20, 30, &c. to 100.
The Back-side of the Sliding-Piece is divided into Inches, Halfs, and Half-quarters, and figured from 12 to 24; so that it may be slid out to 2 Foot, to measure the Length of any thing.
The Lines of Numbers, A, B, or C, being either of them constructed (which see in the Chapter concerning Gunter’s Scale) the Line D, from thence, may easily be constructed.
For having set 4 directly under 1, for the Beginning of the Line; to find where any Division suppose, must be placed, take twice the Distance from 4 to 5, on either of the Radius’s of either of the Lines of Numbers A, B, C, and lay off from 4 that Extent, which will give the Division 5. Proceed thus for all the other Divisions and Subdivisions, by always taking the double of them on the Lines A, B, or C.
Note, For the manner of Notation, on this Rule, see the Line of Numbers on the Carpenters.
Section I. The Use of this Rule in measuring plain Superfices
Use I.To measure a Geometrical Square.
Let there be a Square whose Sides are each 5 Feet; set 1 on the Line B, to 5 on the Line A; then against 5 on the Line B, is 25 Feet the Content of the Square on the Line A.
Use II.To measure a right angled Parallelogram, or Long-Square.
Let there be a Parallelogram, whose longest Side is 18 Feet, and shortest 10; set 1 on the Line B, to 10 on the Line A; then against 18 Feet on the Line B, is 180 Feet the Content on the Line A.
Use III.To measure a Rhombus.
Let the Side of a Rhombus be 12 Feet, and the Length of a Perpendicular let fall from one of the obtuse Angles, to the opposite Side, 9 Feet; set 1 on the Line B, to 12, the Length of the Side, on the Line A: then against 9, the Length of the Perpendicular on the Line B, is 108 Feet the Content.
Use IV.To measure a Rhomboides.
Suppose the Length of either of the longest Sides of a Rhomboides to be 25 Feet, and the Length of the Perpendicular let fall from one of the obtuse Angles to the opposite longest Side, is 8 Feet; let 1 on the Line B, to 25, the Length, on the Line A: then against 8 Feet on the Line B, stands 200 Feet the Content.
Use V.To measure a Triangle.
Let the Base of a Triangle be 7 Feet, and the Length of the Perpendicular let fall from the opposite Angle to the Base, 4 Feet. Set 1 on the Line B, to 7 on the Line A; then against half the Perpendicular, which is 2, on the Line B, is 14 on the Line A, for the Content of the Triangle.
Use VI.The Diameter of a Circle being given, to find it’s Content.
Let the Diameter of a Circle be 3.5 Feet: set 11 on the Girt-Line D, to 95 on the Line C; then against 3.5 Feet on D, is 9.6 on the Line C, which is the Content in Feet of the said Circle.
The Reason of this Operation, is, that As the Square of 11, which is 121, is to 95; So is the Square of the Diameter of any Circle, to it’s Content. Also, from the Nature of the Logarithms, it is manifest, if any Number, taken on a single Line of Numbers (whether whole or broken, in the manner that the Line D is), be set to another Number, taken on a double Line of Numbers of the same Length; that the Square of the Number taken on the single Line of Numbers, will be to the Number it is set against; on the double Line of Numbers, as the Square of any other Number, taken on the single Line of Numbers, to the Number against it on the double Line of Numbers.
Use VII.To find the Content of an Oval or Ellipsis.
Let the Transverse, or longest Diameter, be 9 Feet, and the Conjugate, or shortest Diameter, 4 Feet; to find the Content of this Ellipsis.
Theorem. The Content of every Ellipsis, is a mean Proportional between a Circle, whose Diameter is equal to the longest Diameter of the Ellipsis, and a Circle whose Diameter is equal to the shortest Diameter of the same Ellipsis; as is manifest per Cor. 3. Prop. XI. Lib. 11. of Sturmy’s Mathesis Enucleata.
Therefore a mean Proportional must first be found between 4 and 9, the longest and shortest Diameters; to do which by the Sliding-Rule, set the greater of the two Numbers 9 on the Girt Line, to the same Number on the Line C; then against the lesser Number 4, on the same Line C, is 6 the mean Proportional sought on the Girt-Line. Now we have only the Content of a Circle to find, whose Diameter is 6 Feet; which, when found, will be the Content of the Ellipsis sought: therefore (by the last Problem) set 11 on the Girt-Line D, to 95 on the Line C; then against 6 Feet on the Girt-Line D, stands on the Line C, 28.28 Feet for the Content of the aforesaid Ellipsis.
The Reason of the Operation for finding a mean Proportional between two Numbers, as 4 and 9, is manifest from what I said in the last Use of the Property of a double and single Line of Numbers Hiding by one another. And from this Theorem, viz. That if there are three Numbers continually proportional (as 4, 6, and 9), the Square of the greatest (as 81) is to the greatest (9), as the Square of the middle one (6), or the Rectangle under the Extremes (which is equal to it, per Prop. 20. Lib. 7. Eucl.) is to the lesser Extreme (4).
This Use may be easier solved at one Operation by the Lines A and B, thus; set 1.27 on the Line E, to the transverse Axis 9 Feet, on the Line A: then against the Conjugate Axis 4, on the Line B, stands 28.28 Feet on the Line A, for the Content.
Note, The standing Number 1.27, is the Quotient of 14 divided by 11; also as 14 is to 11, so is the Rectangle under the transverse and conjugate Axes of any Ellipsis to it's Area; whence the Reason of this Operation is easily manifest.
Section II. Of measuring Timber.
Use I.To measure Timber the common Way.
Take the Length in Feet, Half-feet (and if desired), in Quarters; then measure half-way back again, where girt the Tree with a small Cord or Chalk-Line; double this Line twice very even, and this fourth Part of the Girt, or Circumference, which is called the Girt, measure in Inches, Halfs, and Quarters of Inches; but the Length must be given in Feet, and the Girt in Inches. The Dimensions being thus taken, the Tree is to be measured as square Timber, the Girt, or \(\frac{1}{4}\) of the Circumference being taken for the Side of the Square, in the following manner.
Always set 12 on the Girt-Line D, to the Length in Feet on the Line C; then against the Side of the Square, on the Girt-Line D, taken in Inches, you will find on the Line C the Content of the Tree in Feet.
Example I. Suppose the Girt of a Tree, in the middle, be 60 Inches, and the Length 30 Feet, What is the Content? Set 12 on the Girt-Line D, to 30 Feet on the Line C; then against 15, the one fourth of 60, on the Girt-Line D, is 46.8 Feet the Content on the Line C.
Example II. A Piece of Timber is 15 Feet long, and \(\frac{1}{4}\) of the Girt 42 Inches: Set 12 on the Girt-Line D, to 15 on the second Radius of the Line C; then against 42, at the Beginning of the Girt-Line D, is, on the Line C, 184 Feet, the Content sought.
Example III. The Length of a Piece is 9 Inches, and a Quarter of the Girt 35 Inches, What is the Content? Now, because the Length is not a Foot, measure it by your Line of Foot Measure, and see what decimal Part of a Foot it makes, which will be .75; then set 12 on the Girt-Line, to 75 on the first Radius of the Line Cl and against 35 on the Girt-Line D, is 6.4 Feet on the Line C, for the Content.
Example IV. A Rail is 18 Feet long, and the Quarter of the Girt 3 Inches: set 12 on the Girt-Line D, to 18 on the first Radius of the Line C; then against 30, which must be taken for 3, on the Girt-Line D, is just 1.12 Feet for the Content.
The Reason of the Operations of this Use, is manifest from what I said about the Property of the Lines D and C, in Use VI. and this Theorem, viz. that as 144, the square Inches in a Foot, is to the Content of the Square Base of a Parallelopipedon taken in Inches; that is, to the Square of \(\frac{1}{4}\) of the Girt: So is the Length of a Parallelopipedon taken in Feet, to the Solidity of the said Parallelopipedon in Feet.
This Use may be sooner done by taking all the Dimensions in Foot Measure thus, count 10, 20, 30, 40, &c. on the Girt-Line to be 1, 2, 3, 4, &c. and then place 10 on the Girt-Line D (now called 1) to the Length of the Tree on the Line C, and against the Girt, in Foot Measure, on the Girt-Line D, stands the Content on the Line C.
Example I. Let the Length of a Tree be, as in the first Example foregoing, viz. 30 Feet, and the Girt 60 Inches, or 5 Feet, What is the Content? Set 10 (now called 1) on the Girt-Line D, to 30 Feet on the Line C; then against 1.25 Feet, the one fourth of the Girt, on the Girt-Line 13, stands 46.8 Feet on the Line C, for the Content, as before.
Example II. A Piece of Timber is 15 Feet long, and one fourth of the Girt is 42 Inches, or 3.5 Feet, What is the Content?
Set 10 on the Girt-Line, to 15 on the first Radius of the Line C; then against 3.5 Feet on the Girt-Line, is 184 Feet on the Line C, the Content required.
Example III. A Length is 9.75 Feet, and \(\frac{1}{4}\) of the Girt 39 Inches, or 3 Feet \(\frac{25}{100}\) set 10 on the Girt-Line to 9.75 on the Line C; and against 3.25 Feet, on the Girt-Line D, is beyond 100 on the Line C: in this Case take half the Length, and then the Content found must be doubled, as here:
Set 10 on the Girt-Line, to (half of 9.75) 4.87; and then against 3.25 is 51.5; the double or which is 103 Feet, the Content required.
Note, If the Content of any Piece of Timber in Feet, be divided by 50, you have the Content in Loads: but some will have a Load to be bur 40 solid Feet; therefore you may take which of the two is most customary with you.
Use II.To measure Round Timber the true way.
The manner of measuring Round Timber in the last Use, being the common way, but not the true one, as I have already said in speaking of the Carpenter’s Rule: I shall now give you a Point on the Girt-Line D, which must be used instead of 12, which is 10.635, at which there ought to be placed a little Brass Center-Pin: this 10.635 is the Side of a Square, equal to a Circle, whose Diameter is 12 Inches.
Example. Let a Length be (as in the second Example of the last Use) 15 Feet, and the 1 of the Girt 42 Inches: let the said Point 10.635, to 15 the Length; then against 42, at the Beginning of the Gut Line, is 233 Feet for the Content sought: but by the common way, there arises only 184 Feet.
Note, As the Area, or Content of a Circle (in Inches) whose Diameter is 12 Inches, is to the Length of any Cylinder in Feet So is the Square of the Circumference of the Base of the Cylinder, in Indies, to the solid Content of the Cylinder in Feet.
Also the common Measure is to the true Measure, as u is to 14; that is, as the Area, or Content of a Circle, to the Squire of it’s Diameter; which, from hence, will be easily manifest: Call the Diameter of any Circle D, and 4 \(\frac{1}{34}\) the Circumference C; then the Content of the said Circle will be equal to D × C; therefore D × C; is to D × D, as 11 is to 14. But the common Measure (because the Length of the Piece is the same) will be to the true Measure, as C × C, the Square of the Circumference, to D × C the Content of the said Circle; whence D × C must be to Dq, as Cq is to D × C; and by comparing the Rectangles under the Means and Extremes, they will be found equal; therefore what I proposed is true.
If the Girt of a Piece of Timber be taken in Feet, the point for true Measure is .886, or .89, which is the Side of a Square, equal to the Content of a Circle, whose Diameter is Unity And then, for the foregoing Example, the Length being 15 Feet, and \(\frac{1}{4}\) of the Girt 42 Inches; set the aforesaid Point 89 on the Girt-Line, to the Length 15 Feet on the Line C, (in the first Radius) then against 3.5 Feet (which is 35) on the Girt-Line D, is 233 Feet on the Line C, the true Content required.
Use III.To measure a Cube.
Let there be a Cube whose Sides are 6 Feet; to find the Content: set 12 on the Girt-Line D, to 6 on the Line C; then against 72 Inches (the Inches in 6 Feet) on the Girt Line D, is 216 Feet on the Line C, which is the Content required.
Use IV.To measure unequal square Timber; that is, if the Breadth and Depth are not equal.
Measure the Length of the Piece, and the Breadth and Depth (at the End; in Inches; then find a mean Proportional between the breadth and Depth of the Piece; Which mean Proportional is the Side of a Square equal to the End of the Piece: which being found, the Piece may be measured as square Timber.
Example I. Let there be a Piece of Timber whose Length is 13 Feet, the Breadth 23 Inches, and the Depth 13 Inches: set 23 on the Girt-Line D, to 23 on the Line C; men against 13 on the Line C, is 17.35 on the Girt-Line D for the mean Proportional. Now again; letting 12 on the Girt Line D, to 13 Feet, the Length, on the Line C; then against 17.35 on the Girt-Line D, is 27 Feet the Content required.
Example II. Let there be a Piece of Stone 7.4 Feet in Length, 30 Inches in Breadth, and 23.5 Deep: set 30 Inches on the Girt-Line D, to 30 on the Line C; then against 23.5, on the Line C, is 26.5 on the Girt-Line D; then let 12 on the Girt-Line D, to 7.4 on the Line C; and against 26.5, on the Girt Line, is 36 Feet the Content sought.
Use V.To find the Content of a Piece of Timber in Form of a Triangular Prism.
You must first find a mean Proportional between the Base, and half the Perpendicular of the triangular End, or between the Perpendicular and half the Base, both measured in Inches, and that mean Proportional will be the Side of a Square equal to the Triangle.
Then to find the Content, set 12 on the Girt-Line D, to the Length in Feet on the Line of Numbers C; and against the mean Proportional on the Girt-Line D, is the Content on the Line of Numbers C.
But the Dimensions being all taken in Foot-Measure, and the mean Proportional found in the same; then set 1 on the Girt-Line, to the Length on the Line C; and against the mean Proportional in the Girt-Line, is the Content in the Line C.
Example There is a Piece of Timber 19 Feet 6 Inches in Length, the Base of the Triangle at each End 21 Inches, and the Perpendicular 16 Inches: to find the Content.
Set 21 Inches on the Girt-Line D, to 21 on the Line C; then against 8 on the Line C, is 12.95 on the Line D, the mean Proportional; then set 12 on the Line D, to 19.5 Feet the Length, on the Line C; and against 12.95 (the mean Proportional) on the Girt-Line D, is 22.8 Feet the Content on the Line C. Or thus, take all the Dimensions in Foot-Measure, and then the Length 19 Feet 6 Inches, is 19.5, the Base 21 Inches, is 1.75, and the Perpendicular 16 Inches, is 1.33. Now set 1 on the Girt-Line D, to the Length 19.5 on the double Line C; and against 1.08 on the Girt-Line D, is 22.8 Feet on the Line C, for the Content.
Use VI.To measure Taper Timber.
The Length being measured in Feet, note one third of it, which may be found thus: set 3 on the Line A, to the Length on the Line B; then against 1 on the Line A, is the third Part on the Line B: then if the Solid be round, measure the Diameter at each End in Inches, and substract the lesser Diameter from the greater, and add half the Difference to the lesser Diameter, the Sum is the Diameter in the middle of the Piece; then set 13.54 on the Girt-Line D, to the Length on the Line C; and against the Diameter in the middle, on the Girt-Line, is a fourth Number on the Line C. Again; set 13.54 on the Girt-Line, to the third part of the Length on the Line C: then against half the Difference on the Girt-Line, is another fourth Number on the Line C; these two fourth Numbers added together, will give the Content.
Example. Let the Length be 27 Feet (one third of which will be 9), the greater Diameter 22 Inches, and the lesser 18, the Sum of the greater and lesser Diameters will be 40; their Difference 4, half their Difference 2, which added to the lesser Diameter, gives 20 Inches for the Diameter in the middle of the Piece. Now set 13.54 on the Girt-Line D, to 27 on the Line C; and against 20 on the Line D, is 58.9 Feet. Again, set 13.54 on the Girt-Line, to 9 on the Line C; then against 2 on the Girt-Line (represented by 20), is .196 Parts: therefore, by adding 58.9 Feet, to .196 Feet, the Sum is 59.096 Feet the Content. If all the Dimensions are taken in Foot-Measure, then you must add the greater and lesser Diameters together, which in this Example make 3.33 Feet; half of which is the Diameter in the middle of the Piece, viz. 1.67 Feet, the Difference of the Diameters is 0.33 Feet, half of which Difference is 0.17 Feet.
Then set 1.13 on the Girt-Line, to the Length 27 Feet on the Line C; and against 1.67 on the Line D, is 58.9 Feet: then again, set 1.13 on the Line D, to 9 Feet on the Line C; and then against 0.17 on the Line D, is 196 Parts of a Foot, and both added is the Content; that is, 58.9 and .196 added, makes 59.096 Feet as before.
If the Solid is square, and has the same Dimensions; that is, the Length 27 Feet, the Side of the greater End 22 Inches, and the Side of the lesser End 18 Inches, to find the Content in Inch-Measure: set 12 on the Girt-Line, to 27 the Length of the Solid, on the Line C; and against 20 Inches, the Side of the mean Square on the Girt-Line, is 75.4 Feet. Again; set 12 on the Girt-Line, to 9 Feet, one third of the Length, on the Line C; and against 2 Inches, half the Difference of the Sides of the Squares of the Ends, on the Girt-Line, is .25 Parts of a Foot; both together is 75.65 Feet the Content of the Solid or thus, When all the Dimensions are taken in Foot-Measure, set 1 on the Girt-Line, to the Length 27 Feet on the Line C; then against 1.67 Feet, the Side of the middle Square on the Girt-Line, stands 75.4 Feet; and setting 1 on the Girt-Line to 9 Feet, one third of the Length on the Line C, against 0.167, half the Difference of the Sides of the Squares of the Ends on the Girt-Line, is on the Line C, .25 Parts of a Foot; which added to the other, makes 75.65 Feet, as before, for the Content.
Note, The fixed Numbers 13.54, and 1.13 are, the first, the Diameter of a Circle whose Area, or Content is 144; that is, the Number of square Inches in a superficial Foot; and the other, the Diameter of a Circle whose Area is Unity.
Use VII.To find how many Inches in Length will make a Foot-Solid, at any girt, being the Side of a Square not exceeding 40 Inches.
Let the Girt, or Side of the Square, taken upon the Girt-Line, be set to 1 on the Line C: then against 41.57 of the Girt-Line, is the Number of Inches on the Line C, that will make a Solid-Foot.
Example. Let the Side of a Square be 8 Inches: set 8 on the Girt-Line D, to 1 on the Line C; then against 41.57 on the Girt-Line D, is 27 Inches for the Length of one solid Foot. To do this in Foot-Measure; the Side of the Square 8 Inches, in Foot-Measure, is .66 Parts, which taken on the Girt-Line, and being set to 1 on the Line C, against 1 on the Girt-Line, is 2.25 Feet, for the Length to make one Foot of Timber.
Note, 41.57 is the Square-Root of 1728, the Number of Cubic Inches in a solid Foot.
Use VIII.The Diameter of a Circle, or round Piece of Timber, being given: to find the Side of a Square within the Circle; or two know how many Inches the Side of the Square will be, when the round Timber is squared.
Rule. Set 8.5 on the Line A, to 6 on the Line B; then against the Diameter on the Line A, is the Side of the Square on the Line B.
Example. Let the Diameter be 18 Inches: set 8.5 on A, to 6 on B; then against 18 on A, is 12\(\frac{1}{4}\) the Line B, for the Side of a Square within the Circle. The same done in Foot-Measure: the Diameter being 18 Inches, is in Foot-Measure 1.5; then set 1 on the Line A, to .707 on the Line B; and against the Diameter 1.5 on the Line A, is 1.7 on the Line B; that is, 1.7 Foot is the Side of an inscribed Square in a Circle, whose Diameter is 1.5 Foot.
Note, The given Numbers 8.5 and 6, or more exacter, 1 and .707, are, the one the Diameter of a Circle, and the other the Side of a Square inscribed in that Circle.
Use IX.The Girt of a Tree, or round Piece of Timber being given; to find the Side of a Square within.
Rule. Set 10 to 9 on the Lines A and B; then against the Girt on the Line A, are the Inches for the Side of the Square on the Line B.
Let the Girt be 12 Inches; set 10 on the Line A, to 9 on the Line B; then against: 12 on the Line A, is 10.8 on the Line B, for the Side of the Square. By Foot-Measure it is thus; the Girt 12 Inches is one Foot; then set 10 on the Line A, to 9 on the Line B; and against the Girt 1 Foot, on the Line A, is .89 Parts of a Foot for the Side of the Square within.
Note, The Numbers 10 and 9, or 1 and .9, shew when the Square within the Circle is 1 the fourth Part of the Circumference is .9 Parts of the same. Also, by this and the last Use, you may know, before a Piece of Timber be hewn, how many Boards or Planks of any Thickness it will make.
Use X.The fourth Part of the Girt of a round Piece of Timber being given; to find the Side of a Square equal to it.
Rule. Set 1 on the Line A, to 1.128, on the Line B; then against the one fourth of the Girt, on the Line A, is on the Line B, the Side of the Square equal to it.
Example. Let the Girt (that is, one fourth of the whole Girt), be 16 Inches; What is the Side of a Square equal to it? Set 1 to 1.13, on the Lines A and B; then against 16 on the Line A, is 18 on the Line B; which shews, that a Square, whose Side is 18 Inches, is equal to a Circle, whose Girt is 64 Inches, and \(\frac{1}{4}\) of it's Girt 16 Inches.
Use XI.To find the Solidity of a Cone.
Let the Diameter of the Base of a Cone be 12 Feet, and it’s Altitude or Height, 24; to find the Content.
This Use may be solved at one Operation, thus: set 1.95 on the Girt-Line, to the Height of the Cone 24, on the Line C; then against the Diameter of the Base of the Cone 12, on the Girt Line, stands on the Line C, 904.8 Feet, for the Content.
Note, 1.95 is the Square Root of the Quotient of 42 divided by 11: and As the Quotient of 42 divided by 11, is to the Height of any Cone; So is the Square of the Diameter of it’s Base to the solid Content.
Use XII.To find the Solidity of a Square Pyramid.
Suppose the Side of the Base is 8 Inches, and the Height 30, set 7 on the Girt-Line to \(\frac{1}{3}\) of the Length, viz. 10, on the Line C; then against the Side of the Base 8, on the Girt-Line, is 640 Inches, on the Line C, for the Solidity.
Use XIII.To find the Solidity of a Sphere, by having the Circumference given.
Let the Circumference of a Sphere be 22 Inches; to find the Content. As 2904 is to 49, So is the Cube of the Circumference of a Sphere to it’s solid Content: therefore set 53.8 (the Square Root of 2904) on the Girt Line, to 49 on the Line C; then against the Circumference 22 Inches on the Girt-Line, is a fourth Number. viz. 8.09. Again, set 1, on the Line B, to 22 on the Line A; then against 8.09, on the Line C, stands 179.6 on the Line A, for the Content of the said Sphere in solid Inches. If the Diameter had been given, you must have used the fixed Numbers 4.57 and 11, instead of 53.8 and 49, and then have proceeded as before: because As 21 is to 11, So is the Cube of the Diameter of a Sphere to the solid Content thereof.
This Use may be otherwise solved at one Operation, thus: set 7.69 on the Girt-Line D, to the Circumference of the Sphere 22 Inches, on the Line C; then against 22 Inches, on the Girt-Line D, stands, on the Line C, the solid Content 179.6 Inches. If the Diameter be given to find the Solidity at one Operation, you must set 1.38, on the Girt-Line to the Diameter on the Line C; then against the same Diameter, on the Girt-Line, stands, on the Line C, the Content.
Note, 7.69, and 1.38 are, the one, the Square Root of the Quotient of 2904 divided by 49; and the other, the Square Root of the Quotient of 21, divided by 11.
Use XIV.The Circumference of a Sphere being given, to find it’s Superficies.
Suppose the Circumference of a Sphere be 20 Inches, What is the Area of it’s Superficies? Set 4.69 (the Square Root of 22) on the Girt-Line D, to 7 on the Line C; then against 20 Inches on the Girt-Line, stands upon the Line C 136.5, the Area of the Superficies of the Sphere.
The reason of this is, because 22 is to 7; So is the Square of the Circumference of a Sphere to the superficial Area thereof.
Use XV.To find the Solidity of the Segment of a Sphere.
Say, As 21 is to the Sine; So is 11 times the Square of the said Sine, added to 33 times the Square of half the Chord, to the solid Content of the Segment. As suppose the Sine be 10 Inches, and half the Chord 16 Inches; to find the Content: say, As 21 is to 10; So is 9548, the Sum of 11 times the Square of 10, added to 33 times the Square of 16, 10 the Content 4546.6 Inches.
Use XVI.To find the Area of the Convex Superficies of the Segment of a Sphere.
Say, As 14 is to 44 times the Diameter of a Sphere; So is the Length of the Sine of any Segment thereof, to the convex Superficies of the said Segment. Suppose the Sine be 12 Inches, and the Diameter 30; say, As 14 is to 1320; So is 12 to 1131.4 Indies, the Content sought.
