Chapter XXXVI: Part XII: Mathematics (4)

(24·43 × 20) / 2 = 244·3 square inches. Area of the sector.
(22·42 × 16·56) / 2 = 185·6376 square inches. Area of the triangle.
244·3 - 185·6376 = 58·6624 square inches. Area required.

_To find the area of a semicircle._

1. Multiply ¼ of the circumference by the radius, and the product will be the area.

2. Or, multiply the square of the diameter by ·7854, and half the product will be the area.

_Example._—_Rule 2._—Required the area of a semicircle, the diameter being 50 inches.

(50 × 50 × ·7854) / 2 = 981·75 square inches. Area required.

_To find the area of an ellipsis, or oval._

Multiply the longest diameter, or axis, by the shortest, then multiply the product by ·7854 for the area.

_Example._—Required the area of the ellipse, whose diameters are 25 inches, and 18 inches.

25 × 18 × ·7854 = 353·43 square inches. Area required.

_To find the area of a parabola, or its segment._

Multiply the base by the perpendicular height, and take two-thirds of the product for the area.

_Example._—Required the area of a parabola, whose base is 20 feet, and height 12 feet.

20 × 12 = 240
⅔ of 240 = 160 square feet. Area required.

MENSURATION OF SOLIDS.

_A solid_ is a body containing length, breadth, and thickness.

_Solids are measured_ by cubes, whose sides are each an inch, a foot, a yard, &c., and the solidity, capacity, or content of any figure is computed by the number of such cubes as are contained in it.—_Vide Cubic measure, page 276._

_A cube_ is a solid contained by six equal square sides.

_A pyramid_ is a solid whose sides are all triangles meeting together in a point, the base being any plane figure whatever. It is called a triangular pyramid when its base is a triangle; a square pyramid when its base is a square, &c.

_The segment of a pyramid, cone, or any other solid_ is a part of D E F G cut off from the top by a plane D E F, parallel to the base A B C.—_Vide Fig. 21, Plate 2_, HEIGHTS, DISTANCES, and PRACTICAL GEOMETRY.

_A frustrum, or trunk_, is a part A B C D E F, that remains at the bottom after the segment is cut off.

_A cone_ is a round pyramid, of which the base is a circle.

_The axis of a solid_ is a line from the vertex (or point) to the centre of the base, or through the centres of the two ends. When the axis is perpendicular to the base, it is a right prism, pyramid, or cone; otherwise it is oblique.

_A sphere_ is a solid contained under one convex surface, and is described by the revolution of a semicircle about its diameter, which remains fixed.

_The centre of the sphere_ is such a point within the solid as is everywhere equally distant from the convex surface, or circumference of it.

_The diameter_ (_or axis_) _of a sphere_ is a straight line, which passes through the centre, and is terminated by the convex surface.

_A segment of a sphere_ is a part cut off by a plane, the section of which is always a circle, called the _base of the segment_.

_A sector of a sphere_ is that which is composed of a segment (less than an hemisphere) and of a cone.

_A prism_ is a solid, the sides of which are parallelograms, having its ends equal, and similar plane figures.

_Prisms are named_ according to the number of angles in the base.

_A cylinder_ is a solid, the two ends of which are circular; and it is described, or formed, by the revolution of a right-angled parallelogram about one of its sides, which remains fixed.

_To find the superficies of a prism, or cylinder._

Multiply the perimeter of one end of the prism by the length, or height of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the prism, when required.

Or, compute the areas of all the sides, and ends separately, and add them all together.

_Example._—Required the surface of a cube, whose sides are each 5 inches.

5 + 5 + 5 + 5 = 20, perimeter of one end.
20 × 5 = 100, surface of sides.
5 × 5 = 25, area of one end.
100 + 25 + 25 = 150 square inches. Surface of cube.

_To find the surface of a pyramid, or cone._

Multiply the perimeter of the base by the slant height, or length of the side, and half the product will be the surface of the sides; to which add the area of the base when required.

_Example._—Required the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet.

3 + 3 + 3 = 9, perimeter of base.
(9 × 20) / 2 = 90 feet. Surface required.

_To find the surface of the frustrum of a pyramid, or cone._

Add together the perimeters of the two ends, multiply their sum by the slant height, and take half the product.

_Example._—How many square feet are in the surface of the frustrum of a square pyramid, whose slant height is 10 feet, each side of the base 3 feet, and each side of the less end 2 feet.

3 + 3 + 3 + 3 = 12, perimeter of base.
2 + 2 + 2 + 2 = 8, perimeter of less end.
((12 + 8) × 10) / 2 = 100 feet. Surface required.

_To find the solid content of a prism or cylinder._

Find the area of the base, or end, and multiply it by the length of the prism, or cylinder. Fora cube, multiply its side twice by itself; and for a parallelopipedon, multiply the length, breadth, and depth together for the content.

_Example._—Required the solid content of a cube, whose side is 24 inches.

24 × 24 × 24 = 13824 square inches. Content required.

_To find the content of the solid part of a hollow cylinder._

From the content of the whole cylinder considered as a solid, subtract the content of the hollow part, also considered as a solid, and the difference will be the solidity required.

_Example._—Required the content of the solid part of the hollow cylinder whose exterior diameter is 12 inches, the interior diameter 8 inches, and height 20 inches.

12 × 12 × ·7854 = 113·0976, area of base of cylinder.
113·0976 × 20 = 2261·952, solidity of whole cylinder.
8 × 8 × ·7854 = 50·2656, area of base of hollow cylinder.
50·2656 × 20 = 1005·312, content of hollow part.
2261·952 - 1005·312 = 1256·64 cubic inches. Solidity required.

_To find the solidity of the frustrum of a cylinder._

Multiply the area of the base by half the greatest, and the least lengths, and the product will be the solidity.

_Example._—Required the solidity of a frustrum, whose diameter is 24 inches, the greatest length 36 inches, and the least length 20 inches.

24 × 24 = 576. Square of the diameter.
576 × ·7854 = 452·3904. Area of the base.
452·3904 × (36 + 20) / 2 = 12666·9312 Cubic inches.
Solidity required.

_To find the content of a pyramid, or cone._

Find the area of the base, and multiply that area by the perpendicular height, and take ⅓ of the product.

_Example._—Required the solidity of a square pyramid, each side of its base being 30, and its perpendicular height 25.

30 × 30 = 900, area of base.
(900 × 25) / 3 = 7500, solidity required.

_To find the solidity of the frustrum of a cone, or pyramid._

Add into one sum the areas of the two ends, and the mean proportional between them: take ⅓ of that sum for the mean area, which multiply by the perpendicular height, or length of the frustrum.

_Note._—_To find a mean proportional._

As one of the sides of the base is to the homologous, or corresponding side of the other end, so is the area of the base to the mean proportional required.

_Example._—Required the number of solid feet in a piece of timber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also the length of the perpendicular altitude 24 feet.

15 × 15 = 225, area of the base.
6 × 6 = 36, area of the top.
As 15: 6 :: 225: 90, mean proportional.
24 feet = 288 inches.
((225 + 36 + 90) × 288) / 3 = 33696 cubic inches
= 19½ cubic feet.

_To find the surface of a sphere, or any segment._

Multiply the circumference of the sphere by its diameter, which will give the whole surface.

Or, square the diameter, and multiply by 3·1416.
Or, square the circumference, and multiply by ·3183;
or divide by 3·1416.

_Note._—_For the surface of the segment, or frustrum_, multiply the whole circumference of the sphere by the height of the part required.

_Example._—Required the superficies of a globe whose diameter is 24 inches.

24 × 24 × 3·1416 = 1809·5616 square inches.

_To find the solidity of a sphere, or globe._

1. Multiply the surface by the diameter, and take ⅙ of the product.

Or, multiply the square of the diameter by the circumference, and take ⅙ of the product.

2. Cube the diameter, and multiply by ·5236.

3. Cube the circumference, and multiply by ·01688.

_Example._—Required the content of a sphere, whose axis is 12.

12 × 12 × 12 × ·5236 = 904·7808. Content required.

_To find the solidity of an hemisphere._

Find the solidity of the sphere, and half the content will be that of the hemisphere.

_Note 1._—Any sphere, or globe _twice_ the diameter of another contains _four times_ the superficies, or area of the other, and _eight times_ the solid content. Hence the superficies of spheres are as the squares, and the solidity as the cubes of their diameters.

_Note 2._—The cube of the diameter of a sphere in inches, multiplied by ·00188, will give the number of _imperial gallons it will contain_.

_To find the solid content of a spherical segment._

1. From three times the diameter of the sphere, take double the height of the segment; then multiply the remainder by the square of the height, and this product by ·5236.

2. Or, to three times the square of the radius of the segment’s base add the square of its height; then multiply the sum by the height, and the product by ·5236.

_Example._—Required the content of a spherical segment 2 feet in height, cut from a sphere of 8 feet diameter.

(3 × 8) - (2 × 2) = 20
20 × 2^2 × ·5236 = 41·888 cubic feet. Content required.

_To find the diameter of a sphere, its solidity being given._

Divide the solidity by ·5236, and take the cube root of the quotient.

_Example._—The solidity of a sphere being 113·0976 solid inches, what will be its diameter?

113·0976 / ·5236 = 216, the cube root of which is 6 inches,
the diameter required.

_To find the weight of an iron shot, its diameter being given._

Take ⅛ of the cube of the diameter, and ⅛ of that eighth, and the sum of these two quotients will be the weight in pounds.

Or, as 64 is to 9 lb. so is the diameter cubed to its weight.

_Example._—Required the weight of an iron shot whose diameter is 3·5 inches?

3·5 cubed = 42·875, cube of diameter.
42·875 / 8 = 5·359 5·359 / 8 = ·669
5·359 + ·669 = 6·028 pounds. Weight required.

_To find the weight of a leaden ball, its diameter being given._

Take ⅓ of the cube of the diameter, and from it subtract ⅓ of this third, and the remainder will be the weight, nearly.

Or, take 3/14 of the cube of the diameter.

_Example._—What is the weight of a leaden ball whose diameter is 3·3 inches?

3·3 cubed = 35·937, cube of diameter.
35·937 / 3 = 11·979 11·979 / 3 = 3·993
11·979 - 3·993 = 7·986 pounds. Weight required.

_To find the diameter of an iron shot, its weight being given._

Multiply the cube root of the shot’s weight by 1·923 for the diameter.

Pr. Cube root. Diameter.
42 3·4760, &c. { Multiplied } 6·684, &c.
32 3·1748 { by 1·923, } 6·103 ”
24 2·8844 { diameter } 5·545 ”
18 2·6207 { of a } 5·038 ”
12 2·2894 { 1 lb. } 4·401 ”
9 2·0800 { shot. } 3·999 ”
6 1·8171 { } 3·494 ”
3 1·4422 { } 2·772 ”

_To find the diameter of a leaden ball, its weight being given._

To 4 times the weight add half the weight, and 3/100 of half the weight; and the cube root of this sum will be the diameter in inches, nearly.

_Example._—What is the diameter of a leaden ball, whose weight is 8 pounds?

8 x 4 = 32 8 / 2 = 4 3 / 100 of 4 = ·12.
32 + 4 + ·12 = 36·12, of which the cube root is 3·3 inches,
nearly. Diameter required.

_To find the weight of an iron shell, its interior and exterior diameter being given._

Take 9/64 of the difference of the cubes of the external and internal diameters, for the weight of the shell in pounds.

_Example._—What is the weight of a shell whose exterior diameter is 12·85 inches, and interior diameter 8·75 inches?

12·85 cubed = 2121·8241, 8·75 cubed = 669·9218.
2121·8241 - 669·9218 = 1451·9022.
9/64 of 1451·9022 = 204·1737 pounds. Weight required.

_To find the quantity of powder a shell will contain._

Divide the cube of the interior diameter in inches by 57·3, and the quotient will be the weight in pounds.

Or, multiply the cube of the diameter by 11, and divide by 21 for the quantity in half ounces.

_Example._—How much powder will fill a shell, whose internal diameter is 7 inches?

7 cubed = 343.
343 / 57·3 = 6 pounds nearly. Powder required.

_To find the side of a cubical box to contain a given quantity of powder._[57]

Multiply the weight in pounds by 30, and the cube root of the product will be the side of the box in inches.

_Example._—Required the side of a cubical box to hold 50 pounds of powder?

50 × 30 = 1500, the cube root of which is 11·44, which will be
the side of the box in inches.

_To find the quantity of powder to fill the chamber of a mortar, or howitzer._

Multiply the content of the chamber in inches by 55, and divide the product by 1728, and the quotient will be the quantity of powder in pounds.

_Note._—The chamber of a mortar, or howitzer, is formed of a hollow frustrum of a right cone, and of a hollow hemisphere.

_Example._—Required the quantity of powder to fill the chamber of a 13-inch mortar in which the diameter A B is 9·5 inches, the diameter C E 6·5 inches, and the length D G 21·5 inches. _Vide Fig. 22. Plate 2._ HEIGHTS AND DISTANCES, and PRACTICAL GEOMETRY.

The content of the chamber must be found by finding the content of the hollow frustrum of the cone, and that of the hemisphere (_vide preceding rules_): which in this example will be 999·9741875.

Then (999·9741875 × 55) / 1728 = 31 pounds, nearly.

_To find the quantity of powder to fill a rectangular box._

Divide the content (viz., length × breadth × depth) of the box in inches by 30 for the pounds of powder.

_Example._—How much powder will fill a box, the length being 15 inches, the breadth 12, and the depth 10 inches.

15 × 12 × 10 / 30 = 1800 / 30 = 60 pounds. Number required.

_To find the quantity of powder to fill a cylinder._

Multiply the square of the diameter by the length, then divide by 38·2 for the pounds of powder.

_Example._—How much powder will the cylinder contain, whose diameter is 10 inches, and length 20 inches?

(10 × 10 × 20) / 38·2 = 52⅓ pounds, nearly.

_To find the size of a shell, to contain a given weight of powder._

Multiply the pounds of powder by 57·3, and the cube root of the product will be the diameter in inches.

_Example._—Required the diameter of a shell to contain 6 lb. of powder?

6 × 57·3 = 343·8, the cube root of which is 7, the diameter
required, in inches.

_To find what length of a cylinder (or bore of a gun) will be filled by a given weight of powder._

Multiply the weight in pounds by 38·2, and divide the product by the square of the diameter in inches, for the length.

_Example._—What length of a cylinder 8 inches in diameter will be filled with 20 lb. of powder?

(20 × 38·2) / (8 × 8) = 11-15/16 inches.

_To find the content, and weight of a piece of ordnance._

Divide the length of the gun into as many sections as may be found necessary. Find the content of each (_by preceding rules_) and from their sum subtract the content of a cylinder, whose length is equal to that of the bore, and its diameter equal to that of the calibre of the piece; multiply the difference (if it be a brass gun) by 5·0833, (if an iron gun) by 4·2968, and the product will be the weight in ounces.

_Note._—A cubic inch of gun metal weighs 5·0833 ounces.
Ditto of _cast_ iron 4·2968 ounces.

_To find the content of a cask._

Multiply half the sum of the areas of the two interior circles, viz. at the head, and bung, by the interior length, for the content.

Or, to the area of the head add twice the area at the bung, multiply that sum by the length, and take one-third of the product.

_Example._—Required the content of a cask, its greatest interior diameter being 24 inches, its least interior diameter 20 inches, and the interior length 30 inches.

24 × 24 × ·7854 = 452·3904, area of large circle.
20 × 20 × ·7854 = 314·1600, area of small circle.
(452·3904 + 314·1600) / 2 = 383·2752, half sum.

Then 383·2752 × 30 = 11498·256, the content; which being divided by 277¼ (the number of cubic inches in a gallon) will give the number of gallons contained in the cask.

Thus 11498·256 / 277·25 = 41·4725, &c. Number of gallons required.

_Note._-The content of any vessel in cubic feet, multiplied by 6·232 (or if in inches by ·003607) will give the number of _imperial gallons it will contain_.

EPITOME OF MENSURATION.

OF THE CIRCLE, CYLINDER, SPHERE, ETC.

1. The circle contains a greater area than any other plane figure, bounded by an equal perimeter, or outline.

2. The areas of circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other.

3. The diameter of a circle being 1, its circumference equals 3·1416.

4. The diameter of a circle is equal to ·31831 of its circumference.

5. The square of the diameter of a circle being 1, its area equals ·7854.

6. The square root of the area of a circle, multiplied by 1·12837, equals its diameter.

7. The diameter of a circle, multiplied by ·8862, or the circumference multiplied by ·2821, equals the side of a square of equal area.

8. The sum of the squares of half the chord, and versed sine, divided by the versed sine, the quotient equals the diameter of the corresponding circle.

9. The chord of the whole arc of a circle taken from eight times the chord of half the arc, one-third of the remainder equals the length of the arc.

10. Or, the number of degrees contained in the arc of a circle, multiplied by the diameter of the circle, and by ·008727, the product equals the length of the arc in equal terms of unity.

11. The length of the arc of the sector of a circle multiplied by its radius, half the product is the area.

12. The area of the segment of a circle equals the area of the sector, minus the area of a triangle whose vertex is the centre; and base equals the chord of the segment.

13. The sum of the diameters of two concentric circles multiplied by their difference, and by ·7854, equals the area of the ring, or space contained between them.

14. The sum of the thickness, and internal diameter of a cylindric ring multiplied by the square of its thickness, and by 2·4674, equals its solidity.

15. The circumference of a cylinder multiplied by its length, or height, equals its convex surface.

16. The area of the end of a cylinder multiplied by its length, equals its solid content.

17. The area of the internal diameter of a cylinder multiplied by its depth, equals its cubical capacity.

18. The square of the diameter of a cylinder multiplied by its length, and divided by any other required length, the square root of the quotient equals the diameter of the other cylinder of equal solidity, or capacity.

19. The square of the diameter of a sphere multiplied by 3·1416 equals its convex surface.

20. The cube of the diameter of a sphere multiplied by ·5236, equals its solid content.

21. The height of any spherical segment, or zone, multiplied by the diameter of the sphere, of which it is a part, and by 3·1416, equals the area, or convex surface of the segment;

22. Or, the height of the segment multiplied by the circumference of the sphere of which it is a part, equals the area.

23. The solidity of any spherical segment is equal to three times the square of the radius of its base, plus the square of its height, and multiplied by its height, and by ·5236.

24. The solidity of a spherical zone equals the sum of the squares of the radii of its two ends, and one-third the square of its height, multiplied by the height, and by 1·5708.

25. The solidity of the middle zone of a sphere equals the sum of the square of either end, and two-thirds the square of the height, multiplied by the height, and by ·7854.

26. The capacity of a cylinder 1 foot in diameter, and 1 foot in length, equals 4·895 imperial gallons.

27. The capacity of a cylinder 1 inch in diameter, and 1 foot in length, equals ·034 of an imperial gallon.

28. The capacity of a cylinder 1 inch in diameter, and 1 inch in length, equals ·002832 of an imperial gallon.

29. The capacity of a sphere 1 foot in diameter, equals 3·263 imperial gallons.

30. The capacity of a sphere 1 inch in diameter, equals ·001888 of an imperial gallon.

31. Hence the capacity of any other cylinder in imperial gallons is obtained by multiplying the square of its diameter by its length; or the capacity of any other sphere by the cube of its diameter, and by the number of imperial gallons contained as above in the unity of its measurement.

OF THE SQUARE, RECTANGLE, CUBE, ETC.

1. The side of a square equals the square root of its area.

2. The area of a square equals the square of one of its sides.

3. The diagonal of a square equals the square root of twice the square of its side.

4. The side of a square is equal to the square root of half the square of its diagonal.

5. The side of a square, equal to the diagonal of a given square, contains double the area of the given square.

6. The area of a rectangle equals its length multiplied by its breadth.

7. The length of a rectangle equals the area divided by the breadth; or the breadth equals the area divided by the length.

8. The side, or end of a rectangle, equals the square root of the sum of the diagonal, and opposite side to that required, multiplied by their difference.

9. The diagonal in a rectangle equals the square root of the sum of the squares of the base, and perpendicular.

10. The solidity of a cube equals the area of one of its sides multiplied by the length of one of its edges.

11. The edge of a cube equals the cube root of its solidity.

12. The capacity of a 12-inch cube equals 6·232 gallons.

_Surfaces, and solidities of the regular bodies, when the linear edge is 1._

+-------------+--------------+------------+-----------+
|No. of Sides.| Names. | Surfaces. | Solids. |
+-------------+--------------+------------+-----------+
| 4 | Tetrahedron | 1·7320508 | 0·1178513 |
| 6 | Hexahedron | 6· | 1· |
| 8 | Octahedron | 3·4641016 | 0·4714045 |
| 12 | Dodecahedron | 20·6457788 | 7·6631189 |
| 20 | Icosahedron | 8·6602540 | 2·1816950 |
+-------------+--------------+------------+-----------+

The tabular surface multiplied by the square of the linear edge, the product equals the surface required:

Or, the tabular solidity, multiplied by the cube of the linear edge, the product is the solidity required.

OF TRIANGLES, POLYGONS, ETC.

1. The complement of an angle is its defect from a right angle.

2. The supplement of an angle is its defect from two right angles.

3. The sine, tangent, and secant of an angle, are the cosine, cotangent and cosecant of the complement of that angle.

4. The hypothenuse of a right-angled triangle being made radii, its sides become the sines of the opposite angles, or the cosines of the adjacent angles.

5. The three angles of every triangle are equal to two right angles; hence the oblique angles of a right-angled triangle are each other’s complements.

6. The sum of the squares of the two given sides of a right-angled triangle is equal to the square of the hypothenuse.

7. The difference between the square of the hypothenuse, and given side of a right-angled triangle is equal to the square of the required side.

8. The area of a triangle equals half the product of the base multiplied by the perpendicular height;

9. Or, the area of a triangle equals half the product of the two sides, and the natural sine of the contained angle.

10. The side of any regular polygon multiplied by its apothem, or perpendicular, and by the number of its sides, half the product is the area.

_Table of the areas of regular polygons whose sides are unity._

+----------+------+-----------+-----------+------------+------------+
| Name of |No. of| Apothem, | Area, when| Interior | Central |
| polygon. |sides.| or perpen-|side is one| angle. | angle. |
| | | dicular. | or unity. | | |
+----------+------+-----------+-----------+------------+------------+
| | | | | ° ′ | ° ′ |
| Triangle | 3 | 0·2886751 | 0·4330127 | 60 0 | 120 0 |
| Square | 4 | 0·5 | 1· | 90 0 | 90 0 |
| Pentagon | 5 | 0·6881910 | 1·7204774 | 108 0 | 72 0 |
| Hexagon | 6 | 0·8660254 | 2·5980762 | 120 0 | 60 0 |
| Heptagon | 7 | 1·0382607 | 3·6339124 | 128 34-2/7 | 51 25-5/7 |
| Octagon | 8 | 1·2071068 | 4·8284271 | 135 0 | 45 0 |
| Nonagon | 9 | 1·3737387 | 6·1818242 | 140 0 | 40 0 |
| Decagon | 10 | 1·5388418 | 7·6942088 | 144 0 | 36 0 |
| Undecagon| 11 | 1·7028436 | 9·3656399 | 147 16-4/11| 32 43-7/11|
| Dodecagon| 12 | 1·8660254 |11·1961524 | 150 0 | 30 0 |
+----------+------+-----------+-----------+------------+------------+

The tabular area of the corresponding polygon multiplied by the square of the side of the given polygon, equals the area of the given polygon.

OF ELLIPSES, CONES, FRUSTRUMS, ETC.

1. The square root of half the sum of the squares of the two diameters of an ellipse multiplied by 3·1416 equals its circumference.

2. The product of the two axes of an ellipse multiplied by ·7854 equals its area.

3. The curve surface of a cone is equal to half the product of the circumference of its base multiplied by its slant side, to which, if the area of the base be added, the sum is the whole surface.

4. The solidity of a cone equals one-third of the product of its base multiplied by its altitude, or height.

5. The squares of the diameters of the two ends of the frustrum of a cone added to the product of the two diameters, and that sum multiplied by its height, and by ·2618, equals its solidity.

THE END.

LONDON:

PRINTED BY W. CLOWES AND SONS, STAMFORD STREET
AND CHARING CROSS.

FOOTNOTES:

[1] _Vide page_ vi.

[2] The Articles omitted consist chiefly of directions, &c., or are not generally required.

[3] NOTE. _In the “Exercise and Movements.”_

Commander’s Words are printed in SMALL CAPITALS.
Executive Small print.
Directions, &c. _Italics_.

[4] _Note._—Vide “Motion,” “Forces,” &c., Velocity, Gravity, and Amplitude.

[5] Vide “Tables,” “Excentric Shot, Experiments.”

[6] _When a shot is jammed in a gun, and cannot be rammed home to the cartridge_, destroy the charge by pouring water down the vent, and muzzle until the ingredients are dissolved, and cleared out of the bore; then introduce a small quantity of powder through the vent, and blow out the shot.

[7] The recoil of guns on sleighs varies from four to five feet when on rough ground or in deep snow; to twenty or thirty yards when on glare ice. In the latter case it is of course necessary to send the ammunition sleighs further to the rear; but the recoil may be considerably lessened by placing a small chain round each of the runners.

Ice of eight inches thick will bear with safety a weight of 1115 lb. (or nearly half a ton) on the square foot.

[8] Old pattern.

[9] Further information relative to mixing the composition, and filling combustibles, &c., &c., may be obtained from the “Aide Mémoire,” under the head, “Pyrotechny, Military.”

[10] This will be discontinued when Shrapnell Diaphragm shells are generally introduced into the Service.

[11] _Vide_ Practice Tables for Ranges, Elevations, &c.

[12] The composition for French cannon tubes is two parts of fulminate of mercury and two of mealed powder, mixed together: then formed into a paste with distilled water, slightly impregnated with gum arabic.

[13] Extracted from “Instructions and Regulations for Field Battery Exercise and Movements” for the Royal Regiment of Artillery: the Sections, &c., being similarly numbered.

Commander’s Words are printed in SMALL CAPITALS.
Executive Common type.
Directions, &c. _Italics._

[14] The Sections, of which merely the heads are given, consist chiefly of details too long for the limited size of the Manual, and they are therefore necessarily omitted.

[15] When Guns are in action, and “CEASE FIRING” is given, all Guns then loaded are to be fired off, and on no account is a Gun to be limbered up, or to move whilst loaded.

[16] The Commanding officer’s Word of command is always to be repeated by the officers.

[17] From “Field Battery Exercise.”

[18] From “Field Battery Exercise,” &c.

[19] From “Field Battery Exercise,” &c.

[20] In the transport of horses to Turkey (July, 1854,) in the Himalaya and Simla steamers, the distance between the upright posts was 2 feet 1 inch in the clear per horse, and the length 9 feet.

[21] “For the guidance of the Farriers of the Royal Artillery. Suggested by Charles Percival, Veterinary Surgeon; and approved of by the Right Honourable the Master-General, and Honourable Board of Ordnance.”

[22] In administering draughts to horses, the greatest possible care and attention are required; should the horse cough, or make an attempt to do so, his head must be instantly lowered, otherwise a portion of the drink will be apt to find its way into the trachea or windpipe, which will produce most distressing symptoms, and often be followed by death. In lowering the head, a can or vessel of any kind should be held under the mouth to catch the drink as it escapes.

[23] From “INSTRUCTIONS FOR THE SERVICE OF HEAVY ORDNANCE.”—_Article 15._

[24] _Words of command_—SMALL CAPITALS.

[25] From “Instructions, and Regulations for the Service, and Management of Heavy Ordnance, for the Royal Regiment of Artillery.” Fourth edition. The Parts, and Articles are numbered in conformity thereto.

[26] _Words of command_—SMALL CAPITALS.

[27] Vide PART 12, “ARTILLERIST’S MANUAL,” etc., The Mechanical powers. The Lever.

[28] By the ballistic experiment, conducted in May, 1837, it was found that, with a heavy 6-pounder gun, a charge of 1½ lb. gave a velocity of 1740 feet, and a charge of 2 lb. a velocity of 1892 feet per second. The shot employed were of a high gauge, windage only ·078 inch, and the powder was of the strongest quality; the weight of the pendulum fired into was 58 cwt. 3 qrs. 16 lb. A light 6-pounder, two feet shorter than the heavy 6-pounder, with similar charges, gave velocities of about 190 feet less.

[29] Extracted from PART 2, and APPENDIX of General Sir Howard Douglas’ highly valued work, entitled “A TREATISE ON NAVAL GUNNERY.” 3rd edition.

[30] _On wads for Heavy Ordnance._

The presence of a compressible body, between the powder and the ball, is necessary for the preservation of the gun. The results of the experiments at Fere, in 1844; at Ruelle in 1844, and 1846; and at Gavres in 1848; with cast iron 30, and 24-pounders, proved that all the pieces, fired without a thin piece of cork interposed between the powder and the ball, burst before 500 discharges were made; whilst those, with which this precaution was taken, sustained 1800 and 2000 discharges without any damage, except an enlargement of the vent. _Vide_ United Service Magazine, September, 1855.

[31] _Vide_ “TREATISE ON NAVAL GUNNERY.” 3rd Edition. By General Sir H. Douglas.

[32] _In Extreme training of a gun to the Right_: Nos. 3, 5, 7, 11, 13, are placed outside; Nos. 8, 6, inside the tackle. No. 13 keeps the end of the fall coiled up.

_In Extreme training to the Left_: Nos 4, 6, 8, 2, are placed outside; Nos. 13, 7, 5, inside the tackle. No. 2 keeps the end of the fall coiled up.

[33] _In running out the right guns_, Nos. 3, 5, 5, 7, man the left tackle; Nos. 4, 6, 6, 2, man the right tackle.

_In running out the left guns_, Nos. 3, 5, 7, 5, man the left tackle; Nos. 4, 6, 6, 2, man the right tackle.

[34] _Note._—When the direction of the gun is to be altered, the word “Traverse” is to be given, if the gun is in, and “Point,” when the gun is out.

[35] _Vide_ Sir Howard Douglas’s highly-valued publication, entitled “A TREATISE ON NAVAL GUNNERY.” Fourth edition.

[36] _Vide_—“United Service Magazine,” No. CCCVIII.

[37] _Vide_ FIELD FORTIFICATION, pages 246, 247.

[38] _Vide_ Preface.

[39] For a square, the length of the perpendicular is ⅛th the exterior side; for a pentagon ⅐th; for the hexagon, and other polygons, ⅙th.

[40] _Vide Tables of Weights, and Measures._

[41] _Vide Tables_ of Weights, and Measures.

[42] In reducing fractions to a common denominator, and in multiplication of fractions, the work may be considerably diminished by cancelling any figures, which are in all the multiples; or by dividing a figure in each of them by any figure which can divide all without any remainder.

[43] See Note, page 268.

[44] _To multiply decimals by 1, with any number of ciphers, as 10, 100_, &c.—This is done by only removing the decimal point so many places farther to the right hand, as there are ciphers in the multiplier, and subjoining ciphers, if need be.

[45] The best way of doubling the root, to form the new divisor, is by adding the last figure always to the last divisor, as appears in the following example.

After the figures belonging to the given number are all exhausted, the operation may be continued into decimals, by adding any number of periods of ciphers, two in each period.

[46] _This rule is only applicable to the very best-made new cordage. The circumference squared should be divided by 6 instead of 5 for the description of rope generally employed._

[47] When the board is tapering, add the breadths at the two ends together, and take half the sum for the mean breadth. _Or else_, take the mean breadth in the middle.

[48] _To strengthen a beam, &c. which is required to support a great weight over a cavity, or ditch._—Place a prop, or short skid, under the centre of the beam, and pass a strong rope, or chain, over the beam lengthways, and under the skid, hauling it very tight, and making fast.

[49] In Lieut.-Colonel B. Jackson’s scientific “Treatise on Military Surveying, &c., &c., &c.,” _Portable trigonometry without logarithms_, is thus introduced—

“The following useful application of Trigonometry, by means of the natural sines, tangents, &c., is taken from an early number of that valuable periodical, ‘The Mechanics’ Magazine,’ and will be found particularly suited to the purposes of the military surveyor.”

[50] For further information on Surveying, and Reconnoitring, reference should be made to the highly-valued publication, entitled “A TREATISE ON MILITARY SURVEYING, INCLUDING SKETCHING IN THE FIELD, PLAN DRAWING, LEVELLING, MILITARY RECONNOISSANCE, &c.,” by Lieut.-Colonel Basil Jackson, containing a full account of every surveying instrument, and the right adaptation of them.

[51] 1. The Reconnoitring protractor is not intended to supply the place of the Theodolite, or other expensive instruments, when very great accuracy is required in surveying, or in trigonometrical observations; but, in the hands of officers accustomed to the use of it, bearings may be rapidly taken, heights and distances ascertained, roads traversed, &c., &c., with sufficient accuracy for a military survey, or reconnoissance.

The protractor has a tripod, on which it is to be steadily fixed for taking angles, &c.; but the instrument can nevertheless be used without the tripod; and mounted officers may, after a little practice, make a reconnoissance with the protractor alone, especially if they are able to measure, or calculate the distance of base lines, by the length of the paces of their horses.

2. A survey, &c., may be very rapidly taken in the field, by laying drawing-paper on the face of the protractor, under the marginal scale, fixing it firmly by means of drawing-pins in the sides, and using, at the first station, the edge of the index as a ruler to set off on the paper, at once, by observation through the sights, the angles of the objects whose distance is required; drawing a base line parallel to the tube side of the instrument, and also lines at the angles found. At the second station, the paper must be moved a few inches, for a base line to be drawn; at the termination of which (the second station) the index is to be directed to the objects, as before, and lines are to be produced until they intersect those drawn at the first station: thus the position of the objects will be obtained; and, by using the scale on the index for the length drawn for the measured base line, as well as for the lines directed to the objects, their respective distances will be ascertained.

3. The reconnoitring protractor, and all other instruments for surveying, &c., &c. can be readily obtained from Messrs. Elliott, 56, Strand, London.

[52] Or Reconnoitring protractor.

[53] To erect a perpendicular, _vide_ “Practical Geometry.”

[54]

3 inch cube full of air floats 1 lb. in water.
3 inch cube of water weighs 1 lb. in air.
1 cubic foot of water weighs 64 lb. in air.
1 ditto coal ditto 80 - 64 = 16 in water.
1 ditto sand ditto 95 - 64 = 31 in water.

A suit of clothes and a pair of boots, which weigh 7 lb. in air, when well saturated with water, only weigh in water 1 lb.

[55] _Vide also Definitions_—TRIGONOMETRY, page 301.

[56] _Gunter’s chain_ is in length 4 poles = 22 yards = 66 feet, and is divided into 100 links. Each link is therefore 22/100 of a yard, or 66/100 of a foot, or 7·92 inches. _Land is estimated_ in acres, roods, and perches. _An acre_ contains 10 square chains, or as much as 10 chains in length and 1 chain in breadth; or in yards it is 220 × 22 = 4840; or in poles it is 40 × 4 = 160 square poles; or in links it is 1000 X 100 = 100,000 square links. An acre is divided into 4 parts called roods, and a rood into 40 parts called perches, which are square poles, or the square of a pole of 5½ yards long, or the square of a quarter of a chain, or of 25 links, which is 625 links. Thus the divisions of land measure are—

625 square links = 1 pole, or perch.
40 perches = 1 rood.
4 roods = 1 acre.

The length of lines, measured with a chain, should be set down in links as integers, instead of in chains, and decimals. Therefore, after the content is found, it will be in square links.

[57] 57·3 is the number of pounds of powder contained in a cubic foot, when shaken; and 55 pounds when not shaken. According to the first case, one pound of powder will occupy 30 cubic inches; and according to the second case one pound will occupy 31·4182 cubic inches.

TRANSCRIBER’S NOTE

Footnote [37] is referenced six times from page 237;
footnote [52] is referenced twice from page 311;
footnote [54] is referenced four times from pages 317 and page 318.

Obvious typographical errors and punctuation errors have been
corrected after careful comparison with other occurrences within
the text and consultation of external sources.

Some hyphens in words have been silently removed, some added,
when a predominant preference was found in the original book.

Some { bracketing in some tables has been adjusted or removed for
readability.

Except for those changes noted below, all misspellings in the text,
and inconsistent or archaic usage, have been retained.

Pg viii: page number ‘8’ replaced by ‘48’.
Pg xvi: Added new section ‘CONGREVE ROCKETS.’ to the ToC.
Pg xviii: ‘Embrasures’ replaced by ‘Embrazures’.
Pg xxi: ‘312, 313, 314’ replaced by ‘312–4’.
Pg 61: ‘thirty ronnds of’ replaced by ‘thirty rounds of’.
Pg 69: in the table header ‘Fore ... Diameter’ replaced by
‘Fore ... Hind’.
Pg 72: in second column of the table ‘3¾’ replaced by ‘2¾’.
Pg 80: ‘4⅗ inch Mortar’ replaced by ‘4⅖ inch Mortar’.
Pg 85: in the table ‘5½ in c’ replaced by ‘5½ inch’.
Pg 91: in LEVERS section ‘lb. oz.’ replaced by ‘ft. in.’
Pg 171: in the table ‘1’ replaced by ‘10’.
Pg 171: in the table ‘1209’ replaced by ‘1200’.
Pg 176: ‘to facilite the’ replaced by ‘to facilitate the’.
Pg 187: ‘assist 2 at’ replaced by ‘assists 2 at’.
Pg 195: ‘at an elevavation’ replaced by ‘at an elevation’.
Pg 220: the Remarks column has been moved under the table to
conserve table space.
Pg 222: ‘a longe range’ replaced by ‘a long range’.
Pg 226: ‘they ricoched and’ replaced by ‘they ricocheted and’.
Pg 226: the italic markup on the small table has been removed.
Pg 227: ‘the same is in’ replaced by ‘the same as in’.
Pg 234: ‘left betweeen the’ replaced by ‘left between the’.
Pg 235: ‘placed at tho top’ replaced by ‘placed at the top’.
Pg 238: ‘embrasures should be’ replaced by ‘embrazures should be’.
Pg 291: ‘(5 × 1)’ replaced by ‘(5 + 1)’.
Pg 318: the footnote in the original book ‘See note, p. 317’ was
redundant and has been removed.
Pg 320: ‘is is a’ replaced by ‘it is a’.

Footnote [7]: ‘to five eet’ replaced by ‘to five feet’.