By the Gauge-Points.

First find a mean proportional between the breadth and ngth.

Set 20 on C to 20 on D; then against 100 on C will be 4.83 on D, the mean proportional.

Set 47-098 on D to 1 on C; and opposite to 44-83 on D ill be 906 of an imperial malt bushel on C.

Set 5-21 on D to 1 on C; and against 44.83 on D is 74.1 ›s. on C.

2. Required the content, at 1 inch deep, of a parallelogram, › tallow and hard soap, the sides being 96 and 48, and the erpendicular upon the former 36 inches.

Ans. 114-1348 lbs. tallow, 127.34 lbs. hard soap. 3. Required the content, at 1 inch deep, in starch and reen glass, of a triangular vessel, the base being 118 inches, nd the perpendicular upon it 72 inches, and one of the angles 9o. Ans. 122-0688 lbs. starch, 502-1276 lbs. green glass. 4. Required the content, in imperial gallons and bushels, at inch deep, of a vessel in the form of a trapeziod, the parallel ides 68 and 142, and their perpendicular distance 76 inches. Ans. 28 77987 gallons, or 3.5975 bushels.

5. Required the content, in pounds of starch, of a trapezium, of which the diagonal is 78, and the perpendiculars upon it 23 and 15 inches. Ans. 43.1465 lbs. Here the multiplier is 0287356, the divisor 34-8, and the gauge-point 5.899. 23+15.5 38.5.

=

=

Set 34-8 on A to 39 of 78 on B; and against 38.5 on A will be 43.146 lbs. on B.

6. Required the content, in old wine and ale gallons, of a regular octagon, of which the side is 42 inches.

Here the multipliers are 0209023 and 0171221, the divisors are 47.8417 and 58.4041, and the gauge-points 69168 for wine, and 7-6423 for ale gallons.

42 x 421764

1764

017122

47.84 )1764(36.87 wine gal. 1764 58-404)1764(30-203 ale gal. ⚫020902

30-203208 ale gal. 36.871128 wine gal.

By the Gauge-Points.

Set 7.64 on D to 1 on C; then against 42 on D is 30.2 ale gallons on C.

Set 6.92 on D to 1 on C; and at 42 on D is 36.9 wine gallons on C.

7. Required the content, in imperial bushels, of a regular hexagon, of which the side is 138 inches.

Ans. 22.331 bushels.

III. If the inches in any of the gauge-points be laid on a rule, and this distance be divided into 100 equal parts, the dimensions may be taken with that rule, and then the content may be found without using the multipliers or divisors. Thus, if 7.64 inches be divided into 100 equal parts, the side of the octagon in Ex. 6, measured by this rule, would be 5·496 ale gallons; and this, multiplied by itself, would give 30-206 ale gallons for the content.

1. Required the content, in imperial gallons, of a circle, of which the diameter is 40 inches.

Ans. 1600x0028326 = 4·52316 imperial gallons.

By the Gauge-Points.

Set 18.8 on D to 1 on C; then against 40 on D is 4:53 gallons on C.

If 18.8 inches be divided into 100 equal parts, the diameter measured by this scale would be 2.13, which, multiplied by itself, gives 4-5369 imperial gallons for the content.

2. Required the content, in imperial gallons, of a sector of a circle, of which the radius is 42 inches, and the arc 118 inches. Ans. 8-9369 imperial gallons 3. Required the content, in hard soap, of a trapeze, the diagonal being 32 inches, and the perpendiculars upon it from the angles 18 and 14 inches. Ans. 18-865 lbs. 4. Required the content of a quadrant, at 1 inch deep, in plate glass, the radius being 16 inches. Ans. 21-907 lbs.

IV. Some preparation is often necessary before the question can be wrought by the sliding-rule, as in the following examples.

1. Required the content, in imperial gallons, of a segment of a circle, the diameter 50, and the versed sine 10 inches. Ans. 10'000 ÷ 50=200 the tabular versed sine, opposite to which is 1118238 the tabular area; and 1118238 × 50o x0036065 1.00823 imperial gallon.

Set 16.65 on D to 1118 on C; and at 50 on D is 1:01 im. perial gallon on C.

1

2. Required the content, at 1 inch deep, in tallow, of a triangular vessel, of which the sides are 36, 24, and 20 inches.

Here the half sum is 40, and the remainders are 20, 16, and 4. A mean proportional between 20 and 40 is 28-284, and between 16 and 4 is 8. Then,

Set 30-28 on A to 28-284 on B; and against 8 on A is 7.47 lbs. tallow on B.

3. What is the content, in imperial gallons, of an ellipse, of which the axes are 72 and 50 inches?

Ans. 10-17936 imperial gallons.

PROB. X. To gauge solids.

When the depth is greater than one inch, set the gaugepoint to the depth instead of 1.

1. Required the content, in imperial gallons, of a rectangular prism, of which the length is 81, the breadth 26, and the depth 25 inches.

Ans. 81 x 26 x 25 x⚫0036065 = 189.882 imp. gallons.
By the Gauge-Points.

Set 25 on C to 25 on D; and at 81 on C is 45 on D, a mean proportional between 25 and 81. Then,

Set 16-65 on D to 26 on C; and at 45 on D is 190 imperial gallons on C.

2. Required the content, in imperial gallons and bushels, of an octagonal prism, of which the depth is 80 inches, and each side of the base 63 inches.

Ans. 63 x 80 x 01741395529-262 imperial gallons, = 691.158 bushels.

By the Sliding-Rule.

Set 7.578 on D to 80 on C; and at 63 on D is 5529.3 imperial gallons on C.

Set 21-434 on D to 80 on C; and at 63 on D is 691.16. imperial bushels on C.

3. Required the content, in imperial gallons, of a cylindrical vessel, the depth 40 inches, and the diameter of the base 27 inches.

Ans. 27 x 40 x '002832682.599 imperial gallons.

By the Sliding-Rule.

Set 18.8 on D to 40 on C; and at 27 on D is 82.6 imperial gallons on C.

4. Required the content, in imperial gallons, of the frustum of a square pyramid, the depth 24 inches, each side of the lower base 26, and of the higher 34 inches.

Ans. 34+26=60, and (60o — 34 × 26) × 24 ×·0012022 =78.364 imperial gallons.

By the Sliding-Rule.

First set 26 on C to 26 on D; and at 34 on C is 29.72 on D, the mean proportional between 26 and 34. Then,

Set 28.84 on D to 24 on C; and at 60-0 on D is 104-2 on C. .. 28.84 . . 24

29.7.-25.8 . .

78.4 im. gal.

NOTE. These, with most other questions, may be wrought more easily by Prob. XII. of MENSURATION OF SOLIDS; and therefore it is proper to give tables for it, and the rule for working it by the sliding-rule.

TABLE IV.

GAUGE-POINTS TO BE USED WHEN THE MIDDLE AREA IS

Find the squares or products of the sides or diameters at the top and bottom, and of the double of those in the middle: the sixth part of the sum of these, multiplied by the proper multiplier in Table I. or II. will give the content.

By the Sliding-Rule.

Set the gauge-point on D to the length on C; then oppoite to the sides or diameters at the ends, and to twice that in he middle on D, will be found three numbers on C; and hese three, added together, will give the content.

To work the last question by this rule. (26+34 +60°) × 24 ×·0036065 =78.362 imperial gallons.

By the Sliding-Rule.

Set 40.79 on D to 24 on C; then at 60 on D is 51.6 on C.

5. Required the content, in imperial gallons, of a frustum of a rectangular pyramid, the depth of the frustum 100 inches, he sides of the upper base 18 and 8 inches, and the sides of he lower base 27 and 12 inches.

Ans. (18 x8+27 × 12+45 × 20) × 100 ×·0036065 = 32.2282 imperial gallons.

By the Sliding-Rule.

Set 40·79 on D to 100 on C; then at 18 on D is 19:47 on C.

6. Required the content, in imperial gallons, of the frustum of a cone, the depth of the frustum 100 inches, and the diameters of the bases 18 and 12 inches.

Ans. (182+12o +30o) × 100 ×·0028326 = 64-58328 imperial gallons.

Set 46.02 on D to 100 on C; then at 18 on D is 15·3 on C.

7. If the axis of a globe be 100 inches, how many imperial gallons will it contain?

In a sphere, the square of twice the middle diameter is three times the square of the axis.

(10000+30000+0) x 100 x 0028326

Ans. imperial gallons.

1888-4

Set 46-02 on D to 100 on C; then at 200 on D is INNN-7 imperial gallons on C.

8. Required the content, in imperial gallons, of a bowl o