(2.) 7912÷2

=

3956, radius of the sphere.

3956-3956 cos. 23° 28'-the height of the zone. 7912×3.1416= circumference of the sphere. -3.597256

Log. 3956

Log. cos. 23° 28′ =9.962508

=3.559764 Log. 3956 cos. 23° 28'.

..3956--3628.808=327·192, height.

=2.514803

3628.808

Log. 3.1416

= .497151

(3.) The height of the temperate zone is

=

(cos. 23° 28' sin. 23° 28') x 3956, which being multiplied by 7912 × 3·1416, will give the surface required.

cos. 23° 28'

sin. 23° 28'

Difference

= .917292

= .398215

=519077, hence by loga

(4.) The height of the whole sphere is evidently 7912, and hence 7912 × 7912 × 3.1416 the whole surface; which being wrought by logarithms, is as follows:

(2.) 5236×4 × 4×4=33·5 104 cubic inches.

(1.) By Rule II.

PROBLEM X.

(20 x 3-9x2) x 92 x 5236=1781-2872, solidity. (2.) By Rule I.

(3 × 102+92) x9x 5236=1795-4244, solidity. (3.) By Rule I.

(3 × 162+82)x8x-5236=3485-0816, solidity. (4.) By Rule II.

(3 x 30-2x24) x 242 x 5236-12666-9312, sol.

(1.) By Problem XI., Mensuration of Surfaces, find the area of the generating segment as follows:

2+(14) 61

= =21 feet, the radius.

2×11 3

11

Tan. arc==;.. Log. tan. arc=10+ Log. 3

2

– Log. 4 — 10+·477121—602060=9.875061; ··arc=36° 52′ 11′′; hence the whole arc= 147° 28′ 46′′-147-4794°; wherefore

1.018193

21

84849

2.036386

Subtract sin. 147°28′ 46′′ = 268801

Mult. by (2)2

Again, to find the surface, we have as above,

=112°

Leg. 8-Log. 1510+903090-1.1760919-726999 - Log. tan. 28° 4′ 21′′; therefore the whole arc= 17′ 24′′=112·29°; hence,

And 34-4895373 being multiplied by 12:5664,"
Gives 433-40932 inches, the solidity.

==;..Log. tan. arc-10+Log.3

10

-Log. 4=9·875061= Log. tan. 36° 52′ 11′′; hence the

whole arc =147° 28′ 46′′=147·4794°; wherefore,