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

Ds

DO

=8=384615 nat. sin. DOn=22° 37′ 11′′

2

../DOC 45° 14′ 23′′

=45.23972°

The area of the segment DCn will therefore be found as follows, by Problem XI., Mens. of Surf.

Subt. nat. sin. 45° 14′ 23′′ = 3550295

Mult. by R2 (52)1

Carried over,

·0397594 (1.)

2704

1590376

Now since half the chord divided by the radius gives the nat. sine of half the angle at the centre, the same as in the last example, the calculation will be the same, down to the result marked (1.)

1728)6662-0853964 product solid in.

Ans. = 3-85537 solid feet.

PROBLEM XIV.

(1.) 14+6=20×62 =720 × 2.4674,

Gives the solidity

=1776-528,

(14+6)×6×9-8696-1184-352, the surface.

(2.) (31+9)×9×9×2-4674-7994-376, solidity.

(31+9) x9x9.8696

=3553-056, surface.

=986-96, surface.

(3.) (15+5)×5x5 x 2.4674-1233.7, solidity.

(15+5)x5x9.8696

25-455816, solidity.
1.732051
36

62.353836 surface.

(2.) Tabular solidity 1x43 64 solidity. Tabular surface=6x42 = 96 surface.

(3.) Tab. sol.

7-663119 x 33 =206-904213, solidity.

Tab. surf.20.645729 × 32= 185:811561 surface.

G

.

(4.) Tab. sol. 471405 × 103 =471-405, solidity. Tab. surf. 3.464102 × 102=346-4102, surface.

(5.) Tab. sol. 2-181695 x 831117-02784, solidity. Tab. surf. 8-660254 x 82=554.256256, surface.

32-795203 acres, 2 ro., 32.7952 poles. Ans.