MEN
176
MEN
2
for the first term is — in both, and the second term in theo
approximation only differs from that of the proper series bya quantity less than
2 cv
From this Theorem, ■
3
+ —, we may derive c + —
& C u
A -, as in Theorem XI. Rule 7, for the rectification of
c r
curves.
In the sector a d c e a draw the chord A c and e d perpen-dicular to a c, cutting ac at b; let ac' = c, and bd = tjthen be = de — db = r — v; consequently the area of
the triangle ace= —— X c =---; but the area
A M
2 c v <y3
of the segment A d c is - + — : therefore the area of
6 3 2c
. 2 c« , v s cr—cv cv
the sector adceais equal to —5— + --1--— = —
O 4 C m O
+ x- + -a 5 and if this area be divided by the radius, we2 c 2
, . cv v 5 c c cv v 3 , . .
obtain — + -— + o’ or o + IT" + 5— for the half arc >
6 r 2 cr 2 2 6r 2cr
or c + *—• + —, for the whole arc. And if —7-—— be sub-
3 r cr 8v
Theorem XIY. Rule 7, of the areas of plane fig atet>
s 2 + v 3
-, being thq area of the segm ent >
8 c v 1
stituted for r, we obtain c +
3 <? + 12 v 2
the value of the arc in terms of c and v.
8v<
c 3 4- 4 c v
for
11 r* X
1 jl s t ar -sine of the half arc of which is s, the versed sine v, an ^radius r, evidently follows from Theorem X. Rule
multiplying half the arc, X * ” —, by the radi° 9
v 1 0 ’2 11 + 3 J
Theorem XYI. being x \ will easily ^
b \ 3 2 6/ t he
obtained from the following consideration, viz., supp 09ellipsis to be completed, and a circle to be described upe j e jaxis, which bisects the base of the elliptic segment; the ar y,the base of the elliptic segment, to be continued, if nece f e0 itmeet the circumference of the circle on each side; 0 fwill be as the base of the circular segment is to the ba eJ itthe elliptic segment, so is the area of the circular se ° „ a ofto the area of the elliptic segment; therefore, if the a {the circular segment be known, that of the elliptic seg^ ^ ^
will follow; now the area of the circular segment 1 9 3
+ therefore, 6 : l :: ^ ^ : \ X
the area of the elliptic segment.
Areas op Curved Surfaces.
Rule 1, is an approximation ; ® n $6' what dependence is to be placed J
then (d xf X (a-\-x) = adr-{-2adx-{-aa?-{-d‘x-\-and (d — x) X (a — x) ad 1 — 2a dx ax* — <f x +therefore
Theorem VII.following will showresult obtained by calculation
Let a — the axis of the spheroid ;
6 = the diameter of the great circle ; ,
and x = an indefinitely small distance, comparea or d. . • of ^
Now .5236 a d* is the expression for the solidly, .spheroid, being two-thirds of its circumscribing cyl ,n
2 d x* + x? = a d 2 nearly, being in excess ;
2d 3? — 3? — a ct nearly, being in defect;
4adx
2d‘x
+ 2 x 3 will be the solidity of a thin shell, the
tb> c
ness of which is x ; therefore 4 ad + 2d 2 + 2x 3 will be thearea of such a shell, or 4ad + 2d 2 , by leavfhg out the quan-tity 2ar*, which is indefinitely small: therefore .5236 (4 ad+ 2 d 2 ) is the area of a spheroid nearly; it is exactly so whena and d are equal; for then .5236 (4 a d + 2 d 1 ) becomes.5236 X 6 d? = 4 X .7854 <P, but the greater the differencebetween a and d, the more will the error be; because thethin shell varies more in its thickness.
Corollary 1.—Hence the surface of every rectangularprism with a square base is 4 a d + 2 d 2 .
Corollary 2.—When a = nothing, then .5236 (4arf + 2d 2 )becomes .5236 X 2 d? = 1.0472 d 1 , instead of .7854 d 2 , thearea of the base.
Example for the Oblong Spheroid.
11200 = 4 arf+2<P
.6236
11200
1047200
5236
5236
5864.3200 = .5236 (4 a d + 2 <f)
The true answer is 5882.6385, the difference 1 9 ea o^ eonly 18.3185, which is not worth regarding in so l a £S ^her. In a dome which is the half of an oblong s P^ eerror would only be 9.1592 less than the truth.
Example for an Oblate Spheroid.
50 = a
40 =d
40 = a
50
40 -d
40 = d
50 = d
50
2000 = a d
1600 = <r
2000 = a d
2500 s= ^
4
2
4
2
■
_____
_ —
8000 = 4 ad
3200
8000 = 4 ad
5000
3200
5000
13000 = arf+ 2d*