Encyclopaedia Britannica > Volume 3, Anatomy-Astronomy

ARCH.
of the extrados, to find that of the intrados, is the more im- ( )
portant of the two, and more difficult. Its solution xe- x c ( —£ n *. 1
quires the integral calculus ; but the difficulty is no great- y—be^-^Ve c+H)eV Je c^dx—e c J ecXdx\.
er than that of finding the area of a curve whose equa¬
tion is given. This can always be accomplished, it not in
finite terms, at least by infinite series. We shall now give
a general solution of the problem. . . 7
Lao-range has shown (Theorie des Fonctions Analy-
tiques, chap, viii.) that the integration of the equation
d'2y
dy?
To determine the arbitrary constants b and b\ we must
consider that a tangent to the intrados at the vertex is
perpendicular to the vertical DH; therefore, when ;r=:0,
then 0 ; but from the equation just found we get
dx
-£ = x.
(where X denotes an explicit function of the variable x),
can always be accomplished when two particular integrals
of this other equation, viz.
d*y y_ - 0
dx1 (? ~~ ’
X —£
are known. Now y — ye and y — qe c are such inte¬
grals (p and q being arbitrary constants, and e=: 2-7182818,
the number whose Napierian logarithm is unity), as may
be proved by differentiation; therefore, following La¬
grange, to integrate the equation (B), viz.
dry y _ v
we assume
X_ —X_
yzzpec-\-qe c
for the complete integral equation; but now p and 9 aie
to be considered as functions of the variable x. lo de¬
termine these, we differentiate, and get
± = l{pe‘-qe ‘}+e^£+‘
dq
dx'
-dp .
ej+e
ax
^=0,
ax
md then we have
dy If -
dx c j
Again, by differentiating, we find
<Py ij -7I
^ = y-yIK‘+V ■q+cC dx]
= 2!+I{c7^_e“7j}.
(?— c l dx dx)
This result, compared with equation (B), gives
1 f Ldp dq\ v
cl dx dx) c-
From this and the equation
-dp , dq
ecA--\-e c 0,
dx dx
we obtain ^putting X = —
dp _c ~
c X,
dx 2
dq _ c
i=-2etX’
Since p and q are indeterminate functions of x, we may
assume that
^jL = -{be^—b'e +%\e\fe ^Xdx + e “fe^Xdx}.
Now, when «=0, then =1, and e ^ — 1; and by hypo¬
thesis the integrals f ecXdx,fe cXdx in this particu¬
lar case vanish; therefore, when a?=0, the last equation
becomes 0=6— b', so that b'=b. But again, in the gene¬
ral equation (C), when a;=0, then y=d; therefore _we
have also c'=26 and b=\c'. On the whole, the equation
of the extrados is
(D)
^ _ L ee q-e e c Xdx—e CJec Xrfx j-;
the integrals being taken so as to vanish when a;=0.
We have thus brought the solution to depend on the in¬
tegration of the two differentials
£. ——
ec Xdx, e c Xdx,
which in fact will only differ in their sign, because the
branches of the extrados on opposite sides of the vertical
are exactly alike, and therefore the substitution of x
for + x will not change the sign of u nor of X. Now this
integration can always be effected by known methods,
therefore the second problem may be regarded as com¬
pletely resolved.
57. Example. Let us suppose that the extrados is a ho¬
rizontal straight line EF.
The line FT being supposed to touch the curve, let us
as before put
d = DB, c" = EA, s — DE,
x = DQ, y = FQ, 9 = angle TPK.
In this case « = 0 and X = 0, and the equation of the
curve is simply
y=!{*'+*7}.
p — ^Je cXdx-\-b; q— — ^j'ecXdx-\-b
Here b and b' are arbitrary constants, and the integrals
are supposed to be taken so as to vanish when x=0 rru''
complete integral equation is now
vol. m.
This case of the general problem has been resolved in
sect. 25; the equation of the curve is, however, here
given under a different form. We shall now deduce from
ft a formula for logarithmic calculation.
In all curves, tan. <p in the present case
Let ^ be such an angle that
— X
e7 = tan. (45° + £ then e 7 = tan. (45 — £ ^).
Hence, by the arithmetic of sines (Algebra, sect. 244,
(K) and sect. 240, (C) No. 1),
*_ -£ 1 _ 2
ec-j-e <•■ =
The -
ec — e c —
' cos.(45° + ^<//) cos. (4o°-
sin. 4
cos.(45° + cos.(45°-
cos.-4/
_2sin.-4/_
COS.-vJ/
3 Gr
-i+)
=2 sec. \j/
2tan.4/