the great practical question is to determine the conditions of the pressure under those possible circumstances, which are most unfavourable to the stability of the arch; circumstances which manifestly occur in the state bordering upon its rupture. This question necessarily then supposes a direction of the pressure, and therefore of the line of resistance, touching the extrados at the crown, and the intrados at the haunches; and, this being supposed, all those conditions of the equilibrium which depend upon the nearer approach of the voussoirs after the first striking of the center, or which arise from the long continued pressure, or from the influence of changes in the temperature, are eliminated.

Let us then assume that the line of resistance touches the extrados at the crown, so that pr (1 + a) cos 0; by equation (49), P(1+a) cos → - cos¥} = {X+ y2 (¦ a2 + a) (¥ − →)} sin Y

Y

P tan ¥ = X + ro († a2 + a) (¥ − 0) — r2 a2 (} a + §) tan ;

α

Dividing this equation by sin Y, subtracting unity from both sides, and reducing

Substituting this value for the first member of the preceding equation; multiplying by the denominator of the fraction in the second member, and neglecting powers of a above the first,

Hence, by Lagrange's theorem, we have from equation (60), neglecting powers of a above the first,

Where is taken to represent the value of when a = 0,* so

#

By Lagrange's Theorem, if y=z+apy, then, neglecting powers of a above the first, and representing by fy any function of y,

Let y=FY, y=F1Y, ƒy=FY; .. z = FY,, oz = F11, ƒz = F2Y ̧·

If therefore FY = FY2+ aF1Y and F be any other function of Y, then, neglecting powers of a above the first,

↓ (1-tan') (1-tan2) - (tan +tan') (1-0)

2

2

2

By equation (48), omitting powers of a above the first,

-1

also by equation (61), ¥ = 2 tan-1 {} (y + √y2 — sin2 O) sec2

2

- 2 tan

Ꮎ cos2

2

2

{sin

cot

Ꮎ costan

cos → + sin2

2

2

2

Ꮎ +cos2-tan2 2

Ꮎ

cos2

2

2

{}{

Ꮎ

Ꮎ
cos+costan

Ꮎ -cos2 - tan

Whence, by Lagrange's Theorem, neglecting powers of a above the first, we have, by equation (60),

Equations (62), (63), (65) determine the values of P and Y, that is, the pressure upon the key-stone and the positions of the points of rupture, for every condition of loading, and every form of the gothic and circular arch.

Assuming → = 0 and eliminating the value of tan we have, by equations (62), (63), and (65),