of Edinburgh, Session 1882-83. 
177 
represent the curve very closely. It may be remarked that when A 
is 60°, the circle osculating the curve at B has 0 for its centre ; and 
that when the crossing is at right angles, the radius of curvature at 
the vertex is just one-half of the major ordinate. 
Since the whole area ABO is expressed by sin A, and OBPH 
by sin a, we have 
but 
wherefore 
APH = c2(sin A - sin a) 
JPH^ = c2(cos a - cos A) 
2APH : OB^ : : sin A - sin a : cos a - cos A : : 1 : tan 
A -j- Cl 
2 
so that 
2. ABO:OB2::l:tan^ . 
And thus, when the crossing is perpendicularly, as in fig. 3, the 
area of one wave of the curve is just equivalent to the square of the 
major ordinate. 
Equation (4) gives us 
x = c f 2 ^(cos a - cos A) , 
so that the second equation becomes 
dl = 
-^(cos a - cos A) da , 
\/2 
and thus the length BP of the curve is to, be got by the integration 
of this transcendental expression, that is. 
or 
I — / (cos a — cos A) da . 
choA.f2J ^ • ’ 
If we suppose the plane of the paper to be placed vertically, OB 
being directed toward the zenith, and if we imagine ON to represent 
the extreme position of a simple pendulum, the velocity acquired 
in descending from N to % (PI. II. fig. 7) would be proportional to 
^y(cos a - cos A), and thus the element of time in which an incre- 
ment da of the arc is passed over would be (cos a - cos A)~^da, and 
we thus have this remarkably beautiful theorem. 
(cos a- cos A) da 
