zk 6 Mr. Knight on the Attraction of such Solids 
It is plain that, to find C, we need only change h into b\ and 
the reverse, in the last expression, whence 
P y v V + fr* + b y Va z + b z -\-b‘ z + b 
a V' 
Prop. 5. 
To find the attraction of a regular polygon, on a point 
situated perpendicularly over its centre. 
As this figure is composed of isosceles triangles, if we put 
b for the perpendicular from the centre on one of the sides, 
and r for the tangent of half the angle at the centre, subtended 
by one of those sides, we have, by Prop. 1, for a polygon of 
n sides, 
A = 2 n arc (tang. = ~ y / 1 -f- ~r-^ 2 ) — 2 n arc (tang. = --J» 
which, because the last term = [n — 2) v, by Euclid 1.32. 
Cor. 1, is 
A = 2/z arc (tang. = A 1 »f- l —JL ) — ( n — 2 ) it. 
Prop. 6 . 
To find the attraction of a circle, on a point situated per- 
pendicularly over its centre. 
This is only a particular case of the last proposition, when 
n is infinitely great and r infinitely small. 
It is easy to see, that the arc whose tangent is —s/ 1 + -J- b % 
will have for its cosine — r - , if we keep only the first power 
of r; consequently we may put it under the form ~ — arc 
(sine = —=£L=.) = . - —Ad— very nearlv; this multi- 
1 W + W 2 W+6 a ’ J * 
