S5& Mr. Knight on the Attraction of such Solids 
passing through rW, parallel to the base, will be found by 
Prop. 2 , if we put x for a, and mx for b ; and is in the respec- 
tive directions pm, pg, po. 
A = <p (i, m, r) -f $ (i, m, /) — arc (tang. = -Lj — arc 
(tang. = ~) ; 
B = J r x (i,m,r'); C=4/(i ,m,r) — <J/(i 
If we multiply these by x and take the fluents, the actions 
of the pyramid are found to be 
A = X(p ( i , m, r) -j- x(p ( i, m, r') — x arc (tang. = -LJ — x arc 
{tang. = yj ; 
B = x x ( l, m, r) -j- ( i, m, /); C = xfy (i, m, r) — xp 
(i, m, P). 
If the pyramid, whose action we were seeking, had been 
that whose is u'mv, we must have used the other values of 
A, B, C given in Prop. q. 
Prop. 8, 
Let fig. 6 represent any pyramid whatever, whose base 
uvuVu is terminated by right lines ; to find its attraction, on a 
point at the vertex p, both in quantity and direction. 
Let a perpendicular from p meet the base at m, and draw 
lines from this point to all the angles u, v, &c, of the base. It 
is plain, that the solid will thus be divided into such pyramids 
as were considered in the last proposition ; so that the pro- 
blem is already solved. 
Cor. We may apply this to the pyramid whose base is 
a rhombus, and the vertex placed perpendicularly over its 
centre. By proceeding as in the proposition, it will be divided 
