VOL. XXVIII.] PHILOSOPHICAL TRANSACTIONS. 11 



then that part of c will be ■^-~— X ex* .v. Hence, taking the fluent, it will 

 be c = (4a^ + ^^) X -yca^. But the cone itself is a = -^ca^, and co = -fa : 

 Therefore cg = - = . 



A A 80 



And after the same manner the calculation proceeds in other figures, when 

 the ratios of ch to he, and of hz to p are more compounded. 



Exam, 3. To show the manner of calculating the quantity d, let the figure 

 proposed be a parallelopipedon, whose face perpendicular to the horizon, and 

 parallel to the plane of the motion of the centre of gravity, is abd, fig. 10. 

 Draw the diameters ef and hi, and let the altitude of the elements be p, and 

 make tr parallel to hi; putting gp = a, gh = b, gs = x, and sz = v. Then 

 will D = vxxx + ivvv. Hence the part of d in the right line tr will be 

 2bxx^ + 2b^x; and again taking the double of the fluent, it will be d = -f 



(ba^ -f- b^a). But a = 4ab; hence - = ^^-— — = tV^b^. 



Exam. 4. Let the last example be in the sphere, a great circle of which 

 is Btr, its diameter ab, and centre g, fig. 11. Then drawing the lines as 

 appear in the scheme, it will be d = g^"^ X j& + gw^ X p. But the sum of all 

 the Gs'^ X p in the right line tr, is g^^ drawn into the area of the circle on the 

 diameter tr. Also the sum of all the Gm^ X j& in the right line ki, is gw^ X 

 area of the circle on the diameter ki. Hence it easily appears that d is = 4 

 times the fluent g**^ into the area of the circle whose diameter is tr. Let 

 therefore cbe the area of the circle whose radius squared is 1 ; and let ga = a, and 

 gs = X. Then will d = 4xx^ X (ca'^ — cx^) = 4caxx'^ — 4cxx^. Hence, taking 

 the fluent, and making x =: a, it will be d = -^ca^. But a = ^ca^. 

 Hence - = -fa^. 



A ^ 



Because of the affinity of the solution, I shall here add a problem on finding 

 the centre of percussion. 



Prop. 6. Frob. 3. To find the centre of percussion of any body, revolving 

 about a given point; which centre must be such, that striking against an 

 obstacle, and at the same time the body being disengaged from the centre 

 of suspension, it shall not incline to either one side or another. 



First it appears that the point must be found in the plane of motion of the 

 centre of gravity. For if the body be resolved into prismatic elements perpen- 

 dicular to that plane, they will be carried about by a parallel motion; hence 

 the moments on each side of that plane will be equal. Therefore, by the re- 

 sistence made in this plane, no point of the body will be driven out of it. 

 Let that plane then be ab, fig. 12, to which let the body be reduced, by a 

 contraction of the prismatic elements into particles p, situated at the points z, 



c 2 



