involved in the Construction of Artillery. 403 



(o being a constant introduced by integration). But since, when x = P, P=0, the equa- 

 tion will, in that case, become 



c 

 and, eliminating e by means of this equation, we find 



^^ 2 



R_x\ 

 .V r)' 



But when a; = r, P=Fr, therefore Fr = —^l -= ), and eliminating kA between these 



equations, 



(12) 

 also, 



x"- 



and since this is the measure of the extending force on each square inch of the material, it 

 must not exceed a certain limit, T; but it is evident that it is greatest when x is least, that 

 is, when x = r; therefore the greatest pressure which this tube will bear is given by the 

 equation, 



" 11°. Let it be supposed that the pressures of the different shells were originally so pro- 

 portioned, that when the greatest internal pressure is applied, the tension T=^- shall 

 be uniform throughout the tube ; then by equation (9) dP + Tdx = 0, and integrating 



P=T{R-x), (13) 



T 



and the internal pressure F= — (R-r) is greater than the pressure corresponding to the 



same tension in the first case, in the ratio of R' + i^: Rr + r'. 



" IIP. To find the original distribution of pressure which leads to this result, we must 

 suppose the internal pressure F removed, or (which is the same thing) apply a pressure 



-F=-T?-^. 

 r 



Then, by equation (12), if;; be the resulting value of P at any other surface whose radius is 

 X, we have 



