402 Mr. Mallet on the Physical Conditions 



force on each unit of length of the tube, tending to sijlit this sliell into two semi- cylinders, 

 is ifx, and the forces which resist this are 2/' (.r + dx) plus the tenacity of the material. 

 Let this latter force be supposed proportional to the extension, and let 8 bo the increase 

 of the radius .r produced by this extension; then, since the extension of the circumfe- 

 rence, divided by the circumference, is equal to the extension of tlie radius, divided by 

 the radius, the resistance at each extremity of the diameter, along which the tube is sup- 



s 

 posed to spUt, will he k - dx (the value of A is immaterial, as it does not appear in tlie final 



result); and the condition of equilibrium will be 



g 

 2/:r=2/' {x^'dx) + 2k-dx; 



or, if 



fx=P, f {x + dx)==P + dP, 

 the equation becomes 



dP^k-dx = 0. (9) 



" But it must be observed that this shell can only communicate pressure to the next one 

 by virtue of its resistance to compression in the direction of the radius, and that this resis- 

 tance may (according to the common theory) be assumed to be proportional to the com- 

 pression, that is to say, since dx was the original thickness, and d {x+ S) the thickness 



under pressure, the resistance is proportional to - — ; therefore, if A' be the exponent of this 

 ratio, 



f^l^-k'f. (10) 



■' X dx 



" From these two equations Pand S are to be determined; multiplying them, we have 

 PdP = kk' SdS, and integrating, 



P= = Ai'(S^-A=), (11) 



A being the value of g at the surface, which is free from pressure. It is not necessary to 

 suppose k = k', but the calculation will be abridged by the supposition; and eliminating o 

 between (9) and (11), we have, on this supposition, 



dP -/(P' + k'-A^) „ 



-_+ =0; 



dx X 



and, by integration, 



V(P' + A=AO = P+'^. 



