124 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION 



*=y -p-- (97). 



And the practical rule which this equation affords, may be expressed 

 in words in the following manner. 



RULE. Multiply the load on the safety valve by the square 

 of the diameter of the cylinder ; divide the product by the 

 entire pressure on the piston, and the square root of the 

 quotient will give the diameter of the safety valve required. 



132. EXAMPLE 8. The diameter of the safety valve is ^ of an inch, 

 the load upon it 46.875 Ibs., and the entire pressure on the piston 

 of the cylinder is 18750 Ibs. ; what is its diameter? 



Here we have given = J of an inch, w= 46.875 Ibs., and Pzr 18750 

 Ibs. ; consequently, by substitution, we have 



46.875 D*=. 25* X 18750, 

 from which, by division, we shall obtain 

 .25 a X 18750 



46.875 

 and finally, by extracting the square root, we get 



D = -v/ 25 5 inches. 



If both sides of the equation marked (93), be divided by w the 

 weight on the safety valve, we get 



and by extracting the square root, the general expression for the value 

 of D the diameter of the cylinder, becomes 



-y ^' (98). 



And from this equation we derive the following rule. 



RULE. Multiply the entire pressure on the piston of the 

 cylinder by the square of the diameter of the safety valve, 

 divide the product by the weight upon the safety valve, and 

 extract the square root of the quotient for the diameter of the 

 cylinder sought. 



133. EXAMPLE 9. The diameter of the forcing pump is one inch, 

 that of the safety valve is one fourth of an inch, and the power or 

 force with which the plunger descends, is equivalent to 750 Ibs. ; what 

 is the corresponding weight on the safety valve ? 



