Of FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 89 



/ = AC, the length of the chord on which the pressure is a 



maximum, 



p in the pressure on the chord AC, 

 x =. c D, the perpendicular height of the centre of gravity of the 



chord A c above the lower extremity of the diameter, 

 and s ~ the specific gravity of the fluid. 



Then we have o = c? x; CE=z:2a:, and by the property of the 

 circle, the length of the chord becomes 



consequently, the pressure upon it is 



but this, according to the conditions of the problem, is to be a maxi- 

 mum ; therefore, by putting the fluxion of the expression equal to 

 nothing, we obtain 



MX (d* 4dx + 3z 2 ) = ; 



therefore, by omitting the common factors 2dx and transposing, we 

 shall have 



and this quadratic equation being reduced, we get 



x ^d. (48). 



COROL. 1. Consequently, to determine the chord by construction, 

 make BE equal to one third of the vertical diameter BC, and through 

 the point E, draw the straight line EA at right angles to BC, and 

 meeting the circumference in the point A ; then from c, the lower 

 extremity of the diameter, inflect the straight line CA, and the thing 

 is done. Or thus : 



2. Find an angle such, that tan. J^/2". 707 11, which 

 happens when 0zn 35 15' 51" ; therefore, at c the lower extremity 

 of the diameter, make the angle BCA equal to 35 15' 51", and the 

 straight line CA will be the chord required. 



It has been shown above, that according to the property of the 

 circle, the length of the chord is == ^2dx; if, therefore, the value 

 of x as determined in equation (48), be substituted instead of it, in 

 the foregoing value of /, we shall have 



l=ldi/6. (49). 



99. This is the proper form of the expression when adapted for 

 numerical operation, and the practical rule which it supplies, may be 

 expressed as follows. 



