ON SEMICIRCULAR PLANES. THE CHORD OF MAXIMUM PRESSURE. 167 



therefore, by transposing and casting out the common terms, it is * 



cos. 2 0nz2sin. 2 0. 



But according to the principles of Plane Trigonometry, we have 

 2sin. 2 = 2 2cos. 2 <; consequently, by substitution, the above 

 expression becomes 



cos. 2 zz 2 2 cos. 2 ; 



therefore, by transposition and division, we obtain 



cos. 2 0=rf, 

 and by extracting the square root, we get 



cos.0rz:-v/-| ; 



finally, let both sides of this expression be multiplied by r, for the 

 purpose of adapting it to the proper radius, and we shall have 



# = rcos.0:zrr|/!-, the same as above. 



165. The practical rule for reducing this equation, may be expressed 

 in words at length in the following manner. 



RULE. Multiply one third of the radius of the given semi- 

 circular plane by the square root of 6, or by the constant 

 number 2.44947, and the product will give the distance of 

 the point on the vertical axis below the surface of the fluid, 

 through which the chord of maximum pressure passes. 



166. EXAMPLE. The radius of a semi-circular plane, immersed in a 

 fluid agreeably to the conditions of the problem, is 27 inches ; at 

 what distance below the surface of the fluid must a horizontal chord 

 be drawn, so that the pressure which it sustains may be greater than 

 the pressure sustained by any other chord drawn parallel to it? 



By operating according to the rule, we shall obtain 



x 9 ^6*= 9 X2.44947 = 22.04523 inches. 



167. The same example admits of a very simple and elegant geo- 

 metrical construction, which may be effected in the following manner. 



Let ACS be the semi-circular plane, of 

 which the diameter AB is parallel, and the 

 radius D c perpendicular to the horizon ; draw 

 the chord BC, and from c as a centre, with 

 the radius CD, describe the circular arc DKH 

 cutting BC produced in the point H. 



Through the point c draw the tangent c K, 

 and let fall the perpendicular H i meeting c K 

 in i; join DI to intersect the arc A EC in E, and through the point E 



