MAXIMUM PRESSURE ON A SECTION OF A CONICAL VESSEL. 169 



Let ABC be a section passing along the axis of the conical vessel, 

 of which c is the vertex, and AB the diameter of its base. 



Conceive AI to be horizontal, and produce 

 the axis CD to meet the horizontal line AI in 

 the point i ; then is A ic the angle of inclina- 

 tion between the axis and the horizontal line 

 AI. 



Let G be the point in the axis through which 

 the plane of the required section is supposed 

 to pass, arid through G draw the straight line 



EF parallel to AB, and GH perpendicular to AI; then is EF the dia- 

 meter of the section, and GH the perpendicular depth of its centre of 

 gravity below A, the highest particle of the fluid. 



Put Rizr AD, the radius of the base of the conical vessel, 

 H c i), the axis or height, 

 r ZZTEG, the radius of the section on which the pressure is a 



maximum, 



a ~ the area of the section, 



d ~ GH, the perpendicular depth of its centre of gravity, 

 p the pressure perpendicular to its surface, 

 zr: A ic, the angle of inclination between the axis of the cone 



and the horizon, 

 x zr CG, the distance between the section and the vertex of the 



cone, 

 and s z= the specific gravity of the fluid. 



Then, because of the rightangled triangle ADI, and from the prin- 

 ciples of Plane Trigonometry, we have 



R : DI : : tan.0 : 1, 

 and from this, we obtain 



consequently, by adding the axis, we get 



cizz R cot.<j> -f-H, 

 and again by subtraction, it is 



G i ~ R cot.p 4" H x. 



But the triangle GHI, is by construction right angled at H ; there- 

 fore, by Plane Trigonometry, we have 



GHmc?zz:{Rcot.^ -j~ H a?}sin.0. 



Again, the triangles CD A and CGE are similar to one another; 

 therefore, by the property of similar triangles, we have 



