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Re: Volume Calculation for Liquid Inside Cylindrical Container (2 cases)
Posted:
Jan 10, 1999 2:30 AM


Hi, I come up with volume of liquid in the tank as: V = L*(r^2*ACOS((rd)/r)(rd)*SQRT(2*d*rd^2)) where L = length of tank; r = radius of cylindrical tank; d = depth of liquid in the tank (The format should be easily cut and pasted into most BASIC and spreadsheet programs with perhaps minor alteration).
Derived thus:
Imagine the circle area at the end of a horizontal cylinder filled to a given level. The level of the liquid at the edges of the circular form an isosceles triangle with the center of the circle. From the center drop a vertical plumb line (perpendicular to the liquid level) to the bottom of the circle. The altitude a of this triangle is also a common leg of the two mirrorimage congruent right triangles. With a little bit of trigonometry, geometry, and algebra we should be able to demonstrate the derivation of the formula.
r = radius d = depth of liquid a = r  d (the altitude of the isosceles triangle so formed and also the common leg of the 2 right triangles).
AREA OF SECTOR A(sector) = (pi * r^2 * theta)/(2 * pi) = (theta * r^2)/2 Theta is the angle of the sector at the center of the circle. We have: (1/2) * theta = ACOS(a/r) = ACOS((rd)/r) or theta = 2 * ACOS((rd)/r) A(sector) = 2 * ACOS((rd)/r) * (1/2) * r^2 or A(sector) = r^2 * ACOS((rd)/r)
AREA OF TRIANGLE A(triangle) = (1/2) * a * 2 * sqrt(r^2  a^2) = a * sqrt(r^2  a^2) = (r  d) * sqrt((r^2  (r  d)^2)) = (r  d) * sqrt((r^2  (r^2  2dr + d^2))) = (r  d) * sqrt(r^2  r^2 + 2dr  d^2) A(triangle) = (r  d) * sqrt(2dr  d^2)
AREA OF SEGMENT A(segment) = A(sector)  A(triangle) = r^2 * ACOS((rd)/r)  (r  d) * sqrt(2dr  d^2)
Now we can multiply this by L, the length of the cylindrical tank so:
V(liquid in tank) = L*(r^2 * ACOS((r  d)/r)  (r  d) * sqrt(2dr  d^2))
 Bye Best wishes Lee Panteli Poulikakos wrote in message <36945f6b.4514136@news.wenet.net>... >I am in the process of writing a simple program for the company I work >for that calculates the volume of a liquid inside a cylindrical >container. Now if the container is placed vertically, the volume is >simply pi * r^2 * h where r is the radius of the circle and h is the >height of the fill level inside the cylindrical container. However, I >am not sure how to handle the case where the container is placed >horizontally. One might say simply take the container and place it >vertically. Well, that is not really possible since these are >cylindrical tanks with weights in excess of 5,000 lbs. When the >container is placed horizontally, the radius of the circle doesn't >really play a role in the volume. At first I thought that may be some >calculus might help, however there is really no dimension here to >integrate over. For that reason and through talking to some colleagues, > >I am under the impression that this problem involves possibly some >advanced geometry and trig. If anyone has any suggestions on how to go >about solving this problem, I would appreciate it if you could email me >at ppouli1@uic.edu. > >Thanks in advance, > >Panteli Poulikakos >
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