Hi, I come up with volume of liquid in the tank as: V = L*(r^2*ACOS((r-d)/r)-(r-d)*SQRT(2*d*r-d^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).
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 mirror-image 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((r-d)/r) or theta = 2 * ACOS((r-d)/r) A(sector) = 2 * ACOS((r-d)/r) * (1/2) * r^2 or A(sector) = r^2 * ACOS((r-d)/r)
-- Bye Best wishes Lee Panteli Poulikakos wrote in message <email@example.com>... >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 firstname.lastname@example.org. > >Thanks in advance, > >Panteli Poulikakos >
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