Gang, Situation: Horizontal tank with the ends of the tank in the shape of an ellipse. Dimensions: Tank length 35", Narrow width of ellipse 10" and height of ellipse 16". Tank is mounted in a horizontal position with the longest dimension of the ellipse in a vertical position. I found a formula for determining the total tank volume, (about 20 gal) but I would like to make a "dip stick" with gallon markings on it. I know that halfway would be a 10 gallon mark and the markings on the stick would be duplicated from the "bottom half" to the "top half" but I haven't been able to find a formula to calculate how far up the stick the 1 gallon level would be, the 2 gallon level, etc., etc. Anyone know a formula, or where to look. I've checked a bunch of places on the net and looked through all the books, but maybe someone needs to look "outside the box" (or tank in this case) for me. Thanks. Ken.
If you can drain the tank then cheat like a SOB. Drain it, and then add back one gallon at a time, marking your stick as you go. Much easier than the math.
"Rich McCarty" wrote: Integral Calculus, grasshopper ^^^^^^^^^^^^ It should be a fairly easy problem in integral calculus, but I have been out of school for 40 years. If I solve the problem before someone else comes up with the answer, I'll get back to you.
I don't get the "grasshopper" part, though. Maybe it'll show up when I integrate. :-)
Heheheh - do it the dumb-bloke's way (my way) pour in a gallon, put ya stick in and mark it. Add another gallon and repeat, etc. It also makes sure that you don't forget to "carry the 1", or similar :-) Good luck.
Perhaps try a CAD program that will divide the ellipse into slices and then measure the area of the slices. Mine (TurboCad) won't measure the area of the slices. AutoCad, maybe? Gary Brady Austin, TX
"Gary Brady" wrote:(clip) Perhaps try a CAD program that will divide the ellipse into slices and then measure the area of the slices. (clip) ^^^^^^^^^^^ That's basically how I solved the problem, but, not having a CAD program, I wrote the ana;lytical geometry equation for the elli[pse, and did a bunch of arithmetic. I figured the area of the end of each 1" slice, and then converted to gallons.
Ken, I am sending you the figures by separate e-mail so as not to clutter up the newsgroup. If anyone else is interested, (unlikely), I will be happy to share.
Thank you, Thank you very much. If the tank has a constant cross sectional area, the problem would actually only be a double integral. If the tank is truely ellptical, then the calculus is far from trivial. However, I doubt a real-life tank would be truely ellptical. Probably more like two half circles joined by straight sides, in which case the problem can be divided into 3 parts. The double integral of pieces of a circle invloves trig conversions if I remember correctly. I suppose a real man would do it polar coordinates....
Thanks for all the input.... I appreciate the efforts and will do some "oughtin and noughtin" as an experiment. I would have thought tank manufacturers would have some easy formulas for different sizes of tanks they make (like the gasoline storage underground tanks
- which I realize are cylindrical in shape and easier) but I guess this is not a common problem. Thanks again. Ken.
It's easiest to do if you think of the origin in the middle of one end with x to the right and y up. Since, according to your post, the length is constant from end to end, the volume is just the area*length. taking the high to be 2B = 16" and the width to be 2a=8"
The equation of an ellipse is x^2 y^2 --- + --- = 1 a^2 b^2 Looking at the end, consider a horizontal stripe which is 2x wide by dy high and integrate A = 2ab Int'Sqrt(1-u^2)du'from -1 to (h-b)/b where h is the height of the liquid above the _bottom_ of the tank V = AL where L is tank length The answer(s) would be in cubic inches so we need to multiply by
0.004329 to get US gallons.
For the semi-circular case, set b = a = 5 and do the problem for 0 to 5. Then add to that the straight sided part up to 11". Then use bottom answer backwards with a bit of fiddling for the rest
This problem is straight from second semester calculus in any decent college or university.
I have been following this thread and would be ROFLMAO if I wasn't so close to tears. I have taught Calculus, Electronics and Computing at college/university level and guess where the hardest working, most dilligent students, or at least their parents, came from. It sure wasn't the US or Canada (yes, we're almost as bad up north)!
If North Americans were half as interested in Math and Physics as they are in professional sports or watching the Simpsons (like that young fool who thinks this problem needs a triple integral which he can't do) it would be hard for the Asians to compete given the technological lead the US used to have.
It's no wonder jobs are being exported at such a high rate!
The academics that know it, enjoy doing the math. I was happy that I was able to remember the math today for cylindrical volume for a 2-piston air compressor pump without needing to look it up.
The tank's only 35" long.. put a sight gauge along the top and stand it vertical during the filling, heh.
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