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#1
02-22-2006, 01:10 PM
 ReuvenB Charter Member Join Date: Sep 2001 Location: Portland, OR Posts: 757
Liquid in a tilted frustum

So suppose I have a conical frustum, like a drinking glass. I know the measurements of the two radii and the height of the frustum. In the frustum, I have a liquid that sits in the glass. The liquid, having a flat surface, fills up the frustum just enough that it touches the "lip" of the larger end, and the opposite side of the smaller end of the frustum.
What's the volume of the liquid in terms of the two radii and the height?

My math teacher says that this is unsolvable, but I've gotten two equations that should, when integrated and added together, give the volume of the liquid. Unfortunately, they're rather daunting, and I'm wondering whether my math teacher is right, and I should give up now before I begin the really hard work.

Is this a well known problem? Am I missing something? Is there some part of this that I won't be able to integrate?
#2
02-22-2006, 02:57 PM
 Zulema Guest Join Date: Oct 2000 Posts: 1,644
Did he say why he thought it was unsolvable? I don't understand why it wouldn't be.

You are talking about a regular cone shape with the point cut off?
#3
02-22-2006, 02:59 PM
 Zulema Guest Join Date: Oct 2000 Posts: 1,644
#4
02-22-2006, 03:19 PM
 moes lotion Guest Join Date: Nov 2002 Location: North of a Great Lake Posts: 197
I haven't tried to derive a solution to this problem, however I believe SP2263 has misunderstood the situation. This isn't a simple case of what is the volume of a conical frustum, but instead what is the volume of liquid in a frustum that is tipped to one side and filled until the surface of the liquid just touches the opposite edges of the top and bottom.

I suspect that the solution to this problem will involve an integral that has no closed form solution but could be approximated numerically.
#5
02-22-2006, 03:22 PM
 MikeS Charter Member Join Date: Oct 2001 Location: New London, CT Posts: 3,752
It would help to know what you're doing to try to solve this. I'm working through it with the help of Mathematica right now (using spherical coordinates), but it's not giving me anything useful. What kind of bounds of integration are you using?
#6
02-22-2006, 03:27 PM
 Zulema Guest Join Date: Oct 2000 Posts: 1,644
OK, I'm obnoxiously posting three times in a row.

I missed the tilted part in the title.

If both ends of the cone are cut so they are parallel but the whole shape is tilted, the ends are not circular anymore. I'm not sure what you mean by radius. I think this is what you mean by "tilted frustum".

Here is a description of how to figure this out if you have the surface areas of the two planes. The planes have to be parallel. Scroll down a little more than half way.

http://home.att.net/~numericana/answ...mula.htm#tcone
#7
02-22-2006, 03:28 PM
 Omphaloskeptic Guest Join Date: Oct 2001 Posts: 1,263
If I've done it correctly, it does have a closed-form solution. (I worked in Cartesian coordinates, FWIW, finding the area of a differential slice parallel to the endcaps and integrating that along the axis of the frustum.)
#8
02-22-2006, 03:36 PM
 Zulema Guest Join Date: Oct 2000 Posts: 1,644
I finally know what you're talking about. You're tilting a regular frustum.

Give us some numbers to work with, let's see what we all come up with.
#9
02-22-2006, 03:45 PM
 Omphaloskeptic Guest Join Date: Oct 2001 Posts: 1,263
Quote:
 Originally Posted by SP2263 I finally know what you're talking about. You're tilting a regular frustum.
The formula a little farther down on the page you linked to in post #6 can be used here, for a different solution method than I used.
#10
02-22-2006, 04:52 PM
 ReuvenB Charter Member Join Date: Sep 2001 Location: Portland, OR Posts: 757
Quote:
 Originally Posted by Omphaloskeptic If I've done it correctly, it does have a closed-form solution. (I worked in Cartesian coordinates, FWIW, finding the area of a differential slice parallel to the endcaps and integrating that along the axis of the frustum.)
That's what I did, looking at them as a whole collection of circles of increasing (or decreasing, depending on how you look at it) radii, having progressively larger parts sliced off of them at the surface of the liquid. You say there's a closed form solution?

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