Frustum of a Pyramid with a Rectangular Base
Date: 02/20/2002 at 12:42:03 From: Brad Nelson Subject: Frustum of a pyramid with rectangular base Back in 2000 you gave a solution you derived using calculus to someone wanting to know how to figure the volume of a rectangular based frustum of a pyramid see: Volume of a Frustum-Like Structure http://mathforum.org/dr.math/problems/cunningham.5.12.00.html V = a1b1 + a2b2 + (a1b2 + a2b1)/2 x h/3 I am an engineer with a water treatment agency and need to figure the amount of water per foot of elevation in our reservoirs that happen to have the same shape as previously described. To satisfy my curiosity, could you please send me a copy of your derivation? I have had up through differential equations in college and am a bit rusty, so lay it on me. Thanks in advance.
Date: 02/20/2002 at 14:58:30 From: Doctor Peterson Subject: Re: Frustum of a pyramid with rectangular base Hi, Brad. You'll be interested in this later page, where I answered a question like yours about the volume contained by such a shape up to a given depth: Volume of a Trapezoidal Solid http://mathforum.org/dr.math/problems/greg.11.15.00.html It will also be of interest to you that the shape under discussion is a special case of a more general shape I wasn't aware of at the time, called the prismoid, or, even more generally, a prismatoid: MathWorld - Eric Weisstein: http://mathworld.wolfram.com/Prismoid.html http://mathworld.wolfram.com/Prismatoid.html The latter page gives the same formula as in my reference above, for the volume in this much more general case, where you just have two parallel polygonal bases joined to one another by straight edges: V = h/6 (A1 + 4M + A2) where h is the altitude, A1 and A2 are the areas of the bases, and M is the area of the cross section halfway between. Now, my calculus derivation for our special case was based on the fact that the length and width are changing linearly with height. So the dimensions of the rectangular cross-section at height x will be a = a1 + (a2-a1) * x/h b = b1 + (b2-b1) * x/h (To check these, see what they are at x=0 and x=h.) Then I integrated the product ab with respect to x, from 0 to h. It's a pretty easy integration. I later rederived the same formula by dissecting the shape into several prisms and pyramids. That takes a little more visualization and calculation, but no calculus. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 02/21/2002 at 10:08:06 From: Brad Nelson Subject: Frustum of a pyramid with rectangular base Thank you for the information. I was able to integrate it and come up with the same answer. My only other question is what principle or theorem you used for the a and b of the rectangular cross-section at any height x? a = a1 + (a2-a1) * x/h b = b1 + (b2-b1) * x/h Once again, thanks for your prompt response to my inquiry.
Date: 02/21/2002 at 10:21:06 From: Doctor Peterson Subject: Re: Frustum of a pyramid with rectangular base Hi, Brad. You can just draw a side view: a1 +------------+ ------------- / / \ |x | / / \ | | +------------+-----+ ------ |h / / \ | / a1 / a2-a1 \ | +------------+-----------+ ------- a2 You can see that the width at distance x from the top is a1 + (a2-a1)x/h, using similar triangles. (I've got this upside-down from the original labeling, but that doesn't affect the math.) Is that what you wanted? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 02/21/2002 at 11:30:39 From: Brad Nelson Subject: Frustum of a pyramid with rectangular base Aha, I see the light. I guessed you were probably using similar triangles but was having a hard time visualizing it for some reason. Thank you for your time and patience with a math-rusty engineer. You have been most helpful.
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