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Old 13th January 2003, 03:30 AM   #1
kneadle is offline kneadle  United States
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Unhappy isosceles tetrahedron formula

Would someone like to suggest one of the following:

A formula for determining the volume of an isosceles tetrahedron, NOT the formula for determining the volume of a regular tetrahedron NOR for the irregular non-isosceles?

A calculator or modeler of some sort that will do it graphically?

A website that has the actual formula spelled out?

Or solve this problem:

The outside measurements of a 3/4" thick wood base are equilaterally 21". What are the measurements of the 3 isosceles triangles which will give the resulting tetrahedron 1.5 cubic feet (don't neglect the wood thickness)? Show your work so I can steal the formula.

Much appreciated.

Dave
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Old 13th January 2003, 04:36 AM   #2
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Default Thread

So let me get this straight,

The isosceles tetrahedron has an equilateral triangle as a base, with edge length of 21"?

If that's the case then the isosceles triangles should have one edge at 21" and the isosceles edges at 63.325"

I just used a ProE model and changed the height of the tetrahedron until the enclosed volume was the right value, this does incorporate the thickness of the box.

I'm sure there's some formula out there you could modify, but since you're doing it an odd way you might need to do some sort of iterative solution anyhow.
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Old 13th January 2003, 04:41 AM   #3
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Default Re: isosceles tetrahedron formula

Quote:
Originally posted by kneadle
A website that has the actual formula spelled out?
IIRC the volume of any regular solid like this should be 1/6 x the area of the base x the height.

dave
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Old 13th January 2003, 06:02 AM   #4
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The area of an equilateral triangle is A = 0.433 * X^2.

In this case x = 21 => A = 190.95sq.in.

The volume would be 0.433 * (the integral of X^2) from 0 to h, where h = height of the isosceles tetrahedron.

The desired volume is (V) = 1.5cu.ft. = 1728 * 1.5 = 2592cu.in.

Solving for h above

=> 2592 = 0.433 * (h^3/3)
=> h^3 = 17958.43
=> h = 26.187in.

This is the height of the isosceles tetrahedron

By trig, the edge of each side is 28.857in. and 21in. at the base.

Rodd Yamashita
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Old 13th January 2003, 01:19 PM   #5
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Default Re: Re: isosceles tetrahedron formula

Thanks all, but see the bottom of this post because I got two quite different answers. BTW, Here's why I asked in the first place: http://mathforum.org/dr.math/faq/for...trahedron.html Upon seeing this, my eyes glazed over.

Quote:
Originally posted by planet10


IIRC the volume of any regular solid like this should be 1/6 x the area of the base x the height.

dave
I don't think an isosceles tetrahedron is a regular solid. If it is, then I have never seen it described as such. You can see by Roddyama's response above that it is a little more complicated than solving for a regular solid.

Quote:
Originally posted by MITMechE
If that's the case then the isosceles triangles should have one edge at 21" and the isosceles edges at 63.325"
Quote:
Originally posted by roddyama
By trig, the edge of each side is 28.857in. and 21in. at the base.
Ok. Do you two guys wanna wrestle for it?

Dave
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Old 13th January 2003, 02:34 PM   #6
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Default Discrepancy

I think the difference is I account for a box in my calculations. The 1.5 cubic feet is the volume inside the box, and the 21" edge length and 63" edge length are the outside edge lengths of the box.

Using the dimensions of the inside tetrahedron (the dimensions of the volume) with edge lengths 18.14" and 55.57" I double-checked with the matrix formula and my answer is correct unless there's some other mistake.

1.0e+003 *

0 0.3291 0.3291 3.0880 0.0010
0.3291 0 0.3291 3.0880 0.0010
0.3291 0.3291 0 3.0880 0.0010
3.0880 3.0880 3.0880 0 0.0010
0.0010 0.0010 0.0010 0.0010 0

The determinant of that matrix is 1.9350e+009

1.9350e+009 = 288 * V^2

Therefore V = 2.5920e+003 in^3 = 1.5 ft^3
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Old 13th January 2003, 03:18 PM   #7
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Default Re: Discrepancy

Quote:
Originally posted by MITMechE
I think the difference is I account for a box in my calculations. The 1.5 cubic feet is the volume inside the box, and the 21" edge length and 63" edge length are the outside edge lengths of the box.
...and I was worried that I wouldn't have room to properly port the box!

5 ft 3 in? I've got to work on my sales technique to the wifey before I start construction on these mamas.

Thanks tons,

Dave

PS--do you know where the two different calculations are off? Even if Roddyama didn't take into account the inside of the box, his is still much different than yours.
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Old 13th January 2003, 04:15 PM   #8
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The differences most likely resulted from a couple of things:

Using 21" edge length instead of 18.14" for the base
Not accounting for the box itself.

They probably aren't off as much as they look, the truth of the matter is a tall tetrahedron isn't a very efficient way of holding volume.
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Old 13th January 2003, 05:49 PM   #9
kneadle is offline kneadle  United States
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Well, thanks to yours and roddyama's supplied calculations, I can achieve a shorter and broader tetrahedron. In addition, 1.5 cu ft was the absolute maximum for the driver in question. It will still sound good (given a skillful construction) with much less volume. I'll advise as things progress.

Keep a lookout for "tetrahedron DIY project finished" in the next few weeks, and I'll be sure to post photos and credits.

Thanks

Dave
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Old 13th January 2003, 06:44 PM   #10
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Default Re: Re: Discrepancy

Quote:
Originally posted by kneadle
Dave

PS--do you know where the two different calculations are off? Even if Roddyama didn't take into account the inside of the box, his is still much different than yours.
Mine was a guess pulled from my head (from info stored there a long time ago). If Rod solved the integral correctly then i probably just mis-remembered. Maybe the 1/6 should be 1/3?

dave
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