Find the volume of an irregular tetrahedron form its edges:
Suppose you are given the 6 sides of an irregular tetrahedron and you need to find the volume consumed by it.
Let the given sides to be u, v, w, W, V, U. Here, (u, U), (v, V), (w, W) are considered to be opposite edge pairs ( opposite edges means the edges which do not share common vertices ). Now the volume can be found from the following formula:Let:
u′ = v² + w² - U²
v′ = w² + u² - V²
w′ = u² + v² - W²
Now:
volume = 1⁄12 × √(4u²v²w² - u²u′² - v²v′² - w²w′² + u′v′w′)
This formula is derived from the determinant which can be found here for more reading. As the formula is symmetric, the ordering of the pairs won't make any change to the formula.
Read more properties about Tetrahedrons from Wikipedia.
Thanks so much for posting this, I had to do this problem too, and I couldn't find anywhere that did it :D
ReplyDeleteGreat, thanks for calculation this, saves the rest of the world some time :)
ReplyDeleteLode
Hi there,
ReplyDeleteThis formula works as long as the sides are more or less proportional.
I have tried to work out the volume of a tetrahedrom with sides:
40,425, 426,444.9,131.6 and 137.5 and the formula does not work because it becomes a negative number under the square root.
Does ANYONE know how to work out the volume of ANY irregular tetrahedron? I have spent weeks trying to get the answer.
Thanks
That’s really cool!
ReplyDeleteTo Anonymous of May 15, 2012: assign a letter to each of those six numbers, using the diagram above with u, v, w, U, V, and W and post it here. Please make sure you follow the requirements and have u and U each commanding two different, non-overlapping pairs of vertices, and the same with v and V and w and W. I'll check out your math. I'm sure you made a mistake because I've used the formula successfully with tetrahedrons that are similar to yours. I say that I've used it successfully because not only do I not get a negative number under the square root but because I've devised an alternative means of obtaining the volume using trigonometry and integral calculus, and it produces results that are identical to those of the formula.
ReplyDeleteTo Zobayer: Thank you for posting this formula, which enabled me to verify that my alternative method produces perfect results.
ReplyDeleteAbout the formula: I wonder if you have any information regarding its derivation--exactly how it is derived, and who was the clever person who did it. I'm particularly curious to know if Archimedes or one of the other ancients derived it--it seems well beyond their capabilities but I could be wrong.
@roger you might wanna look at this http://mathworld.wolfram.com/Cayley-MengerDeterminant.html
DeleteSeriously, this is brilliant! I found this at some other place too, but totally drove me wrong for hours! Thank you for such simple, intuitive presentation.
ReplyDelete