colonel@gloria.UUCP (Col. G. L. Sicherman) (05/21/85)
Does anybody know of a _compact_ expression for the volume
of a tetrahedron in terms of its edges--something comparable
to Hero's and Brahmagupta's formulas in plane geometry? I
have an expression but it's rather long and hard to remember.
--
Col. G. L. Sicherman
...{rocksvax|decvax}!sunybcs!colonelran@ho95b.UUCP (RANeinast) (05/24/85)
>Does anybody know of a _compact_ expression for the volume >of a tetrahedron in terms of its edges--something comparable >to Hero's and Brahmagupta's formulas in plane geometry? I >have an expression but it's rather long and hard to remember. > >Col. G. L. Sicherman Here's the expression, in the easiest form to remember it: (1/12)*sqrt( a^2*d^2*(-a^2+b^2+c^2-d^2+e^2+f^2) b^2*e^2*(+a^2-b^2+c^2+d^2-e^2+f^2) c^2*f^2*(+a^2+b^2-c^2+d^2+e^2-f^2) -a^2*b^2*f^2-a^2*c^2*e^2-b^2*c^2*d^2-d^2*e^2*f^2). Here's how the tetrahedron's edges are lettered: |`\ |` \ | ` \ | ` \ | ` \ c| `d \e | ` \ | /` \ | b/ `` \ | / f`` \ |/ ``\ ------------- a The secret is that (a, d), (b, e), and (c, f) are opposite, non-touching edges. Keeping that straight helps with the first 3 lines in the formula (note where the minus signs are). The last 4 subtracted terms are just made up of the triangular faces of the tetrahedron. I had played around with this formula in high school, and it took me about 10 minutes to rederive it (I know a lot more fancy tricks now). No, there is no simple factorization as there is for Hero's formula (unless you have some special case like a=d, b=e, c=f [in fact, you can use the special case factoriation to show that a general case factorization does not exist]). -- ". . . and shun the frumious Bandersnatch." Robert Neinast (ihnp4!ho95b!ran) AT&T-Bell Labs
colonel@gloria.UUCP (Col. G. L. Sicherman) (05/24/85)
To save time, here's the old (ugly) formula:
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
432 V = 3[(Aa) (B +b +C +c ) + (Bb) (A +a +C +c ) + (Cc) (A +a +B +b )]
2 2 2 2
- 6[(ABC) + (Abc) + (aBc) + (abC) ]
2 2 2 2 2 2 2 2 2
- [(Aa) (A +a ) + (Bb) (B +b ) + (Cc) (C +c )]
__ __ __ __ __ __
where A = OA, B = OB, C = OC, a = BC, b = AC, c = AB.
--
Col. G. L. Sicherman
...{rocksvax|decvax}!sunybcs!coloneljs2j@mhuxt.UUCP (sonntag) (05/24/85)
> Does anybody know of a _compact_ expression for the volume > of a tetrahedron in terms of its edges--something comparable > to Hero's and Brahmagupta's formulas in plane geometry? I > have an expression but it's rather long and hard to remember. > Col. G. L. Sicherman I just computed it, using techniques from high-school level calculus, and got: V = 1/(6*2^(1/2)) * L^3, where L is the length of a side. Is that compact enough? Just in case anyone's interested, this is how I got that formula: h V = S (1/2)s^2 sin(60) dy 0 Where 'S' is the integral sign, h is the height of the tetrahedron. h=(2/3)^(1/2)*L s is the length of the side of the triangle formed by the intersection of the tetrahedron and y=constant. s = L - y (L/h). The (1/2)s^2 sin(60) term is the area of an equalateral triangle with sides of length s. -- Jeff Sonntag ihnp4!mhuxt!js2j "You can be in my dream if I can be in yours." - Dylan
igor@fisher.UUCP (Igor Rivin) (05/25/85)
> Does anybody know of a _compact_ expression for the volume > of a tetrahedron in terms of its edges--something comparable > to Hero's and Brahmagupta's formulas in plane geometry? I > have an expression but it's rather long and hard to remember. > -- > Col. G. L. Sicherman > ...{rocksvax|decvax}!sunybcs!colonel *** REPLACE THIS LINE WITH YOUR MESSAGE *** Yes, there is a formula. Let Aij be the length of the side connecting the ith and the jth vertices. Than V^2 = | 0 1 1 1 1 | | | | 1 0 A12^2 A13^2 A14^2 | | | 1/(288) * | 1 A12^2 0 A23^2 A24^2 | | | | 1 A13^2 A23^2 0 A34^2 | | | | 1 A14^2 A24^2 A34^2 0 | This formula (with the obvious generalisations) works for all dimensions. The constant, in general, is (-1)^(n-1)/(2^n*(n!)^2), in n dimensions. Igor.
trainor@ucla-cs.UUCP (06/05/85)
More about volumes of tetrahedrons... >>Yeah, but it is easier to remember that the determinant >>of the deformation matrix changes volumes. >Not if you never knew it in the first place. What is the >deformation matrix? How does this give volumes? Is there >a fairly simple way to show it? I can never remember formulas... If you have some object A which encloses a volume V, and a 4x4 deformation matrix M which transforms A, then the volume of the new object is just V' = V * |det(M)| The absolute value is for reflections and the like which have negative determinants. For this particular problem, all you need to find is M, which transforms the standard tetrahedron to whatever tetrahedron you want. For the tetrahedron P,Q,R,S: 1 | [P-S 0]| 1 | [P 1]| V = --- |det[Q-S 0]| = --- |det[Q 1]| 6 | [R-S 0]| 6 | [R 1]| | [ S 1]| | [S 1]| From and old UCLA Math 169 exam: Q: Find the volume of the tetrahedron with vertices [1 1 1], [2 4 8], [3 9 27], [5 25 125] A: 1 | [1 1 1 1]| V = --- |det[2 4 8 1]| 6 | [3 9 27 1]| | [5 25 125 1]| 1 | [1 1 1 1]| = --- |det[1 2 4 8]| 6 | [1 3 9 27]| | [1 5 25 125]| 1 = --- (2-1)(3-1)(5-1)(3-2)(5-2)(5-3) 6 = 8 ARPA: trainor@ucla-locus.arpa --> trainor@locus.ucla.edu UUCP: ...!{sch-loki,silogic,randvax,ihnp4,sdcrdcf,ucbvax}!ucla-cs!trainor SPUD: trainor@russet.spud