Solving an irregular tetrahedron

1. Oct 11, 2011

remettub

I'm dealing with a problem that seems (to my uneducated mind) like it should be more or less straightforward, but for some reason I've been unable to find any help on forums that are geared towards high school and college level math. Please forgive me if the solution is obvious.

If I know the lengths of the three edges of face A on a not-necessarily-regular
tetrahedron, and I know the three angles formed at the vertex opposite
face A (vertex P), how can I determine the other edges and angles of the tetrahedron?

With this information it is simple to determine the angles on face A. After this I am at a loss on how to proceed. I have attempted unsuccessfully to substitute into the sine law equivaletent for tetrahedrons (that the product of the sines of the clockwise angles adjacent to a given face is equal to the product of the sines of the counterclockwise angles), and then solve for one of the unknown variables, however it seems that the unknown variable always cancels itself out.

Any suggestions on how to approach this problem would be appreciated.

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2. Oct 12, 2011

gsal

are you looking for a 100% analytic method? or are you writing a computer program that can have a well defined algorithm that will take you there every time? 'cause I don't know the first one, but have a couple of ideas about the latter one.

3. Oct 12, 2011

AlephZero

If you know the length SR and the angle SPR, then from plane geometry you know that P lies on the arc of a circle, and if the center of the circle is O, angle SOR is twice angle SPR.

So, if you define some parameter (say p1) that gives the position of P on the circle, you can find the lengths SP and RP in terms of p1.

Do the same thing for the other two sides, using parameters p2 and p3.

You now have two expressions for the length of each side QP, RP, SP, and equating them gives you 3 equations for the 3 parameters p1, p2, p3.

4. Oct 12, 2011

remettub

gsal: I'm looking for an analytic method if possible.

AlephZero: thanks for the insight. I'll give it a try.

5. Oct 12, 2011

HYSL

Let the angle at the vertex P that corresponds to SR to be a, and the angle that corresponds to RQ to be b, and the angle that corresponds to QS to c. Let SR=l, RQ=m, QS=n (known), and PS=x, PR=y, PQ=z (the unknowns), then

x2 + y2 - 2 xy cos(a) = l2,.............(1)
y2 + z2 - 2 yz cos(b) = m2,............(2)
z2 + x2 - 2 zx cos(c) = n2,.............(3)

This gives you three equations for the three unknowns, but these are not linear equations, and I think will give you two solutions. I don't know if there is a simple way of solving this.

6. Sep 26, 2012

torchfire

Please, let me know if you have found the analytical solution for the problem.
I'm dealing with the same problem and can't get the solution however I've tried a lot of ways to solve it.
I would use the solution for position, orientation finding for a practical use.

7. Sep 26, 2012

HYSL

I have not been able to come up with an analytical solution; even solving this numerically is a mess.

The only way I can think of is something like this: use eqs.(1) and (2) to eliminate the y2 term, then use this with eq.(3) to elminate the z2 term. You now have a quadratic equation of x2, with coefficients consisting of the constants and y, z, and yz terms. Solve for x and substitute it into either eq.(1) or (3), which gives you another equation in terms of y and z (call this eq(4)). The new eq.(4) and the old eq.(2) give you two equations and two unknowns, y and z. On the other hand, you can express y and z parametrically as

y = $1\over\sqrt{2}$ { $\cos{t}\over\sqrt{ 1 - \cos{b} }$ + $\sin{t}\over\sqrt{ 1 + \cos{b}}$ }
z = $1\over\sqrt{2}$ { $\cos{t}\over\sqrt{ 1 - \cos{b} }$ - $\sin{t}\over\sqrt{ 1 + \cos{b}}$ }

This pair satisfies eq.(2) for any value of t. You now substitute the expressions for y and z into the eq.(4) and try to find the value of t that satisfies eq.(4) (numerically).

As I said, it's still a mess.

8. Sep 27, 2012

torchfire

Maybe you will find murrdpirate0's answers interesting. He found out that the problem was discussed by Mathematicans, it's called Three-dimensional resection. The algebraic and numerical solutions can be used in geodesy for example. If you're further intereseted, read the thread. The suggested book 'Algebraic Geodesy and Geoinformatics' deals with the problem from page 226.

9. Sep 30, 2012

remettub

Torchfire: From the other thread, it looks like you may have all the answers you need... but... I think I had an application very similar to yours, and I eventually decided on a four point system. I believe I found a solution using four coplanar points that was relatively simple. I could dig it up if that would be useful.

However, since I was using real world data, I was dealing with a certain amount of error in the measurements, and I don't think the estimation method I used coincided with the actual "maximum likelihood" scenario. If you've come across anything about that, I'd love to hear about it.

10. Oct 2, 2012

torchfire

I'm interested in your solution. It would be interesting to compare with the general four point system solution, which I found here:

http://perception.inrialpes.fr/Publications/1989/HCLL89/Horaud-CGVIP89.pdf

The coplanar case is presented as a special case.

I have also found an enourmously useful presentation on the problem. It deals with the POSIT algorithm as well:

http://mesh.brown.edu/3DP/pdfs/Grembowietz-2004-3dPoseEstimation.pdf [Broken]

I think this POSIT algorithm maybe very useful in practice. Maybe it's not so sensitive for the measurement errors as the algebraic method you invited and used.

Another wonderful articles in the subject:

http://avss2012.org/2006papers/gjkw/gk2.pdf [Broken]
http://www-users.cs.umn.edu/~joel/_files/Joel_Hesch_ICCV11.pdf
http://haralick.org/journals/three_point_perspective.pdf