Solve for dimensions of rectangular prism given inertia

In summary, the author is having difficulty solving an inverse problem for inertia and is seeking advice. Any suggestions would be helpful.
  • #1
krenzke
2
0
I am trying to solve something of an inverse problem for inertia, and having a real tough time with it. Any suggestions would be helpful.

For a rectangular prism the mass and moments of inertia are:
m = rho*A*B*C
Ixx = m/12*(B^2 + C^2)
Iyy = m/12*(A^2 + C^2)
Izz = m/12*(A^2 + B^2)

Where m is mass, rho density, A width, B length, C height, and Ixx, Iyy, Izz are the principal moments of inertia. My problem involves a hollow rectangular prism, with wall thickness t. So the equations then become:

mOuter = rho*A*B*C
mInner = rho*(A-2*t)*(B-2*t)*(C-2*t)
m = mOuter - mInner
Ixx = mOuter/12*(B^2 + C^2) - mInner/12*((B-2*t)^2 + (C-2*t)^2)
Iyy = mOuter/12*(A^2 + C^2) - mInner/12*((A-2*t)^2 + (C-2*t)^2)
Izz = mOuter/12*(A^2 + B^2) - mInner/12*((A-2*t)^2 + (B-2*t)^2)

To see how these equations work imagine taking a rectangular prism and subtracting out a smaller rectangular prism from the inside. Given length (A), width (B), height(C), thickness (t), and density (rho) I can easily solve for the mass and moments of inertia. What I would like to do is given m, rho, Ixx, Iyy, and Izz solve for A,B,C, and t. I tried doing the hand-calc to solve the 4 equations for the 4 unknowns and quickly ran into a seemingly intractable 15th-degree polynomial in t with tons of unknown coefficients. If anyone has any ideas on how to solve this I would greatly appreciate it. It doesn't have to be a closed form solution (though that would be best), iterative or other methods would work as well.

Any ideas?
 
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  • #2
Welcome to PF!

If you want to pursue a numerical solution coded by yourself (as opposed to using a numerical solver in, say, Matlab or Mathematica), one of the standard tools in that toolbox would be to try solve it using Newton-Raphson [1] on your system of equations. If you want to try that, you should probably write it up as a system of equations of 4 unknowns and 4 knowns, as density is just a scale factor, and note that you can break symmetry in A, B and C by restricting the solution to [itex]A \geq B \geq C > 2t > 0[/itex] and make an initial guess that is asymmetric, like for instance [itex] B = A/2, C = A/4, t = A/16[/itex] which should give the initial guess [itex]A^3 = \frac{512}{43}\frac{m}{\rho}[/itex].

[1] http://en.wikipedia.org/wiki/Newton's_method#Nonlinear_systems_of_equations
 
  • #3
Thanks for the reply! I finally did try a Newton-rhapson method coded in c++, and it seems to work pretty well. I've been using the eigenvalues of the inertia tensor as the initial guess, with some fraction of the smallest value as the thickness. I have noticed that it will periodically converge to a non-physical value (i.e. A,B,C or t less than zero), but in this case taking the absolute value of the non-physical answer and putting that back into the iterative solver seems to come up with a different (and realistic) solution. Not sure if there is a better way to ensure that the solution lands in the positive quadrant...perhaps a boundary function of some sort?
 
  • #4
krenzke said:
I have noticed that it will periodically converge to a non-physical value (i.e. A,B,C or t less than zero), but in this case taking the absolute value of the non-physical answer and putting that back into the iterative solver seems to come up with a different (and realistic) solution. Not sure if there is a better way to ensure that the solution lands in the positive quadrant...perhaps a boundary function of some sort?

I'm probably just stating the obvious here, but anyway:

If the non-physical solutions pop up after a large jump in state (due to a small-magnitude Jacobian giving rise to a large state change) you could try limit the magnitude of the state change.

If the non-physical solution appears during a normal "slow" convergence sequence, then your initial guess must belongs to the attraction basin of that solution and a different initial guess is (as you say) obviously needed, probably one where the value for A, B and C are more equal in magnitude. If non-physical solutions are a real problem a practical "work-around" may be to map out the attraction basins of different initial guesses to see if there is a class or pattern of initial guesses that always converge to the physical solution and then stick to those.

If I recall correctly, there should also be a fair bit of convergence theory which may help you if you want to analyse convergence in a more theoretical way, although I'm not sure how well it will apply to your non-linear system.
 
  • #5


I understand the frustration of trying to solve an inverse problem and encountering difficulties. One suggestion would be to try using a numerical method, such as the Newton-Raphson method, to solve for the unknown variables. This method involves iteratively refining an initial guess until a satisfactory solution is reached. Another option could be to use a computer program or software specifically designed for solving complex equations, which may be able to handle the 15th-degree polynomial you encountered. Additionally, reaching out to colleagues or experts in the field may provide valuable insights and alternate approaches to solving the problem. Good luck with your research!
 

1. What is a rectangular prism?

A rectangular prism is a three-dimensional geometric shape that has six rectangular faces and all angles are right angles.

2. What is inertia?

Inertia is the resistance of an object to change its state of motion. It is directly proportional to the mass of an object.

3. Why is it important to solve for the dimensions of a rectangular prism given inertia?

Knowing the dimensions of a rectangular prism and its inertia can help in understanding its stability, strength, and motion when subjected to external forces.

4. How can I solve for the dimensions of a rectangular prism given inertia?

You can use the formula I = 1/12 * m * (a^2 + b^2) to solve for the dimensions, where I is the inertia, m is the mass, and a and b are the dimensions of the rectangular prism.

5. Can I use this formula for any rectangular prism?

Yes, this formula can be used for any rectangular prism regardless of its size or orientation. However, the measurements used must be consistent (e.g. all in meters or all in centimeters).

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