in reply to Re: [OT] Forces.
in thread [OT] Forces.

Then, you can just ignore the circle and apply the forces directly in the point A.

What I was trying to indicate was that I was aware of the small discrepancy in the final position of the point F; and that because the angles involved are chosen to be very small at each step of the iterations; the affect of that discrepancy is minimal for each step.

However, the forces that will result from the next iteration of the stress tensor, are affected by the rotation of the body from the previous step, hence it cannot be ignored completely.

It is clear that those small errors will accumulate through the iterations; and if once the model is run -- it'll take many days -- if the accumulated error is too large, which will be obvious once I can see the final position of the body -- it should end up back where it started once the assembly has made a full revolution around B which is my terminating condition -- then I will have a handle on what kind of additional calculation (or possibly a fudge factor) I need to incorporate at each step to alleviate the error accumulation.

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Re^3: [OT] Forces.
by salva (Canon) on Feb 15, 2016 at 09:39 UTC
    Then knowing the resulting force is not enough you also need to know the moment.

    The library you are using to integrate the tensor may also provide you some function to calculate its moment around some given point.

[reply]
      The library you are using to integrate the tensor may also provide you some function to calculate its moment around some given point.

      It does -- see Re^3: [OT] Forces. -- but at this point I am unsure how to make use of it?

      All I know about the value returned is this quote (the only one) from the manual:

      Rˆ2 (i.e. Moment of Inertia/density) This integral is used to determine the moment of inertial of the selected blocks. To obtain moment of inertia, the result of this integral must be multiplied by the density of the selected region. For 2D planar problems, the moment of inertia about the z-axis (i.e. about x=0,y=0)

      In terms of my diagram, x=0, y=0 is the point B.

      With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
      Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
      "Science is about questioning the status quo. Questioning authority". I knew I was on the right track :)
      In the absence of evidence, opinion is indistinguishable from prejudice.
[reply]
        The moment of inertia is a different thing (which BTW, you also need to know).

        I was actually referring to the moment of the force or the torque, which given your tensor field F(v) with v the vector (x, y, 0), it is the integral of the torque field defined as the vectorial product F(v) ^ v (supposing, the rotational axis is (0, 0, z)).

        In the same way for linear movement you have the equation F = m * a, for a rotational movement there is T = I * alfa where T is the torque or moment of force, I the moment of inertia, and alfa the angular acceleration.

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