Torque Question

A few comments.

Please give some background for those who haven't read your other posts. Where does your "force" come from, for example?

My first comment would be that if it looks good to you, then perhaps it is working fine and you do not need to worry about the true correctness. But, sadly, your simulation is not yet consistent with our current "laws" of physics!

So, if you are concerned about having a correct, very realistic simulation...there are a few things that don't seem right here.

You correctly wrote that (in 2D) T = F * r and T = I * a (with a being angular acceleration). But from this you came up with T = F * I * r. Which is just wrong, probably due to an algebraic derivation gone wrong. So, your computed torque is wrong. If T = F * r, then it cannot also be T = F * I * r (the extra I makes it an entirely different equation with inconsistent units). What really happens is that T = F * r = I * a. So, since you have F, I, and r, you can use algebra to solve for a = F * r / I (in 2D only---3D isn't this simple).

You are assuming that the turntable applies a constant force to the vinyl, to accelerate it, it appears. But, since you're using radius = 14.6 (the radius of the vinyl), you have applied the turntable force at the outer rim of the disk. This cannot be physically realistic. No real-world turntable connects with the vinyl at exactly the rim (which it would have to do to be able to apply a force at exactly the rim). A real turntable has a surface on which the vinyl sits. The actual applied torque is distributed across the surface. It is perfectly appropriate to represent the distributed torque (torque/radius) if you know what the effective radius is, a reasonable first guess would be the radius of gyration of the vinyl. (Though in real life it'd depend on the contact area and variation in surface properties between the platter and the vinyl). But, even this could be wrong for your simulation.

You haven't mentioned whether or not you are considering the possibility that the vinyl will slip on the turntable, e.g., do you have a friction model or is the vinyl rigidly locked to the turnable and you just want to find the acceleration?

If your vinyl is rigidly locked (as I suspect), then your approach to computing the torque is wrong. What you should do in this case is identify the raw torque output of the motor (should be somewhere on the specs). That torque will already be in the correct units (or at least units of torque not force) and so you don't need to compute T = F * r. You will just have T. And....this T would be applied to the turntable platter plus the vinyl alike. The platter is likely MUCH heavier than the vinyl and so the platter will provide the dominate component of total inertia---the vinyl in this case will be nearly negligible. ....The idea is that the motor output gives you T directly. The I is a sum of the platter I plus the vinyl I. Acceleration is then a = T / I, with T being the motor torque and I being the total I. The platter and vinyl accelerate at the same rate if they are locked together.

Lastly, you do not need to type report all 16 digits of precision! That might be what your code produces, but the #'s certainly aren't that accurate and it makes your post hard to read!

Hope this helps!

Graham

Well since my variables are floats anyhow, those decimals come down to like a few digits anyways.

Yeah I kinda figured it was wrong, but the right way looked wrong. Perhaps it was because I wasn't involving the platter as well. I have three different forces involved. One is the motor that auto rotates the vinyl. Another is 2 kinds of friction (the break for when you hit the stop button, and the other when you cut off the power, which takes longer to stop, both are separataly done.), and the mouse to spin the record, which I believe is applied torque.