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BrowserUk has asked for the wisdom of the Perl Monks concerning the following question:
I'm trying to solve a problem to do with forces as represented by this freebody diagram.
The large circle represents a body, that is free to rotate about axis A, that has a complex of external, non-contact forces acting upon it represented by the wiggly red arrows. The composite of these forces is calculated (by a library call using Stress Tensor integration), and the combined result is returned as (ex.) X: 335.123 N Y: -74.2858 N.
In addition to the body being free to rotate about A, the short (10 units) link arm it is attached to can rotate about the axis B.
The problem is to work out what position the combined mechanism will end up in as a result of the forces acting upon it.
For now, I'm ignoring the possibility that forces acting may change as the assembly rotates. This is part of a complex simulation and the basic idea is that from a known starting point, I use the library to calculate the net force acting at some instance in time; from that work out the position the assembly moves to; adjust the model to reflect that movement, and then re-calculate the stress tensor integration for the next instance. If the time-step is small enough and the iteration count high enough, then it should give a reasonable approximation of the affects of the forces over time.
In keeping with the "what have you got so far" ethic, the second (lower) state diagram in the image shows what I think will happen.
- The short link A-B will rotate anticlockwise about B by an angle (to the horizontal) represented by the purple line that connects B to a point F, which is a point on the periphery of the circle through which the resultant of the X/Y force acts through A in the original position.
- The body will rotate clockwise around A such that the position of the point F on its periphery will end up in the same place(ish) as it was in before the movement takes place.
I realise that the position of F in the resultant is slight displaced from its original position in space, but as shown, the angles involved are greatly exaggerated. The actual actual angles will be very small at each iteration, so the inaccuracy at each step will also be small. In the end, it is the position of the body in space that is important, not its rotation, nor the exact position of any point on it.
I'm fairly sure I can handle coding the trig to resolve the two rotations; what I'm looking for here is any kind of argument for or against my assessment of what will happen?
My thanks to anyone with enough math to tackle the problem and bored enough to bother :)
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Re: [OT] Forces.
by salva (Canon) on Feb 14, 2016 at 12:33 UTC
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In the end, it is the position of the body in space that is important, not its rotation, nor the exact position of any point on it
Then, you can just ignore the circle and apply the forces directly in the point A.
If the only thing that matters to you is the final position (it is not clear from your post), then you can avoid all the calculations. The stable position happens when AB is aligned with F. Though in the absence of any friction, what you have there is a pendulum.
If the force changes with the position, the logic remains the same, though you will have to find the point A where F(A) and AB align.
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Without knowing the masses and the moments involved,
The model can return the area of the body, and from that its volume can be approximated, which combined with the densities can allow an reasonably accurate mass to be determined.
It comes out at 0.75 Kg.
However, this is a 2D finite element analysis, so the model is an infinitely thin radial (cross-axial) slice of the mechanism, which leaves me questioning the relevance of the total mass? Or whether there is any useful subdivision of it that can be used?
The model can return a figure labeled as: "R^2 (ie. Moment of Inertia / density )", and further defined as: "Integral of (x^2 + y^2)".
Which for the body in question gives a value of:9.88017e-8 m^5.
But, at this point, I have no idea what if any relevance or use that has in what I am trying to calculate?
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.
In the absence of evidence, opinion is indistinguishable from prejudice.
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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.
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.
In the absence of evidence, opinion is indistinguishable from prejudice.
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Re: [OT] Forces.
by RichardK (Parson) on Feb 14, 2016 at 12:31 UTC
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You could resolve force F to be 2 vectors one parallel to AB & one at 90 degrees. The parallel force F' can be ignored as it's just trying to stretch/compress the link AB, the other, F'', provides a turning moment around B.
sorry for the ascii graphics ;)
F' <---- A-------B
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/ |
F V
F''
Then as you iterate your solution F'' will reduce to zero, as you rotate AB in proportion to the magnitude of F'' for each step.
It's not entirely obvious to me that the body will rotate about A, it will depend on how the external forces are applied to it. So without more information I don't know if your intuition is correct. | [reply]
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It's not entirely obvious to me that the body will rotate about A, it will depend on how the external forces are applied to it. So without more information I don't know if your intuition is correct.
Yes, it appears that there's missing (or hidden) information.
If the forces from the wiggly red arrows don't have tangential components, then they won't spin up the disk/cylinder around A.
If you aren't concerned about how fast it will get there, you don't need the masses and just need the component of the total force on the cylinder tangential to B (i.e. perpendicular to the A-B line, which will rotate about B). At equilibrium, that force will be zero and all the force will be parallel to the A-B axis (unless the force is strong enough to stretch or break things...).
At risk of messing it up because I'm doing geometry in my head before eating, it looks like F_perp=F_x*sin(theta)+F_y*cos(theta), where theta is the angle measured in the little interior angle that you show in the second pic. When F_perp=0, then all the force is along F_parallel. For completeness that should be F_parallel=F_x*cos(theta)+F_y*sin(theta).
You can check this by looking at the first picture, where theta=0. sin(0)=0 and cos(0)=1, so F_perp=F_x*0+F_y*1. If the linkage makes it all the way to 90 degrees, then F_perp=F_x*1+F_y*0, so I think that's right.
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Yes, it appears that there's missing (or hidden) information.
I'm not hiding anything. At least not deliberately.
If the forces from the wiggly red arrows don't have tangential components, then they won't spin up the disk/cylinder around A.
I'm not party to any tangential forces involved. That is to say, the only knowledge I have of the forces acting is the bland, X-component/Y-component values returned from the integration.
My thinking on why the body will rotate clockwise around A as the force causes A to rotate anticlockwise around B, is simply the conservation of momentum.
That is: if we assume zero friction in the bearing at A (which we must), and an absence of viscosity in the medium (air, water, vacuum) surrounding the body (again, we must since we have no information), then the orientation of the mass of the body will tend to remain the same as the assembly rotates around B, simply because there is no force acting on it to cause it to change that orientation.
The net affect of the body maintaining its orientation with respect to the universe(B, and the rigidly attached "wall"), as the point A rotates anticlockwise around B, is that the mass of the body appears to (and actually does) rotate clockwise with respect to A.
If you aren't concerned about how fast it will get there, you don't need the masses and just need the component of the total force on the cylinder tangential to B (i.e. perpendicular to the A-B line, which will rotate about B). At equilibrium, that force will be zero and all the force will be parallel to the A-B axis (unless the force is strong enough to stretch or break things...).
Agreed. The purpose of the freebody diagram, (and the static link from B to the "wall"), is to isolate this part of the mechanism from the rest of the Universe and so allow you to simplify the overall problem to one that calculates everything with respect B.
At risk of messing it up because I'm doing geometry in my head
I concur with your trig.
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.
In the absence of evidence, opinion is indistinguishable from prejudice.
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You could resolve force F to be 2 vectors one parallel to AB & one at 90 degrees.
The library routine actually provides me with the X & Y components; I have to combine them to derive the resultant vector.
The parallel force F' can be ignored as it's just trying to stretch/compress the link AB, the other, F'', provides a turning moment around B.
The problem with that is it implies that the two components of the force can be applied separately; but that doesn't work. For example, if you applied them in the other order, the turning moment followed by the parallel force, the result is different because after you've applied the turning moment, the parallel force is no longer parallel :)
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.
In the absence of evidence, opinion is indistinguishable from prejudice.
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