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Stewart platform forces?

Discussion in 'DIY Motion Simulator Building Q&A / FAQ' started by Erik Middeldorp, Jun 4, 2020.

  1. Erik Middeldorp

    Erik Middeldorp Member

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    Does anyone know how to calculate the forces in a Stewart platform for various positions? I'm thinking about how the size of the platform would affect the forces, I imagine for the same amount of movement a large platform would more evenly distribute the forces over the actuators vs a small platform. I'm planning to build an actuator to test but I'm wondering how much of the weight of the platform it needs to be able to support. I imagine a small platform with large movements might generate individual actuator forces larger than the weight of the rig.
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  2. Ads Master

    Ads Master

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  3. Gadget999

    Gadget999 Well-Known Member

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    My Motion Simulator:
    2DOF, DC motor, Arduino, 6DOF
    smaller platforms can have larger motion angles

    try simcalc to work out the forces
  4. Erik Middeldorp

    Erik Middeldorp Member

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    As far as I can tell, SimCalc only works for 2 dof rigs.
  5. noorbeast

    noorbeast VR Tassie Devil Staff Member Moderator Race Director

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    Last edited: Jun 5, 2020
  6. Dirty

    Dirty Well-Known Member Gold Contributor

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    Hey Erik,

    if you want to calculate the forces of every actuator, I'd find that incredibly cool :thumbs. The excel sheet I made will not help you much, because I have used only some very rough approximations.

    Between the lines, I read that you have put some deep thoughts into this already. I like that! I agree, the bigger the platform(s), the less the load will shift for a given movement. And yes, you can have constellations where a single actuator might have to carry much, or even all of the load of the entire rig. Theoretically even more. Theoretically the load could even become infinitely large.

    I have considered calculating the precise (static-only!) forces. Here are my thoughts, tell me what you think about them:
    x = lateral y= longitudinal z = vertical
    1. Determine the weight of the upper platform in Kg.
    2. Multiply with 9.81 to get weight forces in N.
    3. Determine the center of gravity of the upper platform . You only need the x-y coordinate (horizontal plane).
    4. Determine the x-y coordinates of all six upper connecting points.
    5. Determine the distribution of the platform weight forces on these six points. That is harder than it sounds.
    6. Determine the angle of each actuator from the horizontal plane.
    7. For each actuator calculate the axial force by multiplying its vertical force share with "cos(angle) / sin(angle)" (I think!)
    Let me know if you want to write a program or something. I'd be happy to help/test/bounce ideas back'n forth/etc...

    Cheers, Dirty :)
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  7. Erik Middeldorp

    Erik Middeldorp Member

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    Hi Dirty,

    That sounds to me like a reasonable way to calculate the static forces. I think to calculate the weight distribution it might work to do it from the side, treating the platform as a simply supported beam with just two supports then just divide the reactions up for the 2 actuators at the back and 4 at the front. I think that would work for a platform in a neutral position at least. If the load wasn't centered sideways I think you could do the same thing from the front to split the load between the two pairs of front actuators.

    I could be wrong though. I previously thought platforms where the actuator pairs don't share an attachment point would bind if the actuator lengths weren't carefully controlled. Reading a thread on here and thinking about it helped me realise that was incorrect.

    I've found a couple of pdf's that cover the inverse kinematics and forces for hexapods. But I don't understand the equations and some of the terminology yet. Hopefully I can find tutorials or explanations to get me up to speed.

    There's also this forward kinematics solver with links to source code and an article which might be useful.

    I'll see if I can figure out those pdfs. If I can, I might have a go at making a program but I am a very novice programmer.
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  8. Dirty

    Dirty Well-Known Member Gold Contributor

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    I think so too. However, for a platform in its neutral position (surge, sway, heave, yaw, pitch and roll all at zero) the load distribution on those points is equal anyways. 1/6th of the total weight force per connecting point. Then multiply with "1/tan(angle)". Not much to calculate there, I think.
    It becomes interesting if the C.G. is offset from the plane of the circle of the upper connecting points and the platform is tilted :) Still considering the platform to be static, no motion.

    First of all, I agree, you CAN consider the six upper connecting points to be three pairs (each pair sharing a point). It will make the calculations easier and change the result only marginally. If you use a computer, you might as well take the six points into account independently but that gets quite a bit more complicated.

    Instead of this:
    IMG_1317.jpeg

    Use this:
    IMG_1319.jpeg

    I'm trying to lay down my thoughts here unfiltered, so let me know if I go wrong somewhere:
    • Since we only care for the distribution of the weight forces and these are by definition perpendicular to the horizontal plane, we can project the platform CG and all three (or six?) connecting points into the horizontal plane simply by ignoring the z-component. No matter the orientation of the platform in space. The resulting points will lay on an ellipse (because the platform was tilted) and the CG will not be in the middle (because it was offset vertically and the tilt moved it off center).
      IMG_1321.jpeg
    • Instead of looking at the platform from the side/behind as you suggested, we will have to take three looks at these points (see above). Each such that two of them align. Then you can consider each of those cases as a simple supported beam. The two aligned points cannot be distinguished, but the load on the single point can be determined.
    • Do that for all three "view directions" and you have the load on each of the three points.
      View 1 determines the load on point A
      View 2 determines the load on point B
      View 3 determines the load on point C
      IMG_1322.jpeg
    :thumbs I think that would be a way to do it. But I'm open for opinions anytime!

    I wouldn't give this assignment to a 10th grader, but I certainly would to a 12th grader who considers to go study mechanical engineering,... or build a Stewart platform :)

    That's no excuse! I am a novice too! The only question is: Does that problem keep you up at night? If it does, you WILL write that program. Period. If it doesn't, then there is no point in writing it.

    In general: Yes, you were! As long as we are talking about real actuators (ones that cannot more than double their length) and platforms of reasonable size and joints with sufficient range of motion, there should always be a valid solution. In fact there can be as many as 24 solutions, but given that you start with a normally oriented platform, you will always keep a normally oriented platform and are able to return to a normally oriented platform. You cannot "break" the platform with a small movement of an actuator. You have six actuators, and six DOFs for the platform to "give way". Creating singularities where an actuator is "in plane" with the platform and/or the horizontal plane is close to impossible!

    Cheers,... Dirty :)
    Last edited: Jun 6, 2020
  9. Dirty

    Dirty Well-Known Member Gold Contributor

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    I just saw in the video you posted at 2:15, you can see that the software they wrote calculates how many solutions there are and you can cycle through all of them :)
  10. Erik Middeldorp

    Erik Middeldorp Member

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    I think what you've described should work. But I think I'd rather try to understand the solution in the previous links since it would allow dynamic forces to be calculated.