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Torque calculation of solar tracker

  1. Feb 20, 2013 #1

    i am currently designing a solar tracking system. i am confused at the point of calculating the torque required to rotate the solar tracker. i have a shaft in the cylinder which is rotating with 4RPM.

    i know that Power= Toque* angular velocity or 2*pi*RPM* Torque

    where Torque is= Force* radius

    radius= should i take the radius of the shaft? ( please see the picture)
    Force= how can i calculate the force?
    progress assembly.JPG

    i know its stupid but i am confused and i need to continue with this

    thank you
  2. jcsd
  3. Feb 20, 2013 #2


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    For gravity, radius is the horizontal displacement of the center of gravity of the system relative to your axis. And force is just the weight of the structure.
  4. Feb 20, 2013 #3
    could you please be more specific? in my case do i take the radius of the shaft or the radius of the cylindrical base attached to the shaft?

    thank you
  5. Feb 20, 2013 #4


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    I think my post is as specific as possible. Do you know the concept of the center of gravity?
    Can you determine the position of your center of gravity?
  6. Feb 21, 2013 #5
    Ummmm.....no. Wrong approach.

    You want torque required to turn the assembly. More specifically, you want the torque required to accelerate that assembly to a desired rotational speed in a desired time. That is

    T = Jα

    Torque = (Mass Moment of Inertia of the Assembly about the shaft) X ( angular acceleration)

    ...and angular acceleration is approximated by ΔAngularVelocity / ΔTime .

    But this is not sufficient. You must size the driver/motor/gearing for maximum peak torque. That includes rotational inertial torque plus every other resistance to rotation that you can dream up (friction, stiction, wind, gravity, cobwebs, yadda yadda yadda). You need to analyze it for slowing down to a stop & holding it steady also. Then awww, what the heck, double that and use that to size your driver/motor/gearing.

    If this assembly is designed with a CAD system of any useful power, then you can use the CAD system to calculate the MMI about the shaft. It can be done manually, but it would sure be tedious.

    The only instance of using "radius" for this type of analysis was if you have a Radius of Gyration of the assembly...but that's too darn much trouble to fool with.
  7. Feb 22, 2013 #6


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    The required rotation speed and acceleration to follow the sun is negligible, but the motor has to overcome the torque due to the weight of the structure.
  8. Feb 23, 2013 #7
    Mass properties of progress assembly ( Assembly Configuration - Default )

    Output coordinate System: -- default --

    The center of mass and the moments of inertia are output in the coordinate system of progress assembly
    Mass = 50454.14 grams

    Volume = 30048791.17 cubic millimeters

    Surface area = 5588654.59 square millimeters

    Center of mass: ( millimeters )
    X = -45.25
    Y = 58.44
    Z = 527.49

    Principal axes of inertia and principal moments of inertia: ( grams * square millimeters )
    Taken at the center of mass.
    Ix = (-0.01, -0.00, 1.00) Px = 3565295302.31
    Iy = (0.51, -0.86, -0.00) Py = 16485736325.98
    Iz = (0.86, 0.51, 0.01) Pz = 17926686404.42

    Moments of inertia: ( grams * square millimeters )
    Taken at the center of mass and aligned with the output coordinate system.
    Lxx = 17554525288.05 Lxy = -630256839.06 Lxz = -72393884.81
    Lyx = -630256839.06 Lyy = 16857393767.64 Lyz = -44632156.42
    Lzx = -72393884.81 Lzy = -44632156.42 Lzz = 3565798977.02

    Moments of inertia: ( grams * square millimeters )
    Taken at the output coordinate system.
    Ixx = 31765453901.78 Ixy = -763673439.14 Ixz = -1276546609.18
    Iyx = -763673439.14 Iyy = 30999270315.94 Iyz = 1510802666.01
    Izx = -1276546609.18 Izy = 1510802666.01 Izz = 3841422169.88

    i got the above results from solid works. i am lost.. moment of inertia should i use?

    thank you
  9. Feb 26, 2013 #8
    Perhaps not. It's a simple calculation to be sure, and a disastrous project if it is not negligible. If one must accelerate a massive object to even a slow speed, significant torque must be applied to get it moving from zero speed. Once the mass is in motion, then the torque requirement will be reduced. Only if the motor must overcome gravity effects of the "weight" and therefore must counteract an induced torque T=WgL. Otherwise it's **inertia**. The wide dimension of this panel structure will produce a significant inertia due to the distributed masses away from the axis of rotation.

    Regarding whether your SolidWorks-calculated inertia is Ix, Iy, Izz, or anything else: You must get the Mass Moment of Inertia of the mass **about the desired rotational axis**.

    This is probably NOT the CAD model coordinate system axes that are given by the values you listed. So you must read the SW help file and (what? don't remember exactly) either have it calculate about a new reference geometry Axis or new reference geometry Coordinate System that you assign at the correct location to give you the answers you need.
  10. Feb 26, 2013 #9


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    So why didn't you do it?

    The system just has to complete one cycle per day. Let's assume it accelerates from rest to ##\omega=\frac{\pi}{24hours}## in 1 second (quite quick!). For a point-mass m with distance d to the axis, this requires a torque of ##md^2\frac{\pi}{86400s^2}##. Required static torque due to its angle relative to the vertical position is ##md\sin(\alpha)g##.

    Assuming d<100m (we don't want to rotate a skyscraper, right?), static torque is larger for ##\alpha>0.00037##. For d=100m, this is a deflection of just 4cm at the top. I am sure the system is designed to operate with angles some orders of magnitude larger. It is easy to extend this analysis to arbitrary shapes, the conclusion stays valid as long as the structure does not exceed 100m (or 1000m, if you remove one 0 from the angle...).
  11. Feb 26, 2013 #10
    thank you very much both of you
  12. Apr 21, 2015 #11
    Acceleration torque is moment of inertia about rotational axis times the angular acceleration.

    but Should we factor ------- Weight of payload X Distance between motor axis and CG of payload as well ???????

    And add both of them?
  13. Apr 21, 2015 #12


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    It also has to overcome wind force on the structures (panels I presume). Worst case wind force is likely an order of magnitude higher than those other forces.

    You need a design wind speed. You also need to rapidly move the panels to the feather position if winds exceed design speed. Alternatively, you would need a clutch to decouple the drive system and design the aerodynamics of the structure to self feather like a wind vane does.
  14. Apr 21, 2015 #13

    jack action

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    You must use the parallel axis theorem to find the proper moment of inertia (including payload) when the center of mass does not coincide with the axis of rotation: 1d0a8cfeb36821096b25a1df551318e5.png
  15. Apr 21, 2015 #14


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    Yes. This has been answered over a month ago in the previous posts.
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