# Torque calculation of solar tracker

 P: 25 hi 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? i know its stupid but i am confused and i need to continue with this thank you
 Mentor P: 11,928 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.
 P: 25 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
 Mentor P: 11,928 Torque calculation of solar tracker 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?
 Mentor P: 11,928 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.
 P: 25 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
P: 157
 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.
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.
Mentor
P: 11,928
 It's a simple calculation to be sure
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...).
 P: 25 thank you very much both of you

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