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Torque calculation of solar tracker 
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#1
Feb2013, 05:44 AM

#2
Feb2013, 08:12 AM

Mentor
P: 11,631

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.



#3
Feb2013, 08:30 AM

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 


#4
Feb2013, 11:19 AM

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P: 11,631

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? 


#5
Feb2113, 08:22 PM

P: 154

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. 


#6
Feb2213, 07:26 AM

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P: 11,631

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.



#7
Feb2313, 08:26 AM

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 


#8
Feb2613, 11:40 AM

P: 154

Regarding whether your SolidWorkscalculated 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. 


#9
Feb2613, 12:43 PM

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P: 11,631

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 pointmass 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...). 


#10
Feb2613, 12:45 PM

P: 25

thank you very much both of you



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