2
$\begingroup$

I'm trying to calculate the torque needed to rotate an array of solar panels. I've found some formulas for calculating torque, but they require moment of inertia. I designed this array in Autodesk Inventor and it gives me multiple options for the moment of inertia calculation. I've tried asking on the autodesk forums but no one will help me. I would be fine calculating this manually if need be also.

The pivot point is in the center of the array and the dimensions are: 117" x 1.5" x 129 1/16". The weight is 470 pounds and the center of gravity is in the center of the axis of rotation. The pivot is on the center of the 117" side. (see drawing below)

enter image description here

Which of these moments of inertia would I want to use in this equation, should I use principal global or center of gravity? I'm currently on the principal tab and it gives me I1, I2, and I3. Are any of these the right one?

The speed of rotation is no more than 0.001 rpm. (0.00003333333 π rad/s) Does anyone know what value I should use for the angular acceleration on a dc motor? I know very little about physics and I'm having trouble.

$\endgroup$
  • $\begingroup$ Does the pivot axis pass through the center of mass ? $\endgroup$ – ja72 Aug 8 '15 at 19:46
  • $\begingroup$ Yes, the pivot axis does pass through the center of mass. The frame is symmetrical both in length and in materials. $\endgroup$ – mberna Aug 8 '15 at 19:58
  • $\begingroup$ @user2201182, what is the purpose of your calculation? If you already have a motor for this application, a calculation would be extraneous. If, on the other hand, you are attempting to select a motor that will perform as intended, that would be useful information. $\endgroup$ – David White Aug 8 '15 at 20:13
  • $\begingroup$ @DavidWhite I would like to find out the smallest motor that could work when geared up. To do this I need to find the necessary torque since I already know the speed. I'm planning on using a worm gearing system so that I can get a very large gear ratio with low speed, high torque, and no back pressure on the motor. I have very little money however, so I need to plan it out pretty well ahead because I can't spend a lot of money to buy a variety of ratios and motors. $\endgroup$ – mberna Aug 8 '15 at 21:42
  • 2
    $\begingroup$ Torque gives you angular acceleration, not rotation speed. You need torque for 2 things: start rotation from a stop, overcome resistance. Given your quite low speed, acceleration is almost irrelevant. You need to overcome resistance. If this panel is exposed to elements, wind load could be many times greater than the forces needed for acceleration. I don't think the moment of inertia of the panel is going to be your constraint for motor size. $\endgroup$ – BowlOfRed Aug 8 '15 at 22:49
1
$\begingroup$

The answer first: given your axis of rotation, which is the x-axis, you are looking for the moment of inertia associated with that axis, which is the first, smallest principal moment ($I_1$). It only becomes that simple because in your setup the coordinate axes and symmetry axes are identical.

Now for some comments: As @BowlOfRed also pointed out in the comments, there is some hints that suggest that the moment of inertia is of no or little significance when dimensioning your motor for this setup:

In a perfectly constructed assembly, that is slowly accelerated to such slow speeds (say you allow about 5-10s to accelerate from zero to final angular velocity), motor torque needed to achieve this acceleration is going to be negligible. In an ideal system, the constant rotation itself needs no torque to be sustained. What you do have to take care of is

  1. System asymmetry due to imperfect construction
  2. Friction
  3. Power of the elements (wind, precipitation, debris)

I expect the additional resistance introduced through those effects to be much higher than the "clean" moment of inertia.

As the forces will be highest at the centre of rotation, you may want to consider any drive mechanism that attaches to the outer area of the panel (using pistons, chains, rope or similar), as it will have to overcome much lower resistance that way. This (pistons attached to the outer area) is in fact the way the solar panels at my workplace are set up.

All the best for your project!

$\endgroup$
0
$\begingroup$

The torque needed at the pivot A to rotationally accelerate an object by $\ddot{\theta}$i s

$$ \tau_A = I_A \ddot{\theta} + c_x W $$ where

  • $I_A = I_{com} + m c^2$ is the mass moment of inertia at the pivot
  • $c$ is the total distance between the pivot and the center of mass
  • $c_x$ the horizontal distance between the pivot and the center of mass
  • $W$ is the weight of the part

In your case if the pivot is at the center of mass ($c=c_x=0$) then all you need is

$$ \tau_A = I_A \ddot{\theta}$$ and you need to measure the mass moment of inertia about the pivot axis. I don't how Inventor handles this, but you should be able to choose which point and what coordinate system you want inertia measured about.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.