Motor for rotating platform

  • Thread starter Mr Zakaria
  • Start date
  • #1
Hi all,

I am planning to build solar tracker which will have a platform that rotates about vertical axis according to sun. The platform should have max tangential speed of 1.03 m/min and min tangential speed of 0.017 m/min.
The platform has a mass of 4 tonnes and radius of 10 m and can be modeled as uniformly distributed object. The precision needed for the tracking is 0.01°
The question is how much torque needed to rotate the platform?
I'm going to use wheels for the rotation mechanism with one of the wheels is driven by a motor
If you have any suggestion about where i can find suitable motor or gear reduction for that purpose, please let me know
Thank you in advance
 

Answers and Replies

  • #2
96
0
The peak torque requirement will be composed of 2 major components, friction and drag forces, and accelerating the mass. The steady state rotation will require less torque, primarily to over come the friction and drag forces. You need to estimate the friction and drag forces based on consideration of the elements in your design. The peak required acceleration would come fromthe operational criteria, how fast it needs to be brought into position or correct a positional error.

I question that driving this with a wheel (friction) drive will provide the positional accuracy required (0.01 degrees, which is about 0.8 mm at the rim). There will need to be some positional error correcting or servo mechanism, coupled with an accurate position feedback and reference, as part of the drive control. You may want to think about a gear and rack drive, or if you only need to track for part of the day a linear actuator and crank mechanism may work.
 
  • #3
Thanks DickL for the answer.
Yes, the tracker will have a control system which will control the position according to sun.
Can you please explain me more how can i calculate the drag forces and accelerating the mass?
 
  • #4
96
0
Figuring the torque required for acceleration is the easy part. The formula is:
T= I*α
Where T= torque (N-m), I is the mass moment of inertia (kg-m^2), and α it the angular acceleration (radians/sec^2). For a disc with evenly distributed mass I = m*r^2/4. For the 10m, 4 tonne platform this gives an I = 25,000 kg-m^2. (But see the additional I elements noted below.)

You will need to develop the required maximum angular acceleration from either the operational criteria or the control system's error correction requirements. The precision you noted (0.01 deg) is the starting point for developing the maximum angular acceleration needed, but you will need to look at how fast you want the platform to be able to respond to commands. Start-up may be the worse case and have the need for the greatest amount of acceleration, however too high or too low acceleration rates can make the control system unstable or unable to correct errors.

You've also given a maximum speed of 1.03 m/min (0.0172 m/s), which is an angular velocity of 0.00343 rad/sec. You might want to start by arbitrarily selecting (guessing) at a time needed to reach full speed, say 10 sec. Using 10 seconds gives an angular acceleration rate of 0.0343 rad/sec^2. Feeding that into T=I*α gives a torque for acceleration of 857.5 N-m. A point of caution here, this arbitrary 10 second time has important consequences to the control system's capability. I would guess you will want a time more in the 1 second range (i.e. 10 times the above torque).

Okay, as I said that is the easy part. Estimating the friction and drag will be less straight forward. For this you will need to work from a conceptual (or better) design. How many wheels will be used (6, more?), what type of wheel (pneumatic, solid urethane...), rolling resistance of the wheels (including their bearings, which may not have significant drag), the drive system (gear units, flex couplings, brakes, etc.). Each of these elements will contribute loses that the motor will need to over come. Some rolling resistance coefficient values for pneumatic tires from an old reference use values of 0.01 – 0.03 of the load. Gear units add resistance as a result of internal friction and losses, think of its efficiency. There will probably be a need for a brake (unless a four quadrant controller is used) and brakes often contribute resistance, even when they are released.

Each of these elements also contributes some additional rotating mass, that should be added to the effective total I for calculating the torque required for acceleration. Adding this additional rotating mass isn't simply adding their individual I's to the total I. The wheels, motor, etc. will rotate at a greater angular velocity than the main platform. To reflect their contribution to the main platform their I's need to be increased by the square of their respective angular velocities – (ω(wheel) / ω(platform))^2. The motor, if running through a gear reduction, may have a surprisingly large contribution to the total I due to the combined wheel and gear ratio adjustment.

The drive wheel needs to have enough load and friction to avoid, or at least minimize, slippage. Pneumatic tires on dry rough concrete can have high friction of traction coefficients. However, when running on polished concrete or metal, that may be dirty, oily or wet, can end up with coefficients below 0.05. Using that value for the traction coefficient and the above torque leads to a drive tire load of approximately 1,800 N. Assuming that all of the above additional effects were to double the required torque, the value is still within the required vertical support load. However, you will probably want to include some form of suspension on the drive wheel to assure it maintains an adequate force against the platform should there be any unevenness to the platforms running surface.

In the above I referred to pneumatic tires, that is because I'm quite familiar with them. For this application I would be more likely to consider urethane tires.

I hope this helps and that I wasn't too long.
 
  • #5
Thanks DickL
This definitely will help me in estimating the torque needed for the platform
 
  • #6
I have just gone through the calculation
Isn't the inertia for a disc I = 0.5 m*r², which will give 200 000 kg m²?

I will use 30 cm diameter wheels for this application and with this precision of 0.01 degrees, the wheel will have about 1.09 RPM at max tangential speed and 0.02 RPM at min tangential speed. My question; Is the torque that i calculated to rotate the platform the same as the torque needed for the motor to rotate the wheel? How can i calculate the torque needed for one motor which will rotate one wheel in case i will use two motors which will drive two wheels in this application?
 

Related Threads on Motor for rotating platform

Replies
2
Views
1K
  • Last Post
Replies
8
Views
936
Replies
1
Views
3K
Replies
3
Views
530
Replies
9
Views
6K
  • Last Post
Replies
7
Views
2K
Replies
18
Views
10K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Top