Need help with motor sizing calculations for solar panel mount - Page 1
Forum section: electronics / beginners

author: cbc02009
author_rank: Regular Contributor
timestamp: September 16, 2018, 04:07:34 pm
content: Hello all,I have a 50W solar panel (Datasheet here:http://grapesolar.com/docs/GS-STAR-50W.pdf) that I am trying to build a solar-tracking mount for, and I am having trouble figuring out how to calculate the power of the motors I need.The solar panel will need two motors, one for movement about the vertical axis, and one for movement about the horizontal axis. I have approximated the moment of inertia for each axis of the panel using the formulaCode:I = (1/12)*m*(h^2+w^2)and I know I can use that in combination with the necessary angular acceleration of the panel to find the required torque of the motor.My problem is, I have literally no idea what kind of acceleration is necessary (or even reasonable) for my project. It really doesn't need to move very fast, since the sun doesn't move terrible fast in the sky, but that's velocity, not acceleration.In fact, I don't know that I care very much about the actual acceleration of the panel at all, it's more that I don't want to buy an under-powered motor and have it stall.Any advice on where to find calculations, or even just where to start my search for the answer would be greatly appreciated.Thanks!

author: IanB
author_rank: Super Contributor
timestamp: September 16, 2018, 04:15:08 pm
content: I think you may be barking up the wrong tree. My guess is that friction in the gears and bearing system is likely to be the most important load on the motor.

author: CM800
author_rank: Frequent Contributor
timestamp: September 16, 2018, 04:21:52 pm
content: Quote from: IanB on September 16, 2018, 04:15:08 pmI'd strongly suggest calculating it all out.Calculate out a 10 second (you don't really want to wait a day to move it to the position you want, should you want to change. move 180*Using a movement curve without constant velocity gives you 1 rev / 20 seconds. (0.05rps)So find a motor that's pretty slow.For the acceleration, it will be twice the velocity (as you need to move half the distance in half the time) so 0.1rps /s (convert that to radians)The torque you will require is going to be the Inertia, multiplied by the Acceleration in Radians.

author: rstofer
author_rank: Super Contributor
timestamp: September 16, 2018, 05:31:04 pm
content: Have you considered the wind load on the panel? Around here we would design for 80 MPH. No, we're not going to hit that number more than a couple of times per year but that's what the building codes require and it seems reasonable to consider it.These tracking projects are all over the Internet. Maybe Google can help.https://www.electronicshub.org/sun-tracking-solar-panel/

author: cbc02009
author_rank: Regular Contributor
timestamp: September 16, 2018, 08:50:04 pm
content: Quote from: CM800 on September 16, 2018, 04:21:52 pmThank you, that's very helpful. That's exactly what I was looking for.Quote from: rstofer on September 16, 2018, 05:31:04 pmThis is a proof-of-concept prototype for a school project, so out of pocket cost is kind of the most important factor. I've seen projects online using servos and stepper motors, but don't those both require constant input from the MCU to maintain their position, thereby reducing the efficiency of the system compared to a regular gearboxed motor? I was hoping to use a wormgear to both increase the gear ratio, and also keep the panel held in place when we stop the motor.

author: rstofer
author_rank: Super Contributor
timestamp: September 17, 2018, 04:08:43 pm
content: Even a regular DC motor is going to require power to hold position. If you want to be able to completely disconnect the power, and you do, you need some way of holding position. A worm gear drive for rotation and perhaps a lead screw drive for tilt might be a good way to go. Something where the mechanics of the system hold position with or without the motor.As you are using LDRs to track the Sun, you don't really need a stepper motor. However, the stepper has the advantage of moving a predictable amount for every step but, if you plan to depower the motor, you need to remember the state of the outputs so that the motor will cog back to the same relationship (assuming the rotor hasn't drifted too far). But this will also be a problem with a DC motor used as a servo as well.It's interesting to use LDRs to try to locate the sun. That won't work quite as well on cloudy days but it may still be ok.Personally, I would use one of the celestial navigation formulas that apply to my Latitude and predict where the Sun is going to be, clouds or not. Telescope mounts do this all the time.Reed's Almanac used to have a section on celestial navigation. One of the neat features was a 5th degree polynomial used to predict the location of the Sun for a given Latitude and Longitude. The equation was derived annually by the Royal Observatory in London. I used to have it programmed into my HP 48GX calculator - back when I was involved with that kind of thing. Alas, I haven't seen it in a long while.There are suitably accurate equations that don't need to be calculated on an annual basis. It's not like you're trying to make landfall.I think you can get quite a bit of cogging out of a stepper motor if you short the windings. Some H-Bridges may include that capability. But you would still need to power the driver, just not the motor.References to Sun tracking are included here:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3355400/

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2018, 06:28:46 pm
content: Quote from: rstofer on September 17, 2018, 04:08:43 pmThis bit I don't get. Surely a high ratio worm drive will be fixed in position if you remove torque from the input shaft? The output shaft will be locked and the only possible movement will be backlash in the gear train?

author: rstofer
author_rank: Super Contributor
timestamp: September 17, 2018, 07:19:31 pm
content: Quote from: IanB on September 17, 2018, 06:28:46 pmYes, a worm gear will work perfect! But there is no mention of this in the OP. The gear ratio also influences the required torque and, ultimately, horse-power (or fractions thereof). Then too, the drive train is also a load on the motor.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2018, 07:28:45 pm
content: Quote from: rstofer on September 17, 2018, 07:19:31 pmSee below:Quote from: cbc02009 on September 16, 2018, 08:50:04 pm