## Design Problems- Solar Panel design

Hi All

I am designing the solar panel deployment mechanism using torsion springs for a satellite. I have a few questions to ask here and appreciate if someone could help:

1. How can I calculate the Force required by torsion spring to open the panel?
( mass of panel= 92.35 grams and dimension 327.5 mm x 63 mm)

2. How can I determine that the lever that will support the panel to open can bear the weight?
( please see the attached zip file containing pictures for details on this)

Kindly let me know if you think the details are less and you need further details.

Thanks and Regards
Attached Files
 solar panel.zip (61.6 KB, 32 views)
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 Recognitions: Gold Member Welcome to PF, Hamzaaaa. I'm not a scientist, but it strikes me that weight-bearing shouldn't be an issue, since the satellite will be in micro-gravity. Likewise, it should take very little effort to deploy the device. The mass will still have to be overcome, but not under any appreciable influence of gravity. The required force would depend upon how quickly you want the thing to move.
 Thank you Danger. I do have the same views but the only thing that is bothering me is that as you might have seen from the figure that the lever-top part to which the panel is attached is only 2 mm thick. I want to know if that could hold the weight and open the panel. I have set the applied force to be 1 N by each spring. And therefore acceleration calculated would be around 11 m/s^2. But the force is an assumption. I want to know if it is possible to calculate the force required to deploy the panel in no gravity condition. is there any such equation? Thanks in advance

Recognitions:
Gold Member

## Design Problems- Solar Panel design

Unfortunately, someone else will have to field that question. I'm sure that there must be some applicable formulae, but I have no idea of what they might be. Sorry.
 Ahah! That is Solid Works :) Good luck!
 Recognitions: Homework Help Science Advisor hamzaaaa: There are two torsional springs per panel. Let's assume the torsional spring arm contacts the lever at a distance c = 7 mm from the axle centerline, applying a force F1 to the lever at the contact point. Let's assume the axle coefficient of friction is mu = 0.20, and the axle diameter is D = 3.0 mm. If (c - 0.5*mu*D) > 0, it indicates the panel will rotate when force F1 is applied. The angular acceleration of the panel, in rad/s^2, will be alpha = 54.080*F1, where F1 = force applied by each of the two torsional springs (N) at contact point c. The moment on each of your four bolts, in units of N*mm, will be M1 = 0.042 908*alpha. Therefore, if F1 = 1 N, then alpha = 54.080 rad/s^2. The tangential acceleration of the tip of the panel would then be r*alpha = (0.0800 m)(54.080 rad/s^2) = 4.326 m/s^2. And the moment on each bolt would be M1 = 0.042 908*54.080 = 2.320 N*mm. Moment M1 produces prying at the bolt. Distribute the M1 moment based on the heel contact (edge) distance at the bolted connection, thereby obtaining the prying force and bolt tensile force. Then you can compute the bending stress on the lever due to this prying force (and you can compute the bolt tensile stress). However, you will also need to determine what causes your panel to stop deploying, and figure out the stopping time (or maybe the stopping distance); and thus compute the panel angular deceleration during stopping. Then you can use the M1 equation, above, to compute the corresponding moment on each bolt during deceleration (impact). The stopping deceleration is likely to be far greater than the deployment acceleration.

 Quote by nvn However, you will also need to determine what causes your panel to stop deploying, and figure out the stopping time (or maybe the stopping distance); and thus compute the panel angular deceleration during stopping. Then you can use the M1 equation, above, to compute the corresponding moment on each bolt during deceleration (impact). The stopping deceleration is likely to be far greater than the deployment acceleration.
Couldn't it stop just because of being geometrically conditionated?
 Manufactures rate all springs according to their size. There are extension springs and compression springs. Look at the manufactures chart rating for the spring tension. You can order springs with any tension you like. Then it becomes a simple math problem of leverage. If a satellite is in outer space with no gravity then weight is not a factor but mass is. http://www.mcmaster.com/#extension-springs/=2pzgya This is not a very good spring chart, it is just basic BS. You need to contact the spring manufacture for some technical data for all the springs they manufacture. It is free just give them a call they are more than happy to supply customers with what ever they need to make an educated guess in placing an order. Be sure to ask for the sales department. http://www.engineersedge.com/spring_extension_calc.htm You need to read about spring technology. A spring will typically extend or compress 1/3 to 1/2 if its length. Larger wire diameter makes a stronger spring. Square wire is stronger than round wire. Larger coil diameter makes an easier to extend or compress spring. The spring gets strongs as it is extended or compressed. Decide what you need then order the approprate spring. http://www.centuryspring.com/ http://www.newcombspring.com/ http://www.katyspring.com/ http://www.mwspring.com/ Most manufactures specialize in a certain type of spring, large springs, small springs, long springs, short springs, strong springs, etc. Here are links to 4 manufactures they may not be what your looking for. Do a web search. Figure out what type spring you need then order a catalog from a manufacture that makes that type spring.
 Dear Nvn and Gary350 Thanks a lot to both of you for your great replies. I will try working out what nvn suggested and then I will further ask. Gary 350 I have one more query regarding torsion springs. When the torsion springs are compressed their internal diameter would also decrease? Suppose I have a 2mm rod on which there will be a torsion spring. So what is the tolerance I should keep for the internal diameter of the spring? Is there any formula or relation I can calculate the decrease in the inner diameter of the spring upon compression, provided I have the length (15mm) of the spring and number of turns(25)? Thanks alot
 You may have to experement to see how it works. I would think .030" would be plenty of clearance.
 Hi I am also looking at designing a torsion spring mechanical deployment system for a solar panel array. I was wondering if anyone knew of a way to dampen the spring on deployment? I have looked at greases but I feel outgassing will be too much of a concern. Thanks