solar project and I got myself into a pickle...I need some help

Ni-Cads may be better in this application, even if they're heavier than NiMH.

A brushless motor will be far more efficient than a brushed motor. Unfortunately the motor controllers available today optimize for lightness and full-throttle efficiency; the motor-controller system isn't terribly efficent at part throttle.

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Tim Wescott
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my project is to build an all solar enduro electric glider. I have figured out the air frame and I need to figure out the electronics. for anyone who knows rc planes I am using the Great planes 400gd electrifly system that is rated 7.2-8.4 volts max continuous amps:22A max internal resistance is .0036( i dont know the unit) I was thinking of using 30 solar cells (15 in series with an additional cell in parallel) 15 cells equal 7.5 volts at 7500ma I was hoping running the second set of 15 cells in parrallel would increase my current so the motor would not overload the cells. each cell is rated 500ma at .5 volts. if this is correct( I doubt I am right about it though lol) I need to construct a powersource that can hold a charge of 5v at 1a for atleast

5minutes. (just incase I lose power from the cells)

another thoery I had was to let the solar cells charge another power source and the motor runs off of that (ofcourse I wouldnt run the reciever and the motor of the same source) so that the solar cells would not be subejected to that type of draw.

please help me in any way shape and form (I can read schematics) feel free to take any info and draw up shematics, because this will be entered into the annual sciece fair if it is successful. thanks alot for the help matt

Reply to
mattsimonton

This makes me think I've got the wrong notions about lift/drag/weight/thrust/etc.

If you really could build a glider that would stay up under solar power on a science fair budget, wouldn't an unmarked black van show up at your door and whisk you away to some undisclosed underground location to develop military weaponry? What am I missing about the feasibility of such a project? mike

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Reply to
mike

That's 22,000 mA

A fairly decent 2 by 4 cm cell produces .4, optimistically .5 volt at

200 mA or maybe a bit more in direct sunlight. That means 10 to 12.5 milliwatts per square centimeter. 7.5 volts at 7500 mA is 56.25 watts, which requires about half a square meter of solar cells. That is if the sun is directly overhead and the sky is clear. Does your airplane have that much area to put these cells on?

First, I think you will get more like .45, maybe .4 volts rather than .5 when drawing a load. Second, putting cells in series adds voltage, and not current. Putting

15 of these cells in series will give .6-.75 volt at 500 mA. To get 7500 mA, you need 15 cells in parallel, which gives .4-.5 volt. To get 15 volts at 7500 mA, you need a series-parallel array of cells, with 15 series strings in parallel, with each series string having somewhere from 15 to 18 cells - totalling 225 to 270 cells.

That smaller amount can be supplied by a cell array of two paralleled series strings, each having 10-12 cells.

Possibly a cell array comprising two paralleled series strings, with each series string having 15-18 maybe 19 cells, could charge a series string of four or five NiMH cells. There are NiMH cells of AA size and 2000 mAH nominal capacity that can survive 7500 mA current (for example "IC3" ones of Ray-O-Vac brand [available at Target] and also available at Radio Shack under their brand). They do get somewhat hot with that much current, but they survive that at least to some usable extent.

You may want to get a NiMH charging IC to detect the cells reaching full charge. CAUTION - detection of full charge can fail if the sunlight weakens when the cells are filling up, which allows for some possibility of overcharge. Also, AA NiMH cells probably have their life expectancies shortened by heating up from passing 7500 mA. Be prepared for batteries to not live as long as they "should".

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

Forget about the 22A drain, pick another motor. Try a low power Astro Colbalt motor for a glider(more efficient). This will only keep the glider air born since you will not harness the power for a 22A motor. Hi-Start lauch, and no batteries. They add weight. Gliders with motors just use the availble power to climb to altitued and thats it. I believe the AMA mag had someones design for a solar glider. The wings were completely covered. Not sure when, 92' or some where around then. Hope this helps.

Cheers

Reply to
Martin Riddle

Throw it away. You need to start from nothing, and design ultralight. You are talking very slow flying (lowers drag), and completely covered in solar cells. Solar cells alone weigh about (for .5mm thick cells) some 1Kg/m^2. At an efficiency of 10% (system), that's 100W/Kg or so. This is a good number, but it rapidly plummets, once you start mounting the cells. Firstly work out where on the structure you're going to mount the solar cells, then work out the area exposed to the sun. Now, you design to that power number.

Reply to
Ian Stirling

I was afraid there was going to be another variable I over looked. (darnet) I am currently designing a 2.5 meter sailplane the only problem is that the motor may be too weak to pull up the weight. When I get the kinks worked out I will get a few sponsors. but going brushless is out of my budget. (since I now need 220-225 cells ughhh) the motor will only be used during a climb and a speed pass, so it will not be used during the whole flight. seems I need to find an alternitive to solar power lol, its to involved for the budget I have, but thanks alot all and I really appreciate it. matt

Reply to
mattsimonton

Graupner used to produce a Solar powered model, the Solar-Uhu, circa 1991.

1.8m span, 20 solar cells. Motor direct drive 7.2V, 9.5A max. rpm 8500 with 8" by 4.5" folding propeller. Flying weight 1.4kg. The solar panels were polycrystalline, rated at 0.5V, 1.2A each (50 by 100 mm, weight 5g). I suspect that that output rating demands a full sun (E = 100 mW/cm^2), but the blurb says that 'currents of 500mA can be generated in overcast or evening sky'.

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Reply to
Dave Garnett

figured

Hmmm, speaking of what I know (paragliding), which is the lightest glider possible for a given payload:

A good glider has about 1m/s best sink rate. The average total flying weight is about 1kN which means you need about 1kW to sustain an horizontal flight in still conditions (neither thermals nor sinks). Using Don's 10mW/cm^2 figure, this is 10m^2. The average paraglider surface loading is about 3.5kg/m^2 and the corresponding surface for 100kg total weight is 28m^2.

This leave you about 1.8kW for an average 1.8m/s climbing rate. Not bad, provided light, flexible cells are available.

It is possible to lower the sink rate by increasing the glider surface, would you further need it. But this will be, of course, to the detriment of your air speed (about 35/40km/h for today gliders).

Now, as good as they are, a paraglider has one of the worst gliding ratio of all the flying objects (8-10), but it has the smallest surface loading, which is probably why you don't see solar powered gliders: they are not on the good side of the power budget crossover point.

And also it is the slowest one, which is probably not very interesting for the military people.

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Reply to
Fred Bartoli

This is the minimum possible amount - it's the number you get if you cut the wings off, and put a winch between it and the body, and raise at

1m/s.

The wings are the most efficient bit, as they move lots of air. The less air you move, the faster you have to move it, and the worse the v^2 term in kE=1/2 *m* v^2 bites you.

To make it even stable, you've got to move a lot of air, more than half that moved by the wings, which is a big problem.

Reply to
Ian Stirling

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