im planning an open field solar system and am trying to find out how much re-enforcment to place under the system to ensure the system doesnt fly away. who knows anything about physics?
2006-08-14 01:09:34 · 6 answers · asked by redirus91 3 in Science & Mathematics ➔ Physics
sorry I forgot about the dimension. its a 30° angle with 9 meters wide by 3 meters tall. this would be of course a non continuous surface. it will have gaps around the panels of about 1-2 fingers wide all around each panel.
2006-08-14 01:52:57 · update #1
tony i think you should wait till your older to fly something tha big.
2006-08-15 21:52:29 · update #2
i'm not much in the physics field, but I do know a bit about wind and construction. I sail. A 20 mph wind can propel my boat (25 foot length, 8 foot beam, 4000 lb displacement and 4 foot draft) at about 9-10 miles per hour under full sail and with a 110% genoa jib on the bow. If you ever manned the jib winches under that kind of strain you would know that you handle the ropes with gloves to prevent painful friction burns and you keep your fingers away from the winch drum or you will lose a finger. That's a pretty healthy wind. I'd advise anchoring your panels to something solid like a concrete foundation with studs set in the concrete. I know this won't give you a physics answer but will give you anidea of what such wind can do.
2006-08-14 01:19:56 · answer #1 · answered by dread pirate lavenderbeard 4 · 0⤊ 0⤋
A 20mph wind is a pretty slow wind!
If your installation is in open field, you must consider winds up to 80 mph! These winds may not "lift" your construction, but they may certainly bend over the structure or wipe them away.
Remember, the force of the wind (that is, the pressure it can exerce on a surface) is a function of the shape of the object "pushed" by the wind times the square of the wind velocity (in m/s) times the surface of the object (F=k*S*v^2)
k is a "coefficient of penetration" equal to 1 for a flat surface, going towards for a very pointy surface (the tip of a needle).
F is more or less equal to 100N for 0.01m^2, flat square, at 40kph.
2006-08-14 08:51:36 · answer #2 · answered by just "JR" 7 · 0⤊ 0⤋
I don't know the physics, sorry, mate but I've recently crushed my shoulder, three ribs and a tooth flying a 3.5 sqr mtr power-kite in a twenty mph wind. I weigh 10.5 stones. Good luck.
2006-08-15 12:00:23 · answer #3 · answered by Tony 1 · 0⤊ 0⤋
horizontal force of wind with the velocity of 11 m/sec would be about 70 newton /m^2
Lift force depends on inclination of panels.
2006-08-14 08:50:11 · answer #4 · answered by Anonymous · 0⤊ 0⤋
It all depends on the angle of the panels.
2006-08-14 08:14:51 · answer #5 · answered by whyme? 5 · 0⤊ 0⤋
Depends size / height of solar panels
Would expected planning permission would be required so detailed plans and drawings would be required !
2006-08-14 08:14:45 · answer #6 · answered by Bob 4 · 0⤊ 0⤋