Quote:
Originally Posted by drkiraco
to keep going:
Thrust = mass flow * velocity
Power = (1/2) mass flow * velocity^2
For a given thrust, we could pick any combination of mass flow and velocity, but to minimize energy used, we want greater mass flow and less velocity.
max power for a single FIRST motor is 480 watts (12V * 40A). call it 500W for round numbers. So for 50 N you could do, for example, 2.5 kg/s air at 20 m/s. Now, to put this in perspective, 20 m/s = 44 mph. Seems a little high to me, but maybe not out of the question. Probably you wouldn't get anything faster than this. Now put 2.5 kg/s of air in perspective. That's 5.5 lb/s. Since a cubic foot of air has a mass of ~0.075 lb at standard temp and pressure, we need 73 ft^3/s. That's 4406 ft^3/min (scfm). That's a rather large number. Maybe not totally out of the question though.
So, unless I messed up the math, I'd guess that 50N of thrust could be achievable without violating the laws of physics or any FIRST rules, but I don't think it would be cheap or easy...
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I did math myself starting with your equations and got the same answer, so you're probably correct.
Now, how big would the swept area on this hypothetical fan be? You've got a column of air 18.5 meters long (for my calcs) long passing through through every second, that contains 2.7kg of air, which is 2.11 cubic meters.
V = pi * r * r * h
2.11 = pi * r * r * 18.5
sqrt(2.11 / (18.5 * pi)) = r
0.19 = r = 19cm radius fan, or 7.5 inches. That's a pretty big fan, but not unusably big. This could be do-able, but the speed out the back and the CFM rating seems pretty insane. I searched for 4000CFM fans, and came up with stuff sold as "whole house fans", and I doubt that those would put out the kind of velocities needed.
Is this a correct calculation?