Quizzes & Puzzles4 mins ago
Help With Power Calculation Please
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I have designed and built a device that harnesses centrifugal force.The power output is in the form of
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).
If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).
If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?
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Best Answer
No best answer has yet been selected by bill barlow. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ."why does it matter"! - the fact that you ask that tells me all I need to know! Forget any calculations, forget any comments above, test the theory. Connect your "engine" to a small alternator, it can turn that can't it? Feed that to a car battery, connect a 12v-240 transformer to the battery and plug the drill into that. Also plug the drill into the mains, split the wires so both sources feed the drill. Turn on the mains and the drill, leave the drill on and turn off the mains. If you are correct the "engine" should run fior ever with no power input. Tada, nobel prize!
@bill
//I have measured the weight these pistons can lift to be 122lbs//
You said 'lift' and 'weight'.
For starters, weight is a force. Mass x gravity
50kg x 9.81 = 490 N
Force x distance (meters)
15mm = 1.5 cm
1.5 cm = 0.015 m
490 x 0.015 = 7.35 Nm
66.666 times per second = 489.9951 NM.s^-1
500W in, 490W out.
98% efficiency is pretty impressive. Nice set of bearings you're using!
490 x 0.015
//I have measured the weight these pistons can lift to be 122lbs//
You said 'lift' and 'weight'.
For starters, weight is a force. Mass x gravity
50kg x 9.81 = 490 N
Force x distance (meters)
15mm = 1.5 cm
1.5 cm = 0.015 m
490 x 0.015 = 7.35 Nm
66.666 times per second = 489.9951 NM.s^-1
500W in, 490W out.
98% efficiency is pretty impressive. Nice set of bearings you're using!
490 x 0.015
I presume that's a joke, Tora. There are any number of ideas like yours, all of which have fallen flat because they have not allowed for losses.
I was talking in the pub to someone who simply couldn't understand why you couldn't connect a motor to an alternator to feed the motor.
Give it up, Tora. It ain't possible.
I was talking in the pub to someone who simply couldn't understand why you couldn't connect a motor to an alternator to feed the motor.
Give it up, Tora. It ain't possible.
why are you telling me to give it up, it's not my post, I know the FLOTD.
http:// en.wiki pedia.o rg/wiki /First_ law_of_ thermod ynamics
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