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calculating motor torque required | |||
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Posted by: tspco ® 12/18/2005, 14:07:58 Author Profile eMail author Edit |
I have a 150 pound machine on the drawing board, it has 4 wheels each side has one motor and gearhead plus a final 2:1 chain drive, the tires are 13" tall, and the machine will move at 10 mph, according to my calculations I need in the neighborhood of 130 in/lbs torque per side, I tried an online calculator to select the gear head and the program responded with a gearhead that only has 20 in/lbs torque, how can this be?
I quess what I want here is a second opinion of my calculations, to verify accuracy any takers? |
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Re: calculating motor torque required | |||
Re: calculating motor torque required -- tspco | Post Reply | Top of thread | Forum |
Posted by: Kelly_Bramble ® Administrator 01/13/2006, 20:45:33 Author Profile eMail author Edit |
Anything here help you? /motors/motor_menu.shtml |
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Re: calculating motor torque required | |||
Re: calculating motor torque required -- tspco | Post Reply | Top of thread | Forum |
Posted by: zekeman ® 12/18/2005, 21:21:07 Author Profile eMail author Edit |
Unless you specify the time it takes to get up to speed , you cnnot get a definitive answer as to the torque required or the motor horsepower.And why do you need two motors to power this machine; one should be sufficient with a differential.
For example, if it takes 5 seconds to come to the speed of 10 mph, the torque on the wheels comes out of the formula F=Mass*acceleration Torque= F*r where r is the wheel radius Assuming uniform acceleration, for the example indicated, Acceleration=10mph*5280/3600/5 sec =2.85 ft/sec^2 And neglecting friction F=150/32* 2.85 = 7 lb and T=7*6.5= 45.5 lb in The max RPM at the wheels is 52800*12/60/13pi= 258 RPM; a total gear ratio of 10, for example would give a max motor RPM of 10*258= 2580 RPM with motor torque of 45.5/10=4.6 lb in . With gear inefficiencies, this could be somewhat higher. If you change the time to come to speed by any factor, you would raise or lower the torque inverse proportionally. Or if you change the max motor RPM, the gear ratio as well as the motor torque would be inversely proportional. |
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Re: calculating motor torque required | |||
Re: Re: calculating motor torque required -- zekeman | Post Reply | Top of thread | Forum |
Posted by: tspco ® 12/19/2005, 07:38:16 Author Profile eMail author Edit |
Got me on that one, but you are close I was figuring 6 seconds, to get up to speed. |
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Re: Re: calculating motor torque required -- tspco | Post Reply | Top of thread | Forum |
Posted by: lifejunkie ® 01/13/2006, 19:35:52 Author Profile eMail author Edit |
You only need enough power to break the coefficient of static friction. F = umg Where F is the friction force, u is the coefficient of static friction (0.9 is ok for tires on pavement), m is the mass, and g is the acceleration due to gravity. After you get this number, multiply it by the radius of the tire, and you will have the maximum torque you need for the vehicle. |
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