Torque Calc: 300 FTLBS@400rpm, 36/18 Gear Ratio

AI Thread Summary
To calculate the torque after gear reduction, a motor with 300 FTLBS of torque at 400 rpm and a gear ratio of 36/18 results in a driven gear rpm of 800, which halves the torque to 150 FTLBS. The discussion also highlights a project involving a 16 hp gas engine powering a pump and hydraulic motor for an urban car, aiming for a top speed of 50 mph without accumulators. Key figures include an engine RPM of 2000, rolling resistance of 0.127 per tire, and an acceleration time of about 20 seconds to reach 50 mph. The user seeks assistance with complex calculations related to drawbar and incline grades due to a lack of math background. The project reflects a desire to innovate in urban vehicle design while addressing efficiency challenges.
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If I have a motor with 300 FTLBS. of torque at 400 rpm with a 36 tooth gear and the driven gear is 18 teeth what is my torque. Thanks' Ray M.
 
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I guess your driven gear would have a rpm of 800 rpm since the no. of teeth are half. So now your rpm is doubled and to keep the transferred power constant the torque would become half of 300 i.e. 150 ftlbs.
 
What is your attempted solution to the problem?
You're welcome.
 
I have been trying to design a hydostatic set up using a 16 hp gas engine to power a pump and hydraulic motor for a 1200 pound
urban car top speed of 50 mph effiecently. using a variable displacement pump and fixed motor. no accumalators. I've Got a few figures like engine RPM 2000 Rolling resistance per tire is .127 and speed 0-50mph in about 20 secs [I think] Max grade of incline 15%. But I don't have the Math background to do the more complex calculations such as drawbar, or grade of incline ect...I just don't want to come acroos as a cracked for wanting to do this project.
 

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