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

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Discussion Overview

The discussion revolves around calculating the torque output of a system involving a motor with specified torque and RPM, along with a gear ratio. Participants explore the implications of gear ratios on torque and RPM, as well as related engineering considerations for a vehicle design project.

Discussion Character

  • Technical explanation, Conceptual clarification, Homework-related

Main Points Raised

  • One participant asks for the torque output when a motor with 300 FTLBS of torque at 400 RPM is connected to a gear system with a 36-tooth driving gear and an 18-tooth driven gear.
  • Another participant suggests that the driven gear would rotate at 800 RPM due to the gear ratio, proposing that to maintain constant power, the torque would be halved to 150 FTLBS.
  • A third participant inquires about the attempted solution to the initial problem posed, indicating a desire for further clarification or elaboration.
  • A fourth participant shares their project involving a 16 hp gas engine powering a hydraulic system for a vehicle, expressing a need for assistance with complex calculations related to the design.

Areas of Agreement / Disagreement

There is no consensus on the torque calculation, as participants present differing views on the implications of the gear ratio and the associated torque output. The discussion remains unresolved regarding the initial torque question.

Contextual Notes

Participants express varying levels of mathematical background, which may affect their ability to engage with the technical aspects of the discussion. Some assumptions about power transfer and efficiency are implied but not explicitly stated.

Who May Find This Useful

Individuals interested in mechanical engineering, gear systems, torque calculations, and vehicle design may find this discussion relevant.

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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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