Torque on a Shaft: 5HP & 10:1 Gear Ratio

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Discussion Overview

The discussion centers on the relationship between horsepower, gear ratios, and torque in a mechanical system. Participants explore how power is transmitted through a gearbox and the implications for torque, particularly in the context of a motor rated at 5 horsepower with a 10:1 gear ratio.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions whether the horsepower remains constant throughout the system or if the gearbox multiplies the power.
  • Another participant asserts that power is constant through a transmission and is defined as the product of torque and angular rate.
  • A subsequent reply emphasizes that power remains constant minus losses due to inefficiencies, suggesting that this is relevant to the original question.
  • Another participant provides equations indicating that power output equals power input multiplied by efficiency, and torque output is related to torque input adjusted by the ratio of input speed to output speed, also factoring in efficiency.

Areas of Agreement / Disagreement

Participants express differing views on the nature of power transmission through the gearbox, with some asserting that power remains constant while others highlight the impact of efficiency losses. The discussion does not reach a consensus on these points.

Contextual Notes

Participants reference efficiency and its effects on power and torque but do not clarify specific assumptions or definitions regarding efficiency or the conditions under which their statements hold true.

blake92
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If a motor has a horsepower of 5hp and a gear box ratio of 10:1 does the hp through out the system remain 5? or does the gear box multiply the power?

how does this affect the torque?
 
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Power is constant through a transmission and is the product of torque and angle rate.
 
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Doug Huffman said:
Power is constant through a transmission and is the product of torque and angle rate.
...minus the loss due to inefficiencies. Not to nitpick on you Doug Huffman, but considering the OP's question I think it is worthy of pointing out.
 
power out = power in * efficiency
torque out = torque in * (input speed / output speed) * efficiency
 

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