Question about torque, RPM and gear ratios...

[Conceptual]

Am I right in assuming that at a given RPM a motor produces a set amount of torque, and if the gear ratio on the output shaft is 3:1, then the torque is devided by 3?

IE: 1000 RPM and 100ft/lbs with a 2:1 gear up you would have 2000 RPM, and 50ft/lbs?

I know this may be super simple to most on here, but I want to be sure that I am not skewed on this fact.

THANKS!

[sinspawn1024]

You're close.

Gear ratios are given as a ratio of the number of teeth on the idler gear (the gear being driven) vs. the number of teeth on the pinion gear (the gear doing the driving).

So if a gear with 10 teeth is driving a gear with 20 teeth, it's gear ratio is 20/10, 2.0/1, or 2:1 (any of these representations can be used).

So in your example, if a torque of 100 lb*ft is delivered at 1000 rpm to a set of gears with an overall ratio of 2:1, the output of the gearset will be 200 lb*ft of torque at 500 rpm (minus whatever power losses due to friction etc.)

[alelanza]

A good read, not precisely about your question but it does answer it as well:

http://eng.kaps.cz/news-kaps/newton_met ... -1017.html

[xpensive]

There's even an easier explanation: Power is always Propulsionforce times vehiclespeed.

With a constant enginepower, propulsionforce will decrease proportionally with speed,
regardless of gear ratios or CVTs.

This also why it's so easy to get wheelspin from a standing start.

[Conceptual]

Thanks for the replies!

And to the poster that worked out my equasion, I guess that mine would have been correct if I used 1:2 ratio?

Thanks again!

[Conceptual]

A good read, not precisely about your question but it does answer it as well:

http://eng.kaps.cz/news-kaps/newton_met ... -1017.html

Dead link?

[alelanza]

A good read, not precisely about your question but it does answer it as well:

http://eng.kaps.cz/news-kaps/newton_met ... -1017.html

Dead link?

Odd, it works for me... what if you go here:

http://eng.kaps.cz/zavodni.php

And then look for what is at the moment i type this the 5th article down?
Again it's mainly about power vs torque, so the other answers on this thread are a lot more effective than mine, but I think it would also answer your question. Long winded though...

[xpensive]

Another way of seeing things is that Power is always Speed times Torque.

In SI-units: Watts equals radians per second (Speed) times Newtonmeters (Torque).

Example: A 720 Hp (529 000 W) at 18000 Rpm F1 engine has an SI-speed of (18000/60)*2pi per second, which gives a torque of 529 000 W over 600*3.1416 radians/second = 281 Nm.

Not much really when you compare with the Torque of a standard vehicle engine.
But the power comes from the ferrocious speed of 18000 Rpm.

You can transfer one for the other, but the end result is always the same.

[sinspawn1024]

Thanks for the replies!

And to the poster that worked out my equasion, I guess that mine would have been correct if I used 1:2 ratio?

Thanks again!

Yeah that's exactly right. The only thing is that most engineers would write it as 0.5:1

Not a big deal technically, but gear ratios in transmissions are just about always put as a decimal:1. That way you can tell at a glance that if it's less than one, it's an overdrive gear, and if it's more than one it's a reducer. Then you can tell just by looking how much it overdrives or reduces. It's basically a convenience thing.

[Scotracer]

Interesting topic to bring me back up to speed with the forum since my vacation. If what I type has already been said I apologize for I am still on vacation and typing from my ipod.

Take an example engine:

750BHP @ 18,000rpm (basically a current F1 unit)

Calculating the torque (since I'm on my iPod I must do it roughly in my head) we get about 215lb.ft @ 18000rpm (crank ang. Velocity). If we have a 3:1 ratio as you suggest then this torque becomes 645lb.ft @ 6,000rpm at the output shaft of the 'box. The rotational speed drops but the torque rises since energy must be conserved.

I don't use the force method (torque) on engines anymore (unless doing tranny work) as it's combersome compared to energy method ( power)

[xpensive]

I always try to see things another way Scot.
If you keep Power the same (minus some friction of course), then Power is always Torque times Angular velocity or Rpm.

You can trade Torque for Rpm if you wish, but power remains the same.

[Scotracer]

I always try to see things another way Scot.
If you keep Power the same (minus some friction of course), then Power is always Torque times Angular velocity or Rpm.

You can trade Torque for Rpm if you wish, but power remains the same.

I know it is. But since torque IS power, effectively I drop one of the terms as soon as possible as it makes no difference to the value. And that's what I mean - Torque analysis or Power analysis of a car gives you the Same end result but with the Power (energy method) you don't have to worry about ratios or mechanical efficiencies, you have a power value; an ultimate thrust calculated from a lot less work. Power is conserved.

But different methods for different engineers

[xpensive]

Torque is NOT Power, there's where you got it completely wrong I'm afraid.