Gear ratio question. ugh i'm confused.

okay. i recently had a forum asking about gear ratio that has me so confused on the concept of ratio that it hurts.

if you have a 40 spline gear and a 20spine gear. and ur power input is on the 40 spline gear. doesnt that mean the 20 spline gear will rotate twice every time the 40 spline gear rotates once?

Averagesupernova
Gold Member
It doesn't matter which shaft is the driven shaft. Whenever a 40 tooth gear meshes with a 20 tooth gear the shaft that the 40 tooth gear is mounted to will make one revolution for every 2 revolutions the 20 tooth gear makes.

It doesn't matter which shaft is the driven shaft. Whenever a 40 tooth gear meshes with a 20 tooth gear the shaft that the 40 tooth gear is mounted to will make one revolution for every 2 revolutions the 20 tooth gear makes.

okay. so with this statement. we could typically say that the 20 spline tooth gear will provide twice the work of the 40 spline tooth gear correct?

Redbelly98
Staff Emeritus
Homework Helper
okay. so with this statement. we could typically say that the 20 spline tooth gear will provide twice the work of the 40 spline tooth gear correct?
No. It will have twice the number of revolutions, and half the torque. So the amount of work would be equal.

No. It will have twice the number of revolutions, and half the torque. So the amount of work would be equal.

What if torque isnt a desired effect but rotation. its for a propeller. would it still be equal work? cuz this could be the difference of 1000lbf and 2000lbf, which would be 100% more efficient and very beneficial.

The faster a propeller spins, the more torque it needs.

So, you can't just slap any small motor onto a gearbox, and ask it to spin the propeller as fast as you want. It simply won't be able to do it. An airplane propeller might only spin at 2000-3000 rpm for example, but it might need thousands of foot-pounds of torque to do so. The result is that, in a lot of cases, the power of the engine is actually put through a REDUCTION gearbox, to decrease the rpm and increase the torque going to the propeller.

Otherwise, the engine simply would not have the strength to spin the thing fast enough.

Redbelly98
Staff Emeritus
Homework Helper
What if torque isnt a desired effect but rotation. its for a propeller. would it still be equal work? cuz this could be the difference of 1000lbf and 2000lbf, which would be 100% more efficient and very beneficial.
It's okay if the amount of rotation is your desired effect, but that would be different than work. Work has a strict definition in physics and engineering: Force-times-Distance, or equivalently Torque-times-Angle for a rotating shaft.

I don't know if you have some application in mind, or this was a hypothetical question. In practice, you can't just "gear up" to an arbitrarily high rotation rate because you need enough torque to drive whatever load is attached.

The faster a propeller spins, the more torque it needs.

So, you can't just slap any small motor onto a gearbox, and ask it to spin the propeller as fast as you want. It simply won't be able to do it. An airplane propeller might only spin at 2000-3000 rpm for example, but it might need thousands of foot-pounds of torque to do so. The result is that, in a lot of cases, the power of the engine is actually put through a REDUCTION gearbox, to decrease the rpm and increase the torque going to the propeller.

Otherwise, the engine simply would not have the strength to spin the thing fast enough.

yea im not an idiot thanks, that was too ur small motor remark.

secondly theres no 1000's of foot pounds of torque. more like 120 foot pounds of torque that i need out of a 310 foot lb torque motor.

and third, you might want to look up how a helicopter works...the goal is to reduce torque...not do a burn out on the highway, oh wait...theres no tires....just a propeller.

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It's okay if the amount of rotation is your desired effect, but that would be different than work. Work has a strict definition in physics and engineering: Force-times-Distance, or equivalently Torque-times-Angle for a rotating shaft.

I don't know if you have some application in mind, or this was a hypothetical question. In practice, you can't just "gear up" to an arbitrarily high rotation rate because you need enough torque to drive whatever load is attached.

thanks for your reply tho, the reason why i ask is because i have an excess amount of torque compared to the rotational weight of the propeller. which seems like a waste because theres no friction application. just aviation.

tuning down the torque to match the weight of the propeller will initially raise the rpms if using this 2:1 gear ratio or something of the sort correct?

i mean obviously if i have too much torque and a moderate amount of rotation for vertical thrust, that seems like a waste.

so i thought this gear ratio for less torque and more rpm would make a more efficient propeller application. which ultimately would use less gas while producing more thrust.

am i right so far on this?

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There is only the amount of torque proportional to the power divided by rotational speed. For example, if the propeller uses power that only requires 120 ft.lbs. of torque and the motor is directly coupled to it, the motor is only producing 120 ft.lbs of torque. No torque is wasted. If the motor is producing an excess of torque, the propeller will speed up until an equilibrium is reached. For efficiency, you try to minimize the power required to produce the required thrust. Unless I am misunderstanding your application, the best way to maximize efficiency would be to choose a gear ratio such that the motor and propeller both are operating at their respective design speeds or speeds where they are most efficient.

There is only the amount of torque proportional to the power divided by rotational speed. For example, if the propeller uses power that only requires 120 ft.lbs. of torque and the motor is directly coupled to it, the motor is only producing 120 ft.lbs of torque. No torque is wasted. If the motor is producing an excess of torque, the propeller will speed up until an equilibrium is reached. For efficiency, you try to minimize the power required to produce the required thrust. Unless I am misunderstanding your application, the best way to maximize efficiency would be to choose a gear ratio such that the motor and propeller both are operating at their respective design speeds or speeds where they are most efficient.

....okay....thats kind of what im asking but thanks for restating my question...

so im guessing that you agree with what i said then, finding the correct gear ratio to reduce this 310ft lb of torque to 120lb of torque while increasing propeller speed for more thrust, is the best application for my flying machine.

jack action
Gold Member
When you want to match a power source to a load, you do not look at the torque but at the power.

Your propeller will need some amount of power to do the job (read http://www.auf.asn.au/groundschool/propeller.html#propellers" [Broken]). So it's the product of torque AND rpm that matters, not the torque alone. A propeller is designed to be efficient at a particular rpm, so the rpm is fixed, you can't modified it at your will.

For a particular propeller with a particular aircraft, you will get a "power required curve", like this one:

[PLAIN]http://www.auf.asn.au/groundschool/powercurve.gif [Broken]

Then you will overlap the power curve of the engine of that curve, like this:

[PLAIN]http://www.auf.asn.au/groundschool/poweravailable.gif [Broken]

Every time both curves are at the same point, it is a stable condition. If the engine power is greater than the power required, than the propeller will accelerate until it reach a stable condition. The power required will never be greater than the engine power.

The gearbox is used to match the maximum rpm of the engine to the maximum rpm of the propeller. No matter the gear ratio, the power stays constant (minus some efficiency losses) between the gearbox input and output.

You can also check this more technical http://continuouswave.com/whaler/reference/propellerPowerCurve.html" [Broken]. It is about propeller for boat propulsion, but the same principle apply.

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yea im not an idiot thanks, that was too ur small motor remark.

All I have to go off of is what you write, and "we could typically say that the 20 spline tooth gear will provide twice the work of the 40 spline tooth gear correct?" was indicative that you didn't know the basics. So I explained it as basically as I could. Second, not knowing something doesn't make one an idiot...especially if you know that you don't know something, and try to remedy it....which is the point of this site. I apologize if my explanation insulted you though....

and third, you might want to look up how a helicopter works...the goal is to reduce torque...not do a burn out on the highway, oh wait...theres no tires....just a propeller.

?
When a helicopter has a turbine engine spinning at 10000rpm and making thousands of horsepower, and the propeller is spinning at a few hundred rpm....well I don't want to say anything that might insult you again. I'm sure you know the formula to derive how much torque that propeller is seeing.

Jakkincorpe, you've been told just how gears work several times now, both by me and others. You continue to ignore what we are saying and respond with flippant remarks.

We know how gears work, you obviously don't. I suggest you listen to what people are saying.

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ur all right i reall dont know how they work. it just seems common sense that 1 rotation that can be multiplied by 2 would produce more power. its fine though, thanks for the replies everyone. imma take a break on all of this and research a little more.

I'm not saying don't ask questions or be put off as it's taking a while, or my bluntness. I had to be blunt to get through that you are sticking with a misconception and not listening to what people ar esaying. You are trying to tackle too many things that you don't have a full understanding of.

There is no shame in not understanding something, we are here to share our knowledge and collectively we do have a very large knowledge base.

Common sense in a very dangerous thing regarding engineering as even a minor error in understanding it can lead to massively wrong conclusions. What you think of as 'more power' is acutally 'more torque'. The power is always constant.

If you can come up with a clear idea of what you want to do in schematic form. This means the specifics such as motor size, gear ratio and proellor size are all not needed.

EG: Motor -> Gearbox -> Propellor

With the aim of lifting you vertically (if I recall correctly).

We can take it from the motor one step at a time of how the powr is transferred and why no new power is available.

stewartcs
ur all right i reall dont know how they work. it just seems common sense that 1 rotation that can be multiplied by 2 would produce more power. its fine though, thanks for the replies everyone. imma take a break on all of this and research a little more.

Just always remember one thing: The power in, must equal the power out minus any losses, and you can't go wrong.

CS

I'm not saying don't ask questions or be put off as it's taking a while, or my bluntness. I had to be blunt to get through that you are sticking with a misconception and not listening to what people ar esaying. You are trying to tackle too many things that you don't have a full understanding of.

There is no shame in not understanding something, we are here to share our knowledge and collectively we do have a very large knowledge base.

Common sense in a very dangerous thing regarding engineering as even a minor error in understanding it can lead to massively wrong conclusions. What you think of as 'more power' is acutally 'more torque'. The power is always constant.

If you can come up with a clear idea of what you want to do in schematic form. This means the specifics such as motor size, gear ratio and proellor size are all not needed.

EG: Motor -> Gearbox -> Propellor

With the aim of lifting you vertically (if I recall correctly).

We can take it from the motor one step at a time of how the powr is transferred and why no new power is available.

wait, is all this confusion happening because i'm saying "power" instead of "torque"?

like as in i know the motor power will never change, but the smaller gear would produce more torque to the turbine than the motors power would alone. is that right?

russ_watters
Mentor
wait, is all this confusion happening because i'm saying "power" instead of "torque"?
The confusion is that you are still confusing power and torque with each other. You aren't clear on how they are different and how they can be related to each other mathematically. You must start treating them as the completely separate entities that they are and paying attention to the mathematical relationships. Ie:
like as in i know the motor power will never change, but the smaller gear would produce more torque to the turbine than the motors power would alone. is that right?
"more torque...than the motor's power would alone" is a jumble of words that doesn't have any meaning because power isn't torque. Knowing the power of a motor tells you very little about what the torque at the motor might be.

The motor has produces a certain torque at a certain rpm. Multiply them together and you get power. Apply a gear ratio and the rpm at the drive shaft goes down while the torque goes up (or vice versa) and the powe stays constant....and the rpm and torque at the motor stay constant.

Note, however, that depending on the type of motor and the application, the motor may simply fail to do what you want it to do. In the case of a synchronous AC motor driving a fan, for example, the rpm of the motor is fixed, but the power is not. If you gear-up the fan to spin faster, the rpm of the motor stays constant, the rpm of the fan goes up, the torque at the fan goes up, and the torque at the motor goes up a lot. The net result for a fan is that doubling the speed of the fan increases the motor power by a factor of 8. And a synchronous AC motor will successfully run at pretty much whatever power you demand from it.....until it burns itself up.

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Wow. This is what was confusing me. I knew what I was talking about the whole time I just wasn't wording it correctly. Jesus what a waste of time. Now that I'm over that freaking hump. I'm relieved. so I'm not to big on a flying machine right now.. My family and I saw this weird *** plane with a huge light source just floating from miles away. It looked like a planet but it moved slightly in every direction. Plus through a telescope it looked as if it was spinning. Then we saw a jet fly by towards the thing so we were all like... Aliens? Haha idk it was weird. So flying is a no cuz I don't wanna be confronted by a B2 ever in my life. However. This gear ratio realization has led me to thinking about a car with amazing gas mileage by just using the gears and keeping a lightweight design. I'm on my iPhone right now so I'll go in further detail tomo. Night.

russ_watters
Mentor
Wow. This is what was confusing me. I knew what I was talking about the whole time I just wasn't wording it correctly.
In your earlier posts, your math was wrong, so this clearly was not a matter of wording it was a matter of conceptual understanding. The problems continue:
This gear ratio realization has led me to thinking about a car with amazing gas mileage by just using the gears and keeping a lightweight design.
There's no magic with gear ratios on cars. As others mentioned, they are already designed for optimum fuel efficiency, matching the most efficient operating point with the required output at a certain cruise speed. You cannot increase fuel economy by further changes in gear ratio unless you are adjusting for different operating conditions.

Also, fuel efficiency is not much of a function of weight for a car at cruise speed. Fuel efficiency is a function of drag.

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I didn't say anything before, but I know exactly what you idea is (I think) as I've seen it before. I'll try to be clairvoyant. You were going to suggest that you could use a gear that kept an engine at idle but allowed the wheels to turn at many many rpm giving high road speed.

There are 2 ways of describing what is wrong with this one method using torque the other using power (in reality they describe the same thing).

1) You will use a gearing that increases the rpm and so reduces torque at the wheel, meaning it won't have the necessary thrust to attain the speed you want. The drag will just overpower the car and not allow it to accelerate. Ie you don't have enough wheel torque. OR 2) You don't have enough power at idle to push a cartain car with drag 'x' at a certain speed. You don't have enough engine power.

In the case above at idle rpm, you dont have the torque necessary (and therefore the power necessary) to propel the car at high speeds. No matter what the gearing.

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I didn't say anything before, but I know exactly what you idea is (I think) as I've seen it before. I'll try to be clairvoyant. You were going to suggest that you could use a gear that kept an engine at idle but allowed the wheels to turn at many many rpm giving high road speed.

There are 2 ways of describing what is wrong with this one method using torque the other using power (in reality they describe the same thing).

1) You will use a gearing that increases the rpm and so reduces torque at the wheel, meaning it won't have the necessary thrust to attain the speed you want. The drag will just overpower the car and not allow it to accelerate. Ie you don't have enough wheel torque. OR 2) You don't have enough power at idle to push a cartain car with drag 'x' at a certain speed. You don't have enough engine power.

In the case above at idle rpm, you dont have the torque necessary (and therefore the power necessary) to propel the car at high speeds. No matter what the gearing.

the idle thing was for the vertical thrust. my calcs of the thust issue showed me at idle the propellers would be producing enough thrust to lift the vehicle. which the equivelent was like 2400lb of thrust while the vehicle only weighed 2000lbs.

but in this about the gearing of an actual car u are correct lol there is no way i'm going to be able to produce enough torque to go 60mph at idle haha.

In your earlier posts, your math was wrong, so this clearly was not a matter of wording it was a matter of conceptual understanding. The problems continue:
There's no magic with gear ratios on cars. As others mentioned, they are already designed for optimum fuel efficiency, matching the most efficient operating point with the required output at a certain cruise speed. You cannot increase fuel economy by further changes in gear ratio unless you are adjusting for different operating conditions.

Also, fuel efficiency is not much of a function of weight for a car at cruise speed. Fuel efficiency is a function of drag.

yes, there is no magic in gear ratios. but if you have a low enough idle gear at a lower rpm than factory standard with a more efficient engine you can double your mpg. i'm thinking about a diesel motor so heres what i came up with using a 96-98 jetta turbodiesel

with a 24" wheel diameter, it will take 220 rotations to travel one mile. at 60mph the engine would be at 1900rpm, meaning the tranny gear would be 2.33:1 and the differential would be 3.7:1. i used 1900rpm because thats the engines maximum torque (149ft/t). so at this speed with the ratios the torque to the wheels is 1,284ft/t at 220rpm.

this is a very excessive amount of torque for highway cruising. my 2000 mustang only used 730ft/t at 220rpm and it weighed 200lbs more than the jetta.

thats what i'm saying. if you cut the rotation of the engine in half, while still producing enough torque to travel at 60mph. you'll gain mpg.

about the weight thing as well with the fuel efficiency russ, its a proven fact that if you can cut 50lbs off your cars weight you'll see a 6-10% gain in mpg. the GM ultralite factored that into producing a car that only weighed 420lbs, met USA speed standards, size, and got 80mpg. the thing was that the car was way to expensive to produce. but thats the problem with alot of things these days. just not enough ppl manufacturing the products to create the parts.