Gear ratio question. ugh i'm confused.

jakksincorpse

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

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.

jakksincorpse

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

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

jakksincorpse

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.

Lsos

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

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.

jakksincorpse

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.

jakksincorpse

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?

Turbodog

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.

jakksincorpse

....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

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 this). 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:



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



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 link. It is about propeller for boat propulsion, but the same principle apply.

Lsos

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....

?
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.

xxChrisxx

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.

jakksincorpse

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.

xxChrisxx

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

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

CS