Gear Ratio's - How to properly show them?

I feel like this is a really stupid question. But it seems like it see it both ways all over the internet and these forums.

Drive : Driven (Input:Output)

    or

Driven : Drive (Output:Input)

E.g. is a direct drive wheel with high speed internal gearing considered to be 1:1.6 or 1.6:1

I feel like it’s supposed to be Drive : Driven (Input:Output), therefore with the above example it should be 1:1.6

It’s not a stupid question. I find myself asking the same thing all the time. Since it’s one of those things people do differently, like which side of the road they drive on in different countries, I always try to specify exactly what I mean.

It’s like clockwise vs. counter-clockwise. It really matters from which direction you look at it.

Depends on if it’s a speed ratio or torque ratio. I would tend to prefix the ratio by “geared up” (increase in speed) or “geared down” (increase in torque). The car guys always express in terms of input/output, eg. the rear axle has a 3.43:1 ratio, meaning input turns 3.43 revolutions for each turn of the output.

Edit: I fixed my dumb mistake.

1 Like

Actually it gets even more complicated. Because the argument in some cases isn’t output:input or vice versa but actually between weither its gear teath(my preference) or angular velocity

WPI teaches a motor powering a 12 tooth and geared to a 60 tooth as 1:5

Input:output
12:60

This meaning gear ratio( ratio of the gears) and then say torque ratio to mean ratio of the torques(they also say that torque ratio should include the efficiency but thats more of a practice to ensure students don’t forget 37 stage gear ratios are bad)

Actually, in gearhead speak (at least in relation to cars and trucks), it is input-output. A 3.43 rear axle ratio is a higher speed, lower torque ratio than a 4.10. It would be 3.43:1 or 4.10:1, input-output. A “lower” rear axle ratio (higher torque, lower speed) is numerically higher than a “taller” ratio (lower torque, higher speed).

Doh, of course you are correct, serves me right for typing that post quickly as I was leaving the house today.

1 Like

But I guess that’s kind of the point though…if you have a standard way of showing it, then it does not have to be prefaced with descriptions.

For example if it is always (input : output) then

1:3 implies speed
3:1 implies torque

This method seems to make sense to me because it can be read (for 1:3) as “When the motor turns ONCE the output turns THREE times” There is logic to it…I think.

It seems like there has to be a correct/standard way of doing this.

Gear ratio is an ambiguous term in my opinion. People can take it to mean tooth ratio, speed ratio, or torque ratio. You are better off using one of the latter two when referring to your gear train. I looked back in my textbooks and couldn’t find “gear ratio” anywhere.
The second issue is that it is not clear what goes first, input or output. In my experience it has always been the latter (output:input).
So a speed ratio of 3 has the output turning faster than a speed ratio of 2.

This is the crux of my question. With all the engineering in this world, I find it hard to believe there is not a standard.

Good to know. This makes sense to me. I would be awesome if everyone on these forums could start using a common language regarding gear ratios.

Well in general we use gear trains to increase either torque or speed. Rarely is our intent to decrease the speed or torque (even then we call it a reduction ratio), these are just consequenses. So it would make sense to me that a ratio that expresses our desired change would have the output on top so you can see by what factor your parameter has increased. Does that make sense? I know it is sort of poorly phrased.

But if you increase speed you decrease torque. It is a relationship. Not quite sure if I’m understanding what you mean.

My point is that a person should not have to state what they are talking about. If everyone uses the same format, it will be implied whether it is a speed or torque gearing.

If it is indeed input : output, then 1:3 automatically means you are gearing for more speed and 3:1 automatically means you are gearing for more torque.

I may be thinking about this all wrong though.

I think you’re accidentally reversing this.

If it is input : output, it would be for example:

12 tooth driving : 60 tooth being driven =
1 : 5 for torque (because 60 tooth gear makes 1/5 of a revolution for every revolution of the 12 tooth gear)

(You were saying 1:3 means speed and 3:1 means torque)

A quick search of the term “gear ratio” on Google immediately finds this article. By reading it, and several other articles, it became apparent to me that the convention is to refer to a gear ratio in terms of speed (and that it is synonymous with the terms speed ratio or velocity ratio), and of the form input : output. Therefore, if you are discussing a gear ratio to increase the speed of your drive by a factor of 2, you would be talking about a 1:2 gear ratio. For every 1 rotation of the input gear, you get 2 rotations of the output gear.

Another option to solve this confusion would be to stop using the term “gear ratio,” and instead use “velocity ratio” or “torque ratio,” depending on the type of ratio being discussed.

Oh ok 4149G’s post makes sense then he was probably talking about 1:3 in terms of revolutions rather than in terms of teeth #.

I was but your previous point a good point though. What you said makes me realize that I’m probably not thinking about it correctly.

I was thinking 1:3 would mean for every one turn of the motor the output turns three times. But my logic falls apart then when applying that, as you said. If a 12 tooth gear is driving a 36 tooth gear (1:3), that is actually a torque gearing, not speed.

Thank you for pointing that out…I don’t know what this simple topic confuses me so much.

I think this is how my brain has been interpreting all of this as well. As you said 1:3 would would be read as: “for every one turn of the input there are 3 turns of the output” (speed).

This does fall apart if using number of teeth for calculating though (as I said in the previous post). 12 tooth driving 36 teeth would reduce to 1:3, but would need to be shown as 3:1 if you want to use the “speed” or “velocity” idea of gear ratio.

uhg. I’m confusing myself.

to avoid confusion i just use fractions

1/3 ratio means output speed = 1/3 input speed

though where i think most of the confusion comes from is assuming x:y means x/y

In the “speed” gearing, the 36 tooth gear is the driven pinion, and the 12 tooth gear is the output gear. So it does simplify to 3:1.

I believe the “proper” way to do it is [input speed]:[output speed], which is the same as [output torque]:[input torque] or [output number of teeth]:[input number of teeth].

In a vex context it’s usually not that important to get it right, because the context should tell you. Arms are always geared down, and drives are always (except in extreme cases) geared up.

We’ve found that for vex, especially for drives, the most useful way to talk about gearing is this:

First you combine all the elements of the drive (internal gearing, external gearing, wheel size, wheel orientation) and compare that effective gear ratio to a reference of tank drive with torque motors geared 1:1 on 4" wheels. Then you divide both sides of the ratio by the smaller side to give a ratio in the form 1:[x].*

McChicken’s drive used speed motors geared 1:1 on an X drive with 2.75" wheels. This works out to be the same as a tank drive with 4" wheels and torque motors geared up by 1.6:1, so McChicken had an “effective” drive ratio of 1:1.6. It doesn’t matter whether this is expressed as 1:1.6 or 1.6:1, because you know that unless otherwise stated all drives will be geared up.

This way you can compare any two drives to each other without needing to state the internal reduction, the external reduction and the wheel size separately. Of course since the reference gearing is based on specific vex motors and wheels, it doesn’t generalise well to other domains.

*The reason for using 4" wheels and torque motors as the reference is that not so long ago torque motors were the only ones available and 4" wheels were the only size that was practical for competition, because that was the size that dual-roller omniwheels came in.