simple gearing - I have confused myself

[Tony Ennis]

If I have two shafts and I want a 7.2:1 ratio between them, what gearing would I use? Now, 72:10 is pretty obvious. But what if I wanted to accomplish the task using a compound gear? I've confused myself to the point I keep going in circles.

[Tony Ennis]

After more effort, I got this:


shaft1 ==== 10
compound == 40 10
shaft2 ======= 18


which I believe is 10 / 40 * 10 / 18 = .14, which is 1:7.2

I brute-forced that. I don't know a systematic way, though the ratio of the compound and the shaft2 gear implies a method.

[LES A W HARRIS]

90/50 X 80/20 = 7.2:1

cHEERS,

[The Artful Bodger]

After more effort, I got this:


shaft1 ==== 10
compound == 40 10
shaft2 ======= 18


which I believe is 10 / 40 * 10 / 18 = .14, which is 1:7.2

I brute-forced that. I don't know a systematic way, though the ratio of the compound and the shaft2 gear implies a method.

10 is a fairly small pinion, how about 20/80 and 20/36, I am assuming that gives the same ratio..

[Forrest Addy]

OK you got a ratio (7.2:1) you want to convert to change gears. If the ratio doesn't work out to a single reduction (5 to 1 is about the limit) you need to go to a compound reduction.

Generally you start with the square root of the starting ratio but just by looking you can see that 7.2 is divisible by 3.0 producing compound ratios of 3.0 and 2.4.

So 3.0 x 20 = 20/60 and 2.4 x 20 = 20 x 48. There is your compound gearing: 20/60:20/48.

Or 22/66:30/72 or whatever works with the available tooth counts. There's about a jillion alternative combinations.

[Astronowanabe]

If I have two shafts and I want a 7.2:1 ratio between them, what gearing would I use? Now, 72:10 is pretty obvious. But what if I wanted to accomplish the task using a compound gear? I've confused myself to the point I keep going in circles.

prime factors of 10 are (2*5)
prime factors of 72 are (3*3*2*2*2)

you can drop a 2 they have in common 5 : (3*3*2*2)

one of your gears needs to be a multiple of 5; say n
the other(s) will need to be some combination of n*(3*3*2*2)

how you end up with that will depend on the gears at hand


Edited to correct a typo changed a 3 to a 2.

[RB211]

prime factors of 10 are (2*5)
prime factors of 72 are (3*3*2*2*3)

you can drop a 2 they have in common 5 : (3*3*2*2)

one of your gears needs to be a multiple of 5; say n
the other(s) will need to be some combination of n*(3*3*2*2)

how you end up with that will depend on the gears at hand

Don't you mean 5: (3*3*3*2)?

[rohart]

I can proceed to work out as many combinations as I like, and even to do that with my limited selection of gears.

What I can't get a good handle on, is how to restrict my gearings to those that I will actually be able to set up without gears that should not be meshing clashing with each other.

The extent of this difficulty is worse the wider the gear spindles are.

For example, take the 10:40, 10:18 solution that Tony offered to his own question. With the first 10 tooth gear on the input shaft, driving the 40 tooth gear, compounded with the second 10 tooth gear, the output 18 tooth gear cannot be mounted because its spindle is obstructed by the 40 tooth gear.

And this is not the only type of clash that rears its ugly head.

You can set up strict allowable relations between adjacent gears, but I find the problem becomes mind-boggling, and I have to resort to trial and error.

[Astronowanabe]

Don't you mean 5: (3*3*3*2)?

No, that would be 5:54, OP needs 5:36 which is 10:72 which is 1:7.2

Edit: I see the mistake that I wrote that made to you think that.
where I wrote 5: (3*3*2*2*3) the last three is a typo it should be a two

[Astronowanabe]

I can proceed to work out as many combinations as I like, and even to do that with my limited selection of gears.

What I can't get a good handle on, is how to restrict my gearings to those that I will actually be able to set up without gears that should not be meshing clashing with each other.

The extent of this difficulty is worse the wider the gear spindles are.

For example, take the 10:40, 10:18 solution that Tony offered to his own question. With the first 10 tooth gear on the input shaft, driving the 40 tooth gear, compounded with the second 10 tooth gear, the output 18 tooth gear cannot be mounted because its spindle is obstructed by the 40 tooth gear.

And this is not the only type of clash that rears its ugly head.

You can set up strict allowable relations between adjacent gears, but I find the problem becomes mind-boggling, and I have to resort to trial and error.

that is a more interesting problem, which so far I have not run into, probably because I have a QC box. wish I could say I would look at it
but I'm in up over my eyeballs these days.