Engineers: How to design a gear train

[friedo]

Let’s say I want to make a gear train with a 12:1 ratio. There are an infinite number of combinations for the sizes of gears that I can use.

But, let’s say there’s two additional requirements:

1. The first and last gear of the train must rotate in the same direction (therefore the number of gears must be even.)

2. The first and last gear of the train must have the same axis of rotation.

In case you haven’t figured it out yet, these are design requirements for a clock.

Is there a formula or table or something that I can use to find a solution? Picking an even number of gears to make 12:1 is easy, but I need to pick gears with the exact correct diameter to get them to stack up right in order to make this device.

[treis]

The best way to go about this would be to determine what size gears you can get, and then from those pick the combination that uses the smallest number of gears, and is easiest to assemble.

[friedo]

For the purposes of this (theoretical, not practical) discussion, assume I am making all the gears from scratch, so there are no pre-set sizes to choose from.

[gotpasswords]

@friedo

1. The first and last gear of the train must rotate in the same direction (therefore the number of gears must be even.)

Don’t you mean the number must be odd?

2. The first and last gear of the train must have the same axis of rotation.
So, nothing oddball like bevel or worm gears. Just a batch of nice simple straight-cut gears.

these are design requirements for a clock.
Aren’t clockworks already pretty well worked out?

[friedo]

@gotpasswords

Don’t you mean the number must be odd?

No, I mean even, because a gear train has two gears per axle except for the first and the last gears.

So, nothing oddball like bevel or worm gears. Just a batch of nice simple straight-cut gears.

Right.

Aren’t clockworks already pretty well worked out?

Yes, but that’s not the point. I’m trying to figure out how people figured this out.

[Santo_Rugger]

Set up an equation, putting the driving gears’ numbers of teeth (N) divided by the driven gears’ numbers of teeth (n), and manipulate the equation such that it equals 12.

N1/1 * N2/n2 * N3/n3 * 1/n4

I believe this is what you’re asking, but if you’d like more detail, let me know.

IANA gear expert, but my senior design project was designing a sun and planetary gear system, in tandem with a CVT, to try and produce a tractor transmission that was infinitly variable.

[Oslo_Ostragoth]

Dammit, as a ME in college, I knew this stuff cold. I hate that I no longer recall how to figure this stuff out.

[Dervorin]

Well, for what it’s worth, Richard Feynman was given the following advice when he encountered a similar problem:

@Surely You're Joking

Second, when you have a gear ratio, say 2 to 1, and you are wondering whether you should make it 10 to 5 or 24 to 12 or 48 to 24, here’s how to decide: You look in the Boston Gear Catalogue, and select those gears that are in the middle of the list. The ones at the high end have so many teeth they’re hard to make, if they could make gears with even finer teeth, they’d have made the list go even higher. The gears at the low end of the list have so few teeth they break easy. So the best design uses gears from the middle of the list.

[casdave]

Looks like you will have to find a way of using internal barrel gears to get an even number of elements with input and output moving in the same direction.At least one transfer will have to work this way.

[Kevbo]

the least number of reductions, given the common axis of input and output, is 2.

Integer pairs of factors for 12 are:

6 and 2

3 and 4.

12 and 1.

The 3-4 pair lets each pair of gears do about half the job. . If we want the gears to all have the same pitch, the sum of the teeth of the two sets must be the same. This gives us 3 equations:

i/j=3
m/n=4
i+j=m+n

Simple gears need at least 12 teeth. For this reason, clocks often use lantern pinons which will work well with as few as 6 teeth. So start with J=6 and see if the above equations have an integer solution set . If not, try 7,8,9 etc.

In this way, you are likely to find a solution. If not, you might consider making the two gear sets with different pitches, though this might lead to depthing trouble if not precisely done. Non-metric standard gears are specified by “diametrical pitch”. Using this, the total number of teeth gives you the axle spacing. Metric gears use a modulus system I’ve never had to deal with.
Gear reduction ratios needn’t be integers though. 12 has an infinite number of pairs of non integer, rational factors. The one equation that the gears must satisfy is:

im/jn=12

As I said before, you want the two ratios i/j and m/n to be fairly close to equal. So start picking i/j ratios (rational) between 3 and 4, and see what you get for m/n. Then substitute these for the first two of the group of three equations above and proceed.

For non-prime reductions, there is generally a solution. Prime reductions require at least one gear with the prime number of teeth. For example, to obtain the 127/100 ratio needed to cut metric threads on an imperial lathe you must have a 127 tooth gear. (though 47/37 is pretty darned close)

[zut]

@friedo

Is there a formula or table or something that I can use to find a solution? Picking an even number of gears to make 12:1 is easy, but I need to pick gears with the exact correct diameter to get them to stack up right in order to make this device.

But there’s still an infinite number of solutions to this. You’d have to add in a few more requirements to get down to one exact solution.

In order to simplify, let’s add one more requirement: you want to do this the simplest way possible, which is exactly four gears: one on the driven shaft (of radius r[sub]1[/sub]), two on an idler shaft (of radii r[sub]2[/sub] and r[sub]3[/sub]), and the last on the driven shaft (of radius r[sub]4[/sub]). Gears 1 and 2 are meshed, as are gears 3 and 4.

Now, for those four radii, you have only two requirements. The first is that the gear ratio must be 12:
(r[sub]1[/sub]*r[sub]3[/sub])/(r[sub]2[/sub]*r[sub]4[/sub]) = 1/12
The second is that the centerline of the driving and driven shaft must be the same, which means the distance from the driving/driven shaft to the idler shaft must be the same:
r[sub]1[/sub] + r[sub]2[/sub] = r[sub]3[/sub] + r[sub]4[/sub]

That’s two equations and four unknowns. Can’t get there from here. [On preview I see Kevbo beat me to the punch, but let me keep going.] However, each gear reduction must be a rational fraction, to be able to set the number of teeth (in other words, you couldn’t have radii ratios of SQRT(12), for example).

Let’s arbitrarily choose speed ratios of 1:3 and 1:4. This eliminates two unknowns, because r[sub]2[/sub]/r[sub]1[/sub] = 3 and r[sub]4[/sub]/r[sub]3[/sub] = 4. However, it also eliminates an equation. Now let’s arbitrarily set a centerline distance–say 1-1/2 inches. Now the centerline equation breaks into two parts:
r[sub]1[/sub] + r[sub]2[/sub] = 1.5
r[sub]3[/sub] + r[sub]4[/sub] = 1.5

Substituting in, we get r[sub]1[/sub] = 0.5, r[sub]2[/sub] = 1.0, r[sub]3[/sub] = 0.375, and r[sub]4[/sub] = 1.125. Any meshing gears with stated diameters will work for your application. If you’re making them yourself, remember the pitch has to be the same on meshing gears, so that the ratio of the number of teeth is the same as the ratio of the diameters.

You can reset your required centerline distance and recalculate, or reset the speed ratios and recalculate.

[friedo]

Thanks, everyone! This is some really useful information.

Merry Christmas!