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Many amateur metalworkers seem to be confused about how one goes about calculating gear ratios. It's a sufficiently recurrent source of questions from my website that I wrote the following article in an attempt to help people understand the underlying 'theory'.
We can sum up almost all of gear ratio theory into one simple relationship that's worth memorizing...
IF TWO GEARS ARE IN MESH, THEN THE PRODUCT OF SPEED AND TEETH IS CONSERVED.
Let's put this in terms of usable math. Let's say that we have two gears in mesh. Gear 1 (we'll call it the driver) is turning at speed S1 rpm and has T1 teeth. Gear 2 (the driven gear) is turning at speed S2 and has T2 teeth.
The number of teeth per minute arriving at the mesh point of the gears must be equal. [If this isn't true, you're going to have a pile of gear teeth on the floor.]
For gear 1 the number of teeth per minute arriving at the mesh is S1*T1 and for gear 2 it's S2*T2
Then our relationship above says that:
S1 * T1 = S2 * T2
We can use this simple equation to solve for whatever (single) value is unknown. Consider a simple example where:
S1 = 100 rpm
T1 = 30 teeth
S2 = ?
T2 = 40 teeth
Solving the equation above for S2, we have:
S2 = (T1/T2) * S1 = (30/40) * 100 = 75 rpm
Let's add a third gear to the train. Assume gear 2 drives gear 3 and gear 3 has T3 = 50 teeth. What's the speed of gear 3? Well, since gears 2 and 3 are in mesh, our conservation law says that:
S2 * T2 = S3 * T3
We could do the arithmetic (S3 = (T2/T3) * S2 = (40/50) * 75 = 60 rpm) to find S3. Or, we could note that, since both S1*T1 and S3*T3 are equal to S2*T2, they must be equal to each other.
S1 * T1 = S3 * T3
so,
S3 = (T1/T3) * S1 = (30/50) * 100 = 60 rpm.
What we've just proved mathematically is what every true gearhead knows already...
AN IDLER GEAR BETWEEN A DRIVER AND DRIVEN GEAR HAS NO EFFECT ON THE OVERALL GEAR RATIO, REGARDLESS OF HOW MANY TEETH IT HAS.
(Note that T2 never entered into our computation in the last equation.)
Suppose now that we add a fourth gear with T4 = 60 teeth to our developing gear train. Its speed must be S4 = (T3/T4) * S3 = (50/60) * 60 = 50 rpm. But again, by use of the conservation principle, we have:
S4 = (T1/T4) * S1 = (30/60) * 100 = 50 rpm.
I can continue like this indefinitely but by now you should get the idea...
NO MATTER HOW MANY GEARS ARE BETWEEN THE DRIVER AND (FINAL) DRIVE GEAR, THE OVERALL RATIO DEPENDS ONLY ON THE TOOTH COUNT OF THE DRIVER AND FINAL DRIVEN GEAR.
(Put another way, all the intervening gears are idlers and don't contribute to the overall ratio.)
In fact, if we interchange gears 2 and 3, so that:
S1 = 100
T1 = 30
T2 = 50
T3 = 40
T4 = 60
Then:
S2 = 30/50 * 100 = 60
S3 = 50/40 * 60 = 75
S4 = 40/60 * 75 = 50
the same result we got from:
S4 = (T1/T4) * S1 = (30/60) * 100 = 50 rpm.
*** MAJOR CAVEAT ***
Note that everything said to this point assumes that each of the gears in the gear train is on its own, separate shaft. Sometimes gears are 'ganged' by keying or otherwise welding them together and both gears turn as a unit on the same shaft. This complicates the computation of the gear ratio, but not horribly. Suppose gears 2 and 3 are keyed together into a single compound gear we'll designate g (g for ganged). Assuming S1 and S2 are in mesh, it's still true that:
S1 * T1 = Sg * T2
Sg = (T1/T2) * S1
If gears 3 and 4 are in mesh,
Sg * T3 = S4 * T4
(Remember, S3 turns at the same speed as S2 because they're physically joined and we're calling their shared speed Sg.)
Therefore,
S4 = (T3/T4) * Sg = (T3/T4)*(T1/T2) * S1
So the end-to-end gear ratio is (T1*T3)/(T2*T4) and it *does* depend on the intermediate gears, unlike the previous case when each gear could turn on its own separate axis. Note that the resultant gear ratio is just the product of the two separate gear ratios - (T1/T2)*(T3/T4).
One of the most common places where the hobbyist must grapple with gear ratios is the case of a lathe that uses change gears to cut various threads (as opposed to a lathe equipped with a quick change gear box). In this situation, one has to be concerned about the pitch of the leadscrew as well, since this enters into the calculation.
The easiest way to think about the problem is in terms of carriage motion. If my lathe has a Lead Screw Pitch (LSP) of L tpi, then if the spindle is connected to the lead screw with 1:1 gearing, one revolution of the spindle will move the carriage 1/L inches. Let's assume I want to cut a thread of pitch D tpi ('D' for Desired). Therefore I want the carriage to move 1/D inches for each revolution of the spindle. Clearly, I need a gear ratio (R) determined by:
1/L in/lsrot * R lsrot/srot = 1/D in/srot
where lsrot = leadscrew rotation, srot = spindle rotation.
or:
R = L/D lsrot/srot
For each spindle rotation, the leadscrew has to make L/D rotations. As a simple example, consider cutting D = 16 tpi on a lathe with a L = 8 tpi leadscrew. For each spindle rotation, the carriage must move 1/16 inch. To do this it must turn R = L/D = 8/16 = 1/2 as fast as the spindle so a gear ratio of 1/2 is needed.
It was easy to calculate the resulting gear ratio when we knew which gears were involved but here we have the opposite problem. We've calculated the ratio we need (R) but now need to know which gears to use in what configuration to obtain that ratio. Normally, the gear setups for common thread pitches are covered in the documentation that comes with the lathe. But, when faced with making a replacement 5/16 BSF bolt for a piece of British tooling, you suddenly need to cut (D =) 22 tpi and it isn't shown on the index.
R = 8/22 = 0.363636...
Guessing the gear setup to obtain that ratio may not be easy. More practically, you'd like to make the setup from the gears you have available and not have to buy/make new gears everytime you need to cut some oddball thread. Fortunately, the CHANGE program on my page will do the dirty work for you. (Many combinations are possible for 8/22 but the simplest is a 20 tooth gear driving a 55 tooth gear.)
Compound (i.e., ganged) gears are used often in setting up change gear lathes because they offer the opportunity to achieve a much larger selection of ratios than would be possible with only simple gearing. In simple gearing, as we saw, the ratio depends only on the driver and driven gear - any gears between these can't affect the overall ratio. To obtain a given ratio, you would need to have two gears, each with precisely the right number of teeth to establish the ratio. By compounding, a much smaller set of gears can be used to achieve a large number of ratios.
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