Last time we saw how a gear reduction does as its name implies, reduces the speed of the driven gear with respect to the driving gear within a gear train. Today we’ll see how to work the problem in reverse, so to speak, by determining how many teeth a driven gear must have within a given gear train to operate at a particular desired revolutions per minute (RPM). For our example we’ll use a gear train whose driving gear has 18 teeth. It’s mounted on an alternating current (AC) motor turning at 3600 (RPM). The equipment it’s attached to requires a speed of 1800 RPM to operate correctly. What number of teeth must the driven gear have in order to pull this off? If you’ve identified this to be a word problem, you’re correct. Let’s first review the gear ratio formulas introduced in my previous two articles:
n (1)_{Driven }
N (2)_{Driving }Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required:
n3600 RPM ÷ 1800 RPM = 2_{Driven} = This equation tells us that to reduce the speed of the 3600 RPM motor to the required 1800 RPM, we need a gear train with a gear ratio of 2:1. Stated another way, for every two revolutions of the driving gear, we must have one revolution of the driven gear. Now that we know the required gear ratio,
N_{Driving}2 =
The driven gear requires 36 teeth to allow the gear train to operate equipment properly, that is to say, enable the gear train it’s attached to provide a speed reduction of 1800 RPM, down from the 3600 RPM that is being put out from the driving gear. But gear ratio isn’t just about changing speeds of the driven gear relative to the driving gear. Next time we’ll see how it works together with the concept of
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## Posts Tagged ‘gear reduction’

### Gear Reduction Worked Backwards

Sunday, March 9th, 2014### Gear Reduction

Wednesday, March 5th, 2014
Last time we learned there are two formulas used to calculate gear ratio, If you’ll recall from last time, our formulas to determine gear ratio are:
N (1)_{Driving }
n (2)_{Driven }Now let’s apply them to this example gear train to see how a gear reduction works. Here we have a driven gear with 23 teeth, while the driving gear has 18. For our example the electric motor connected to the driving gear causes it to turn at a speed, n._{Driven}First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other:
N_{Driving}
In gear design nomenclature, the gear train is said to have a Interestingly, the Since we have already determined that the
n_{Driven} Now all we need is one more numerical value to solve Formula (2)’s equation. We know that the speed at which the driving gear is rotating, n:_{Driven }1.27 = 3600 RPM ÷
Based on our calculations, the driven gear is turning at a speed that is slower than the driving gear. To determine exactly how much slower we’ll calculate the difference between their speeds:
– n= 3600 RPM – 2834.65 RPM ≈ 765 RPM_{Driven} So in this gear reduction the driven gear turns approximately 765 RPM slower than the driving gear. Next time we’ll apply a gear reduction to a gear train and see how to arrive at a particular desired output speed. _______________________________________ |