By

Could anyone please advise me the speed ratio of crank shaft to generator? Thanks in advance.

John (UK)

By

By Jim Thode Chehalis Washington on Saturday, March 30, 2013 - 08:00 pm:

Dean,

The crank has 24 teeth and the generator gear has 16 teeth. So the generator turns 1.5 (24/16) times as fast as the motor. If the motor is turning at 1000 rpm the the generator is turning at 1500 rpm.

Jim

By

Thank you Jim.

By

Doug, What role does the cam gear play in working out the gen. speed?

Allan from down under.

By

the generator run off the cam gear it turns 1/2 the speed of the crank adding that information to Doug's calculations the generator would run 750 rpm.

By

I don't think so, let's compute this:

Say the 24 tooth crank gear is turning 1000 rpm, then the 48 tooth cam gear is turning 500 rpm.

1000 x 24 / 48 = 500

Now, with the 48 tooth cam gear turning 500 rpm, the 16 tooth generator gear is turning:

500 x 48 / 16 = 1500 rpm.

So the original answer was correct.

By

Doug is correct. In any three gear setup such as this, the cam gear only serves as a idler (so to speak) and does not contribute to the overhaul gear ratio. The crank gear serves as the drive gear and the generator gear is the driven gear. Also, the cam gear turns the opposite direction while the generator gear turns the same direction as the crank. to do the math, divide 24 by 16 and the answer if 1.5 to one.

By

I concur. It doesn't matter how many gears are in between, only the first and last come into play.

By

I don't think that's right, Hal.

By

Mike,

Think about it, Hal is correct. The number or size of intermediate gears have no effect other then changing the direction. An odd number of intermediate gears and the first and last will turn the same direction. An even number (or zero) of intermediate gears and the first and last will turn in opposite directions.

For an intermediate gear to effect the speed it would have to be a combination gear with a large and small gear on the same shaft.

Jim

By

As long as the gears in between don't have variable number of teeth.

By

Kerry,

It does not matter. You could have all big or all small or any combination of intermediate gears between the 1st and last. And still the ratio of speed of the first to last is controlled only by the ratio of teeth in the first and last.

Jim

By

Then my lathe is in trouble!! I've only got one speed??

By

You mean your lathe doesnt have a change gear set?

By

Guys, the rule doesn't apply if there are two gears on the same shaft. But if it is just single gears, one turning the next, so on down the line, then the final ratio is the ratio of the first and last. A set of gears can be modeled as a set of friction wheels with diameters the same as the pitch diameter of the gears. Assume no slippage (Pretty good assumption on gears). While various diameter friction wheels (Or gears) will turn at various rpm's, the surface speed of their outer diameters is constant. No matter what the size, the outer diameter is moving the same speed. (Or in the case of gears, same number of teeth per minute) Bigger ones turn slower in rpm. Smaller ones turn faster in rpm, but their outer diameters are all moving the same speed. (Or same number of teeth per minute) Therefore, the final ratio of the entire set is nothing more than the ratio of the diameters of the first and final friction wheel (Or the ratio of the number of teeth of first and last gear).

It's just like Mark said above. 24 tooth crank gear running 1000 rpm turning a 48 tooth cam gear, makes it turn 500 rpm. 48/24 is a 2:1 ratio. 1000/2 is 500. The 500 rpm 48 tooth cam gear turns the 16 tooth generator gear at 1500 rpm. 48/16 is a 3:1 ratio. 500 x 3 is 1500.

Now let's take the cam out of it. crank to generator ratio is 24/16 or 1.5:1. 1000 x 1.5 is 1500.

Same either way.

By

Hal -- It looks as if you were correct all along. It works with only one gear between the two, but it didn't seem to me that it would work with more than one until I tried this: If the 24-tooth crank gear is turning at 1,000 rpm, the 48-tooth cam gear turns at 500. Check. If that gear turns a 12-tooth idler gear, that shaft will be turning 2,000. If that 12-tooth gear turns the 16-tooth generator gear, the gen. shaft would turn 12/16 or 3/4 of 2,000, which would be 1,500. Amazing. I never knew that. Thanks.

As I've said before, I learn something new here every day.

By

I was waiting for someone to mention the cam gear, :>)

It's true that the intermediate gears do not affect the final ratio, but only in the case of a simple gear train, where one gear drives another, and that gear drives the next, and so on. But, more to Kerry's point, in more complex gear trains, the intermediate gears DO affect the final drive ratio, as in a manual transmission.

Look at the sketch below. The lower right is a simple gear train. The upper left is double reduction gear train that does not follow the "first & last" final drive ratio formula. Each gear train uses the same gears.

By

I see Hal basically described the same thing in his posting. It would pay to read all the posts before I reply :>)

By

Another way to wrap ones head around this is to think about a serpentine belt set up. The belt is running at a constant velocity (number of feet/seconds) regardless of how many or what diameter pulleys are in the loop. Since the belt velocity is determined by the surface speed of the drive pulley, the rotational speed of any other pulley in the system is a ratio of the drive pulley diameter vs the driven pulley diameter. The diameters of any other pulleys in the system is completely irrelevant.

By

Jim, yes you right but only if every idler gear in between after the first ratio is set, is the same number of teeth, I said variable #'s

By

I'm confused, Kerry. I'm pretty sure you are not saying there are gears with a variable number of teeth. Are you saying that for the formula to work, all intermediate idlers must have the same number of teeth? That is not the case. As long as we are talking about a simple gear train like pictured in the lower half of Jerry's sketch above, you could have 100 intermediate idlers with 100 different tooth counts and the final ratio is the # of teeth of the first gear divided by the # of teeth of the last gear.

By

Thanks Hal, my gears are sometimes a little slow, it's been a long time since I left school!!