I wnat to make a bike with a engine. Now correct me if im wrong...

Im going to run a 8 tooth sprocket on the engine, going to a jackshaft with a 60 tooth on it, on the other end of the jackshaft, there will be another 8 tooth. that 8 tooth will be going to another 60 tooth on the crank of the bike. Then on the other side of the crank there will be a 28 tooth, going to a 28 tooth on my rear wheel.

So my calculations are the reduction intotal wil be 14:1

that will give me 428RPM at the crank of the bike, will that give me enough reduction for a 3HP 2 stroke?

8 to 60 is a reduction of 13.3. Two in series is 177.7 times. Assuming a redline of 4000RPM, that's 22RPM at the wheel, for a 40cm wheel that's about 30m/min, or 1mph.

You didn't submit in your calculation the RPM of the motor, but it is at a 56.25:1 ratio, not 14.5:1 . Round it off to 60 to 1 for the moment. In other words, it would take almost

60 Revolutions of the motor's 8 tooth sprocket to turn the bike crank one turn. Not 14.5. And you also didn't give the size of the bicycle wheel, which will also be necessary to determine speed, and also no weight's to determine Horsepower required, and no target speeds. The gearing from 28 to 28 is just an idler and creates no ratio change.

But assumptions on my part...... for a bike with 26" wheels (pretty standard size giving a circumference of 6.8 feet) to attain a speed of

10 Miles per Hour, the wheels have to make almost 13 RPM. So working backwards, your motor would have to run at 780 RPM (13 X 60). That's pretty slow for a small engine. Stick in one more 7.5:1 to get 420:1 (motor to wheels) and you would need 5800RPM at the motor, which would be a pretty good set-up I think.

But good luck anyway. Let us know what/where/when/how if you get it d>I wnat to make a bike with a engine. Now correct me if im wrong... >

Take out one stage of reduction, and you wind up with a somewhat more reasonable 6.68 mph, assuming that your redline is correct. With a really small engine, it could have a much higher redline, and might actually be more reasonable with the two stage reduction.

snipped-for-privacy@aol.comma (Dave Baker) wrote in news: snipped-for-privacy@mb-m14.aol.com:

Because this is rec.crafts.metalworking not sci.engr.mech. Of course, there is no way to know whether someone who answers on sci.engr.mech is qualified either although, presumably, more readers there are actual engineers and would jump on obliviously wrong answers more quickly and harder.

Ok... I want the bike to get around 20MPH, faster maybe, but for a base 20MPH. I have 26" wheels.

Now I would REALLY like to know how to calculate all of this. I was told in anopther post, that if i have a 10:1 ratio and another ration of 15:1 to just combine them, to get 25:1.

So what are the formulas to determine gear ratios, and speed?

I believe you said you had a 3 hp engine. That is plenty of power for 30 mph if you gear it right. If you will be riding it on moderately hilly roads, gearing it for your 20 mph target will be a happier combination. That's also a much better speed for a bicycle chassis and brakes.

To combine ratios of reductions in series, you multiply them.

Think it through like this: If the first stage has 12 teeth on the input and 60 teeth on the output, then the output turns 1 turn for every 5 turns of the input. A 5:1 ratio.

If the second stage is the same ratio, then the input to the second stage has to turn 5 times to get the second stage output to turn 1 time.

To make the input to the second stage to turn 5 times, you have to turn the input of the 1st stage 25 times. A 25:1 ratio.

Who needs a website? It's your basic "no-brainer":

The length of the gear-train can be extended to infinity (or the limit of practicality, whichever you hit first), yet the calculation remains the same: For a gear-train consisting of N gear ratios, TotalRatio = Ratio1 * Ratio2 * Ratio3 ... * RatioN

So, to use an example with ratio numbers that are pulled out of the clear blue sky, let's assume you've got a 15:1, followed by a 12:1, followed by a 2:1, then a series of four 3:1 gearsets. What's the total ratio? 15 * 12 * 2 * 3 * 3 * 3 * 3 = 29160:1 - It'll take almost 30K revs on the input end to get the last gear in the series to turn 1 revolution. And good luck trying to *STOP* that last gear when the input end is turning... The available torque on it is going to be nothing short of astronomical - So high that it's entirely possible that the gear-train won't be usable under load - Unless it's built *EXTREMELY* heavy-duty, it may very well literally rip itself to pieces, possibly generating quite a large amount of shrapnel in the process, within seconds after power is applied.

On 19 Oct 2003 16:23:03 -0700, snipped-for-privacy@yahoo.com (TJ Poseno) wrote something ......and in reply I say!:

OKIMAPRICK. No acronym intended.

IIRC, we have been here before, TJ. A lot of people tried to explain. You still have not got it right. It is very simple.

I assume you really want to try this. In this case, you REALLY need to grasp some very simple principles, and read what was already posted.

Each shaft on which the gears are commonly captive can only turn at one speed, so on each shaft the ratio is 1:1. Each gear ratio _between_ shafts should be _multiplied_ by the previous one. You need to state recvs and diameters to get speeds........

#1 : 60 / 8 = 7.5

shaft #1 cannot twist

#2 : 60 / 8 = 7.5

shaft #2 cannot twist

So far we have 7.5 * 7.5 = 56.25:1

NOT 14:1 (???) (or even 15 : 1 which would _add_ the reductions)

The 28 tooth to 28 tooth on your bike gives 1:1.

How fast does the 2-stroke run? How fast do you want to go? What final wheel size on the bike?

Listen to what you have been told. Answer the questions asked.

_Think_ about it.

Save the next toke for me.

Sorry.

****************************************************************************************** Whenever you have to prove to yourself that you are not something, you probably are.

Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!

snipped-for-privacy@yahoo.com (TJ Poseno) wrote in news: snipped-for-privacy@posting.google.com:

Just start with first principles. You have an 8 tooth on the motor driving a 60 tooth on the shaft (If I remember correctly). When your motor goes 1 revolution, it goes 8 teeth. If it goes 8 teeth, the driven gear must go 8 teeth. Since the driven gear has 60 teeth for 1 revolution, it has gone 8/60th of a revolution. 8/60 = .133333 of a revolution. 1 divided by .13333333 is 7.5 which is your reduction. Note, you can get to the final answer just by dividing 60/8 which = 7.5

Now on the other end of your shaft, you have another 8 tooth driving a 60 tooth. We've already done the math so we know that is a second reduction of 7.5.

Now, how do you add the two reductions. Well again start with 1 revolution of the motor, you have 8/60 revolutions of the shaft. You calculate the revolutions of the shaft by multiplying the number of revolutions of the motor by the gear ratio. In this case that is 1 times

8/60 (or 1 times .133333 or 1 divided by 7.5, they are all the same calculation). But the shaft has only gone 8/60th of a revolution and the final driven gear produces another 8/60th of a revolution for each 1 input revolution.

So in the shaft case you multiply 8/60 (the number of input revolutions just like above) by the gear ratio which is also 8/60, and that =

0.017777777. So for 1 revolution of the motor, you get 0.01777777777 revolutions at the driven gear. 1 divided by 0.017777777777 is 56.25 so your overall reduction is 56.25. Notice that 7.5 times 7.5 = 56.25.

So the answer to your question is you multiply the reductions. This is the power of transmissions (so to speak). It doesn't take very many gears to produce very large reductions.

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