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?

Err, no.
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.

Hey TJ,
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
done.
Brian Lawson,
Bothwell, Ontario.
XXXXXXXXXXXXXXXXXXX

?????
How about 7.5:1 ?
210:1
19.02 RPM
23.91 Meters/min, or about 0.891 mph
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.
Enjoy,
DoN.

You multiply the successive reductions, not add them.
8/60 is more like 7.66.
--
Bob May
Losing weight is easy! If you ever want to lose weight, eat and drink less.
Works evevery time it is tried!

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?
Is there a website that can explain this?

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.

In North America, football is cool math is not. The result is a
basically inumerate population. Pity. It lets politicians get away
with a lot of lies.
Ted

[ ... ]
And *I* managed to screw up the next stage of the calculation
(and all that followed it) while wondering how you got 13.3:1. :-)
And I *still* don't know how I came up with the figure that I
did. :-)
Enjoy,
DoN.

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.