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32 cc speed

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matman55

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Hey whats up,i need some help calculating how fasti can possible go on a 32 cc weed eater engine.my bike weighs about 40 pounds and im 230 i heard somewere 1 cc can pull 10 pund at 3 miles and hour?im not really math inclined so i need some help on this.:roflol:


:censored::oops:
 

landuse

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It has everything to do about gearing. You can have a 100HP engine, and it will only go 3MPH is geared to go that speed. You will have to give us some more info like sprocket tooth counts, tyre diameter, etc
 

cumminsbayou4x4

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depends on what set up if your looking for speed dont use a frictoin drive set up but if you want speed get a chain saw that will make a world of differece it woundnt be bogging down you could make it up hills to the fastest a weed eater would go i coundnt say cause if you gear it to high it will bogg down gear it to low it wont go but like 5
 

matman55

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ok,weel,its friction drive so its not the most efficient thing in the world,and the tire is 26 inch,hopethat helps
 

augidog

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what's "fast" ?

25-30mph is plenty fast on bicycle wheels & hubs, and that's about all you'll get with that 32cc & friction drive.

btw-if you're planning on riding on-the-road, you should know florida's getting tough on gas-powered bicycles...there's been too much abuse of the limits, and states everywhere are tightening up because of it.
 

matman55

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oh,i thought when you said i wont be going fast at all i thought you ment like limited to 10 mph, but if im going 25-30 on a beach bike im good!i am a bit of a speed demon though.ad fast for me on a bike is 50-80
 

landuse

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I disagree. It all factors together.

"With all other drive types, tire size does matter. Here's why tire size doesn't matter with a friction drive: The speed of the tire rotation in RPM is related to bike speed, but, it is really the speed of the circumference of the tire, in miles per hour, which is directly related to bike speed. After all, the tire is in physical contact with the road, and if there is no slippage, the tire circumference speed and the bike speed HAVE to be equal. Likewise, if there is no slippage, the tire circumference speed and the roller circumference speed MUST be equal, because THEY are in direct contact. Think of it this way - Essentially, the tire is just a transfer roller (or idler wheel,) between the drive roller and the road... A smaller tire would spin faster than a larger tire, but, since the circumference of the smaller tire is proportionally less, (by exactly the same ratio as the tire diameter, and the RPM increase,) there is no difference in bike speed."

This quote was taken from the www.motoredbikes.com sticky on friction drives found at http://www.motoredbikes.com/showthread.php?t=23779. It is very helpful and will answer all the questions that you have regarding friction drives
 

Doc Sprocket

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I think I'm having one of those stupid moments, because I'm not getting this. In my mind, this is akin to a vehicle that has a rear sprocket equal to the tire size. In such a case, that size still factors in to overall vehicle speed, as WRPM still equates to overall vehicle speed as the circumference ultimately translates into a set distance per rotation. Therefore, a larger wheel (or sprocket) travels further per rotation than a smaller wheel would at the same WRPM... No? Forgive me- I'm not being belligerent- I'm just not understanding your argument.
 

brandongeiger2

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I do think landuse is right if you have a 26 inch tire and a 3 inch roller it would take more turns of the roller to make a complete rotation if you are using a 20 inch tire those rotations would be less correct?? but if you had a 5 inch roller it would move more of the tire in a complete rotation the tire size makes no difference it just transfers motion?
 

landuse

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I think I'm having one of those stupid moments, because I'm not getting this. In my mind, this is akin to a vehicle that has a rear sprocket equal to the tire size. In such a case, that size still factors in to overall vehicle speed, as WRPM still equates to overall vehicle speed as the circumference ultimately translates into a set distance per rotation. Therefore, a larger wheel (or sprocket) travels further per rotation than a smaller wheel would at the same WRPM... No? Forgive me- I'm not being belligerent- I'm just not understanding your argument.

You could also think about it this way. If you put the roller right on the ground, it would travel x amount of distance in x amount of time at x rpm. Put the roller against a tire, and the surface of the tire will travel the same distance with the same numbers, no matter the size of the tire.

If the roller on the ground moves 10 inches in one revolution, it will also move the tire surface 10 inches in one revolution.
 

landuse

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whether chain, belt, or friction, (assuming the same tire in all cases) the overal reduction-ratio from engine-crank to rear-hub is all that matters...

the thing that can confuse, until you look more closely, is that with friction drive the power is transferred along the tire's circumference, so the equation appears different in our mind.

but simply put, the math's exactly the same...with friction-drive, tire-diameter factors in twice...once as the "rear sprocket" and once as "tire diameter."

and while the roller's surface-speed IS the same as the vehicle's surface-speed, that fact's irrelevant to the math that's used to calculate ratio and speed.

with a small 2-stroke and a single-speed, the best bet is to pursue it's potential for high-rpm's...to a certain point, a lower gear will result in a faster vehicle than if you ask the engine to power it's way through a higher gear. it's my experience that a 32cc 2-stroke can pull a 26" bicycle best at about 20-22:1 overall-reduction. with a friction-drive, that very-roughly translates to a 1 1/8" to 1 1/4" roller.

i'll say it again, florida's getting tough on (on-road) gas-bicycles, so research is a good idea:
Gas Motorized Bicycles Illegal in Florida?

I think my head is getting sore. :ack2: I am going to have another good read before I can comment again
 
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