Does tyre size (diameter) influence friction drive speed/gearing

landuse

landuse

New Member
Hi there

I was wondering about the above.I read on another forum that the tyre diameter doesn't matter. This is what the forum said:

"Wait a Minute! Doesn't tire size play a factor in calculating the top speed? With friction drives, No, it does not. 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."

Surely the tyre diameter has something to do with gearing using friction drive, and not only the roller diameter? Could someone explain this to me?
 
Think about it. 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.
 
Thanks wayne, I think i am getting my head around all this. It is quite confusing for me who comes from a chain and sprocket background
 
What about this statement that I got from someone (who knows a lot) off another forum:

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

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 "rear sprocket size" and once as "tire diameter." obviously, the roller is the "front sprocket." on the calculator, use inches in diameter (of roller & tire) as number of teeth, move the decimal point (the same number of places for both values) to use whole numbers if necessary.

note: a 26" bicycle rim is 22", add the tire size (1.75, 2.0, 2.25, etc) twice to get actual tire-diameter.

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" (x 2.0") bicycle best at about 20-22:1 overall-reduction. with a friction-drive, that very-roughly translates to a 1.125" to 1.250" roller.
 
It's basically saying the same thing.. the ratio is 1:1 so doesn't matter. if you are doing direct friction drive just multiply the roller circumference X RPM and convert to your desired unit of speed (mph, kpm, feet per second, etc).
 
I have a 32cc Robin on a 26 inch bike with a dax friction drive and a one inch roller. I find the combination perfect for the robin engine. Very nice.
 
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