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

Discussion in 'Friction Drive' started by landuse, Apr 24, 2012.

1. ### landuseMember

Hi there

I was wondering about the above. I had a read about this in the sticky in the friction drive section, but it still didn't make sense to me. This is what the sticky says:

"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?

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2. ### Cavi MikeMember

Only if the bike weighs more. Other than that, no, tire diameter means nothing. The bike can only move as fast as the surface speed of the roller.

*edit* I'll try to describe it a little bit better than him.

I assume you're thinking about how gearing works, right? Well what you've forgot to add into the equation is that when you change your gearing, you haven't changed tire size. Now, if you would have changed your tire size the same % as you've changed your gearing, the bikes speed will stay the same. That's how it works with friction-drives. The tire is the gear so every time you change the tire, the gearing changes with it, cancelling each other out.

Last edited: Apr 24, 2012
3. ### landuseMember

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.

4. ### Richard H.Member

What about it? It's a somewhat wordy and convoluted way of stating tire size doesn't matter.

5. ### landuseMember

I am trying to figure out if size does matter or not. On this forum there is a sticky which says that it doesn't matter, and other people are saying that it does.

I need a bit of clarification here.

6. ### Happy ValleyActive Member

For direct friction drive, the size of the tire will not effect the final gear ratio at all, only the size of the drive spindle does.

If you read this carefully Lou explains the physics clearly in the sticky with the following:

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7. ### landuseMember

Yup, that is the quote I used initially.

Last edited: Apr 26, 2012
8. ### Happy ValleyActive Member

Good, glad it helped.

9. ### HeadSmessWell-Known Member

make a bike with a 1" wheel and direct drive it from the engine

10. ### loquinActive Member

Here's another way to think about it:

Suppose you had a 1 inch drive roller, and two different wheels you're thinking about using. A 26 inch wheel, and a 13 inch wheel.

The 13 inch wheel would spin exactly twice as fast as the 26 inch wheel, BUT, since each tire rotation takes you half as far as the 26 inch wheel, it has to spin around twice to go the same distance... So, at a given engine RPM, the bike would travel at exactly the same speed with either wheel (all other things being equal)

So, long story short... the tire diameter has NOTHING to do with the speed that the bike will go, with a friction drive. The wheel is nothing except a transfer roller. The ONLY two factors in calculating the potential speed of a FD are (1)RPM and (2)roller diameter.

Last edited: Apr 26, 2012
11. ### DeathProofMember

lmfao its just a tire just slap it on and go! some people just over think to much and complicate things

12. ### Cavi MikeMember

How many of us need to tell you it doesn't matter before you finally stop looking for "clarification?" This isn't a matter of opinion but obviously you're treating it as such. I think you just need to go test it out yourself.

13. ### DeathProofMember

lol thats cavimike dont listen to him hes like the biggest negative in mbc he will cut down ppl like crazy here and is jus plain miserable ask anyone hes our oscar the grouch of mbc lol we jus close the lid and let him mumble to himself anyways dude jus put a tire on who cares if it has resistance does it really matter? a thick tire a thin tire who cares jus go ride have fun!

14. ### FabianWell-Known Member

Tyre size certainly does influence friction, hence the reason why high end road racing bikes use very narrow tyres, to minimise wind resistance.
The larger the sectional profile, the greater the wind resistance. At the high end of professional competition, any advantage that can be taken, will be taken.

Technically a larger diameter tyre will also reduce rolling resistance as axle to bearing speed is reduced.

All these frictional and heat losses add up, but it needs to be put in context of the goal that is trying to be achieved.

Last edited: Apr 27, 2012
15. ### GreybeardMember

With a friction drive, roller size alone determines the potental speed. A 1" roller will move the bike forward 3.14" per revolution. A 1.5" roller will move the bike forward 4.71" per engine revolution. Be it a small diameter tire or a large diameter, they both can only move forward by the circumference of the roller on the friction drive.

16. ### DeathProofMember

imagine a car wheel on a bike yes weight and friction matter!

17. ### loquinActive Member

Yes they do.

But, in the context of a BICYCLE, with a small engine mounted, the weight of the bike will fall in a relatively small range, and the tire size will also fall into a relatively small range, and within these ranges, the only truly appreciable determining factor in calculating the maximum theoretical speed of the bike/motor, is the roller diameter and engine speed.

True, you can obtain small improvements in the attainable bike speed, (which cannot exceed the maximum theoretical speed,) by fiddling with tire width, air pressure, rolling friction, etc.

What I am getting at is that the roller diameter and engine speed determine the maximum theoretical speed obtainable. Period. Any other factor simply reduces the obtainable speed of the bike.

From a practical viewpoint, you can fairly easily get to within an MPH or two of the max theoretical speed using standard, relatively inexpensive equipment. (I got a full MPH by replacing the stock tire with a slick, low rolling friction tire, for instance.) Or, you could end up spending thousands to get that final MPH or so. At some point, there is a point of vanishing returns, where you spend more and more to get less and less.

18. ### FabianWell-Known Member

Have to agree with you Lou.

At some point the cost of improving engine performance becomes less costly than reducing rolling resistance or aerodynamic drag, unless you are competing in a control category.

Last edited: May 1, 2012
19. ### loquinActive Member

Back to the OP (landuse) questions. The second reference from the other site tries to equate a "gear reduction ratio" to friction drives... but, that approach is meaningless, in regard to friction drives. As the other person mentions, changing the tire size cancels out the 'gear ratio' change.

The other site poster said:
For a friction drive system, Surface Speed is ALL that matters, and ratio is meaningless. The surface speed is the speed of a point on the circumference (outer surface) of the drive roller. Assuming that there is no slipping between drive roller and tire, or between tire and road:

The surface speed of the spinning roller is EXACTLY the same speed as the surface speed of the spinning tire, NO MATTER THE DIAMETER, because the roller and the tire are in constant contact.

Likewise, the surface speed of the spinning tire is EXACTLY the same as the surface speed of the tire against the asphalt, (or, the road moving past the tire,) because THEY are in contact.

In a system with no slip, the surface speed of the roller is therefore IDENTICAL to the speed of the bike, no matter the tire diameter, and to calculate the maximum theoretical bike speed, you only need engine RPM and roller diameter.

Even though the RPM of the wheel is much less than the RPM of the drive roller, (By the ratio of Drive roller diameter divided by tire diameter) the linear speed of a point on the surface of BOTH is the same. It HAS TO BE if there's no slipping. They are in contact.

Assuming a 1 inch roller and a 26 inch wheel, the roller RPM is 26 times the wheel RPM. Your roller would HAVE to spin 26 times for the bike to travel 81.64 inches (pi * 26 inch)

If you cut the wheel diameter in half, the roller RPM would now only be 13 times faster than the wheel RPM. But, since each wheel rotation will only take you half as far (40.82 inches,) as with the 26 inch wheel, so your wheel would have to spin twice, and the drive roller would STILL need to spin 26 times (13 * 2 times) to make the bike move the same 81.64 inches as with one rotation of a 26 inch tire.

Remember, with a chain/belt drive, all the power is directed to the drive AXLE, to make it spin at a given RPM. In that case, you NEED to use the wheel diameter to calculate the final speed.

But, with a friction drive, all the power is directed against the outer surface of the wheel, bypassing the axle entirely (From a power transfer point of view.) Since the power is already applied AT the tire surface, the tire diameter is irrelevant.

Now, you COULD go through the RPM calculations at each point, but, it's POINTLESS. Because the RPM of the spinning tire varies in direct relationship to the tire diameter, AND the distance traveled with each rotation of the tire also varies in direct relationship to tire diameter, any change introduced by a tire diameter change is exactly and completely canceled out.

Last edited: May 2, 2012
20. ### landuseMember

Thanks everyone. I appreciate the help