Whinning Tire Noise From Friction Drives

Those are close in that they have a flat profile but, the ones I remember (on the rich kid's bikes, not mine) had no tread on them at all...a real slick. Now, in today's sue happy society, maybe you can't offer a real slick because some kid might ride in the rain and get hurt. Is my memory correct about the one's in the 60's being real slicks? I could easily be wrong about this...what do you guys around my age (51) remember?

This flat profile, in my opinion, would work better with the friction drive units out there. Either that, or they need to make a drive roller that has the same radius profile of the tire being used. Either way you get more contact area with less down pressure needed to drive without slippage.

Bill
 
I remember them all to well. They were completely SLICK. No tread what so ever. Not worth a flip on wet ground of any kind, nor loose debris such as sand, gravel or loose dirt. They were great for slide outs though and leaving black marks on concrete drive ways and side walks ! Hee ! Hee !
 
Bill - I'm not aware of any flat slick 26 inch tires. :(

But... you do NOT want a roller that has a concave surface to 'match' the radius of the tire. At all.

Here's the problem. If you have a roller that is 1.5 inch at the widest part, and 1.25 inch at the narrowest part, and it's placed in contact with the tire:
the 1.5 inch portion is trying to make the tire rotate faster than the 1.25 inch portion of the roller at the same engine RPM ... about 20 percent faster!!!

So, either the 1.25 inch part of the roller will be grinding away at the crown of the tire, or the 1.5 inch portion will be grinding away towards the sidewalls of the tire, or BOTH will be grinding away at the tire.

In any case, your tire will be ground away soon, AND you'll be wasting a huge portion of your available power. Not a good situation.
 
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all other factors being equal, slick tires are the best on just about any kind of surface, wet or dry, for traction. They cannot hydroplane on a bicycle (unless you're running them at 15 psi or so.) On a BICYCLE, tread is NOT needed to 'channel water.' About the only time you would want tread is with snow (because as it compacts the snow, some will be forced into the tread, providing more traction.)
 
Bill - I'm not aware of any flat slick 26 inch tires. :(

But... you do NOT want a roller that has a concave surface to 'match' the radius of the tire. At all.

Here's the problem. If you have a roller that is 1.5 inch at the widest part, and 1.25 inch at the narrowest part, and it's placed in contact with the tire:
the 1.5 inch portion is trying to make the tire rotate faster than the 1.25 inch portion of the roller at the same engine RPM ... about 20 percent faster!!!

So, either the 1.25 inch part of the roller will be grinding away at the crown of the tire, or the 1.5 inch portion will be grinding away towards the sidewalls of the tire, or BOTH will be grinding away at the tire.

In any case, your tire will be ground away soon, AND you'll be wasting a huge portion of your available power. Not a good situation.

I see what you are saying but..I respectfully disagree. You are correct about everything you said except the part about them wanting to go different speeds. The tangent point of the largest diameter of the tire will be in contact with the smallest diameter of the drive roller, in your example, 1.250". The largest diameter of the drive roller, in your example 1.5" will be in contact with the area along the radius of the tire closer to the center point of rotation thereby making a smaller diameter tire. (effectively)

I have not run the math on this but I think you will find that the reduced effective diameter of the rotating tire will be exactly compensated (equilibrium) by the larger diameter drive roller and will then be running at the exact same speed.

As I see it, the real fly in the ointment here would be the unknown radius the compression of the tire, which we would need some for traction, that would be formed. In other words, it may not be possible to make it match exactly because you might push yours down just a little harder than me, or visa versa and I believe this would then alter the required radius.

All in all it would probably not be worth the effort. And, you are right, if there is not a very good match with the cross-sectional radius of the tire, it would chew that tire up pretty fast.

Bill
 
No, it won't be, Bill.

With a friction drive, the diameter of the tire has ZERO to do with the final speed of the bike. It is essentially a transfer roller, between the friction drive roller and the ground... You could have a 1 inch diameter tire, or a 100 inch tire. The 1 inch diameter tire would be spinning at a furious rate, while the 100 inch diameter tire would be spinning at 1/100th the speed, but, the GROUND speed of the two would be identical. Speed calcs for friction drive systems depend upon the roller diameter and the RPM. And, that's all.
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Suppose you have a roller that's concave. (saddle-shaped) And, it's an extreme case, with the narrowest part at 1 inch diameter, and the largest diameter of the 'saddle' is 1.5 inches.

You're running at 7000 RPM.

The roller is spinning at the same RPM for both the two points on the roller, correct?

This means that the linear (tangential) speed of the roller at the narrow point is
7000Rev/Min * 1 inch * 3.14 * 60 Min/Hr * 1/5280 ft/Mile * 1/12 Inch/Ft = 20.8144 MPH

And, for the wide part of the roller, at exactly the same RPM...
7000Rev/Min * 1.5 inch * 3.14 * 60 Min/Hr * 1/5280 ft/Mile * 1/12 Inch/Ft = 31.2216 MPH
 
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If you take this argument to the extreme, assume a concave roller that's 1 inch in diameter at the thin part, and 26 inches in diameter at the wide part ... contacting a wheel that's 26 inches in diameter. And, 7000 RPM on the friction roller.

The narrowest part of the roller is in contact with the outer surface of the tire, whereas the widest part of the roller would be in contact with the 'tire' 1 inch away from the hub center.


The outer part of the tire would be spun at a linear speed of 20.8144 MPH. (and, at 269 RPM)

But, the wide part of this roller would have a linear speed of
7000Rev/Min * 26 inch * 3.14 * 60 Min/Hr * 1/5280 ft/Mile * 1/12 Inch/Ft = 541.17 MPH !!!

So, assuming that a roller that's 26 inches in diameter, spinning at 7000 RPM, is trying to transfer that speed to a portion of the tire that's 1 inch in diameter. This means that the tire is going to be forced to spin at 26 times this RPM, or 182000 RPM.

But, it's obvious that the tire can't be forced to spin at two different RPMs at the same point in time.



I ignored the difference in the earlier discussion, as the difference in linear speeds between 1.5 inch and 1 inch rollers is 50%, but the reduction in tire diameter from 26 inches to 25.5 inches is only 1.9% less. The 50% difference in speed 'trumps' the 1.9% difference in speed.

If the roller and the tire had the same diameters, and the narrow part of the roller had the better friction... The outer part of the tire would be spinning at the same tangential speed as the roller, but the outer part of the roller would be rubbing against a part of the tire that CAN'T move (radially in relation to the roller,) as it would be the hub...

The picture below is a gearing analogy...
 

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Lou:

You make some really good arguments here and I will have to go back over your examples and think about this a while.

I agree 100% that if your bike and mine are identical, same engines, rollers, rpms, etc. except yours has a 26" wheel and mine has a 20" wheel that we will still go the same speed. (This alone dispels my prior argument)

The information and experience I was drawing upon was my 20 years in the precision machining industry. One of the many machines my company owned was an ID grinder that had a radius on the drive motor roller (It was a large radius maybe 8") but the roller was only 1" in diameter. It used a flat belt that drove a flat pulley for the grinding tool and it would spin at over 12,000 rpm. So, here we have a radius shape on a flat surface (the belt, 2" wide) and also a flat surface on the driven pulley and, that belt did not wear out or slip in any way.

I am having trouble reconciling this personal knowledge of mine with your statements, all of which I agree with at this point. I will give it some more thought.

It may all be a moot point in that, due to the flexible nature of the tire, the drive roller "sees" the tire as a mating flat surface anyway. In other words, if the tire were solid steel and still had the same shape, we would then have very little contact area between the drive roller and the "tire". But since the tire flattens out, as it does on the road surface, my idea for the radius matching roller amounts to nothing.

I appreciate your responses to my posts and your explanations which I will re-read. My brain does not work as good as it used to, ha ha, but I am really getting into the MB stuff and I love it.

Take care and thanks,

Bill
 
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