Friction drive gear/roller ratios

Happy Valley

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Help me understand something. For example, would it be correct to say the ratio is 26/1 with a 26" bike wheel and a 1" friction roller?

For some reason, with a friction drive set-up I'm having a hard time getting my head around this issue. I was talking with Dave Staton about a friction drive set-up with the Robin EHO35 and asked him which was harder on the engine, smaller or larger rollers. He said the larger the roller the more it worked the engine.

I'm trying to understand the torque/hp elements for this set-up and the variables like roller size that will effect the way I use it.

Thanks for any insights.
 
Hi Happy,

Discounting slippage, bicycle A has a 1" roller, which spins a 26" tire at a 26:1 ratio. Bicycle B has a 2" roller/26" tire with a 13:1 ratio. What Dave meant is that bicycle B would have less mechanical advantage and has to work harder to move the bike at every speed.

Say that an engine has peak torque at 4500 rpm. That's where it feels most powerful. At 26:1 ratio, Bike A feels this power at 13 mph. At 13 mph, Bike B spins at 2250 rpm, or idle speed. However, it would not have the low end power to reach 13 mph, much like a car being in 4th gear. You would have to pedal faster than 13mph before Bike B could weakly push itself forward.

At 9000 rpm, Bike A is screaming at 25 mph, while Bike B is at peak torque at 4500 rpm. At 9,000 rpm, Bike B will be screaming at 50 mph.

So the larger diameter the roller, the more the engine has to struggle at low speed, but the bike would be faster, but not quicker to 50 mph. The smaller spindle will peak at a much lower speed, and will feel perkier until it begs for mercy at 25 mph.

hope this helps.

5-7
 
5-7

Thanks so much.....again.

I couldn't have hoped for a better or more thorough explantion. I think I got it.
 
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One interesting point is that a friction drive is not affected by the size of the wheel. Essentially the friction drive is moving at the speed of the size of the roller as if it were being used directly to the ground. In other words if you had it behind the bike and the roller were touching the ground pushing the bike the diameter would dictate the speed of the bike, the larger the roller the more ground it would cover in each rotation. The way to change the ratio of wheel is to move into more the center, to a smaller diameter and take advantage of the outside being larger making a longer distance in each revolution. This is a little long but sometimes it is a little hard to see that this is real time motion the tire moves at the ground directly in proportion to the circumference of the roller. Have fun, Dave
 
Thanks Egor, good analogy, I'm finally understanding, I think, what I want to know.

0.7/8" diameter = 2.748" circumference
1. 00" diameter = 3.141" circumference
1 1/8" diameter = 3.534" circumference
1 1/4" diameter = 3.926" circumference
1 3/8" diameter = 4.319" circumference
1 1/2" diameter = 4.712" circumference

1/8" increments in roller diameter equals a touch over 3/8" increments in circumfernce.

The 5/8" difference in diameter, for example, between the 7/8" and the 1 1/2" rollers is almost a 2" spread in circumference! Multiply that times rpm.
 
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Dave Station makes it all pretty simple when someone places an order. He asked me which roller I wanted -- I didn't have a clue at the time. He asked my weight (195 lbs). How many hills in our area - we have soom hills here in San Diego. He asked me if I could peddle up the hills without walking the bike - yes - but may be very slow on some.. He sent the 1" roller - has worked perfect for the layman. Happy Riding from - Mountainman
 
So to make it easy with a 4" circumference roller at ten thousand RPM's you are traveling at 40,000 inches a minute. A mile has 63,360 inches in it so you are traveling about 3/4 of a mile a minute, so I think that equates to about 40 MPH. Have fun, Dave
 
Hi, all
Just to chime in with a real world experience I just had on this topic...
I had a 1.3" OD BMX foot peg as a drive roller on my 20" wheel on the Rans recumbent and it was pretty fast with the 25cc Ryobi 4 stroke engine at about 25 mph but it dogged on the hills quite a bit.
So, I switched to a 1" OD BMX foot peg drive roller this morning and found that my top speed isn't much less but the hill climbing and take-off ability is much better.
Also, the smaller spindle results in a lower speed before the engine bogs down and dies which is very nice for making tighter turns at slower speeds without killing the engine (it's a friction drive with no centrifugal clutch so I kill it by putting the brakes on).
In case anyone is looking for the BMX foot pegs, I found a nice pair online that have a 3/8" bore which works on most of my Ryobi engines and one end of the peg is threaded with 24T and the other with 26T. I needed 24T. Very Clever. Here's the web site:

bicyclebuys.com

The SKU is 0255552 and the Product Name is: "Pyramid Pegs 26/24T Black. The cost was $20 with shipping.


Steve G.
Grants Pass, Oregon
 
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not a "ratio"

you already got it i know, but just remember that there is no ratio to worry about when using a drive roller (spindle) friction drive. a good number to remember when calculating speeds is how to go from inches per minute to miles per hour. the formula : in/min / 1056 = miles/hour...happy motoring!
 
7-30-08
Hi, again
Just a final note in case anyone is still reading this thread...
I took a 16 mile trip today with a 1" drive roller instead of the old 1.3" roller on my friction drive, home made, MB.
The smaller roller made a BIG difference in hill climbing. I was able, for the first time, to go into town and back, up a big hill, with no pedaling.
Also, I got 256 mpg which is the most I've ever gotten.
FWIW, I suspect the smaller roller kept the 25cc Ryobi 4 stroke engine in its optimum power band better.
It's hard to believe you can get this kind of performance from a home built bike and that, at least around here, no one even seems aware of MB...

Steve G.
Grants Pass, Oregon
 
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