loquin
Well-Known Member
On a friction drive, the tire does nothing more than to transfer the circumferential velocity of the drive roller to the ground. It's an idler wheel, essentially. The drive roller transfers IT's speed to the tire, which in turn, transfers its speed to the ground.
Now, the ANGULAR velocity (RPMs) DO decrease, but, in a friction system, unlike a sprocket system, the ratios of wheel size to drive roller size have zero effect. If you had a system that could place the drive roller directly against the ground, it would push you along at the same speed as if the drive roller operates through a wheel of ANY size.
Disregarding frictional losses. the speed of the roller (circumference speed, in miles per hour) is exactly the same as the ground speed. And, the circumference speed is calculated by multiplying the RPM times 60, to get revolutions per hour, then multiplying by the roller diameter and by PI, to get the number of inches per hour, then you divide by 12 inches per feet, and by 5280 feet per mile. The result is the speed in miles per hour. The wheel diameter doesn't even enter in to the picture. If you wanted, you COULD calculate the wheel RPM, but it wouldn't make any difference.
Now, if you have a smaller wheel, its RPMs will be higher, but the ground speed will be the same as with a larger wheel. IF you keep carrying this analogy down using smaller and smaller wheels, a 2.6 inch diameter tire would spin 10 times faster than a 26 inch diameter tier, but, because it's 10 times smaller, it HAS to spin 10 times faster, just to stay at the same speed. A 1 inch diameter tire (the same size as the roller) would spin at exactly the same RPM as the drive roller. And, it would be running at the same ground speed as a 26 inch tire, a 20 inch tire, a 10 inch tire or a 2.6 inch tire...
Now, the ANGULAR velocity (RPMs) DO decrease, but, in a friction system, unlike a sprocket system, the ratios of wheel size to drive roller size have zero effect. If you had a system that could place the drive roller directly against the ground, it would push you along at the same speed as if the drive roller operates through a wheel of ANY size.
Disregarding frictional losses. the speed of the roller (circumference speed, in miles per hour) is exactly the same as the ground speed. And, the circumference speed is calculated by multiplying the RPM times 60, to get revolutions per hour, then multiplying by the roller diameter and by PI, to get the number of inches per hour, then you divide by 12 inches per feet, and by 5280 feet per mile. The result is the speed in miles per hour. The wheel diameter doesn't even enter in to the picture. If you wanted, you COULD calculate the wheel RPM, but it wouldn't make any difference.
Now, if you have a smaller wheel, its RPMs will be higher, but the ground speed will be the same as with a larger wheel. IF you keep carrying this analogy down using smaller and smaller wheels, a 2.6 inch diameter tire would spin 10 times faster than a 26 inch diameter tier, but, because it's 10 times smaller, it HAS to spin 10 times faster, just to stay at the same speed. A 1 inch diameter tire (the same size as the roller) would spin at exactly the same RPM as the drive roller. And, it would be running at the same ground speed as a 26 inch tire, a 20 inch tire, a 10 inch tire or a 2.6 inch tire...
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