gas/engine speed/roller/speed/rear wheel speed relationship

mifletz

Member
Local time
4:36 PM
Joined
Aug 25, 2009
Messages
104
Sorry to display my simplicity, but what is the relationship between throttle setting/engine speed/roller speed/rear tyre speed?

e.g. why can a 7/8" roller only achieve 22mph, but a 1.5" can do 30mph, on the flat?

What will be the difference in the throttle setting, engine speed, roller speed and rear wheel speed between a 7/8" and a 1.5" in this case?
 
Last edited:
Sorry to display my simplicity, but what is the relationship between throttle setting/engine speed/roller speed/rear tyre speed?

e.g. why can a 7/8" roller only achieve 22mph, but a 1.5" can do 30mph, on the flat?

What will be the difference in the throttle setting, engine speed, roller speed and rear wheel speed between a 7/8" and a 1.5" in this case?

Unsure what you mean by throttle setting differences.

Simply put, the 7/8" roller would be like first gear in a manual transmission. The bike will only go 22mph before the engine winds out. The 1.5" roller is like fourth gear. The engine will wind out at a higher speed.

Someone here will download a spindle/rpm/speed calculator here soon.
 
You may want to take a look at the Friction Drive FAQ.

There's usually a centrifugal clutch between the engine and the roller. So, the engine has to be above a certain RPM before the clutch engages. That's normally in the 2500 to 3000 RPM range. But, for 4-strokes, the clutch engagement RPM may be lower.

When the clutch is engaged, the roller RPM is the same as the engine RPM (unless the clutch is slipping.)

As the roller (or, for that matter, any wheel) spins, a point on the outside of the roller is moving at some speed that is dependent on the distance from the center of the spinning roller, and the RPM of the roller.

That speed can be calculated as follows:

Velocity (in MPH) equals Roller Diameter (in inches) times RPM times pi times 60 divided by 5280 divided by 12

V = (60*pi*d*RPM) / (5280 * 12)

The whole thing can be simplified to

V = d * RPM * 0.002975

or

V = d * RPM / 336.14

Note - For the metric system, with wheel diameter in cm and speed in Km/Hr, the constants are different:
V = d * RPM * 0,001885
or
V = d * RPM / 530,5

So, a 1 inch roller, at 7500 RPM, could push you along at up to 22.3 mph and a 1.5 inch roller at the same RPM would 33.5 MPH. Likewise, a 7/8 inch roller would give you a 19.5 MPH max speed

Note that this calculation results in the maximum speed you could attain - it does not account for slope, rider weight, wind resistance, engine power, roller/tire resistance, tire rolling resistance, air pressure, or any other variables that could impact your final upper speed.

Here's the link to the gearing/friction drive calculator I wrote a while back.
 
Last edited:
Think about it this way. If you have a wheel turning, a point on the tire face is moving around the axle at the wheel RPM. When the wheel turns 1 complete revolution, the distance that the wheel moved along the ground is equal to the circumference of the wheel. The distance around the wheel. That distance AROUND the wheel is proportional to the wheel diameter. A wheel that is double the diameter has double the circumference. (pi * d is the actual calculation used to find the circumference for a given diameter.) A wheel with half the diameter would move 1/2 as far.

Now, when you have a wheel spinning at some RPM (which stands for Revolutions Per Minute,) you multiply the circumference by the angular velocity (# of revs per minute) to calculate how far the wheel moves in one minute. And, multiply be 60 to get the distance, in an hour, at that speed.

The funny thing is this: For a friction drive, the WHEEL diameter doesn't matter. The ONLY diameter that matters is the drive roller diameter. That's because the wheel & tire is essentially an idler wheel - all it does is to transfer the circumference speed of the roller to the ground. The circumference speed of the roller is exactly the same circumference speed as the tire, because they are in contact. And, the circumference speed of the tire is exactly the same as the ground speed of the bike, because the tire is also in contact with the ground.

Ref the attached sketch. The second tire is twice the diameter as the first, But, the drive rollers are the same diameter. And, for the same engine RPM, both wheels will travel at the same speed. The smaller wheel will spin twice as fast, but, because it is half the diameter, and half the circumference, the ground speed is identical...
 

Attachments

  • Drive-Wheels.JPG
    Drive-Wheels.JPG
    18.8 KB · Views: 852
Last edited:
Bravo Lou for taking the effort to nail that one more time! Those replies need to go into the FAQ for reference 'cause it'll come up again.
 
Lou:

This was a great explanation. So, that means I can use your formulas to calculate my engine rpm with my 1.250" roller hitting 36 mph on the flats right?

I always knew this engine was really screaming but now I will know exactly how fast it has to be going given those parameters.

Thanks again,

Bill
 
Yes, you could, Bill. Just be sure to use the actual, 'loaded' diameter. (calculate the diameter by moving the bike, under load, one measured wheel circumference, then, divide that length by pi.)

The rough estimate is, for a 1.25 inch roller at 36 MPH, 9680 RPM
 
Lou:

Thanks man. Math was always one of my weaker subjects, ha ha. Just by ear I was guessing around 9,000 rpm or so and these numbers are helpful. I will do the loaded measurements as you suggested to get an exact figure but I knew this little engine was really turning.

I had a nice ride today. I love these machines.

Thanks,

Bill
 
Lou:

Yes, you were the one that explained that to me, ha ha. I had thought you meant check my loaded circumference for programming my speedometer in order to get an accurate mph reading so I could then calculate my rpms. I did not do this loaded but I did check both the diameter and a circumference measurements and it all worked out as it should be.

Thanks for your help.

Bill
 
Back
Top