Ok math wizards
Posted September 28, 2006  04:19 AM
Stock gearing is 1449.
Posted September 28, 2006  04:49 AM
I have converted my 2002 426 to Supermoto trim. How do i get back to "stock" gearing ratio?? The wheel have been down sized to 17's. My question is what sprocket combonation would get me back to where I was before the wheel swap?
Stock gearing is 1449.
If the stock wheel was 18 I'd say a 14/46 is the closest to stock.
Or if you prefer to change the front sprocket a 15/49 will do.
Posted September 28, 2006  07:59 AM
Then take your final drive ratio of 3.27:1 (49/15) and multiply by the ratio you got in the first step. In the example, that's 3.27/.93=3.04. Multiply the resulting final drive ratio by 14 to see what rear sprocket you need (3.04x14=42.56), and you see that you need 42 or 43 rear. Multiply by 15 instead, and that gives you 45.6 instead, so you then choose between a 45 and a 46.
Posted September 28, 2006  08:11 AM
So what I would do is pick an arbitrary speed and calculate tire rpm with stock tire.
(feet per minute/circumfrence in feet=tire rpm)
tire rpmXfinal gear ratio=engine RPM
Figure tire RPM at the same speed with new shorter tire(it will be higher)
engine RPM/new tire RPM=new final gear ratio
teeth count front sprocketXnew final gear ratio= new rear sprocket
May be the same as Grayracer, just a different way of doing it.
Posted September 28, 2006  08:28 AM
http://4strokes.com/...earing_calc.xls (Select Open or Save) It is an Excel spreadsheet that does most if not all the math for you. You may have do some of the measurements yourself. Enter the values in Pink and it will autoupdate the rest.
 RCD
Posted September 28, 2006  09:20 AM
True, but it follows that if the diameter is .93 of the original, the circumference will be, too. If you are that concerned with accuracy, then measure the radius instead, and measure from the axle to the ground, or "rolling radius". If that isn't accurate enough, you can mark the tire and the ground, then roll the loaded bike forward ten (or any larger number) revolutions and measure the distance traveled, and divide by 10 (or whatever) for the circumference. That, of course, won't take into account forward motion lost to internal tire flex under power, either.Close but it is circumfrence not diameter that affects gearing. What you are trying to match is engine revolutions vs tire revolutions. I am not sure reducing diameter to .93% of original will give you .93% as many tire revolutions.
The point is that the original question only went to the gear ratio that would compensate for the difference in wheel size. The factors you introduced will change the result by 1% or less.
Try it both ways, and see if it doesn't get you the same result quicker.So what I would do is pick an arbitrary speed and calculate tire rpm with stock tire.
(feet per minute/circumfrence in feet=tire rpm)
tire rpmXfinal gear ratio=engine RPM
Figure tire RPM at the same speed with new shorter tire(it will be higher)
engine RPM/new tire RPM=new final gear ratio
teeth count front sprocketXnew final gear ratio= new rear sprocket
May be the same as Grayracer, just a different way of doing it.
To accomplish the entire calculation for any RPM/Gearing/Wheel size:
((RPMe/OAR)xCw)/1056=Speed in MPH at a given engine rpm and gearing
...where:
RPMe=engine speed
OAR=overall ratio (primary ratio x trans ratio x final drive ratio)
Cw=circumference of the rear wheel
1056 is a conversion factor that yields MPH from inches/minute.
Posted September 28, 2006  09:58 AM
Wow...and don't forget to use a microscope so you can tell the difference between the fly s**t and pepper! Gray, I absolutely love your posts!That, of course, won't take into account forward motion lost to internal tire flex under power, either.
Posted September 28, 2006  10:29 AM
since C = pi x D, then the change in C = pi x the change in D.
whether you represent the change as a percentage or an actual number does not matter.
Now a quiz:
If you wrap a string around the world it will be approximately 25,000 miles long. If you have another string suspended 1" above the ground, all the way around the earth, how much longer will it be?
Posted September 28, 2006  11:16 AM
Gray is absolutely correct (as per usual).
since C = pi x D, then the change in C = pi x the change in D.
whether you represent the change as a percentage or an actual number does not matter.
Now a quiz:
If you wrap a string around the world it will be approximately 25,000 miles long. If you have another string suspended 1" above the ground, all the way around the earth, how much longer will it be?
3.14159265359' longer
Posted September 28, 2006  11:16 AM
If you wrap a string around the world it will be approximately 25,000 miles long. If you have another string suspended 1" above the ground, all the way around the earth, how much longer will it be?
Just to answer the question. It is 3.14159..." longer which equals PI.
Posted September 28, 2006  11:35 AM
Posted September 28, 2006  11:40 AM
Aha, you caught it. The diameter of the circle created by the string 1" above the equator would be D(Earth) + 2". It increases the radius by one, the diameter by two.Since the formula 2*Pi*r is used to get C, then shouldn't the answer be 6.2832" (2*pi)?
BTW, the formula is simply (pi)D. 2(pi)r, (pi)2r, or r2(pi) are all the same thing, but undersimplified.
Posted September 28, 2006  11:41 AM
We have a winner.Since the formula 2*Pi*r is used to get C, then shouldn't the answer be 6.2832" (2*pi)?
The condensed formula was in my original post.
You're looking for the change in C, so pi x the change in D is it.
Most people think the number is going to be huge because you're dealing with a huge number in the first place, but it's really just as simple as delta C = pi x delta D.
Posted September 28, 2006  11:50 AM
I say, if you want to spilt hairs, let's split them fine!Wow...and don't forget to use a microscope so you can tell the difference between the fly s**t and pepper! Gray, I absolutely love your posts!
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