# Ok math wizards

15 replies to this topic
• # Ztard

Posted September 28, 2006 - 04:19 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 14-49.

• # Ulysses

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 14-49.

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.

• # grayracer513

Posted September 28, 2006 - 07:59 AM

Take the diameter of your SM wheel/tire assembly and divide it by the diameter of your original, larger wheel/tire. That will give you a number such as 0.93. That's the amount you need to raise your gearing (lower numerically)

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.

• # scott367

Posted September 28, 2006 - 08:11 AM

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.

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.

• # 00YZ426FMRCD

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

• # grayracer513

Posted September 28, 2006 - 09:20 AM

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.

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.

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.

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.

Try it both ways, and see if it doesn't get you the same result quicker.

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.

• # YZ426F Rider

Posted September 28, 2006 - 09:58 AM

That, of course, won't take into account forward motion lost to internal tire flex under power, either.

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!

• # bg10459

Posted September 28, 2006 - 10:29 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?

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• # gjones

Posted September 28, 2006 - 10:43 AM

Last I checked Circumference was related to diameter!

• # scott367

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

• # Yama4Stroker

Posted September 28, 2006 - 11:16 AM

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?

Just to answer the question. It is 3.14159..." longer which equals PI.

• # YZ426F Rider

Posted September 28, 2006 - 11:35 AM

Since the formula 2*Pi*r is used to get C, then shouldn't the answer be 6.2832" (2*pi)?

• # grayracer513

Posted September 28, 2006 - 11:40 AM

Since the formula 2*Pi*r is used to get C, then shouldn't the answer be 6.2832" (2*pi)?

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.

BTW, the formula is simply (pi)D. 2(pi)r, (pi)2r, or r2(pi) are all the same thing, but undersimplified.

• # bg10459

Posted September 28, 2006 - 11:41 AM

Since the formula 2*Pi*r is used to get C, then shouldn't the answer be 6.2832" (2*pi)?

We have a winner.

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.

• # grayracer513

Posted September 28, 2006 - 11:50 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!

I say, if you want to spilt hairs, let's split them fine!

• # Ztard

Posted September 28, 2006 - 03:29 PM

OK, thanks I think...

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