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# Gearing Help.

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On my current bike I am running 13/51 gearing for mostly hillclimbing and super technical trail riding. I really like this gearing but I wish I could have the same gear ratio with smaller sprocket. On BMX bikes I know that for every 3 added on the front equals 1 in the back. So would the same be 12/49 or is it different because the small one is in the front?

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On my current bike I am running 13/51 gearing for mostly hillclimbing and super technical trail riding. I really like this gearing but I wish I could have the same gear ratio with smaller sprocket. On BMX bikes I know that for every 3 added on the front equals 1 in the back. So would the same be 12/49 or is it different because the small one is in the front?

Just convert the ratio to a percentage by dividing the front into the rear 13/51 = 25.5%....try other gear combinations to get the same quotient.  12/49 = 24.5%...so not the same.  12/47 would be VERY close.....

At least this is how I understand it...

Edited by Fattonz

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You are about correct that one tooth on the front equals 3 on the rear.  But to make 1nd gear feel like 2st gear, you'll have to go up probably two teeth in the front (or six in the rear, or one in front three in rear).

Do you have a close or wide ratio tranny?

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try this another way.  Your gear ratio right now is 51/13 = 3.92.  So for every 3.92 rotations of the smaller countershaft sprocket, you have 1 turn of the rear sprocket or rear wheel.

With 49/12 your gear ratio is 49/12 =4.08..or the countershaft sprocket now turns 4.08 revolutions for every 1 revolution of the rear sprocket or rear wheel.  The 12/49 combo is 'lower geared" and requires more engine rpm to go the same speed as the 51/13 combo.

If you want the same gear ratio use 47/12 = 3.92.  Note that a smaller countershaft sprocket can increase the wear across the plastic slider on the top of the swingarm, increase wear on the chain due to tighter radius at the countershaft sprocket, and your chain may be too long requiring you to remove links (unless you have a lot of room to adjust the rear wheel further toward the back).

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I had to think to remember this one from my Power Engineering classes but you can use the law of means and extremes.

So I'll a write a random ratio like this:

3 : 4  ::  6 : 8

In the above equation the 3 and the 8 are the "extremes", and the 4 and the 6 are the "means"

Now the rule states "the product of the two extremes divided by either mean will give the other mean"  and that "the product of the two means divided by either extreme will give the other extreme"

So in other words:

Using the two extremes:

3 x 8 = 24 then 24 / 6 = 4 or 24 / 6  = 6

Using the two means:

4 x 6 = 24 then 24 / 3 = 8 or 24 / 8 = 3

So knowing the rule we can now apply it to gear selection:

You are using 13/51 and let's say you want to go to a 12T front sprocket and maintain the same relationship:

We could write it as follows where there is one unknown variable:

13 : 51 :: 12 : _

Then we apply the rule:

51 x 12 = 612 then 612 / 13 = 47.08.....but since we can't have .08 teeth, the closest is a 47T sprocket.

So not identical gearing, but very close.  Clear as mud?

Edited by Fattonz

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So 13/51 is actually pretty tall geared? How does it compare to a 12/49? Like the power delivery, does it pull better on the bottom with the 12/49 and pull better on the top with a 13/51?

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A 12 tooth front is not often recommended, the chain sees quite a lot of wear.

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So 13/51 is actually pretty tall geared? How does it compare to a 12/49? Like the power delivery, does it pull better on the bottom with the 12/49 and pull better on the top with a 13/51?

I definitely would not call 13/51 tall gearing...quite the opposite actually.  However it is taller than 12/49...  I could go all the way to 14/38 on my bike, now that's tall...

We've established that 13/51 is like 12/47....so going to a 49T rear sprocket is going to give you more low rpm torque and less top end.....it is "lower" gearing...as someone else put it, you'll need more engine speed to have the same rotation of the rear wheel....

Not sure why you would want to run the 12T though...

Edited by Fattonz

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I would stay away from a 12 cs  , I run a 15/51

Edited by capt justin

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So 13/51 is actually pretty tall geared? How does it compare to a 12/49? Like the power delivery, does it pull better on the bottom with the 12/49 and pull better on the top with a 13/51?

Didnt say 13/51 was 'tall geared', just that 12/49 is 'geared lower' than 13/51.  To be exact, the engine RPM would be 4% higher at a given rear wheel speed using 12/49 vs. 13/51.  This is strictly a mathematical exercise (4.08/3.92 = 104% or 4% higher rpm).  What bike are we talking about?...I dont know how it was equipped from the factory and cant say whether 13/51 is higher or lower geared than the stock configuration.

Ultimately, what determines how fast the rear wheel turns compared to the engine rpms are 3 different things:

• Gear ratio for the particular gear you are in
• The primary drive ratio (within the gearbox), and
• The secondary drive ratio (front countershaft sprocket & rear wheel sprocket).

The only thing we are talking about here is the 3rd item....secondary drive ratio.

Let me give you a concrete example:  WR450F.....1st gear ratio is 2.417.  Primary Drive ratio = 2.652 and Secondary Drive ratio = 3.846 (this is determined by a 50 tooth rear sprocket and 13 tooth front sprocket or 50/13 = 3.846).  So final driveline speed is ultimately reduced by 2.417 x 2.652 x 3.846 = 24.65.  This means for every 24.65 rotations of the engine crankshaft (e.g. rpms) the rear wheel will turn one revolution.  So at 2000 rpm....rear wheel will turn 81.13 times in a minute.

Let's turn those wheel revolutions into something we understand.  Let's say the outside wheel diameter is 22".  We know that circumference is Pi x Diameter or 3.14 x 22" = 69.08 inches traveled for every revolution of the rear tire.  Translating to feet we divide 69.08 inches by 12 which gives us 5.76 ft traveled per every rear wheel revolution.  We multiply by the 81.13 revolutions of the rear tire per minute we calculated above and we get 467.04 feet per minute traveled x 60 minutes to get 28,022 ft traveled per hour and then we divide by 5,280 ft per mi....and we get to something we can relate to....5.30 miles per hour at a constant 2,000 rpm engine speed when we are in first gear. (assuming absolutely no clutch slippage or rear tire slippage).

On your question of how it pulls etc.....you will have to try it out to see how that gearing/rpm work with your particular motor and the power curve of that motor.  Generally speaking, with 12/47, you will be able to ride it slower (a neglible amount) at the same RPM as you used to ride.  At the top end, you will be scrubbing off some top speed (e.g. it will go 4% slower at maximum RPM compared to 51/13).  These are fixed ratios....so there is no way to lower the gearing in 1st gear and up the gearing in 5th gear without changing the actual ratios in the gear box.

Here is a tool that might be useful for you:

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