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Posted
Just surfin the forums and I noticed that Miles shows to have an expired membership?

 

:think:

 

Yeah, he disappeared several weeks ago. Not sure what happened, he just seemed to have faded away...

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Posted
Miles is correct in that the sensor mounts to the pinion gear. But it takes 33 pinion ring gear meshes for the ring gear to go around one time. That means 33 pinion teeth must pass under the sensor for each ring gear revolution with either rear end. As both have a 33 tooth ring gear. So in both rear ends the speed sensor will see 33 pulses for each wheel revolution. The speedometer should be unaffected.

 

Mike

 

Your math is interesting but I think slightly off.

A 10 tooth pinion and a 33 tooth ring gear gives an effective ratio of 3.30:1, or 3.3 turns of pinion for one turn of ring.

 

A 9 tooth pinion and a 33 tooth ring gear gives you 3.66:1 ratio, or 3.666 turns of pinion to one turn of ring.

 

If the sensor is looking at the pinion for its pulse count and not the ring then it will see an increase in pulses per mile with 9 teeth.

If the sensor is looking at the ring, and the ring only then I'm lost and you are correct.

Having never looked at one with the electronic pickup I have no idea on where or how it sees a pulse. Mine has a cable, and it aint right either.

Posted
So as to the actual change in where we pick up on the power band, is it correct to say that the rpm should increase by 10% at any given speed with the diff change?? If so, at 80 miles per hour would change the rpm from about 3600 to about 4000 and so on and so on??

Is that what the GenII run for rpm at that given speed? No wonder the First gen will run away from them, they are geared lower, mine turns 4k at 70 mph

Posted
Your math is interesting but I think slightly off.

A 10 tooth pinion and a 33 tooth ring gear gives an effective ratio of 3.30:1, or 3.3 turns of pinion for one turn of ring.

 

A 9 tooth pinion and a 33 tooth ring gear gives you 3.66:1 ratio, or 3.666 turns of pinion to one turn of ring.

 

If the sensor is looking at the pinion for its pulse count and not the ring then it will see an increase in pulses per mile with 9 teeth.

If the sensor is looking at the ring, and the ring only then I'm lost and you are correct.

Having never looked at one with the electronic pickup I have no idea on where or how it sees a pulse. Mine has a cable, and it aint right either.

 

 

Try again. You've almost got it. It takes 33 teeth for the ring gear to make one revolution. And that is with either gear set.

 

Mike

Posted
Try again. You've almost got it. It takes 33 teeth for the ring gear to make one revolution. And that is with either gear set.

 

Mike

 

What is the point there? 33 teeth on both rings, make one revolution, the difference is at what rpm the pinion turns. From what I read above, the sensor reads the pinion. So that being you could remove the ring and the sensor would still count the pinion turns.

Posted
What is the point there? 33 teeth on both rings, make one revolution, the difference is at what rpm the pinion turns. From what I read above, the sensor reads the pinion. So that being you could remove the ring and the sensor would still count the pinion turns.

 

Your close but need to go one step farther.

 

But we have to agree on a few things;

 

First we must agree that with either gear set the the bike moves the same distance per revolution of the rear tire.

 

Then we must agree that the sensor is counting the number of pinion teeth that go by and not the RPMs of the pinion.

 

Then we must agree that the electronics on the bike are doing the math to figure the distance traveled by one revolution of the rear tire compared to how many teeth went by the sensor in the time it took the rear wheel to make one revolution.

 

Then we must agree that with the 3.667 gear ratio the 9 tooth pinion turns 3.667 revolutions per revolution of the rear tire.

 

Then we must agree that with the 3.300 gear ratio the 10 tooth pinion turns 3.300 revolutions per revolution of the rear tire.

 

Here comes the math test;

 

If the 9 tooth pinion is turning 3.667 revolutions per revolution of the rear tire then the sensor is seeing 9 teeth X 3.667 revolutions per each revolution of the rear tire, or 33 teeth.

 

If the 10 tooth pinion is turning 3.300 revolutions per revolution of the rear tire then the sensor is seeing 10 teeth X 3.300 revolutions per each revolution of the rear tire, or 33 teeth.

 

Now where have I seen that 33 number before:think::whistling:

 

So based on the math, both the 9 and 10 tooth pinions will see 33 teeth per revolution of the rear tire. Thus 33 pulses per distance traveled by the rear wheel in one revolution.

 

I know it is hard to wrap your head around but the math don't lie.

 

OK maybe it was more than one step.......

Posted
Your close but need to go one step farther.

 

So based on the math, both the 9 and 10 tooth pinions will see 33 teeth per revolution of the rear tire. Thus 33 pulses per distance traveled by the rear wheel in one revolution.

 

I know it is hard to wrap your head around but the math don't lie.

 

OK maybe it was more than one step.......

 

You obviously know more about this than you are telling us.

You keep saying the sensor is counting 33 teeth. That is the ring gear.

I read above that it counts the pinion gear revolutions(9 or 10 depending on which gearset is used), what it uses for a trigger was never explained. With this in mind that it is counting the pinion revolutions and not the ring revolutions then the 33 is not even considered to be relevant in the counting process, and indicated speed would change for a given number of ring revolutions.

Posted
What is the point there? 33 teeth on both rings, make one revolution, the difference is at what rpm the pinion turns. From what I read above, the sensor reads the pinion. So that being you could remove the ring and the sensor would still count the pinion turns.

 

 

 

So true. But it still takes 33 pinion teeth to make the ring gear go around one full revolution. Thus 33 pulses to the speed sensor per wheel revolution for either gear set.

 

Mike

Posted
You obviously know more about this than you are telling us.

You keep saying the sensor is counting 33 teeth. That is the ring gear.

I read above that it counts the pinion gear revolutions(9 or 10 depending on which gearset is used), what it uses for a trigger was never explained. With this in mind that it is counting the pinion revolutions and not the ring revolutions then the 33 is not even considered to be relevant in the counting process, and indicated speed would change for a given number of ring revolutions.

 

Herin lies the issue.

The sensor is not counting pinion revolutions, it is counting the number of pinion teeth that go by. Look at the position of the sensor, it is located centered over the pinion teeth.

In order for the sensor to count pinion revolutions, there would have to be a feature on the pinion that only goes by the sensor once per revolution, and the sensor would have to be located over that feature.

 

There were pics posted somewhere looking into the sensor hole to show this clearly, I can not find them right now. Maybe someone has or can take pics of this.

Posted
Herin lies the issue.

The sensor is not counting pinion revolutions, it is counting the number of pinion teeth that go by. Look at the position of the sensor, it is located centered over the pinion teeth.

In order for the sensor to count pinion revolutions, there would have to be a feature on the pinion that only goes by the sensor once per revolution, and the sensor would have to be located over that feature.

 

There were pics posted somewhere looking into the sensor hole to show this clearly, I can not find them right now. Maybe someone has or can take pics of this.

Maybe pictures in 16th post in thread below.

 

http://www.venturerider.org/forum/showthread.php?threadid=75566&page=2

 

Gary

Posted

without a sensor in place to see what it is in close proximity to I am going to take the facts given as true, the sensor does not count the ring gear teeth, nor does it count the pinion revolutions, it instead counts the pinion teeth. Am I on the correct path there Flyinfool?

 

So that means that Mike and I both are wrong, right?

Posted (edited)

Sorry to say on this issue I am not wrong.

 

The sensor counts pulses caused by the the pinion teeth passing beneath it. The type of sensor is a Hall Effect. It creates one digital duty cycle pulse each time a pinion tooth passes. The output is a modified square wave which is read by the speedo unit. It takes 33 pinion teeth for the ring gear to make one revolution. The wheel is connected to the ring gear therefore it takes 33 pinion teeth to drive the wheel one revolution. So there are 33 sensor pulses per wheel revolution. It does not matter how many teeth the pinion has so long as the ring gear has 33 teeth there will be 33 pulses per wheel revolution.

 

The sensor does not count revolutions. It only counts pulses. The speedo converts this to mph based on the calculated wheel diameter, the calculated number of pulses per revolution and the number of revolutions per minute. The formula for this is built into the speedo.

 

Mike

Edited by MikeWa
Posted
without a sensor in place to see what it is in close proximity to I am going to take the facts given as true, the sensor does not count the ring gear teeth, nor does it count the pinion revolutions, it instead counts the pinion teeth. Am I on the correct path there Flyinfool?

 

So that means that Mike and I both are wrong, right?

 

You are on the right path.

the next step, unless the gears are stripped, pinion teeth must equal ring gear teeth for one revolution of the tire.

Posted

I agree with Bob on this one. I can't, for the life of me, see where the 33 teeth on the ring gear has anything to do with the number of teeth the sensor sees on the pinion, since the ring gear is the DRIVEN gear, not the drive gear.

Posted
I agree with Bob on this one. I can't, for the life of me, see where the 33 teeth on the ring gear has anything to do with the number of teeth the sensor sees on the pinion, since the ring gear is the DRIVEN gear, not the drive gear.

 

It does not matter whether the pinion gear has 9 teeth, 10 teeth, 5 teeth or 20 teeth, the speedometer will read the same if the ring gear stays the same, 33 teeth. In one revolution of the wheel, there are 33 RING gear teeth that will pass a given point. For EACH ring gear tooth, there is ONE pinion gear tooth that matches. Therefore, for each RING gear tooth that advances, one pinion tooth advances. So, for the ring gear to make one revolution, the pinion will have 33 teeth that passes the sensor.

The sensor does not read the number of revolutions the pinion makes, it reads the number of teeth that passes by and that is ONE tooth on the pinion for ONE tooth on the ring gear.

It does not matter which gear is the drive gear or the driven gear. It is simply counting teeth.

RandyA

Posted

Counting teeth is easy for me after catching handlebars in mouth but this is giving me a headache. I have a V-Max rear and a FJR rear. Any difference??Shock mount on some of them? I had them put away and am trying to organize things. Pinion is 9 teeth where it meets diveshaft??

Posted

Venturous is correct. And so is Flyinfool. The speedo reading is based on the revolutions of the rear wheel. All the sensor does is make pulses as the pinion, ring gear and wheel turn. For the Speedometer to maintain the same reading it would need to see the same number pulses for each revolution of the rear wheel. So long as the ring gear has the same number of teeth for each gear set (33) the speedo will see the same number of pulses for each revolution of the rear wheel and the speedo will remain unchanged.

 

Mike

Posted
Counting teeth is easy for me after catching handlebars in mouth but this is giving me a headache. I have a V-Max rear and a FJR rear. Any difference??Shock mount on some of them? I had them put away and am trying to organize things. Pinion is 9 teeth where it meets diveshaft??

 

OK, are you talking about the number of splines in the sleeve? To put the V-Max rear on your MKII 1st gen, you will need a different plastic dust cover that goes on the wheel. Other than that, it should be a direct bolt up and you will use your MKII driveshaft. I will see if I still have the plastic bag with the part number on it for the dust cover.

It is fun with the V-Max rear as long as your passenger does not mind the front wheel coming off the ground in 2nd gear. Did that with Linda yesterday accelerating to match traffic and was not getting on it real hard.

RandyA

Posted
Counting teeth is easy for me after catching handlebars in mouth but this is giving me a headache. I have a V-Max rear and a FJR rear. Any difference??Shock mount on some of them? I had them put away and am trying to organize things. Pinion is 9 teeth where it meets diveshaft??

 

The FJR final drives do not have the shock boss drilled out, VMax does have shock boss drilled. Make sure you have the 3" (approx.) long spacer in the FJR drive when mounted. It is a little different than a VMax rear. I struggled with this when I 1st put FJR rear on Tweety. Without the spacer, the final drive/wheel will bind up long before axle is fully torqued.

 

The drive shaft splines are the same in an MKI thru an RSV including VMax's. The length of the drive shaft is different. If you put a VMax or FJR rear end on a 1st gen, an MKII drive shaft must be used. MKII is about an inch longer than an MKI shaft. I am not sure about RSV's but I believe stock RSV shaft works with either rear end. FJR drive shaft is to long to be used on any Venture.

 

Something I haven't seen mentioned with the above discussion on counting pinion teeth VS. ring teeth is that the power is derived from the fact that with less teeth on the pinion gear (9 teeth), the motor has to rev about 9% higher to achieve the same speed as a 10 tooth pinion. This puts the motor farther up into the power band. It helps power, but usually will decrease MPG. This is not always the case, since some riders tend to run motor at to low an RPM for optimal efficiency.

 

Gary

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