Driver size effect on gearing?

[Van Epps]

I've just been starting to dig into getting to know my sled and how things work. Been wondering what effect the driver size / tooth count has on gearing?

Does a higher tooth count driver equate to lower or higher gearing? Does the tooth count have anything to do with it? Is it the OD of the driver that counts? Help me understand it

I would guess the larger the diameter of the driver, the lower the gearing?

 

[vdo1948]

The larger the diameter, the larger the circumference. The circumference determines (roughly) how much ground you will cover in one revolution of your drive shaft. Therefore, the large the diameter, the more ground you will cover in one revolution, thus larger diameter is like higher ratio gears. More teeth means larger diameter so long as you are comparing drivers with the same pitch.

 

[skibreeze]

The larger the diameter, the larger the circumference. The circumference determines (roughly) how much ground you will cover in one revolution of your drive shaft. Therefore, the large the diameter, the more ground you will cover in one revolution, thus larger diameter is like higher ratio gears. More teeth means larger diameter so long as you are comparing drivers with the same pitch.

Not all correct, the larger diameter will make a lower ratio. This is why you put higher ratio gears in when you put on larger tires on a Jeep. The tires by themselves lowered the overall ratio.

 

[snowman429]

Not all correct, the larger diameter will make a lower ratio. This is why you put higher ratio gears in when you put on larger tires on a Jeep. The tires by themselves lowered the overall ratio.

Actually you put in lower gears in when you go to a larger tire. The higher the number the lower the ratio. Ie: 3:73 gears is a higher gear ratio then 4:10 gears

 

[skibreeze]

Actually you put in lower gears in when you go to a larger tire. The higher the number the lower the ratio. Ie: 3:73 gears is a higher gear ratio then 4:10 gears

I know that you put lower gears in, but they are numerically higher. I forgot that you have to convert it.

 

[Van Epps]

OK, thanks ....... I guess?

So the smaller the OD or "circumference" of the driver, the lower the gearing?

 

[joshkoltes]

apples and oranges

its hard for me to find similarities between drive and driven sled parts and straight gearing in the automobiles.
i try to figure wich is which in my drive system such as drive clutch drives driven, driven drives top gear, top drives bottom, bottom drives drivers. so thinking that way, the smaller driver would take more torque to turn and more track speed right?
there are three different types of current drivers also 2.52, 2.86, and 3 inch pitch so a 8 tooth 2.52 is going to be less diameter then a 8 tooth 3" pitch.

 

[Van Epps]

OK, thinking about it like that.... the bottom gear driving the driver, by putting larger outside diameter drivers on you would effectively be lowering your gearing. Right?

Another question: how do I know the pitch of my drivers? I think I may have 2.86 pitch drivers with a 3 pitch track. Not sure how to verify the driver pitch? My drivers are Wahl anti ratchets and Im not sure if they even make a 3 pitch 8 tooth driver?

 

[theultrarider]

The smaller the driver you have, the less track speed you will have and better bottom end. It would be like going from 3:73 gears in a vehicle to 4:10's. Also, since you made this change down stream from the speedo drive, your speedo will now be off. It will appear that you have the same topend as you did before the swap. But that is not the case. It is just like putting a set of 38's on a truck... down stream of the speedo but in the case of using smaller drivers it has the oposite affect as a larger tire does so therefore if you are acutally going 60mph your speedo will show more like 65-70.

 

[manowar]

Yes your on the right track. Gearing is more than just your chain case. It is the sum of your driver circumfrence, chain case ratio, and cluch ratio. So for example lets say you need a top track speed of 100 mph . And you have 9 tooth 2.52 drivers .

 

[Dootimes2]

Look at it this way. If you have a smaller driver it has to make more revolutions ( rpm's ) to turn the track 1 full revolution as compared to a larger driver. This equates to more rpm's with less ground covered = less speed. Relates to a vehicle in a low gear running wfo. Lots of rpm with no speed. More torque and pulling power but no speed. 1 tooth is not that dramitic but the same theory.

 

[Van Epps]

Yes your on the right track. Gearing is more than just your chain case. It is the sum of your driver circumfrence, chain case ratio, and cluch ratio. So for example lets say you need a top track speed of 100 mph . And you have 9 tooth 2.52 drivers .

Yes........ keep going with that thought! I know there are certain ratios that work best. I don't know what they are or how to calculate them (formulas). Any help here?

Look at it this way. If you have a smaller driver it has to make more revolutions ( rpm's ) to turn the track 1 full revolution as compared to a larger driver. This equates to more rpm's with less ground covered = less speed. Relates to a vehicle in a low gear running wfo. Lots of rpm with no speed. More torque and pulling power but no speed. 1 tooth is not that dramitic but the same theory.

I understand what you are saying and it makes sense to me.

 

[manowar]

Crap. That post was way longer than that. I gott look at them after i post.

 

[ottar88]

Use this. It will tell you theoretical top speed.

And if you change drivers, you can adjust gearing to match load sensing.
(with smaller drivers you'll need taller gears to match top speed)
If you use this correctly, it's a great tool. IMO.


Formula to calculate MPH:
RPM / GR x SC / 12 x 60 / 5280 = MPH @ 1:1 clutch ratio
GR = gear ratio (divide top sprocket into bottom sprocket)
SC = sprocket circumference

7T - 2.52" Pitch = 17.64 SC
8T - 2.52" Pitch = 20.16 SC
9T - 2.52" Pitch (std. OEM size) = 22.68 SC
10T - 2.52" Pitch = 25.2 SC
7T Convolute - 3.29" Pitch = 23.03 SC

 

[manowar]

ok here it goes . take your max rpm / your cluch ratio / chain case ratio = driver rpm.
so if 8000 rpm and .75 over drive cluch at full shift then 19-38 in the chain case.
=8000/.75/2.0=5333.333 driver rpm
next if you have 9 tooth 2.52 pitch drivers you cover 22.68'' per revolution.
or if you have 8 tooth 2.52 drivers you cover 20.16'' per revolution.
so you take your driver rpm- 5333 x 22.68= 120952'' per min.then / by 12 = 10079ft per min. then /5280 (ft in a mi.) =1.90mi per min then x 60 to get mph = 114 .

if you do the same with 8 tooht =102 mph.

driver circumfrence is drive pitch x number of teeth.

to figure the diffrence take the driver circumfrence / 5280 x (the old driver / 5280)- 5280 this gives you the feet you need to make up then / the new driver by the diffrence to get how many more driver rpm is needed.
then use the chain case, cluch and rpm to get the rest of your gearing .
if you need more pm me .

 

[manowar]

And if you change drivers, you can adjust gearing to match load sensing.
(with smaller drivers you'll need taller gears to match top speed)
If you use this correctly, it's a great tool. IMO.


Formula to calculate MPH:
RPM / GR x SC / 12 x 60 / 5280 = MPH @ 1:1 clutch ratio
GR = gear ratio (divide top sprocket into bottom sprocket)
SC = sprocket circumference

7T - 2.52" Pitch = 17.64 SC
8T - 2.52" Pitch = 20.16 SC
9T - 2.52" Pitch (std. OEM size) = 22.68 SC
10T - 2.52" Pitch = 25.2 SC
7T Convolute - 3.29" Pitch = 23.03 SC

yea that in less words lol you beat me to it wile i was writing a book.