[fox87stang]

This is not the usual gear question thats been posted a thousand times. After the gears are out of the box, is there any way to tell 3.73 from 4.10s? Are there more (ridges) in the ring or pinion. Im just afraid when i buy 4.10s, I might get sent 3.73s or something.

[HMUSN]

the pinion gear will have either "4:10" or 3:73" on them. Also they differ in size.
the smaller the size of the gear the lower the ratio (higher the number)

[fox87stang]

ok thanks

[ReverendDexter]

Originally Posted by HMUSN the pinion gear will have either "4:10" or 3:73" on them. Also they differ in size. the smaller the size of the gear the lower the ratio (higher the number)

Wait... isn't that backwards? I thought that the ring gear on a 3.73 would be physically smaller than the ring gear for 4.10s, much how the higher gear on the rear casette on a bike is a physically smaller gear. I mean, it has to have more effective "distance" as it goes around the ring gear (9.9% more), right?

Or am I totally misthinking this?

[HMUSN]

the higher the number( 4:10) the smaller in size the gear will physically be. the ratio is lower being lower pertains to the actual number.

I know it sounds backwards but it isn't.

[Modified88Dreamer]

Any gear that ends in 0 is simple to find out what ratio it is. Example 4.10 would be 41/10 41 teeth on the ring and 10 on the pinion. 3.73 is trickier. Let's assume that your mustang is currently equipped with 2.73 gears and that you're planning to install a 3.73 ring and pinion. If you didn't change speedometer gears, actual vehicle speed would 73.1 percent of indicated speed (2.73/3.73 = .731). That being the case, what's required is speedometer gearing that will result in the driven gear spinning at 73.1 percent of its current speed. The speedo gear combination for a 2.73 rear (assuming 29-inch diameter tires) is a 18-tooth drive gear and 35-tooth driven gear. Switching to a 45-tooth driven gear (the most teeth available) would translate to the vehicle traveling at 78 percent of indicated speed - close but not perfect. Obviously, the drive gear must be changed to one with 15 teeth if the 73.1 percent ratio is to be achieved. Going back to the original 35-tooth driven gear for illustration purposes, the 15-tooth drive gear would result in the vehicle traveling at 83 percent of indicated speed - that seems like a step in the wrong direction, until the driven gear is changed. (This is where it gets a little tricky because you have to work with a percentage of a percentage). Matching the 15-tooth drive speedometer gear up with a 40-tooth driven gear brings the drive ratio to 72.8 percent (15/18=.83; 35/40=.88; .83 x .88 = .7304, which rounds to .73, or 73 percent). There you go.

[ReverendDexter]

Originally Posted by HMUSN the higher the number( 4:10) the smaller in size the gear will physically be. the ratio is lower being lower pertains to the actual number. I know it sounds backwards but it isn't.

I'm confused as to how that's possible. Here's my logic, please correct where I'm wrong:

To reduce the gear, you have to increase the effective distance traveled releative to the actual distance, which is done be increasing the number of teeth on the receiving gear (in this case, the ring gear), or decreasing the teeth on the powered/sending gear (in this case, the pinion gear). assuming that teeth remain the same size, as measured by degrees.

However, this isn't like a bike where the mesh is done against a chain, the gears have to mesh against each other, so tooth count alone is meaningless; if you double the tooth count on the ring gear, you'd have to double the tooth count on the pinion gear, or they would no longer mesh (assuming they each retained their original diameter).

This means that to affect the gear ration, you have to change degrees each tooth is worth, without changing the physical size of the tooth.

Going back, we have to increase the "distance" it takes to travel all the way around the ring gear. And being that we cant change the size of the teeth without loosing mesh, we have to change the size of the ring gear. To make each tooth worth less degrees at the same size, we have to increase the size of the ring gear, thus fitting more same sized teeth, and thus each tooth being worth less of the available 360 degrees of rotation.

Again, I'm just trying to learn, I'm guessing that I'm making a faliable assumption somewhere?

[fox87stang]

Originally Posted by Modified88Dreamer Any gear that ends in 0 is simple to find out what ratio it is. Example 4.10 would be 41/10 41 teeth on the ring and 10 on the pinion. 3.73 is trickier. Let's assume that your mustang is currently equipped with 2.73 gears and that you're planning to install a 3.73 ring and pinion. If you didn't change speedometer gears, actual vehicle speed would 73.1 percent of indicated speed (2.73/3.73 = .731). That being the case, what's required is speedometer gearing that will result in the driven gear spinning at 73.1 percent of its current speed. The speedo gear combination for a 2.73 rear (assuming 29-inch diameter tires) is a 18-tooth drive gear and 35-tooth driven gear. Switching to a 45-tooth driven gear (the most teeth available) would translate to the vehicle traveling at 78 percent of indicated speed - close but not perfect. Obviously, the drive gear must be changed to one with 15 teeth if the 73.1 percent ratio is to be achieved. Going back to the original 35-tooth driven gear for illustration purposes, the 15-tooth drive gear would result in the vehicle traveling at 83 percent of indicated speed - that seems like a step in the wrong direction, until the driven gear is changed. (This is where it gets a little tricky because you have to work with a percentage of a percentage). Matching the 15-tooth drive speedometer gear up with a 40-tooth driven gear brings the drive ratio to 72.8 percent (15/18=.83; 35/40=.88; .83 x .88 = .7304, which rounds to .73, or 73 percent). There you go.

Did you major in physics or something, thats waaay to much for me, but the first couple of sentences helped.