Bus 4 spd trans, Anybody done this Options

[jwalters]

(IMG:http://www.914world.com/bbs2/html/emoticons/confused24.gif)

Thought of an interesting topic---has anybody put a beefed up bus or t1 trans in the teener----

I loved the gear spacing on my old superbug--and first gear for my car,,well,,,sucks ass--and second is too tall for a good start off--

Would rather spend 650 bucks on a pro-streeter WITH a rhino case and super diff than thousands upgrading my 901--

Sand rail mid engine cable shifter should be able to be adapted rather easily and the trans install is a non-event...

What you think???? (IMG:http://www.914world.com/bbs2/html/emoticons/confused24.gif)

[jwalters]

(IMG:http://www.914world.com/bbs2/html/emoticons/biggrin.gif) Andys---I do admit I have not done this kind of math in a long time, and I do not have my cheat book auto math in front of me--but to kinda look at this from the other side of the coin:

Lets say that high gear is 1:1 + a 4.429:1 R&P , net ratio = 4.429

A .71 and .84 high gear is less than one--we are taking away--hence, subtraction, I do remember that much from the pre algebra course I took ( I am not a mathematician, and do not claim to be, I use cheat sheets all the time)

So this is not a multiplication problem----A 4.429 R&P is being turned slightly more than per engine rotation, or, .29 times more or put into a % --29% more if the input shaft = 1, and it does. Due to the overdriven .71 gearset

With high gear we have taken away tq multiplication--we are now subtracting tq from a 1:1 constant

So now plug this into the equation--

1 - .84 = .16
1 - .71 = .29 These are the difference of our 1:1 constant

We then take these away from our R&P : Because it is being overdriven by this much with our constant of 1 :

3.88 - .16 = 3.72 Final drive The engine is turning 3.72 times per full revolution of the wheel.

4.429 - .29 = 4.139 final drive The engine is turning 4.139 times per full revolution of the wheel.

This is a difference of .419 between the two, not much on paper, but a tremendous amount in practice.

Remember, the high gear is overdriving the input shaft to the R&P, 1:1 is our base constant--if the high gear set was in fact 1:1, but because of the overdriven high gear we must take this difference away from the R&P.
only when the cogs are of a greater value than 1 do they become a Division equation of underdrive, hence, a 12 tooth input meshed with a 27 tooth output you get a ratio, or division of 2.25:1, hence the 12 tooth one will turn 2.25 times for one turn of the 27 toothed one. We are discussing less than one, .71 & .84 This is a division equation of overdrive to be subtracted from the R&P. All underdrive equations are added to the R&P.

The gear sets themselve only multiply torque, this is a theoretical value-they do not multiply themslves, they divide speed for speed.

I hope this makes sense, the term torque multiplication is just that--theoretical tq--the gears are meshed in a multiplicative manner--whatever gear is driving the other it has a constant of 1, no matter how many teeth it has, it is still 1-but by mulitplying 1 you only get the same constant--- therefore-they always divide speed for speed. A division of a ratio of 12:27 is 2.25:1, this does in fact mean that the work output is multiplied by that same constant, 1 x 2.25 = 2.25, or 100 ftlb of tq x 2.25 = 225 ftlb of tq. But work is not tangible, you cannot hold horspower in your hands ( the numerical value) so this stays in theory only.

But I also do admit that I could be wrong--as I stated I am no mathematician, and even I am staring at my avatar a little too much (IMG:http://www.914world.com/bbs2/html/emoticons/wub.gif)