Sprockets
 

Dunno if anyone can explain or help me out here, I'll attach a diagram and go from there..

A poor diagram I know but, I have a buggy, just wondering how I work out the ratio considering the 17 drops to a 14 on the same shaft, and it also drops in chain pitch.. Don't think I have enough low end power, having to ride he clutch a little while also pushing a few revs just to take off in first. Its running a 6speed VFR bike engine. Also does tyre size effect the ratio?

I'm not good at working out ratios (my brain doesn't like doing the flip-flop of math conversions) LOL but to answer the last part, yes, tire size does factor in. The larger the OD of the tire, the less "bottom end" power you will have. smaller tires, a larger final sprocket, or a larger initial sprocket on the engine will all help get quicker response.

This is a brain teaser.
Somebody slap me with a trout if I'm wrong but it would seem that what ever happens on the jack shaft is irrelevant. It's the motor turning in relationship to the axle that is the final ratio. If I'm correct in this then you have a final drive ratio of 2.47/1. That's pretty tall gearing and would explain your lack of low end power. Going up on the axle sprocket would give a more favorable ratio for low ends.

And while you at it... some stuff to really get the brain cells cooking!
Effects of different size tires......
In order to get a true picture of gear ratio one must use tire circumference in the equation. If you run the same gears as someone else but you have 24" rear tire circumference and someone else has 22", the true ratios are not equal. To figure the true ratio you will want to use IPR (inches per revolution) which simply means the number of inches a buggy will travel per each revolution of engine.
To figure IPR use this equation: Rear Tire Circumference (x) time’s clutch gear (Y) divided by rear axle gear (z) = IPR. (X*Y)/Z =IPR. For example if one buggy (A) was running a 16t trans gear with a 39t rear axle sprocket gear and their tires were 24" the equation would be 24 x 16 / 39 = 9.84
9.84 is the IPR. This means the buggy would travel 9.84 inches per revolution of engine. If another buggy (B) had the same gears but had 22" tires, look what happens. 22 x 16 / 39 = 9.02. Approximately ¾ of an inch for every engine revolution. Don’t confuse engine revolutions with motor rpm’s. we are talking revolutions at the trans sprocket here! Even though both these buggy’s were using the same gears they were not running an equal set-up. Buggy [A] would be faster on the straights, Buggy [B] would be faster coming off the corners
. So what if buggy [B] wanted to run the same IPR as buggy [A] but still use the same tires? To figure that we can replace buggy B's 39 with buggy A's IPR. 24 x 16 / 9.84 = 39.024 This means that for things to be equal, buggy B would need to change their rear axle gear to a 40 or 41 .When the numbers fall into a fraction go to the next highest tooth count.

That's why i can't figure ratio's too well when there's more than an initial and final gear. I think i may have to slap you with a trout ... :dunno: :rofl::biggrin:

With his initial drive-jackshaft it's going 1 to 1. (17-17) if he changed from a 17 on the jackshaft, to a larger sprocket--let's say 34 to keep my brain from melting ... then he's cut the jackshaft revolutions in half. the adjoining jackshaft sprocket going to the final sprocket would also be turning half as many times per engine rev. so he'd be revving higher to get up to speed thus a bottom end boost.

In a two sprocket system--big on the engine means top speed, big on the final means bottom end and vice versa. the jackshaft sort of messes this up of course. i get why it's there, especially in this instance, is to get sprocket alignment and probably useable sprockets on the axle in relation to the motorcycle engine because of the different pitches. been looking forward to figuring this out myself when i decide to do something with the quad's 250 sitting out back. have to run the same setup.

Cheers for the input guys. Yeah I figured because of the jack shaft having two different sprockets the ratio would be altered, as they obviously don't do a revolution at the same speed.. Ad it's running a Vfr750, the rear axle is in two, and joined into a bearing housing containing my 42t sprocket. The problem was it had a diff in it with fwd neutral reverse, but the ratio was like 6:1 and therefor I'd get into 6th ad it would still be pulling and revving too hard.. So we removed the diff, kept the housing and used the extra sprockets. But I reckon I need the ratio 3.5 : 1 at least

Have any pics I'm in the middle of a vfr 750 build my self

A couple sites to check out with calculators in case you don't want to melt your brain.

http://www.bcot1.com/karting/
http://www.compgoparts.com/Technical...Calculator.asp
http://www.fasttrackraceway.com/calc/index.php

Nice Math CKAU, it has become a lost art with the computer.