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Anyone ever try a 7 tooth front sprocket?

Lets See

E=MC` + longitute * Latitude = C the sum of B

100mph

I would guess

Sorry were C = Counter and B = Bull****

[ April 22, 2002: Message edited by: EgoAhole ]

Count on EgoHole to respond

I just started running a 13t front in the woods. I love it. You will loose some top end but in the single tracks I dont think I will miss it

HELLO..................anyone for pie or errr PI.

Pi are round not square and last time I looked its 3.14 still+- a couple of decimal points. So actually dropping 1 in front is like 3.14 in the rear. Do your math. 13/49 combo yeilds a 3.769 final drive. 14/49 yields a 3.5 final drive. NOW less seee what we get with say 14/52...3.71 pretty dang close aint it?

IM such the gear head.

Shawns Gotta Braaaaiin

Shawns Got a Braaaaaain

NaNa-NaNa-Na-Na

Ya know thats true Pie Vs pi, I myslef love Boston Creme Pi, next to me moms lemon Lime of course.

But I really never met a Pi I did not like

[ April 22, 2002: Message edited by: EgoAhole ]

No...actually it has nothing to do with PI (or pie). It is simply the ratio between the front and back sprockets.

You lose roughly 5 miles an hour off of your top speed.

Loose top end? You should gain top end by going down a tooth in the front. Also the rule a thumb changing one tooth in the front is like changing 2 (or is it 3) on the rear. So going down to a 13 on the front would be like dropping to a 47 (0r 46) on the rear from stock. You're bike should haul on Top end but I would imagine you're going to suffer a lot on bottom end.

Going down one tooth in the front is like going **up** three teeth (roughly) in the rear.

13/14 x 88mph = 81.7 mph

This only applies if if the top speed of a stock 426 actually is 88 mph, and you're hitting the rev limiter with either sprocket.

[ April 22, 2002: Message edited by: holeshot ]

Kfrosty.......

I am not being an a hole but......

If you drop a tooth on the front it is actually like adding a couple to the back.(depending on your front-back ratio currently, it is usually somewhere between 2 and 3)

Just think of it this way if you disaggree.

If the front sprocket is smaller it will need to turn more times to cover the same distance it is current doing........Right?

On the back it is the opposite.....

If you have a 51 tooth and drop it to 45t , your wheel would turn quite a bit more if you pulled the chain around 1 complete revelution right?

Get out your mountain bike and look to see which sprockets you are using to go as SLOW as you can and as FAST as you can if you don't believe me.

Whenever I motox or woods ride I always drop to a 13t front and 15 when it is a high speed open DEZ race.

Thumpin Rock Hucker

I am the Ahole here Oppps I mean EgoAhole

You are correct, Raising Counter 1 = 2 to the back

and Visa Versa.......

All I know is that I lost enough speed in every gear to screw up any whoop timing skills I ever had...

KTMinOZ is exactly right, the best way to figure it out is with ratios. A 13/49 (.2653) is approximatly the same ratio as a 14/52 (.2692)....

You could use pi and the ratio of the circumferences, but it would be a pain in the ass.

John

yeah, just divide the counter sprocket by the rear sprocket and you get a 'point something number'

14/48 (mine) = .2917

.2917 is the amount of rotations the rear wheel will do for every rotation of the counter shaft. if you know the internal gear reductions from your crank, you can figure out unloaded, uninhibited (no wind resistance or friction) top speed. just compare that with a different sprocket size, and you can figure out how much of a drop or increase in top speed you will have.

hmmm....that almost sounds like it would work!

.... and if you divide 48 by 14, you get 3.428, which is the number of rotations the countershaft will make for every rotation of the rear sprocket. Right????

and if you take 35 x 10, you realize i spend 350 dollars a year on hare scramble racing.

plus, 15 x 15 for practice at rausch = 225

plus 30 x 3 for another place = 90

plus 2 sets of tires a year @ 115/pair = 330

plus oil @ 5.50/quart x 1.25 quart per change x 14 changes = 96.25

plus repairs or dumb things totaling around 800

EQUALS

$1891.25 a year in riding. assuming thats IT.

I could be so much more well off without that bike.

If you take the number of revolutions per second that the countershaft turns, multiply this by the gear ratio (14/49) you will have the number of revolutions per second for the rear sprocket/wheel.

If you then multiply this number times the circumference of the rear wheel (with tire) you will get the linear speed of the bike.

The circumference of the rear wheel, and the maximum revolutions per second of the countershaft sprocket don't change.

Therefore, we can set up a direct variation; Speed varies directly with the gear ratio, or speed = k(ratio) where k represents the constants; max rps and wheel cir......

Since 88 is the supposed top speed with 14/49 gearing, k = 308 and you can figure out your theoretical top speed with different gearing.

I have 14/51 gearing, so my top speed should be (14/51)*308 which is 84.5 mph. So I lose about 3.5 mph for 2 teeth.

Just something to think about.

John

P.S. I was trying to stay awake this morning at 3:14 while feeding my 6 week old son, and this is the crap that I ended up thinking about. :-) I think when I saw the time, it got me on the subject.

Or you can take the shortcut -

49/51 X 88 mph = 84.5 mph

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Ok, maybe it didn't start out as a 7 tooth sprocket, but it was yesterday morning...

Went to unload my bike for the Hare Scrambles, and my friend pointed out I was "a tooth or two". Turned out I was missing EVERY OTHER TOOTH! Problem was, I could only find a 13 tooth at the porta-store. Got the holeshot, but than people passed me on the straightaway like I was tied down! Could keep the whoops timed either. Bike wouldn't stall though...

Does anybody know how much top end you lose by dropping one tooth up front??