Rotational torque

If you have made a graph based on your engine's torque curve, add the new gearset, and you should be able to see it too.
 
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For those unable, or unwilling, to do the math, here is a simple proof of concept experiment.

Set up an axle with 2 gear hubs on it.

Place identical axles either side capable of holding drive gears. Attach a 1 foot lever on each, facing away from the driven axle. use the same length chain between them.
Install identical gear sets to each side to check the balance of the system. Attach a 5 pound weight to the end of each lever. Adjust to make balance perfectly.
Now change one side to a larger driver gear with the accompanying driven gear to make the same speed ratio.
Swap gearsets side to side to confirm the results.
Go back to the original gearsets to make sure nothing changed.
 
I was thinking a beam torque wrench on the engine and one on the axle .
would be a reasonable experiment .
 
I was thinking a beam torque wrench on the engine and one on the axle .
would be a reasonable experiment .
I see 2 potential problems.

Resolution of 2 torque wrenches
Introduction of too many variables, including the worst.

The human variable.
 
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For those unable, or unwilling, to do the math, here is a simple proof of concept experiment.

Set up an axle with 2 gear hubs on it.

Place identical axles either side capable of holding drive gears. Attach a 1 foot lever on each, facing away from the driven axle. use the same length chain between them.
Install identical gear sets to each side to check the balance of the system. Attach a 5 pound weight to the end of each lever. Adjust to make balance perfectly.
Now change one side to a larger driver gear with the accompanying driven gear to make the same speed ratio.
Swap gearsets side to side to confirm the results.
Go back to the original gearsets to make sure nothing changed.
would you also need levers on the driven to effect the balance ?
I believe I see the concept .
 
Someone should set their dyno up and do 2 separate pulls with "the same ratio" but differing teeth and post the pics.... Make sure the engine is either already fully warm, or allowed to cool, so that variable is gone... Or do 10 pulls each and average them...
 
would you also need levers on the driven to effect the balance ?
I believe I see the concept .
nope.

The weight of the sprockets are the same on both sides of the axle (side facing the drive gear vs side facing away from the drive gear)
With no chain attached, only sprockets, driven axle should be balanced by itself..

Levers on the drive sprocket axle would need to both be level at the start.
If not balanced for the same ratio, move weights in or out on the levers to effect the balance.
 
Maybe if you could do a real simple drawing of your concept, I'm having trouble understanding.
"Rotational torque"? Is there another kind of torque?
Al your are quite right technically.When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia. This relation can be thought of as Newton's Second Law for rotation.
Any force applied in a straight line is measured in force or pressure. And since no rotation occurs, there is no torque.
 
Someone should set their dyno up and do 2 separate pulls with "the same ratio" but differing teeth and post the pics.... Make sure the engine is either already fully warm, or allowed to cool, so that variable is gone... Or do 10 pulls each and average them...
Great idea Ted .
This may be more real world .
Set up kart and dyno with axle connected to dyno input shaft. (Water brake , hydraulic or torque arm style) .
 
The beauty of the simple experiment is it rules out the usual claims of momentum, weight of chain on sprockets, dyno conditions, track changing, etc
Straight up, is a gear ratio a gear ratio?

Or, if you want to be more specific, Is the speed ratio also the torque ratio?
 
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As most with kids will already know, by the time their age reaches double digits, the answer to the question "why" of "because I said so" ceases to be acceptable.

Our system is not only a speed reducer, it is also a torque multiplier.
While it is generally acceptable to use the speed ratio as the torque ratio in most engineering calculations with an overload protection, or safety factor often exceeding 100%, to see small changes which result in a difficult to explain phenomenon, we must use the actual torque ratio.

As the tooth counts get larger, or the speed ratio gets smaller, the difference becomes less noticeable.

Notice I said tooth counts. The phenomena hold true regardless of chain pitch.

Incidentally, I had to switch my stock answer to "To make you ask questions............... And it's working".

I've heard her use it several times. Lol
 
Here is an additional tidbit to get you thinking.
As we transfer less torque with smaller driver gear sets of the same speed ratio, we need to remember that the demand on the engine is less.
If torque demand is less, engine and kart acceleration speeds up.

Maybe why smaller driver is quicker off the corner.
 
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This is also why I would dispute the bigger gearset, same speed ratio, less friction theory.

If that were the case, when we needed better acceleration, we would simply put a rusty chain on the same sprockets and gain that acceleration.
 
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At a given RPM, at full throttle, the engine produces a certain amount of torque, and it doesn't make any difference what size driver is on the crankshaft. You can take that to the bank!

With a 15/60 gear ratio, (4 – 1) if the engine is producing 10 foot-pounds of torque, that torque is multiplied by the gear ratio, and you would measure 40 foot-pounds of torque at the axle. (In a theoretically perfect no loss system) Let's see, if the engine is turning 4000 RPM, we have 10 X 4000 / 5252.1 = 7.63 HP at the crankshaft. If the engine is turning 4000 RPM (no clutch) with that 4-1gear ratio, the axle would be turning 1000 RPM. 40 x 1000 / 5252.1 = 7.63 HP. So, in this theoretical system, we see that the HP is exactly the same at the engine and the axle.

Now, go back and do the exact same math with a 14 driver, same ratio, (14/56 = 4 – 1)

Now go back and do the exact same math with a 15 driver but a 61 tooth axle sprocket. You'll find that the HP increases at the axle. That's why you get more acceleration, more HP at the axle. The trade-off is, you'll need to turn about 100 more RPM on the top, to be going the same speed. Not always possible because of the HP drop off at peak RPM. In this scenario, we've traded an increase in acceleration at the expense of top speed.
 
At a given RPM, at full throttle, the engine produces a certain amount of torque, and it doesn't make any difference what size driver is on the crankshaft. You can take that to the bank!

With a 15/60 gear ratio, (4 – 1) if the engine is producing 10 foot-pounds of torque, that torque is multiplied by the gear ratio, and you would measure 40 foot-pounds of torque at the axle. (In a theoretically perfect no loss system) Let's see, if the engine is turning 4000 RPM, we have 10 X 4000 / 5252.1 = 7.63 HP at the crankshaft. If the engine is turning 4000 RPM (no clutch) with that 4-1gear ratio, the axle would be turning 1000 RPM. 40 x 1000 / 5252.1 = 7.63 HP. So, in this theoretical system, we see that the HP is exactly the same at the engine and the axle.

Now, go back and do the exact same math with a 14 driver, same ratio, (14/56 = 4 – 1)

Now go back and do the exact same math with a 15 driver but a 61 tooth axle sprocket. You'll find that the HP increases at the axle. That's why you get more acceleration, more HP at the axle. The trade-off is, you'll need to turn about 100 more RPM on the top, to be going the same speed. Not always possible because of the HP drop off at peak RPM. In this scenario, we've traded an increase in acceleration at the expense of top speed.
May need to go back and read the entire thread????
 
Dang now ya'll caused me to reread the whole thing . Click , study and scratch my head .
Its a teaching moment .
 
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