Originally Posted by Mech33

Hi Ryan, it's not that straightforward if you want to try and explain things with torque and lever arms, unfortunately. Yes, smaller radius means more tire patch force for the same torque. But is the torque the same as before you changed the tire radius (at a given speed)? No, it's not, because the smaller tire radius also reduced the engine RPM at the same time, and generally the engine torque is reducing as RPM is increasing, and when it gets high enough you're forced to shift to the next gear with an even lower numeric gear ratio. So did you end with a net positive change, or not?

Crunch all the numbers and you'll see that it's easier to just compare the engine power outputs rather than mess with engine torque * gear ratio * final drive ratio / tire radius to try and arrive at wheel force. But either method will arrive at the same answer.

For example, here is an analysis I did for the GT4 comparing the force to the ground with the stock 3.93 final drive vs. an aftermarket 4.62 that is available in each gear:

[img]https://dl.dropbox.com/s/0hmyruqm3s675uk/GT4%20Final%20Drive%20Comparison%20-%20Force%20vs%20Gear.png?

As you can see, at a given speed, sometimes the force available is higher with the old stock 3.93 final drive, and sometimes it's higher with the new 4.62 final drive. Can you explain why, at 55 mph, the stock 3.93 car is putting out more wheel force (and will there force accelerate more quickly at that time) than the modified 4.62 car?

The explanation is straightforward: since the 4.62 forces the engine speed to be higher at a given vehicle speed, sometimes the 4.62 car is forced to be in the next gear (e.g., 4th gear) while the 3.93 car is still in the previous gear (e.g., 3rd gear). In general, the 4.62 car always makes more force to the ground at very low speeds, but once you get up to track speeds, which car makes more force depends on which speed you're at. Here's a plot of the *difference* between the curves:

[img]https://dl.dropbox.com/s/wxeu4v702iyaask/GT4%20Final%20Drive%20Comparison%20-%20Available%20Force%20vs%20Speed.png?

Or you could simply compare the power vs. speed curves and arrive at the same relative answer. Here is the same plot, but showing the difference in engine power output vs. speed:

[img]https://dl.dropbox.com/s/8hcvohefx7yjvdj/GT4%20Final%20Drive%20Comparison%20-%20Available%20Power%20vs%20Speed.png?

You'll notice the zero-crossing points are the same for the force vs. speed and power vs. speed plots. That's why I find it easier to just talk about power... since Force = Power / velocity, it's just easier to calculate than Force = Torque * gear ratio * final drive ratio / tire radius.

Originally Posted by Brian Himmelman

Originally Posted by Riz

Originally Posted by reidry

I believe there are some embedded assumptions in your analysis. You're looking at instantaneous acceleration rather than the volume under the curve. For instance, take your first diagram and note that the force under the curve for each gear is higher though the position of the curve on the X (velocity axis) is shifted to the left (lower) as I mentioned before the top speed available in each gear is decreased as the effective final drive ratio increases.

Drag race guys have been doing this for years, higher numerical ratio to get more force at the pavement combined with wider, stickier tires (drag radials) to minimize wheel spin and maximize the amount of force converted to forward acceleration.

For the road racer, choosing the correct final drive ratio for a given track is more complicated and would be unique for each car, engine, tire and environmental conditions.

If you'd like to discuss it offline, shoot me a PM.

Ryan

Originally Posted by Raucky

Originally Posted by mafoofan