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This is certainly an area where dyno operators can mess up the results.
Onward, thrust is torque modified with gear ratios and tire radius.
Thrust at the rear tires (along with traction) is what should be used
to determine performance. There is a limit to how much thrust the tire
can apply to the road and excessive thrust will give wheel spin.
I've raced several "muscle cars" with mega torque that spun their tires
while I left them behind. Too much torque at low rpm.
Sticky tires give better results.
I plotted out the thrust curves for my modified S2000.
For interest I added the thrust peaks in each gear for the
Dodge Viper SRT-10 2003 model and the 350Z 6 speed 2006 model.
The Viper developed 482 ft lbs rw torque at 3700 rpm and dropped off after that.
The 350Z developed 226 ft lbs rw torque at 4900 rpm and dropped off after that.
My S2000 developed 264 ft lbs rw torque at 8200 rpm.
Using the equation: Thrust (lbs) = torque (lb ft) x gear ratio x 12 inches/tire radius (ft) inches.
I created the following chart. Only the torque peak in gears are plotted for the Viper and 350Z.
(double click on image for larger view)
Notice at 128mph the Viper thrust peak in 5th gear is very close to mine.
The viper thrust is dropping at that point and mine is increasing.
Almost twice the torque for the Viper but with gearing it results in
equal thrust at 128 mph.
If I knew the aerodynamic drag I could calculate the max speed obtainable.
If I had a salt flat I could find top speed and calculate the drag.
Where the aerodynamic drag curve crosses the thrust curve is max speed.
Vipers are well know to not race in 6th and it is obvious why.
The 713 pounds of thrust max in 6th will not over take the aerodynamic drag at those speeds.
My personal take on this:
1. Less thrust initially gives better traction similar to traction control.
2. Torque in the higher rpms with proper gearing gives thrust where you want it like F-1 cars.
Enjoy.
Please note every car has different torque values and these were just two I found dynos for on the internet. They may not be accurate in describing the vehicles and were included for discussion purposes only.
Good driver can modulate throttle at launch to keep tires from spinning. For a given peak power (and weight), a flatter torque curve will always be better and faster.
Good driver can modulate throttle at launch to keep tires from spinning. For a given peak power (and weight), a flatter torque curve will always be better and faster.
Some people call cars like the S2000 (cars with narrow power bands) "inertia cars." The name fits, because once you get them into their power band in low gear you don't want to drop out of the power band until you're ready to stop. With this type of car the fastest acceleration through first gear (to get to the power band) usually requires tire spin. The time to accelerate to the power band without spinning the tires can be much longer than the time lost due some controlled tire spin. The quickest launch with the S2000 has the tires getting grip right at point where the car is at the base of VTEC (6k). If the motor drops much below 6k during the launch times will increase. That's just the nature of inertia cars.
When it comes to cars with fatter torque curves the situation is different, and (at least with anything other than drag tires on the car) a good driver can get the quickest acceleration possible by keeping the tires operating at their optimal percentage of slip (which is usually very small, around 5%).
A fatter torque curve with equal peak power means that the engine with the fatter torque curve is making more power everywhere except at the peak, so of course it's going to be faster. What if vehicle A has greater peak power but vehicle B has a larger area under the torque curve? Then which would be quicker and which would be faster?
Apparently I wrote the equation out wrong in the post. The number was less than one with a tire radius larger than 12 inches. Good pickup. I'll change it. Bigger tire radius results in less thrust all else being equal.
Another way to interpret RED MX5's comment is this:
Torque is expressed in ft-lbs (i.e., lbs * ft). Thrust, at least in the way it's used in engineering (as in NASA) is a pure force, for which lbs is appropriate. So a simple, consistent way to do this calculation is
thrust = torque * gearing / diameter
(because gearing is unitless). Multiplying by diameter gets you lbs * ft^2, an odd combination.
It would also make sense to use overall tire radius instead of diameter, but that's merely a consistent factor of 2 everywhere so comparisons are valid.
If you think about extremes, this becomes clear: for a given power train (engine and gearing), a very small diameter wheel will spin more easily than a very large wheel, given the same coefficient of friction for the tires. By extension, if you assume the tires won't slip the small wheel will generate more thrust, i.e., acceleration. Ultimately, this is why first gear is shorter than sixth. HPH
05-05-2006, 10:32 AM
Torque, smorke. Let's talk about THRUST
What if vehicle A has greater peak power but vehicle B has a larger area under the torque curve? Then which would be quicker and which would be faster? 
