|
For what it is worth, I posted the following on this forum a few months ago:
Certainly, HP=(T*RPM)/5252. And it is true that torque is what accelerates because torque is a force acting on a moment arm (hence the units of foot-pounds for you Americans, newton-metres for most everyone else :-)). But there is more to the story than that. Torque at the flywheel is one thing. It is the torque at the drive wheel that does the acceleration.
In a typical engine, torque is developed primarily during the burn time in a cylinder + a bit after due to PV expansion. Because of combustion burn time, valving and whatnot, a typical engine has a certain piston speed that generates the most force. At piston speeds (ie revs) slower or faster than that, combustion is not optimum and the generated force is less. Now that piston is acting on the crank via a moment arm, giving a time averaged force acting on the moment arm, ie torque at the flywheel. It is a function of engine speed. Let's say it is
100 ft-lb at 1500 rmp,
200 ft-lb at 3000 rpm, and
100 ft-lb at 6000 rpm.
Run that through a transmission and a differential and you have the torque at the drive wheel. If the tranny is in top gear (not overdrive), the ratio is 1:1. If the rear end is 4:1, then the torque at the drive wheel 4 times the flywheel torque at 1/4 the rotational speed, ie
400 ft-lb at 375 rpm,
800 ft-lb at 750 rpm, and
400 ft-lb at 1500 rpm (rpm at the drive wheel).
If you were in a lower gear so that the tranny was 2:1, then double all those torque numbers and halve the drive wheel rpm. So if you were running along at a given speed at 3000 engine rpm in top gear, 800 ft-lb is the torque you have available for acceleration. Your road speed would be the drive wheel rpm x the wheel radius. If the drive wheel has a 2 foot radius, the force at the wheel is (from T = F x distance)
F = T / distance = 800/2 = 400 lb.
Using Force = Mass x Acceleration, or F=ma/g, where g = lbm ft/(lbf s^2) (to get the units right), we get for a 2000 lb car,
acceleration = 400 / 2000 * 32 = 6.4 ft/s^2.
Of course, in real life the force also has to overcome friction (air and rolling) so net available force is reduced by that much.
If you droped down in gearing to 2:1, then for the same road speed, you would be at 6000 engine rpm at 100 ft-lb engine torque and 800 ft-lb at the drive wheel (2x multiplier from the tranny and 4x multiplier from the rear end). So at the same road speed, the torque at the drive wheel is the same at 3000 engine rpm as it is at 6000 engine rpm - for this engine with the given torque characteristic. So acceleration would be the same! If the torque curve were flatter, say 150 ft-lb at 6000 rpm, then the torque at the drive wheel would be 2 x 4 x 150 = 1200 ft-lb, the force at that 2 ft wheel would be 600 lb and the acceleration would be 600 / 2000 * 32 = 9.6 ft/s^2.
Anyway, the point of all this is (a) to show how torque translates to acceleration and (b) to show that the shape of the torque curve is what determines the best shift points, etc. The trick is to match the gearbox to the engine characteristics for the task at hand. Personally, I liked my old Volvo 340 with a continuously variable tranny. It hung around 3500 rpm most of the time and really scooted, given the small engine it had.
You don't really need to know horsepower. All you need is the torque curve, the gear ratios and the wheel radius. Given that, and some friction and mass data, you should be able to calculate acceleration.
Now, although we can do without a horsepower rating (since we can always calculate it from the torque curve), it does come in handy because it is a measure of POWER (ie work per unit time). At a high speed when we are not accelerating much any more, overcoming friction and drag is the issue. You could determine top speed by calculating a force balance but it is simpler to do an power balance. But it is the same thing really; just a matter of taste.
Bottom line: neither a horsepower number nor a torque number is sufficient to determine which is the faster car. You need the CURVE (either torque or hp) and you need the gear ratios, wheel diameter, mass of the car, friction, time lost in shifting ....etc.
I did this from memory (harking back to engineering school) so don't shoot me if I have slipped a digit or something.
Bill
--
Volvo Info Site 1990 745GLE 16valve
|
