Draco | 08-07-2009 06:53 PM | Quote:
Originally Posted by brianau9
(Post 6537679)
of course the acceleration is higher when its at 6000rpm, if torque is constant, hp increase linearly with increase in RPM. And acceleration is proportion to power.
F = ma
P= m*a*d/t something like that.. | Wrong, despite the fact you posted the correct equation (F=MA). acceleration is determined by the force delivered to the ground. This force can be calculated by starting with the torque at the crank, multiplied by the transmission and final drive gear ratios.
Unless we live in a world where the laws of physics cease to apply, acceleration within a gear is directly proportional to torque (after subtracting for other factors such as wind and rolling resistance) and will in fact follow the torque curve of the vehicle. If torque within a gear is constant, the acceleration will also be constant.
In the scenario given, Car B will accelerate faster at that exact point in time. The reason for this is that despite having lower engine torque at 8000 RPM, it will have greater torque delivered to the ground through the drivetrain. As someone said earlier, Car B is "in the right gear" and is taking better advantage of the torque being output by its engine thanks to the gearing multipliers.
This is why sportbikes with 30-40 lb/ft of torque crap all over cars with 10x more peak torque, despite the car weighing only 6x the bike+rider. they take advantage of gearing to spin the engine up (thereby generating insane HP), deliver massive relative force (when compared to car weight) to the ground, and therefore accelerate better. |