Torque vs Horsepower

[brianau9]

Originally Posted by Ferra

what's with all the OT repliess??

let's rethink another scenario: This time, only with ONE car, and assume this car has a COMPLETELY FLAT torque of 300 ft-lbs from 2000rpm up to 7000rpm.

Is the car gonna accelerate
a) much faster
b) bit faster
c) at the same level
when it is running at 6000rpm vs when it is running at 2000rpm? (same gear, no wind drag)

(acceleration measured in increase km/hr per seconds)

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..

[JL9000]

The OP's question is very simple, and in this scenario Car B will accelerate faster than Car A, period.

[freakshow]

Given the OP's example, Car B will accelerate faster.

[Draco]

Originally Posted by brianau9

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.

[cococly]

^It's turning into a Physics lecture.

[!Aznboi128]

Originally Posted by d1

God damn engrish.

engrish for da win

[brianau9]

Originally Posted by Draco

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.

I have been talking about the torque numbers throughout rpm range on the torque curve, but you' re talking about the resulting torque applied to the drive shaft. 2 different subjects. Torque numbers on torque curve have to been looked at together with the rpm, even the torque curve is a flat horizontal line, the acceleration isn't constant like you said. Torque per crank might be the same at 1000rpm and 9000rpm, but the rate of work being done at 9000rpm is 9times higher. So what you are saying is you don't need skills in launching if the torque curve is flat and just dump clutch at 1k rpm?

And yes "acceleration within a gear is directly proportional to torque ", but that's only implies for the torque applied at the wheels. Not taking into friction losses from drivetrain and drag etc, acceleration should be go parallel to the power curve.The rate of acceleration would go parallel to the torque curve.

[yellowpower]

You frickin retards, include the stretch of road and how long its gunna be.

[ToyotaPowah]

IF CAR A SPENT THE EXTRA MONEY IN MODS HE'D BE FASTER. FUCK CAR B

[BNR32_Coupe]

Originally Posted by Draco

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.

this post just KO'd this thread. end of discussion.