# Power and velociy

1. Dec 31, 2003

### kishtik

I cannot imagine P=F v cosa alhough I know its mathematical derivation.

Think a car speeding up from 0 to 30. It cannot have the same acceleration when it continues to 60 km/h. It seems nonsense.

Can you help me understand this better?

Any help is appreciated.

2. Dec 31, 2003

### krab

Why? It seems perfectly sensible to me. Maybe you are confused by the fact that with an internal combustion engine, power does not come on till you are at a sufficiently high engine speed. Try this: With a gear chosen so that you are near maximum power output for your car at so 50km/h, hit the throttle suddenly and note how hard it presses you back into your seat. That's a measure of acceleration. Now do the same at 100km/h. You will notice considerably less force pressing you into your seat at the higher speed.

3. Dec 31, 2003

### Staff: Mentor

Its much easier to think about it with an electric motor as it has a much flatter torque curve as krab implied.

4. Dec 31, 2003

### pallidin

Just a comment, but I think of it this way:
If 30 to 60 mph acceleration only required the same energy as 0 to 30 mph, than space travel would be far, far easier, and anyone could do it in their own backyard.

5. Dec 31, 2003

### Arcon

The torque delivered by the engine is a function of R.P.M. As the RPM increases there is a power peak range where there is a maximum power being delivered by the engine. As the rpm increases past that point the power starts to drop off.

There are some nice diagrams which illustrate this concept for model engines --

http://www.mh-aerotools.de/airfoils/engingeperformance.htm

As for the relationship P = Fv this applies in this case by noting what the force is. The torque on the wheels is being delivered the engine which exerting a torque by applying a force to the crankshaft through a lever arm. It can then be shown the relationship is

P = T*omega

where T = torque and omega = angular frequency.

Why the power/rpm curve has the shape that it does is a whole different and very interesting question. I've never looked into this myself. Perhaps someone has a good answer.

6. Dec 31, 2003

### Bob3141592

LOL. Still, that doesn't stop some of us from trying!

7. Jan 1, 2004

### Staff: Mentor

Not really. For low speed travel, a=f/m. A rocket actually accelerates FASTER the faster it gets because its mass is decreasing.

8. Jan 2, 2004

### Arcon

In which case one uses dp/dt = d(mv)/dt = m*dv/dt + v*dm/dt

9. Jan 20, 2004

### kishtik

Lots of thanks.