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Originally Posted by ArrestMeRed99Z28 F=ma and a=f/m are the exact same equation, same as 6=2x3 and 3=6/2 is the same equation. I tend to believe that drivetrain loss is a constant, delta. The mass of the rotating objects in your drivetrain does not change, so neither does the force required to turn them, but they should accelerate a whole hell of a lot faster.
The reason that the 10-15% percent loss stays the same is that when you have a 500hp car you need a beefier (heavier, or more massive) drivetrain so that you can actually put that power to the ground. That's why the percentages are used, for guesstimation purposes.
AJ |
There is a lot of confusion on this topic. You are correct about 2 things:
F=ma and the mass of the rotating objects does not change.
But that is not the whole story. The drivetrain losses are not due to acceleration alone. There are drivetrain lossses even at a constant speed - i.e zero acceleration. If drivetrain losses were entirely due to acceleration, then the max amount of power (at a fixed rpm, say 6000) put down at the rear wheels would be the same whether you had one transmission behind the engine or 1000 transmissions in series behind the engine. Granted it would take a lot longer to accelerate all the moving parts of 1000 transmissions up to speed than it would take to accelerate one transmission's parts up to speed, but once up to a contatnt speed there is no acceleration and therefore not force due to acceleration (f = ma)
What you are overlooking is friction.
Friction is proportional to the normal force between two rubbing surfaces and the coefficient of friction ( a constant for a given material / lube combination)
As power increases even at a constant speed, the normal forces between all rubbing surfaces in the engine / drive train (bearings, gears, etc.) increases therefore the total friction forces increase. This may be non intuitive at first. As people have said, you could easily turn the drive train by hand so it seems that it doesn't take much force. This is equivalent to saying I could easily tie a rope to a 2x2 sheet of plywood and drag it across a concrete parking lot, but if I put a 1000 lb block of lead on top of it will be just as easy to drag at a constant speed. It won't! Given F=ma, it will require a lot more force to accelerate the plywood with the block of lead on it which you would expect, but it will also require a lot more force just to drag it at a constant speed (zero acceleration), once it has been accelerated up to a given speed, due to the increase in the normal force.
I do have a degree in mechanical engineering - not saying this to brag in any way, but rather to show I didn't just make this stuff up...