Back from hols!
Ringo wrote:Though realistically constant power [Machin addition; from an internal combustion engine] is not possible.
Yep totally agree, I just wanted the theory to be set straight. I'm glad we got there in the end.
I was quite pleased with this "energy principles" explanation of determining vehicle acceleration from the other thread, and now I've managed to convince you of the principles I wanted to include it here as a "quick reference" for other people referring to this thread in the future (who might not have read the other thread):-
Machin wrote:People often talk about brake power and torque as if they are two completely separate aspects of an engine. They are not. They are two different ways of discussing the engine's ability to do useful work (move our car)
The engine takes fuel, burns it and converts the chemical energy stored in it to useful energy at the flywheel. We can express the output rate of this "useful energy" in three basic ways:
kJ per second
Power
Torque x Speed
75 kJ per second is about 100bhp, or 100lbft @ 5250rpm, or 200lbft at 2625rpm, or 50lbft at 10500rpm.
I'm going to avoid using the terms power and torque since they're often misunderstood, I'm going to stick with the energy per second description.
We all know that energy cannot be created or destroyed, only converted from one form to another.
The typical engine can only achieve its highest energy per second output at one engine speed. When we attach this to the differential and wheels this means the car will only be able to transmit this peak energy to the road at one road speed. We can change the road speed by changing the differential ratio or wheel size, but we can't change the amount of energy -it can't be created or destroyed remember.
A refinement to the system is that we can add selectable gear ratios between the engine and the differential. For each additional selectable ratio we add we allow the engine to transmit its peak energy output to the road wheels at an additional road speed. But we don't increase or decrease the energy output rate (because remember energy can't be created or destroyed). If the engine puts out 200 kJ/sec at the flywheel its still 200kJ/sec at the road wheels (neglecting any that has turned into heat along the way), regardless of the gear ratio or wheelsize in between.
This explains why the performance of a typical car is improved by having lots of ratios; it can have its peak energy output rate at lots of different road speeds -at low speeds for good acceleration, and at high speeds for good top speeds, but the gearbox doesn't increase or decrease the energy output rate.
But, if the energy output rate is the same why does my car accelerate quicker in low gears than it does in high gears? Since Kinetic Energy =(1/2).(mv^2) it takes a lot more energy to change the road speed of our vehicle at high road speeds than it does at low road speeds -even if we neglect air resistance (add in air resistance and it takes even more energy!). Therefore our car will accelerate quicker at low speeds than at high speeds for the same energy output rate. [It is also worth noting at this point that a small amount of energy (about 10% on a typical performance car) is also required to increase the rotational speed of rotating parts; gears, wheels, crank shaft, etc).
Since energy output rate and power are directly proportional we can put this another way; 200bhp will accelerate our car faster at low speeds than it will at high speeds. That's just the laws of physics. Remember; What the gearbox has done is allowed us to have the same power output at a low road speed (hence high acceleration at low road speeds) as well as the same power output at high road speeds (hence high top speed), but it hasn't increased/decreased the power or the energy.
The other way to work out the acceleration is the "forces" method, as explained here:-
Machin wrote:F=mA right?
Lets assume that your car has lots of grip (its not traction limited), and that aero resistance (force) is zero (e.g. low speeds). You'll just have to accept that we don't have any rolling resistance in this eaxmple. (EDIT: and that the drive train is 100% efficient -please bear with it).
The mass (and the small amount of rotational inertia of rotating parts) is fixed. In order to achieve a given level of acceleration we need to provide a motive Force at the tyre contact patch (the "F" in F=mA).
In order to determine the motive Force at the contact patch at a given speed you take the engine torque at the engine speed necessary to achieve the road speed, (you can work out the torque if you know the power and the speed at which that power is made since power =torque x speed) you multiply that torque by the gear ratio and the final drive ratio. You multiply this result by the wheel radius -that gives you the motive force at the tyre contact patch.
Yes, the force at the tyre contact patch is increased with a higher gear ratio, but since you've also decreased the speed of rotation (and hence decreased the road speed for a given engine speed), the power remains the same since power =torque x speed
Either method gives the same results if you apply them properly. Hope that all helps.
