What we see here is a linear progression of the HP value. As the TQ is constant and the RPM value obviously cannot skip numbers to a higher value, the graph is straight forward. HP=TQxRPM/5252. This is the power and tq as an engine that operates at 100% VE (see my first post in this thread) throughout its rev band.

This measurement would happen at our perfect engine's crankshaft, or on the dyno of a car that stays in a single gear.

Remember now, that this crankshaft couples to the rest of the powertrain.



This graph is representative of the forces seen at the contact patch. If the RPM value is thought of as how fast our wheel is spinning, and we know that the HP value does not change (half the shaft speed, double the TQ). Whats the problem with this graph? The engine in this example is running at the same RPM, here it is our transmission that is magical or CVT, I can't tell.

Let's apply KISS (keep it simple stupid) here. F=m*a. From the above TQ numbers we find the force by dividing the torque by the distance from the axis of the wheel to its outer most edge, a constant. We now have a force, and our theroretical vehicle's mass is constant. From this it is easy to calc our accleration, for this example our mass is 1200 and the radius is 1.5.



Now, here is where it gets interesting, in the first example, we had an engine, in the second, an engine operating at constant RPM indicating a decreasing TQ number at the wheels. Let's combine the 2.

*thinks for a few minutes on how to do this*

Lets take the HP numbers generated by the first graph, and arbitraraly select two final drive ratio's (FDR), 3 and 4, or a low gear and a high gear. Our formula looks like this then

Here is the graph of those 2



From the equation F=m*a we can determine accleration as previously calculated, and for a constant value of torque, we would have constant acceleration.


Did I screw it up? Probably, but at least I gave it a shot.