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Well I did some back-of-the-envelope thinking about drivetrain loss while watching a Scrubs re-run this weekend.
Unfortunately I can't scan it in so you'll have to bear with me and try to follow along.
Let's set up the system as a mass-spring-damper 2 node system with an input and output torque.
To node 1, we have the input torque from the engine (Te) and the output torque through the damper (b) and inertia (J) to ground as well as the output toqrue throug the spring (Tk) to the second node. From the second node, we have the output torque (Tf) which is essentially the power-to-the-ground.
Te = Tb + Tj + Tk
Tf = Tk
Now if we subititute Te-Tb-Tj for Tk, we get Tf = Te - Tb - Tj.
This makes sense, right? The torque to the wheels is the torque from the engine minus the torque lost though the bearings and inertia. The spring factor disappears because it would only be a factor of the time-response of the system.
Now if we substitute in for Tb and Tj their equivalent torque loss ratings as a function of the rotaional velocity, we get:
Tf = Te - b(Omega1g) - J (dOmega1g/dt)
As you can see, the torque to the wheels equals the torque from the engine minus the friction loss times the rotational velocity of the drivetrain minus the mass of the drivetrain times the acceleration of the drivetrain.
I'll stop short of the laplace domain to say:
And it is proven - drivetrain loss is not a constant number. It is a function of the total coefficient friction throught the drivetrain, the total mass of the rotating drivetrain, and the acceleration/velocity of the drivetrain.
I apologize for my lack of graphics and hope you were able to follow along.
-Ryan
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