Thread: M3 Dynotest
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      09-08-2007, 09:50 PM   #22

Drives: 2004 330ci
Join Date: Nov 2005
Location: Mich

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Originally Posted by chonko View Post
An AWD has more losses becuase of the number of lossy interfaces, gear interfaces and bearings, etc, as well as the larger rotating mass involved
I was trying to make the distinction between using a fixed percentage of drive-train torque losses as opposed to a given torque loss number.

For the same coefficient of friction, the torque loss at 3000 rpm is the same as the torque loss at 6000 rpm, however your engine torque at both rpm is different. Since the enumerator is the same, with different denominator, it will erroneous to assume that your percentage losses are the same.
You can conversely use the fixed percentage loss and apply it to the various torques at the different rpm, which will give you different Parasitic torque loss. Unfortunately, your torque losses for a given coefficient of friction is not a variable loss, but fixed hence you will be liable to erroneous answers adopting this approach.

With regards to what I wrote as a the formula for frictional losses- it is what we use in the industry, and it forms the basis of managing torque losses.

Torque losses= side force * coefficient of friction*radius of the rotating mass

Side force barring external forces due to bending and curvature on the rotating mass= Normal reaction.
Normal reaction= Weight perpendicular to the plane
Weight perpendicular to the plane(horizontal plane)= Mass * Acceleration due to gravity. (Forgive me for using mass and force loosely didn't realise I was in a physics class: but a mass under the influence of gravity does exert force on a plane)
Side Force*radius = torque.

Torque generally speaking is not a function of rpm, but a function of Force and the radius arm.

Engine Torque does however rely on rpm only because the force generated by the engine is a function of rpm
The engine force= mean effective pressure*cross sectional area of the piston*effective strokes per second*number of cylinders
Torque= Force * Radius.

I can't remember stating Parasitic Power loss, I believe I said "Parasitic Torque losses.
Power represents the ability to do work per unit time.
Torque which I am sure you know very well represents the rotating action of force.

If you have two surfaces in contact, I believe you will have friction.
So like you rightly pointed out, the cumulative frictional coefficient across the drive-train is a lot more than a RWD system. And the rotating mass is more than what is applicable to a RWD system.

..... but this one goes to eleven!