Quote:
Originally Posted by sperry
3) THE CENTER DIFF USES A PLANETARY GEAR. THE RATIO IS FIXED. NO AMOUNT OF LOCKING OF A CLUTCH CAN CHANGE THE NUMBER OF TEETH IN THE GEAR. This of it this way: if the motor makes a maximum of 300ftlbs of torque, you will *never* get more than 105ftlbs to the front wheels and never more than 195ftlbs to the rear wheels.
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Scott, I realize Alex's posts were rambling and while your above statement may be true in ideal conditions on paper as was my "Pin locking brake system"
, perhaps you should read your referenced article again...
A differential allows for differing speeds on each of the output shafts. The ratio of the gears in a planetary differential related to output torque is only absolute when it is fully open.
As the article clearly points out, under partial to full locking, the actual amount of torque at each end is a function of a number of different factors.
Quote:
What is the torque split when the LSD clutch is fully engaged and the driveshafts are turning at the same speed? Note: This question only applies to a clutch style LS mechanism. A viscous LS mechanism cannot transfer torque without a speed difference.
The answer to this question is: It depends. The answer is neither “35:65” nor “always 50:50”. Often literature (Subaru’s and others) will mention something like "equal torque distribution" but this is an inaccurate simplification of a much more complex process. A more considered answer would be:
It depends on:
1. The front/rear weight distribution of the car
2. Any dynamic loading of the axles (due to acceleration or deceleration)
3. The available friction at the tires, which depends on the surface under the tire (ice snow gravel tarmac) and the tire itself
4. The amount of lock (the clutch engagement force)
5. The torque being output by the engine
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