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Originally Posted by GtiGyver
Now, in order to equalize the speed of both halfshafts, the LSD will internally increase friction, potentially all the way to locking the shafts together.
Any friction here is applied to the system that is already in rotation.
Minimal losses.
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Ah, you're saying the friction is applied between the two half-shafts. Not the half-shaft and the housing. That would make a bigger difference, of course. Which LSD types work this way? Is it mostly the clutch types?
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he diff is open, we agree that the total power on both wheels must equal the input, and this is why locking one wheel causes the other to spin twice as fast
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Yes. Agreed.
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This friction on the brakes is a pure loss, since it will only result in heat, but in no way will propel the car forward, and thus a loss of power.
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But this would contradict what you said above. First of all, any friction is a pure energy loss. The heat generated by a mechanical differential is equally a loss of energy as applying friction through the brakes is. This isn't to say that a mechanical differential would have less friction losses because it is more efficient. But no matter the efficiency, energy is lost.
The bigger issue here is the statement that applying the brakes is pure loss and does not propel the car forward. This is the part that contradicts the statement above.
If we assume a 100% efficient open diff with a 1:1 gear ratio, we seem to both agree on this chart:
Input = 100rpm | left = 0rpm | right = 100rpm
Input = 100rpm | left = 50rpm | right = 50rpm
Input = 100rpm | left = 100rpm | right = 0rpm
If the input shaft in this scenario does 100rpm and you brake the left wheel down to 95rpm using the brakes on that output shaft, would you not agree that the right wheel will spin at 5rpm?
Btw. I'm really enjoying this informative exchange on the efficiencies of mechanical vs. eLSD.
Please let me know if anything in the first post needs fixing/correcting as that is the main purpose of this thread.