[torsen] LSD in a rear wheel drive
QSHIPQ at aol.com
QSHIPQ at aol.com
Fri Sep 26 18:17:16 EDT 2003
An open diff will distribute torque 50/50 always regardless of tractive
force. If the lowest wheel tractive force is 10lb/ft, then the other wheel will
only produce a maximum 10lb/ft regardless of *actual* traction (traction
In your one wheel on ice scenario, the wheel on pavement will only provide
equal torque (tractive force) to the wheel on ice. IOW, if the wheel on ice is
providing 10lb/ft of tractive force, then the wheel on pavement also provides
10lb/ft of tractive force.
In the torsen with a 3:1TBR if the lowest wheel tractive force is 10lb/ft,
the torsen can distribute torque up so up 3 times that force to the apposing
wheel, or put another way, one wheel can support 30lb/ft of torque, with the
lowest tractive wheel supporting 10lb/ft. Remember too tho, that a Torsen is a
50/50 split diff (just like open or locked) up to wheel slip, so torque *shift*
is only 20lb/ft in a rwd car.
In your one wheel on ice scenario, 10lb/ft is supported by the wheel on ice,
30lb/ft is supported by the wheel on pavement.
WRT torque ring gear vs T1/T2, this list has come to the conclusion that any
torque in excess of tractive ability goes to spinning up the wheel with the
least traction. T1 + T2 = Trg where T1 = tractive force + slip ((+ gearing) -
frictional losses)) and T2 = tractive force + slip ((+ gearing - frictional
So in the scenarios above, the open diff will support 20 lb/ft of torque,
then the wheel on ice will spin up. The torsen will support 40lb/ft of torque,
then the wheel on ice will spin up.
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