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Locked differential and front wheels on ice: As in the previous example, the torque on the front driveshaft that will cause the wheels to spin is 10ft-lbs. Just like in the first example, that is the max amount of torque that can be applied to the front driveshaft. Why? Because the conditions under the front wheels are still the same, the ice will only "push back" with 10 ft-lbs, so the front wheels can only "push" with 10 ft-lbs. This is very important to understand. You can't push against something if it won't push back. But since the driveshafts are locked together, the engine is free to apply more than 10 ft-lbs to its connection at front driveshaft. In fact, think of it as just one large driveshaft, with the engine twisting the shaft at two places (side by side), one with 35% of its effort, the other 65%. Fig 3, the locked case, shows this concept. If the engine applies 100 ft-lbs, what happens? 35 ft-lbs on the "front twist", 65 on the "back twist", but remember the front axle can only accept 10 ft-lbs. If the rear can take 90 ft-lbs before slip, then rest of the engine torque (90 ft-lbs) goes to the rear axle, which is (let us assume) enough to move the car. In this case the torque split is 10%F/90%R (I ignored the case where the rear can take more than 90 ft-lbs we’ll get to that below.)
In this case, what determined what the torque split turned out to be? It was the amount of traction at the axles. But what determined that value? It was the icy surface and the type of tire and the weight the axles. With more weight on the axle, more traction could be achieved (which is why pickup owners put sandbags in their truck beds in the winter). With winter tires, more traction could be achieved as well. So you see how torque split depends on how much torque each driveshaft "can accept" which goes to the first three items in that list: weight, dynamic loading, and friction. Here is another example to make that point:
gravel express
