Montreal Subaru Club

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:

Kolia a écrit :Jolis article.

T'as lu ce paragraphe ?

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:

Front wheels on dirt and rear wheels on tarmac DCCD at 10% lock: Same premises as before but here it will take 50 ft-lbs to spin the front wheels and 200 ft-lbs to spin the rear wheels. Also let us set the DCCD so that it is only 10% engaged and so will slip at 5lbs. In this situation the point at which the clutch can’t transfer torque will occur before the point at which the front tires will slip. After that point any additional torque is divided in the 35/65 ratio, same as the open diff. That changeover point comes at an engine torque of 33.5 ft/lbs, at which point the actual torque at the axles is 6.7ft-lbs F/28.7ft-lbs R. If we increase engine torque to 100 ft-lbs, then the rest (66.5 ft-lbs) is split up 35/65 so we end up with a total actual torque split of 30 ft-lbs F/70 ft-lbs rear. If we then change the lock to 100%, the clutch can transfer all the torque difference (15 ft-lbs) and our “virtual driveshaft” analogy holds and so the torque difference is 20 ft-lbs F/80 ft-lbs R. So by changing the lock % we can change the torque distribution, even though there was no slip in the system.

Bacon a écrit :Je viens de lire ça en diagonale un peu. homer a l'air de bien comprendre comment le diff dccd fonctionne mais a pas l'air d'être très bon pour l'expliquer.

Le ratio de distribution de torque du diff est fixe 65/35 comme il dit. Mais s'il y a un des 2 essieux qui patine, c'est là que le dccd va faire sa job de lsd et va envoyer le torque à l'essieu qui ne patine pas, la quantité de torque transférée étant en fonction du % de lock du dccd.

Greddy a écrit :Tu parle de quantité de torque transféré quand tu dit au début qu'il est fixe. J'suis encore plus fourré moi là

Bacon a écrit :

Greddy a écrit :Tu parle de quantité de torque transféré quand tu dit au début qu'il est fixe. J'suis encore plus fourré moi là

ok, si on sépare le dccd en 2 morceaux. 1er morceau: un diff mécanique open avec un ratio fixe de 65/35. 2e morceau: un pack LSD avec un facteur de lock variable. Ce pack LSD là peux prendre une partie du 65% qui va en arrière et l'envoyer en avant et vice versa. Est-ce que c'est plus clair?

Kolia a écrit :Mon router a planter mausus... :p

Et ceci:

Front wheels on dirt and rear wheels on tarmac DCCD at 10% lock: Same premises as before but here it will take 50 ft-lbs to spin the front wheels and 200 ft-lbs to spin the rear wheels. Also let us set the DCCD so that it is only 10% engaged and so will slip at 5lbs. In this situation the point at which the clutch can’t transfer torque will occur before the point at which the front tires will slip. After that point any additional torque is divided in the 35/65 ratio, same as the open diff. That changeover point comes at an engine torque of 33.5 ft/lbs, at which point the actual torque at the axles is 6.7ft-lbs F/28.7ft-lbs R. If we increase engine torque to 100 ft-lbs, then the rest (66.5 ft-lbs) is split up 35/65 so we end up with a total actual torque split of 30 ft-lbs F/70 ft-lbs rear. If we then change the lock to 100%, the clutch can transfer all the torque difference (15 ft-lbs) and our “virtual driveshaft” analogy holds and so the torque difference is 20 ft-lbs F/80 ft-lbs R. So by changing the lock % we can change the torque distribution, even though there was no slip in the system.

Marzzochi a écrit :ben moi j'ai tout lus...
pis la chu meler en tabar#$%

trop de monde trop d'expliquation...