When understanding open and locked differential, torque transfer is not entirely intuitive. I created this video to hopefully clear up some confusion on how torque transfer works in both open and locked differentials.
This video is one way of advertising cars and their specific features and its advantage over others. It is possible for marketing people to create similar videos like this to simplify technical terms especially those who will buy cars in Long Island New York.
Excuse me I found something confuse me and I would like to show my opinion. The max force(static friction force) created on pavement is 1000(lbf)(1000*1), and the one created on snow is 300(lbf)(1000*0.3), so the total torque is 1000+300=1300(lbf),according to the video. However, when the lock differential works, both tires rotate in the same speed. The max static friction force of the tire on pavement is larger than the one on snow, so the wheel on pavement may rotate and the other(on snow) may spin. Conclusively, the wheel on snow may create kinetic friction force, which may decrease the friction coefficient from 0.3 to 0.2. It is my calculation that the total torque will be 1000*1 + 100*0.2=1200. Please take a look at it and give me some instruction. Thanks a lot!!
This video is one way of advertising cars and their specific features and its advantage over others. It is possible for marketing people to create similar videos like this to simplify technical terms especially those who will buy cars in Long Island New York.
ReplyDeleteExcuse me
ReplyDeleteI found something confuse me and I would like to show my opinion.
The max force(static friction force) created on pavement is 1000(lbf)(1000*1), and the one created on snow is 300(lbf)(1000*0.3), so the total torque is 1000+300=1300(lbf),according to the video. However, when the lock differential works, both tires rotate in the same speed. The max static friction force of the tire on pavement is larger than the one on snow, so the wheel on pavement may rotate and the other(on snow) may spin. Conclusively, the wheel on snow may create kinetic friction force, which may decrease the friction coefficient from 0.3 to 0.2. It is my calculation that the total torque will be 1000*1 + 100*0.2=1200.
Please take a look at it and give me some instruction.
Thanks a lot!!