Limited slip differential, how it works (Revised Edition 2.0)

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  • เผยแพร่เมื่อ 21 พ.ย. 2024

ความคิดเห็น • 12

  • @amegatron07
    @amegatron07 2 หลายเดือนก่อน +1

    Thanks! Quite a useful explanation!

    • @sikasabumotorchannel
      @sikasabumotorchannel  2 หลายเดือนก่อน

      Thank you for watching the video and for your comment.
      If you liked it, please check out my other videos as well!!!

  • @TuanNguyen-ix1tk
    @TuanNguyen-ix1tk 2 หลายเดือนก่อน +1

    Hey bro, What software do you simulate it with?

    • @sikasabumotorchannel
      @sikasabumotorchannel  2 หลายเดือนก่อน

      Thank you for watching the video and for your comment.
      If you liked it, please check out my other videos as well.
      I use a very old 3D modeling application.

  • @blindobserver
    @blindobserver 2 หลายเดือนก่อน +2

    Actually the least misleading video on differentials on youtube that I've found so far. I have been trying to find a video explaining how a clutch type LSD overcomes the zero grip situation compared to gear type differentials, but I have found nothing. I know both can be preloaded, and that preload seems to be the appliable torque to the non slipping wheel, but there is also a general consensus that clutch diffs are not hampered by this problem whereas gear diffs are. I don't understand this and I cannot find proof.

    • @sikasabumotorchannel
      @sikasabumotorchannel  2 หลายเดือนก่อน +1

      Thank you for watching the video and for your comment.
      As you mentioned, I also couldn't find any videos or websites explaining the preload on gear-type LSDs.
      So, from here on, this is just my guess.
      I think preload differs depending on the intended use of the LSD.
      The preload on gear-type LSDs is set during the manufacturer's development, and as far as I know, there are no gear-type LSDs where the preload can be adjusted.
      Since gear-type LSDs are installed in mass-production cars for public roads, the preload is set to either zero or very low in order to ensure comfort and safety. However, in exchange, the response is not good.
      On the other hand, clutch type LSDs are designed for use on racetracks, so I believe they improve response at the expense of comfort and safety.
      Thank you.

    • @blindobserver
      @blindobserver 2 หลายเดือนก่อน +1

      @@sikasabumotorchannel th-cam.com/video/CIH0V8pg8jk/w-d-xo.htmlsi=ZFU884M9SfH2lWsr&t=177
      This video also showcases that typical clutch limited slip differential cannot reliably overcome the zero grip situation. This is likely due to the hill applying too much reverse torque to overcome the preload of the clutch LSD...
      If this is generally the case, what's the benefit of a clutch LSD compared to a gear LSD? How much preload is needed to overcome hurdles like this, and does that make the clutch LSD too uncomfortable for road use? Does the same apply to gear LSD? 😅

    • @sikasabumotorchannel
      @sikasabumotorchannel  2 หลายเดือนก่อน

      Your issue is probably not preload but the locking ratio.
      Around the 3:00 mark in the video you shared, the differential locking ratio of the dark green BMW is shown as 25%. This likely means that only up to 25% of the engine's torque is being transferred to the not slipping wheel.
      As a result, the car cannot climb steep if 25% of the engine's torque isn't enough.
      The silver BMW, on the other hand, uses an electronically controlled LSD, which can adjust to a higher locking ratio as needed, allowing it to get going.
      Some race-spec clutch-type LSDs can have a 100% locking ratio.
      Gear-type LSDs, however, cannot increase the locking ratio in the same way as clutch-type LSDs.
      Thank you.

    • @blindobserver
      @blindobserver 2 หลายเดือนก่อน +1

      ​@@sikasabumotorchannel The car (BMW M5) has 500hp and 520Nm of torque. The engine torque is definitely enough to climb this hill. It may be that the driver would have to get on the accelerator more to get the LSD to start to lock.
      However, in this case, the LSD is working like an open differential, as if the car was rotating around the rear wheel with traction. All the power is going into the right rear wheel. The LSD has to encounter resistance to apply enough torque to the pinion and thus engage the clutch packs, especially if there is no preload. This is my understanding thus far, but I still don't understand the percentages properly.

    • @sikasabumotorchannel
      @sikasabumotorchannel  2 หลายเดือนก่อน +1

      Sorry, my last comment didn’t seem to answer your question. I’ll correct myself.
      It’s a bit complicated, so please watch my video 'Limited Slip Differential, How it Works (Revised Edition 2.0)' while reading this.
      Please pause the video around 4:00.
      We assume the preload of this LSD is 0.
      Because the right tire is slipping completely, the right side-shaft is rotating while the left side shaft is stationary.
      Since the preload is 0, the four pinions rotate and absorb the difference in rotation between the left and right side-shafts. It operates the same as an open differential.
      When the driver presses the accelerator and the engine torque increases, a rotational force is applied to the pressure ring in the direction of the red arrows, but since the right-side tire is completely slipping, no reaction force from the road is applied to the pinion shaft. In other words, the force in the direction of the brown arrow does not occur.
      As a result, the pinion shaft is pushed and rotated by the pressure ring without resistance, so the pressure ring does not expand, and the clutch does not engage, meaning the engine torque is not transmitted to the left side-shaft. Therefore, the car cannot move.
      If a certain amount of preload is set, the clutch will slightly engage, and engine torque will be applied to the left side-shaft, generating the force in the direction of the brown arrow. This will cause the pressure ring to expand, and the clutch will engage.
      However, I don't know the preload setting for the M5 in the video you shared. But it is likely to be a small preload value.
      Does this answer your question?
      Let me know your thoughts.
      Thank you.