Timeline for When a car with an open differential turns, why does the inner wheel slow down and the outer wheel speed up?
Current License: CC BY-SA 4.0
21 events
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| Jan 10, 2022 at 19:15 | comment | added | Reese | In the case of a coasting car, what I mean when I say that both wheels are braked by equal amounts is to demonstrate that both wheels will decelerate by equal amounts unless disturbed by an external force. This is also true for a carriage with independent wheels, whether it is horse-drawn or coasting. It is also the same case as on a car when both wheels are driven by the engine, as the diff simply enables the wheels to acquire different speeds while driven. In all of these cases, conservation of momentum is obeyed. Ergo, in all of these cases, external forces must be involved. | |
| Jan 10, 2022 at 18:55 | history | edited | NMech | CC BY-SA 4.0 |
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| Jan 10, 2022 at 18:40 | history | edited | NMech | CC BY-SA 4.0 |
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| Jan 10, 2022 at 17:09 | comment | added | NMech | @Reese also are you considering differential or you are looking for the generic case e.g. for a carriage with independent wheels? Because somehow I managed to include that when it was probably not your intention. | |
| Jan 10, 2022 at 17:07 | comment | added | NMech | @Reese Apologies for wasting further your time, just trying to understand. Did the last three sentences that you feel made abundantly clear that the problem is a problem of kinetics started with "If the car is coasting, then both wheels are braked by rolling resistance by equal amounts."? | |
| Jan 10, 2022 at 17:07 | history | edited | NMech | CC BY-SA 4.0 |
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| Jan 10, 2022 at 16:06 | comment | added | NMech | Apologies for wasting further your time, just trying to understand. Did the last three sentences that you feel made abundantly clear that the problem is a problem of kinetics started with "If the car is coasting, then both wheels are braked by rolling resistance by equal amounts."? | |
| Jan 10, 2022 at 15:12 | comment | added | Reese | That part I understood, but it didn't address my question, hence why I did not pursue the matter or delve into kinematics. My question is, and always has been, about why the difference in speed occurs, which is the kinetics part. I feel this was made abundantly clear in the last 3 sentences of the original question I posted, but this was effectively ignored by the vast majority of answerers in favour of repeating the premise stated at the start of the question. Regarding accepting the other poster's answer - I'd like for his confirmation that I'm not barking up the wrong tree before doing so. | |
| Jan 10, 2022 at 13:38 | comment | added | NMech | @reese Anyway since you have found an explanation in another answer then you should accept it. | |
| Jan 10, 2022 at 13:35 | comment | added | NMech | First of all I think we are viewing the hammer example from a different perspective. The premise is a uniform circular motion (i.e. there is no acceleration). What I was trying to explain is that if two different uniform circular motions of different velocities, then the forces would be different (and still have a uniform circular motion). The contrast example was that if the velocity remained constant and the force changed then the trajectory would no longer be uniform. (I was not trying to explain the redistribution of forces). | |
| Jan 10, 2022 at 13:14 | comment | added | Reese | In the hammer example, I am asking why the velocity increases, not why the tension (centripetal force) increases when the velocity increases. The main thing that didn't not sit right with me is that, for horse-drawn carriages where there is no differential, only an independent axle for each rear wheel, then an increase in rolling resistance at the outer wheel in a turn implies that the outer wheel is inclined to slow down, not speed up. This could also be demonstrated by towing a pair of wheels, without any load at all on top of them. I think kamran's edited answer is most relevant. | |
| Jan 10, 2022 at 13:02 | comment | added | NMech | @reese regarding the rolling resistance there are many factor that can affect it. The vertical force (or load) on the wheel, the slip angle, the pressure in the tires (due to the increased weight)... There are many factors that can affect the rolling resistance. | |
| Jan 10, 2022 at 13:00 | comment | added | NMech | @reese I feel I need to reply to your open question with another question, and hopefully that will help you pinpoint what doesn't feel correct in my answer. In the example with the hammer that I've added, why does the force on the string increase when the velocity increase? And from another perspective would the velocity increase if the man applied more force on the string, or would the circular path change to something else? (I am trying to demonstrate that the kinematics in this scenario take precedence, which I feel you still doubt). | |
| Jan 10, 2022 at 12:39 | comment | added | Reese | I see. What's the cause of the change in the rolling resistance? | |
| Jan 9, 2022 at 20:52 | history | edited | NMech | CC BY-SA 4.0 |
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| Jan 9, 2022 at 14:11 | comment | added | NMech | During coasting there is no additional torque from the the engine. The net sum is zero. When in straight line, the small rolling resistance forces generate opposite torque on each axle which balancce out. However when turning, the rolling resistance changes between inner and outer wheels. | |
| Jan 9, 2022 at 14:09 | comment | added | NMech | @reese I've modified my answer (although I think it will still require some more). I specifically decided to focus on the coasting case, so that there is no tangential acceleration/deceleration. The reason I did that was your comment that "the right wheel must accelerate forward and the left wheel accelerates in reverse" (somehow it implied that the engine offers this torque. | |
| Jan 9, 2022 at 14:00 | history | edited | NMech | CC BY-SA 4.0 |
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| Jan 9, 2022 at 13:05 | comment | added | Reese | I was only wondering about the relevance of vertical forces. My train of thought is: In the diff, torque is delivered equally to both wheels and the reaction torque is the same from both axles in a straight line. In a left turn, the right wheel must accelerate forward and the left wheel accelerates in reverse. Acceleration in either direction requires a net force. The diff delivers the same torque (twisting force) to both axles in the same direction, yet the two axles accelerate in different directions. The only explanation for a change in the net force that the reaction force changes. | |
| Jan 9, 2022 at 11:20 | comment | added | Reese | I thank you for the effort you put into this answer. There are still some unanswered questions, though. Firstly, you said that the diff allows the wheels to distribute the torque requirements, but all sources are unanimous that a differential always splits the torque evenly between the wheels, and only power can differ, due to different rotating speeds. Are you referring to the resistance torque? Do you mean that, in the case of a left turn, the right rear wheel experiences more friction, and this force opposes the forward rotation of the left wheel via the rotating pinion in the diff? | |
| Jan 9, 2022 at 10:06 | history | answered | NMech | CC BY-SA 4.0 |