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Is it normal wheels go out of true after a fall?

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Is it normal wheels go out of true after a fall?

Old 10-06-22, 11:19 AM
  #101  
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Originally Posted by Yan View Post
You think we are going off topic? We are not off topic, we just dipped down to lower grade level. How are you going to understand grade 8 physics when you don't know grade 7 physics? Now you are complaining about wanting to go back up to grade 8? Is this a joke?
I've tried to child-explain these concepts to you six different ways to no avail. When at first you didn't get it, I dropped down to lower grade level material. Now I have realized you don't know the lower grade material either. Do I have to start dipping into elementary school territory?

Yeah, I'm done. As I said to Sonofamechanic above, if you don't get it any now, you're never going to get it. Unfortunately you are just not there mentally. Dialogue with you is like trying to SAT tutor a kid who skipped grades 7 to 11. He is simply not equipped. Yet somehow he is convinced he is college material. You are a total farce.

Maybe check out some adult re-education courses on www.coursera.org. Since it's a robot taught web course with no human behind it, it will have infinite patience with you, unlike me. Good luck. You're a chemist? Do us all a favor and stay with chemistry.
​​​​
There are so many childish come backs that I could use but, unlike you, I follow the Forum guidelines and (try) to remain respectful. I will not “get” your point because your point is wrong. I assure you that I have had a fine education and I’m a smart guy. I’ll even admit there are times when I’ve been wrong on some things. This just doesn’t happen to be one.

This article is about car cornering but it does use a two wheel “bicycle” model for discussing the lateral forces on the wheel. It discusses “lateral forces” on the wheel extensively throughout the paper.

Here’s a Slow Twitch article that illustrated exactly what I’ve been saying all along. This graph in particular illustrates my point




Notice the big red arrow marked “ground lateral force”? The article is discussing climbing out of saddle but the same physics applies in a corner.

Here’s a calculator that will tell you have much lateral load a wheel will take.

I look forward to you telling me how stupid I am.
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Old 10-06-22, 01:05 PM
  #102  
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….but under the two wheel modeling it clarifies “The model will not consider explicitly the effects of roll movement and camber change”….and (to Yan’s point) it is exactly camber (what we’d call ‘lean angle’) that cancels out lateral force (as the angle of the wheel ‘re-orients’ the forces to be parallel to the rotation of the wheel, or ‘downward’ relative to the wheel. There is no lean in a 4 (or 3) wheel system—in those cases there is absolutely lateral force on a wheel when cornering.
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Old 10-06-22, 03:50 PM
  #103  
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Originally Posted by Sonofamechanic View Post
….but under the two wheel modeling it clarifies “The model will not consider explicitly the effects of roll movement and camber change”….and (to Yan’s point) it is exactly camber (what we’d call ‘lean angle’) that cancels out lateral force (as the angle of the wheel ‘re-orients’ the forces to be parallel to the rotation of the wheel, or ‘downward’ relative to the wheel.
Again, I have to disagree. Ion Preda, the author of the article seems to disagree as well.

hysteresis (indentation), that means the appearance of a force in the contact patch due to the supplementary energy loss in the rubber as it deforms to cover the road irregularities (in the range of 10 μm to 1 mm) when a (driving or braking) torque or a lateral force are applied to the tire or when the wheel is steered (changing its orientation with respect to its displacement direction);
In the discussion of Figure 4 he says]

In the tire–road contact surfaces of the front and rear wheels act the tangential forces Xf and Xr, the lateral forces Yf and Yr, and the normal forces to the path Zf and Zr.
Further down he says

In the hypothesis of a non-violent cornering (with the transversal acceleration an smaller as 4 m/s2), it can consider that the tires’ lateral forces are proportional to the corresponding slip angles:
Yf =Cf αf Yr =Cr αr.
And in the discussion of Figure 6, a model of an avoidance maneuver, he says

Figure 6 shows such an example. The upper plot presents, versus time (s), the steering angles (deg) of the driver wheel (blue) and of the front wheels (green) – in this case, the simulation took into account the compliances of the steering shaft and linkage. The middle plot presents the lateral acceleration (m/s2) versus time (s) – the two plots correspond to the driver’s steering input presented previously (25° max. angle) and to a weaker input (20° max. angle). The lower plot shows the vehicle’s successive positions at equal time intervals (in this example, with the presented driver action on the steering wheel, the avoidance maneuver fails).
There are 20 uses of the word “lateral” in the paper, many of which are referring to lateral force on the tire (and vehicle). Additionally, he discusses slip angle extensively which is a result of the lateral force on the tire.

In the fictional film world, you can see examples of lateral forces on the tires in numerous…actually countless…movies and TV shows. Any time a car slides a corner, spins, slides off the road, etc. so dramatically (!), that is an example of the loss of friction due to the lateral force on the tires. In the real bicycle world, numerous people have experience this limit of friction due to the lateral force and have the road rash and bruises to prove there is lateral force during cornering.
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Old 10-06-22, 10:00 PM
  #104  
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First, we are all in agreement there is lateral force on a wheel in a ‘4 wheel’ configuration where there is no lean angle, or ‘camber’ (in the author’s words). This is what the study is covering. For the few paragraphs of the study that discuss his calculation tool applied to a two wheel application, author specifically says that he is NOT including any factor of camber, so , yes, there will be lateral wheel force in that circumstance as well (although purely in theory because as Yan pointed out much earlier, ANY uncompensated (or matched) lateral force in a two wheel application is not possible…for long…because you fall over. If it helps, think of the bike rider—going into a heavy corner they lean themselves into the turn so that the force of gravity matches the lateral centrifugal (lateral) force….so all the rider feels is a downward pressure parallel to the bikes lean angle….if the rider had ANY unbalanced (unmatched) LATERAL force they would fall in that direction (left or right). If that makes some experiential sense to you (not saying it will) then think that the same is true of the bike…and the wheels as well. Leaning the bike over lets gravity’s downward pull effect a lateral counteracting (“matching”) force to the lateral centrifugal force of a vehicle in a turn. Its why bikes and motorcycles must lean into turns where cars (and tricycles) do not.
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Old 10-06-22, 10:04 PM
  #105  
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Originally Posted by cyccommute View Post

In the fictional film world, you can see examples of lateral forces on the tires in numerous…actually countless…movies and TV shows. Any time a car slides a corner, spins, slides off the road, etc. so dramatically (!), that is an example of the loss of friction due to the lateral force on the tires. In the real bicycle world, numerous people have experience this limit of friction due to the lateral force and have the road rash and bruises to prove there is lateral force during cornering.
So would it be accurate to say that leaning the bike to the inside of a turn reduces but does not eliminate the lateral force on the wheel? I'm envisioning it vectoring some of that lateral force into vertical force.
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Old 10-06-22, 10:27 PM
  #106  
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Originally Posted by oldbobcat View Post
So would it be accurate to say that leaning the bike to the inside of a turn reduces but does not eliminate the lateral force on the wheel? I'm envisioning it vectoring some of that lateral force into vertical force.
The lean is the lateral force. There is also a bit of twisting of the rim upwards because of how the rim/spoke interface works which is a small amount of vectoring of the lateral force into vertical force but, overall, the force of the turn is a lateral force. Take that lateral force away and the bike goes back to a straight line or, alternatively, don’t apply a lateral force and the bike stays on a straight line. Basic first law of motion stuff: “A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.”
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Old 10-06-22, 10:40 PM
  #107  
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I should have specified, vertical relative to the axis of the wheel. Which is partially lateral relative to the ground.
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Old 10-06-22, 11:00 PM
  #108  
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Originally Posted by Sonofamechanic View Post
First, we are all in agreement there is lateral force on a wheel in a ‘4 wheel’ configuration where there is no lean angle, or ‘camber’ (in the author’s words). This is what the study is covering. For the few paragraphs of the study that discuss his calculation tool applied to a two wheel application, author specifically says that he is NOT including any factor of camber, so , yes, there will be lateral wheel force in that circumstance as well (although purely in theory because as Yan pointed out much earlier, ANY uncompensated (or matched) lateral force in a two wheel application is not possible…for long…because you fall over.
Just to be clear, I have never implied that there is any unmatched forces here. His comment was that there is “no lateral force” on the wheel.

If it helps, think of the bike rider—going into a heavy corner they lean themselves into the turn so that the force of gravity matches the lateral centrifugal (lateral) force….so all the rider feels is a downward pressure parallel to the bikes lean angle…
I have zero issue with that explanation. The downward pressure that the rider feels (and bike experiences) is at an angle to the surface and is the result of the sum of the vertical and horizontal forces. We know who is stating that the only force is that force angled to the surface without understanding why it is angled…as it if “just happens”. I understand what the system experiences but I also understand…and am backed up by many, many references…that there is a lateral force acting on the wheels.

if the rider had ANY unbalanced (unmatched) LATERAL force they would fall in that direction (left or right).
Of course. This can happen when the cornering exceeds the tire friction. Lean over too far (or get the center of gravity to a point where it is unsupported) or have the surface conditions change and the wheels lose traction and the whole system becomes unbalanced.

If that makes some experiential sense to you (not saying it will) then think that the same is true of the bike…and the wheels as well. Leaning the bike over lets gravity’s downward pull effect a lateral counteracting (“matching”) force to the lateral centrifugal force of a vehicle in a turn.
Centripetal force, please. But the point here is that there is a lateral force acting on the bicycle wheel. It can’t be neglected. That lateral force is what causes a slight bending of the rim above the contact patch which is analogous to the deflection of the rim when riding without the lateral force. Granted, it is likely only a small deflection as is the deflection when riding the wheel under vertical compression but it is not zero.

By the way, I find these kinds of discussions (mostly) useful. Although some think I’m dumb as a hammer and totally uneducable, I do learn things from these discussion. If nothing else, it makes me do some research. It’s kind of fun to blow out some of the cobwebs.
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Old 10-06-22, 11:03 PM
  #109  
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Originally Posted by oldbobcat View Post
I should have specified, vertical relative to the axis of the wheel. Which is partially lateral relative to the ground.
That is what I mean by the rim bending upward during a corner as the center of gravity leverages the contact point. The graph I included from slow twitch illustrates what I’m saying, although they are talking about side-to-side movement of the rim during out of saddle effort. There wouldn’t be an oscillation of the rim like in that example but there is nothing to preclude the rim from responding in the same way while cornering.
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Old 10-07-22, 06:03 AM
  #110  
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Originally Posted by oldbobcat View Post
So would it be accurate to say that leaning the bike to the inside of a turn reduces but does not eliminate the lateral force on the wheel? I'm envisioning it vectoring some of that lateral force into vertical force.
Exactly. “Vertical” parallel to the lean angle of the bike. And exactly— there is some remaining lateral force (centrifugal I’m referring to as it “pushes” the bike “outward”) which is what the friction of the tire on the road then counteracts—much less than a car tire which has no compensating lean/camber.
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Old 10-07-22, 07:31 AM
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To be clear, the exact same lateral centrifugal force is in play, undiminished, but by angling the bike that becomes more vertical (relative to the bike and, importantly, to the wheel structure.) Its a similar concept to building banking into the curve of a road or even a plane “banking” into a turn. In fact a plane is a great example—the structure of a plane does not have to be engineered to handle significant lateral force nor is it even constructed with sufficient vertical structure to get enough “traction” in the air to make a corner…because it can instead angle into a turn and utilize its structure and surface area to in effect take that corner “vertically” relative to the plane. In an extreme corner you’ll put a small plane almost fully ”over” to 90 degrees and take the full lateral force of a turn as a vertical force. Bicycle wheels are structured similarly in that they rely in a lean to re-orient the bulk of the lateral force of a turn into a ‘vertical’.
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Old 10-07-22, 09:09 AM
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Originally Posted by Kapusta View Post
This thread need less arguing and more spaghetti and bikes with arrows drawn on them.
Originally Posted by cyccommute View Post



.
OK, that will do
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Old 10-07-22, 12:30 PM
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So Ev0lutionz , are you satisfied with an answer yet?
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