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 Turning causes leaning?
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TonicBIA
Male Senior Member
382 Posts


Arlington, Va
USA

Triumph

Sprint ST

Posted - 03/28/2012 :  1:00 PM
James,

I believe you and this poster are working under different models. You're assuming a fixed radius for his examples, he's not. I believe if his argument was presented as well as yours that might be clearer. Of course if I'm wrong on this I'd echo your last sentence.
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shaddix
Ex-Member

Posted - 03/28/2012 :  1:11 PM
I'm not trying to disregard any facts.. just trying to understand..

Let's keep this simple.

Statement: Decreasing speed in a fixed radius turn will decrease lean angle.

Question: Why? What force moves the mass of the bike to the outside of the turn more in line with the contact patch of the tires?
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/28/2012 :  1:16 PM Follow poster on Twitter  Join poster on Facebook as Friend  
"Decreasing speed in a fixed radius turn will decrease lean angle."

First, do you agree with that statement?
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shaddix
Ex-Member

Posted - 03/28/2012 :  1:22 PM
quote:
Originally posted by James R. Davis

"Decreasing speed in a fixed radius turn will decrease lean angle."

First, do you agree with that statement?



Not necessarily. Here's why:
Decrease in speed in a fixed radius turn means that there is a decrease in lateral acceleration towards the inside of the turn. Unless the bike's lean angle decreases, the bike will fall to the inside of the curve as gravity pulling it down on the inside overwhelms the decreasing centrifugal force pushing it towards the outside.
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/28/2012 :  1:39 PM Follow poster on Twitter  Join poster on Facebook as Friend  
That's what I thought was going on in your mind.

I will attempt to answer your question because I think you are asking it in good faith rather than simply as an argument.

When you are in a turn of any kind, your front wheel is always out-tracking.



That means that your wheels are trying to ride out from under the CG of the bike. If you increase your speed without changing your steering angle, that out-tracking angle continues to cause the front wheel to drive out from under the CG - only faster.

The resulting lean angle increases.

If you want a 'model' to hold in your head that does not confuse, try this: assume that centrifugal force is felt more at the contact patches than at your head. In fact, your contact patches are traveling faster than is your head. As that centrifugal force increases because of increased speed, your contact patches are pushed even harder toward the outside of the turn than is your head - resulting in more, not less, lean.
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shaddix
Ex-Member

Posted - 03/28/2012 :  2:03 PM
Ok that's new and weird, I can kind of understand that. Thanks for the assistance, even though I think I'm more confused now than I was in the beginning XD
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shaddix
Ex-Member

Posted - 03/28/2012 :  7:30 PM
Ok so with this in mind then, you would in fact NOT fall down if you could somehow lock the handlebars mid-turn and accelerated or decelerated. You would simply be unable to change direction, but the bike would still lean further if you sped up and lean less if you slowed down.

I'm still not exactly sure if this means the contact patch moves under/away from the mass of the bike or if the bike's mass moves over/away from the contact patch though.
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/28/2012 :  7:48 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Correct. In fact, so long as your tires have traction, the bike CANNOT fall down at counter-steering speed.


quote:
I'm still not exactly sure if this means the contact patch moves under/away from the mass of the bike or if the bike's mass moves over/away from the contact patch though.


I'm not sure why this concept makes any difference in the real world.

In any event, an argument can be made that says that both happen.

Since the tires are not sliding, only the body of the bike is free to move laterally relative to those contact patches. However, since centrifugal force is merely an inertial manifestation, it is also true that the body of the bike tries to continue moving in a straight line and, thus, tries to stay in place (laterally) while the contact patches literally drive away from it.
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shaddix
Ex-Member

Posted - 03/29/2012 :  7:38 AM
quote:
Originally posted by James R. Davis

That's what I thought was going on in your mind.

I will attempt to answer your question because I think you are asking it in good faith rather than simply as an argument.

When you are in a turn of any kind, your front wheel is always out-tracking.



That means that your wheels are trying to ride out from under the CG of the bike. If you increase your speed without changing your steering angle, that out-tracking angle continues to cause the front wheel to drive out from under the CG - only faster.

The resulting lean angle increases.

If you want a 'model' to hold in your head that does not confuse, try this: assume that centrifugal force is felt more at the contact patches than at your head. In fact, your contact patches are traveling faster than is your head. As that centrifugal force increases because of increased speed, your contact patches are pushed even harder toward the outside of the turn than is your head - resulting in more, not less, lean.



question: If the wheels are driving out from the CoG, then why doesn't the bike continue to lean. The lateral acceleration imparted on the bike by the tires is constant with a particular radius and speed. Out tracking or not, the front wheel is still following a line.

Considering that centrifugal force is simply inertia, does not the sprung weight of the rider/bike mass have more force than the wheels?


quote:
Originally posted by James R. Davis

Correct. In fact, so long as your tires have traction, the bike CANNOT fall down at counter-steering speed.


quote:
I'm still not exactly sure if this means the contact patch moves under/away from the mass of the bike or if the bike's mass moves over/away from the contact patch though.


I'm not sure why this concept makes any difference in the real world.

In any event, an argument can be made that says that both happen.

Since the tires are not sliding, only the body of the bike is free to move laterally relative to those contact patches. However, since centrifugal force is merely an inertial manifestation, it is also true that the body of the bike tries to continue moving in a straight line and, thus, tries to stay in place (laterally) while the contact patches literally drive away from it.



Can the contact patches drive away from the mass of the bike, increasing lean angle, without the handlebars turning?

I see what you're saying about the inertia driving the wheels out from under the bike, even though the bike has more inertia. However, when going in the opposite direction, what force drives the wheels back under the mass of the bike, or what force drives the mass of the bike back on top of the contact patches, if not steering?
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/29/2012 :  8:05 AM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
Can the contact patches drive away from the mass of the bike, increasing lean angle, without the handlebars turning?

I see what you're saying about the inertia driving the wheels out from under the bike, even though the bike has more inertia. However, when going in the opposite direction, what force drives the wheels back under the mass of the bike, or what force drives the mass of the bike back on top of the contact patches, if not steering?


I'm detecting a reluctant but positive change in your 'model' of motorcycle dynamics. Bravo!

The answer to both your questions is centrifugal force.

Imagine a broomstick held vertically and balanced on the palm of your hand. Now move your hand laterally (simulating the tire contact patches) and you will observe that the top of the broomstick 'appears' to move in the opposite direction as your hand. In other words, you have just witnessed bike lean in response to out-tracking.

Now if your hand was moved in a circular path (with the leaning broomstick within the circle) at just the right speed, without any additional lateral movement (meaning that you maintain the radius), that broomstick would maintain its lean angle as the top of the stick follows your hand in the circle - and instead of simply falling, the centrifugal force of your spin is holding it up.

For any given spin rate the broom will have and maintain a given lean angle.

Without changing the path traveled by your hand, just changing its speed of rotation, you can observe that the lean angle of that broomstick changes accordingly.

If you slow down the rotation speed (assuming you are maintaining a speed of some consequence), the broomstick stand taller, just like your motorcycle. You can change the lean angle by simply changing the speed.

(Bad example as it lacks out-tracking - JRD)

We agreed early on in this discussion that the lean angle is a function of speed squared and radius and that the functions are independent. You can change either speed or radius, or both, to affect lean angle. By absolutely no coincidence whatever, centrifugal force is also a function of speed squared and radius. And so you see that lean angle is just a manifestation of centrifugal force.

The formula? F = mv2 / r

Said differently,

the 'Second Law of Motion' states that F = ma
where 'a' is acceleration.

And your bike's lean angle is simply: TAN-1(a)
where 'a' is the lateral acceleration experienced in the turn.

These formula prove that your bike's lean angle is simply proportional to speed squared and inversely proportional to the radius of the turn - independently.

That is, you can change just your speed and the lean angle will change (predictably) to a new angle, and will maintain that angle so long as you do not change the radius of the turn (steering).
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/30/2012 :  6:59 AM Follow poster on Twitter  Join poster on Facebook as Friend  
Whenever I post a formula, however trivial, I receive Private Messages suggesting that 'some of us' (readers) don't have the math skills to deal with the subject at hand, and isn't there an easier way to explain things.

I do not apologize for posting simple math formulas. For some, they are the last word as far as proof is concerned, and for others, the surrounding discussion text is sufficient. We have a broad range of interests and skills in our reader base.

Last night I was asked why I insist on using examples using a fixed radius when discussing lean angles. While the poster agreed that simply changing speed would change the lean angle of the bike, in the real world people also change the lateral force - usually at the same time as they change speed - in order to affect lean angle.

Just so all understand ... of course I neither believe nor do I insist that a fixed radius path must be used in a curve. However, I do insist that at counter-steering speeds the rider controls only his or her speed and the path of travel of the bike. There is no 'change the lateral acceleration' or 'change the lean angle' control on a motorcycle.

This is fundamentally consistent with the theme of this thread - that turning causes leaning, not the other way around.
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shaddix
Ex-Member

Posted - 03/30/2012 :  11:27 PM
James!
I found this post by you at the top of page 5 on this thread:
http://www.msgroup.org/forums/mtt/t...&whichpage=5

Now this makes perfect sense. You said "braking is a COUNTER-STEERING input as a result of weight transfer."

"Said differently, braking in a LEFT turn is the equivalent of pushing forward on your RIGHT grip."

I completely understand the transition from leaning the front wheel to TURNING the front wheel when the bike's wheelbase shortens. So EVEN with holding the bars perfectly still in the turn, you'll still get some turn-in with the suspension compression from using the front brake.

Now there's one thing I don't understand. That post makes it looks like you're disagreeing with yourself and that just can't be right.

I think it's just a chicken and egg thing perhaps. You're saying countersteering lets you choose direction, and my understanding of countersteering is that it leans the bike.

Going straight ahead, the bike wants to continue in that direction, when you press on the handlebars, it torques the tire causing it to push the wheel out from under the mass of the bike that is still traveling straight. Now I would call this leaning the bike at this point. And now in order to avoid continuing to fall to the ground, the tire must find an equilibrium where it's forcing enough lateral acceleration to counteract the gravitational pull. So you let up on the press a bit and the bike stabilizes but not so much that the wheel can turn all the way in and stand you back up...

I can't understand how you are saying that the turning comes BEFORE the leaning, as to me in my mind it makes sense that the leaning is what is giving you the ability to turn. Without the lean, you couldn't turn at higher speed because you would just flop to the outside. Now I also understand that turning is required for leaning! You can't lean the bike unless there's a lateral force counteracting gravity!

This is all very confusing, though I guess it's irrelevant how confused I am... just a mental exercise really. When it comes to riding the bike, I know what to do and how it works is irrelevant..

Edited by - shaddix on 03/30/2012 11:39 PM
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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 03/31/2012 :  2:46 AM
quote:
Originally posted by James R. Davis

If you slow down the rotation speed (assuming you are maintaining a speed of some consequence), the broomstick stand taller , just like your motorcycle. You can change the lean angle by simply changing the speed.



I understand this as standing-up of the broomstick as a consequence of the deceleration. It seemed strange to me, so I performed a simple experiment. I went on a playground and sat on a whirl holding a stick in a leant position. A friend of mine helped me rotating the wirl and we did two video clips after few experiments.

First one, I was supporting the stisk from the inside - it stood taller/vertical when we accelerated :
https://picasaweb.google.com/117334...627251385890

Second one , we initially increased the speed and I was supporting the stick from the outside - it fell to the inside when we decelerated:
https://picasaweb.google.com/117334...826622729506

The conditions were not perfect, but hope, we achieved a decent result to show how broomstick in such conditions behaves exactly the opposite to the motorcycle in a turn while accelerating/decelerating.

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rayg50
Male Moderator
2082 Posts
[Mentor]


NYC, NY
USA

Honda

Shadow Spirit 750DC

Posted - 03/31/2012 :  7:22 AM
quote:
The conditions were not perfect, but hope, we achieved a decent result to show how broomstick in such conditions behaves exactly the opposite to the motorcycle in a turn while accelerating/decelerating.

A motorcycle behaves in 2 apparently opposite ways. Below a speed believed to be 10-12 mph if you decelerate while leaned the bike will fall. Above that speed if you decelerate the bike will stand more. I believe that is why the parenthetical comment by Mr. Davis (assuming you are maintaining a speed of some consequence) was included.

In Coney Island we had a ride where you (and maybe 30 other people) walked into a large cylinder and stood against the cylinder's inside wall. The cylinder would begin to spin faster and faster. At some point the floor would drop away and you would not fall with it because the force pushing you out against the wall was stronger than the force of gravity pulling you down.

The speed of travel is a key piece that you need to account for in your experiment (as well as wind and inertia).

BTW, I applaud your initiative. +1.

Ray

Edited to add this link. I hope it does not muddy the water. I would have preferred to link to a video of the ride itself.

http://answers.yahoo.com/question/i...54607AA6h6PS


Edited by - rayg50 on 03/31/2012 7:32 AM
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/31/2012 :  7:29 AM Follow poster on Twitter  Join poster on Facebook as Friend  
Ah. The example fell apart.

The broomstick behaves just as your motorcycle does - at low speed.

That is, it behaves the way it does because there is such low centrifugal force at low speeds that gravity always prevails.

When I said "assuming you are maintaining a speed of some consequence", I meant that you maintained a counter-steering speed.

The broomstick example was apt when trying to show the that contact patches move away from the bike rather than the bike leaning away from the contact patch when initiating a turn (and is then quickly followed by the bike leaning away from the contact patch) - which was in answer to the question about which happens. It was a poor example to then use to demonstrate lean angle being determined by speed and radius in that you cannot attain and maintain the speed necessary to have adequate centrifugal force for the demonstration.

Or, I've been smoking funny green cigs.
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/31/2012 :  7:56 AM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
I think it's just a chicken and egg thing perhaps. You're saying countersteering lets you choose direction, and my understanding of countersteering is that it leans the bike.


This is the fundamental issue being discussed in this thread.

Steering affects direction of travel - whether its direct-steering at low speeds, or counter-steering at higher speeds. It does not determine lean angle. In other words, steering affects the radius of the curve of your path.

Your speed of travel along with the radius of the curve traveled establish a force commonly known as centrifugal. It is that force that determines lean angle.

It is possible to steer the bike in a different path without affecting the bike's lean angle. Steer a tighter path while at the same time slowing the bike down and the resulting centrifugal force can be the same before and after the maneuver which means that the bike's lean angle will not have changed.

But you cannot change the centrifugal force without that changing the bike's lean angle.

It follows than the speed and radius are the independent variables and that centrifugal force is the result - is the effect of speed and/or direction changes, which are the cause.

To some, the idea that steering causes a change in the path traveled is not intuitively obvious. I cannot explain why not. In a private message I had here yesterday, a member insisted that centrifugal force (described as lateral force) changes the radius of a turn, not steering. When I pointed out that never, not once in the history of the universe, did centrifugal force attempt to shorten a radius in that it is merely an inertial force that ALWAYS attempts to increase the radius of a turn to infinity - making the bike travel in a straight line tangential to the curve it was following, he simply would have none of it.

No matter if you are a math guru or not, experience has demonstrated to you that steering changes the path travel. Changing the path traveled (can) change centrifugal force. Centrifugal force determines lean angle. A precedes B which preceded C.

Finally, in a turn *YOU* do not lean your bike - physics does that for you. (Of course I mean the combined CG including the bike and yourself on it.) There is no 'bike lean' control on your motorcycle. At any speed greater than about 10 MPH, you can control only the bike's speed and direction of travel.

But that's enough to control everything else about the bike's behavior.

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rayg50
Male Moderator
2082 Posts
[Mentor]


NYC, NY
USA

Honda

Shadow Spirit 750DC

Posted - 03/31/2012 :  8:21 AM
quote:
But you cannot change the centrifugal force without changing the bike's lean angle.
I take this to mean that if you change the centrifugal force (by changing radius or speed) it will result in a change in lean angle.

The way it reads, I could also take it to mean the opposite.


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shaddix
Ex-Member

Posted - 03/31/2012 :  8:57 AM
A broomstick can't turn it's front wheel in to react to the decreasing lateral acceleration(the bike naturally wants to turn in and stand up). If you wanted to use a broomstick to simulate a bike, you have to use your hand as the front wheel, and move it towards the center of the curve as you slow down, to move the bottom of the broom closer underneath the CoG.

For acceleration, you have no front wheel to turn to the outside to counteract the increasing centrifugal force(push on the inside grip). In order to simulate this with the broom, you would have to move your hand further out away from the center of the turn in order to balance the broom's lean angle with the lateral acceleration.

Now of course, since you're on a whirl, you cannot choose direction. You are choosing lean angle as a result of inputs at the bottom of the broom that determine it's relationship to the mass of the broom further up. This is the same as changing the position of the contact patches in relation to the mass of the bike.


Somehow... I get the sense if you had a light bike rolling along at 9mph with no rider, but you had a remote control for the handlebars and you turned those handlebars to the right, the bike would fall over to the left. In fact, if you had a bike that had no left/right weight change when you turned the handlebars, it would always fall to the opposite direction that you turned the wheel if it was moving at any speed, from 1 to 100..
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/31/2012 :  9:26 AM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
Originally posted by rayg50

quote:
But you cannot change the centrifugal force without changing the bike's lean angle.
I take this to mean that if you change the centrifugal force (by changing radius or speed) it will result in a change in lean angle.

The way it reads, I could also take it to mean the opposite.



Correct. I added the word 'that' to the original post to try to make that clear.

Thanks.
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James R. Davis
Male Administrator
17295 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/31/2012 :  9:33 AM Follow poster on Twitter  Join poster on Facebook as Friend  
The use of a broomstick in the example was simply not a good analogy since I could not simulate out-tracking.
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