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 Turning causes leaning?
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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 03/31/2012 :  10:51 AM
quote:
Originally posted by James R. Davis

The use of a broomstick in the example was simply not a good analogy since I could not simulate out-tracking.



Thank you for this acknowledgement Mr.Davis
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JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 03/31/2012 :  11:29 AM
Thanks for the experiment. What James neglected to mention is that for the broomstick analogy to work, you also need to move the support point towards the centre of the circle.

[ADDED:] I wrote the reply just after reading the original message, but only sent it after all the replies had been posted.

Edited by - JanK on 03/31/2012 1:24 PM
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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 03/31/2012 :  3:11 PM
quote:
Originally posted by James R. Davis


This is the fundamental issue being discussed in this thread.


By this issue, I suppose you mean ''turning causes leaning''. I agree here, but despite it's simplicity there are few interpretations, some of which are misleading and confusing for the reader. Which turning do we mean here, handlebar turning or bike turning? It depends what we choose for turning and leaning. If we speak for handlebar turning then ''turning causes leaning'' is correct. If we speak for bike turning then the opposite ''leaning causes turning'' is also correct. To avoid this ambiguity and questions, it is a good idea to leave simplicity aside and to use more explainatory words to describe this fundamental issue. Here is my understanding : Usually TURNING handlebars is more easy and effective thing to do ( compared to shifting weigths, using trottle, using brakes, road features etc. ) in way to CAUSE LEANING ( initial counter leaning, lean angle amendments or equilibrium lean angle establishment ). Once there is bike LEANING it will CAUSE bike TURNING or deviate from his straight path. Of course, this explainatory chain is difficult to remember, but it makes wording clear for me. Might be there is another shorter and clear wording for this fundament, you folks can propose it of course.

quote:

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.


okay, steering affects the direction/radius of the curve, but how?
Is it direct link ( we turn and straight here we go where our front is pointing ) or indirect? I believe it is indirect influence. Here is my point: Steering input rotates the front tire first. On the other hand due to friction, front tire is placed on the road via it's contact patch. Turning the tire without lifting it means twisting of this contact, rubber distortion. This distortion is the cause for the forces being born which are responsible to push the front tire to the side, to produce track deviation/cornering and instant cornering radius. This radius is product from the push of these forces, you can call them anyhow ( lateral or centripetal ), but most of them are the reason for the deviation and the radius of curvature to be prodused.
Here we can use the popular 'V-e squared then divaded to R-radius equals centripetal acceleration' either way to derive any component from the other two, but that does not change the Newtonian phisics of it, there is a force first as a cause, and then radius as a motion property and result. If there is no force there would be no deviation/radius - imagine your front tire stepping on icy patch, you rotate handlebar but the contact is slipping, no twisting, no centripetal force, no turning, no radius. Based on all above I believe the steering input is responsible for the production of the cornering forces first, and then indirectly the radius. Or said otherwise THE STEERING CONTROL IS POWERFUL CORNERING FORCES CONTROL indeed. If in doubt for this contact patch distortion mechanism, you can read here for example :
http://dinamoto.it/ , follow the links 'PAPERS ON-LINE' , 'The Motorcycle tires' , there is simply explained how steering can produce
cornering forces ( slip mechanism ) and how leaning can produce cornering forces ( camber mechanism ).

quote:

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.


To have a force acting, we should have a reason for it. For example gravity, spring elastic structure, electromagnetic field, body interaction etc. Speaking for the centrifugal force, there is no reason causing it. It is not real existing force which could act over bodies, IT IS FICTOUS, INERTIAL, IMAGINARY FORCE which we accept as existing in order to use Neuton Second Law F=ma in rotational systems.
Otherwise the motion in such sistems does not correspond to the acting forces. What I mean: Let's look at the whirl used to rotate the above used broomstick. Let's look at the whirl rotating, and the broomstick placed inclined on it in balance position. How is this balance established? Gravity pulls down the stick, but the stick lower end rests on the whirl surface, so it can not go down. But the stick mass center is well above and displased from the heel vertical due to the inclination. Why there is no rotation or falling of the stick? It is in ballance, because the stick lower end has a good grip over the whirl surface( or our hand ) and due to that it follows the whirls rotation. Phisics says when there is rotational motion there is a centripetal force causing it, pointing to the rotation center. This force (acceleration V*V/R )counteracts completely the attempt of gravity to pull down, there are equal and opposed moments trying to produce rotation of the broomstick. No centrifugal force here of course, the balance is kept by gravity force and centripetal force ( lateral to the momentary speed of stick heel). These assumptions are valid if we are not standing on the whirl, but instead of that we look from the side. If we step on the whirl, as we all know, we are entering a rotational system for observation. Here the whirl tries to rotate all objects on it, causing centripetal acceleration over them. But we are also under this influence and it looks like objects we observe are trying to fly outwards ( indeed they are simply not flying inwards as everything strongly connected to the whirl). So Newton Second Law is not vallid here??? What to do? How to create link between motions and forces, how to draw conclusions for some particulars based on others ( accelerations, speeds, corner radius etc.)? Answer is to CREATE and PRETEND THAT SOME FORCE EXISTS, AND WE ARE CALLING IT CENTRIFUGAL AS IF THIS FORCE IS THE REASON FOR THE OBSERVED FLYING OUT OF OBJECTS IN THE ROTATIONAL SYSTEM. If we judge this force application point, direction and magnitude, we can place it among others and build correct equations where the Newtonian Second Law is valid. So here we can draw conclusions for the motion properties from such others succesfully. So, what is the direction of this force? Direction is outwards ( also lateral ) because bodies with zero forces are moving there. And what is the magnitude of it's aacceleration? It is also V*V/R, same as centripetal ones, because this is the rate used by centripetal acceleration pulls inside, same is the rate for the objects to fly bizzare away for no reason if not engaged somehow by the rotation. So, let's go back to our case with the broomstick over the whirl. System is rotating, we assign the IMAGINARY centrifugal force to build the observed balance - gravity is pulling down, centrifugal is pulling out the mass center of the stick, these forces has leverage from the mass center to the heel of the stick, we see those opposed moments try rotation in opposite diirection and canceling each other. We have the same balance, can not be otherway. If someone is more interested in formulas or the math/phisics of it, here you can start:
http://en.wikipedia.org/wiki/Rotati...erence_frame

Okay then, why was this all mess about, what was the problem to pretend we are on board, rotating, and building force balance and conclusions from centrifugal force presence? There is no problem of course, results are the same if all is considered correctly. But if we are trying to understand motion, its cause and effect ( or A, B, C as Mr.Davis says ) we can not pretend that IMAGINARY force arises from speed and radius whithout changing lateral forces ( REAL ONES RESPONSIBLE FOR THE MOTION ), and is responsible for the lean balance. This could be only the way for calculating figures from such others. The way to explain and understand cause-result chains could be nothing but REAL FORCE FIRST - REAL MOTION/CONSEQUENCE after. Back to bikes here is how it starts - steering input - twisted ccontact rubber patch - centripetal forces - curvative path followed by the front tire and radius of curvature observed or measured.( Here I omit the influence of camber forces being born when we have lean angle not zero, but I believe you follow the links above and see why ).

quote:
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.


I can't agree here also. The radius is a result, not an independant variable, it is motion property product from real acting centripetal forces. It could be independent variable only if known from observations and used in equations to find out it's cause(s). Centrifugal force is a result indeed, result from our imagination, tricky enough to enter rotational system, build balance there and ONLY CALCULATE other motion properties of interest - for example the approximate value of the lean angle ( most of passionate readers on this board know the equation ... )

quote:
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.

I was that reader Mr.Davis writes here. I have an understanding how steering causes change of path travelled. Yes, I described centrifugal force as lateral, because it is perpendicular to speed vector. No, I did not claim this force changes radius, I meant it is the centripetal force which is also lateral, as I explained above. I saw you think of
centrifugal when I mean centripetal and tried to explain you, but you did not accept that. Sorry for that, we are humans, each of us has limited ability to understand the other. You can, if you like and think that it would be for good use here, quote our PM correspondence here. I am OK with that. I can only explain my understanding of your belief that fictious force used to ease calculations and brief descriptions of well known facts can be RESPONSIBLE for whatever REAL MOVEMENTS or TRACK DEVIATIONS. I did it above several times already, as much as I can. Sorry, that'all I can do


quote:
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.

Changing the path travelled is a result from forces pushing the bike into corner, centripetal forces, derived from contacts with the road. These forces are exact measure for our imaginary centripetal force, used sometimes as convinience. So these centripetal forces are a cause, both path traveled and centrifugal forces are results. First is real, the second is imaginary.

quote:

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.

Finally, my contrary opinion in brief: I or you whatsoever, do participate in leaning the bike, rider is a part of this common physics, we are CPU on top of the controls you know. There are bike lean controls on every motorcycle, and the most important and powerful of these at our disposal is the steering input, directly manipulating the centripetal ( and the value of your preffered imaginary centrifugal ) forces. Here well stands the basic fundamental defended : TURNING CAUSES LEANING.

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

Hope it is enough, let this wish be for all of us
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JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 03/31/2012 :  6:43 PM
quote:
Originally posted by James R. Davis

The use of a broomstick in the example was simply not a good analogy since I could not simulate out-tracking.



Even if it could simulate it, that would not have made any difference. Out-tracking cannot be the complete explanation why a bike would lean more if the speed increased.

quote:
Originally posted by James R. Davis

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.



In reality not all bikes' front wheels out-track! http://books.google.si/books?id=rJT...lter&f=false
http://dinamoto.it/dinamoto/8_on-li...ng/Tires.htm
If the front tire is softer than the rear tire by a certain amount, the bike exhibits understeer. In such cases the front wheel's slip angle is positive. The actual direction of travel points to the outside of the front wheel direction - the red and blue arrows and their labels in the Figure are exchanged.

What were to happen if front wheel in-tracked? According to this explanation, the in-tracking angle would cause the lean angle to decrease if the speed increased since it would cause the front wheel to drive in from under the CG - only faster. The combined effects of increased speed and decreased lean angle would cause something similar to a high-side. Such a bike would probably be unrideable, except perhaps by a specially trained rider.

I've read that understeering bikes are unpleasant to ride, but I have never heard that the cornering behaviour changes so drastically. Based on that I can only conclude that something else must be going on.
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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 03/31/2012 :  7:17 PM Follow poster on Twitter  Join poster on Facebook as Friend  
I have never heard of a production motorcycle constructed in such a way as to result in under-steer. The slip angle of the rear tire is always somewhat more than that of the front tire. That results in an over-steer vehicle. The example you referenced is NOT one of under-steer, it is one of extreme over-steer which is usually caused not by improper selection of rubber, but by overloading (weight) of the front tire.

Though I have the highest regard for Mr. Cosslater (I have studied much of his work), the example shows extreme over-steer resulting from the lean of the bike (its roll angle) causing a negative slip angle for the front tire. The analysis is meant to show equilibrium of forces (lateral and vertical).

However, out-tracking is not the same as positive slip angle.

Out-tracking is pointing the front tire slightly toward the outside of the actual path of travel in order to CAUSE centrifugal force to lean the bike toward the inside of that path.

In other discussion we have shown that in certain RARE cases there can be neutral steering, but that in virtually all real-world scenarios it requires positive forward pressure on the inside grip to maintain a given lean angle. That positive forward pressure results in OUT-TRACKING.

Still, if you believe that out-tracking is not what accounts for maintenance of lean angles, what do you believe does this?

quote:
If the front tire is softer than the rear tire by a certain amount, the bike exhibits understeer. In such cases the front wheel's slip angle is positive. The actual direction of travel points to the outside of the front wheel direction - the red and blue arrows and their labels in the Figure are exchanged.

The red and blue arrows and their labels in the figure are NOT reversed without resulting in an opening up of the path traveled. That is, I agree with you that you can destroy the stability and functionality of a motorcycle's front-end from a steering point of view with thoughtless modifications and choices of tire rubber, but we are here talking about production (unmodified) motorcycles with reasonable choices for tires.

I was initially led by your post to consider slip angles to be the equivalent of out-tracking, but that quickly changed as I thought about it.
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Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 04/06/2012 :  4:47 PM
Sorry that I'm so late to the party, and I hope I am not covering something already addressed. I tried to read through all the posts carefully, but I might have missed something.

1. shaddix hypothesized several times about locking the steering of a bike in a turn, and I didn't see anyone say that it would immediately become unstable and crash. I agree with James, of course, that the lean angle is set by turn radius and forward speed, but that only works if the steering is free to make the adjustments necessary, and this happens automatically in any well-proportioned bike.

2. The primary input to a bike is steer torque, and the actual steer angle is a complex function of that input torque, forward speed, geometry, and mass distribution. Sharp reports that attempting to control a bike by inputting steer angle is "problematic". I can't tell if shaddix finally got things straight in his head, but perhaps this is a detail that would help. In his example of a bike with "remote control for the handlebars", he is correct that forcing the bars to steer one way will cause the bike to fall in the other direction. If only a torque is applied instead, and the angle remains free, then a stable bike will enter a turn in the other direction and not crash. That's what the first image in the wikipedia counter-steering article is trying to show. Thus, as James said: "so long as your tires have traction, the bike CANNOT fall down." We disagree about how to describe the speeds at which this is true, but that is a separate matter.

3. The out-tracking that James describes is explained by Cossalter, I believe, as kinematic steer angle on pages 26-29 of his book, Motorcycle Dynamics. As James explained, it is completely separate from tire slip angle, which Cossalter explains on pages 59-60.

4. In a paper by Cossalter et. al. on "Steady turning of motorcycles", published in 2007 in Proc. IMechE Vol. 221 Part D: J. Automobile Engineering, they tested four different bikes. The sport bike, the street bike, and the cruiser always over-steered, as James suggested, but the forth, a stock scooter, consistently under-steered.

I can see why it would seem counter intuitive for slowing down in a turn to decrease the lean angle. What pulls the bike upright, as shaddix asked? It is not as spooky as orbital mechanics, however, where using reverse thrust to slow down eventually results in actually going faster. Instead, for me, the crucial detail is that, on a stable bike, the steering angle automatically adjusts in response to the change in speed and continuously modifies the lean angle to match the speed.
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Robsalvv
Male Standard Member
204 Posts


Melbourne, Vic
Australia

Kawasaki

ZX9R

Posted - 04/10/2012 :  2:19 AM
I was embroiled in a similar topic in the forum I moderate... several riders were utterly convinced that they destabalised the bike with a steering input, ultimately choosing a lean angle as a result of that input and believing that the radius was a by product of that lean angle. This is despite being able to lean a bike over while still going in a straight line and a bunch of other reasonable objections.

The "we choose a lean angle" scenario doesn't make sense to me but I can understand why riders believe it. A rider is the centre of their universe so they see their hand pushing the bar and an almost instantanous lean angle developing in response to that push. They see a correlation and assume it causal. It's hard to dissuade them of that view.

It's far more sensible that lean angle results as a function of the speed and curved path. Even if I don't understand the myriad of processes that occur from countersteer to lean angle, that big picture view makes sense and was reinforced by a recent experience.

I was watching my nephew (17 yo) play soccer. He is a very active centre / wing type player and in this particular game was sprinting in bursts, but he had cause to change direction often, sometimes even during mid sprint. I noticed that his body was leaning when running in a curved path. More lean for tighter curves, less for more open curves. Despite the obvious differences between a motorbike and a running human body, that leaning observation couldn't be a fluke.

He already thinks I'm a bit odd, so when I asked him about how he changed directions, he looked at me quizzically and shrugged his shoulders, "I just do". I asked him if he knew he was leaning when he ran a curved path... and the answer was, "I don't know, what's the point?". Well, I should be grateful, at least it was more than a grunt.

The point was, he wasn't dialling up a lean angle to obtain a certain path. He manipulated his feet and arms and legs to conspire to get him onto a particular path. His body however, leaned as a result of the chosen path and speed. That to me should seal it for anyone who still thinks they lean or choose the lean angle to turn.

Just having a look at some American football running back pictures, I see many examples of the same thing:



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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 04/11/2012 :  6:08 AM
Nice and funny :)
We all observe the same phenomena, exploit it well,
and just argue with the explanations :)

Here is an example: body travelling on a constant velocity. If no forces disturb this motion, it continues straight. No curve travelled, no curvature radius. But if we assign a constant force pulling/or pushing sideways, perpendicularly to its velocity, the path travelled will deviate from the straight direction of the velocity, and there will be curvature and radius. The equation that links things is of course the famous here A=V*V/R. 'A' is the acceleration provided by the applied lateral force ( A= Force acting divided on mass being acted on , if we push with 10 Newtons force a 10kg mass, the acceleration produced is 10/10= 1 meter per squared second ). 'V' is the magnitude of the velocity, and 'R' is the radius of the curve traveled. If we do units correctly, it is easy to plug this in a MS excel worksheet, and knowing two of these three ( A,V,R ) we can find the third one. Knowing V and R, we can solve for A, knowing A and V we can solve for R, knowing A and R we can solve for V. Never mind the way we solve, it is simple logic that speed and force ( acceleration sideways ) are the causes, and the path or radius traveled is the result. It is not wrong to calculate the cause ( A ) using speed ( V ) and radius ( R ) if we know them, but that does not mean the radius or curved path is the reason for the lateral acceleration.

So, it is same for bikes cornering, inverted brooms rotated on a carousels, baseball players running in a curve. The more speed ahead ( given the constant sideways force ) gives more straightened curve and less radius as a result. The more sideways force pushing ( given the constant speed ) gives more tight curve and less radius as a result. The opposite is also true - The less speed ahead ( given the constant sideways force ) gives less straightened curve and less radius as a result. The less sideways force pushing ( given the constant speed ) gives less tight curve and bigger radius as a result.

If you say to me 'I choose path' I strongly believe you did it by manipulating speed or lateral forces pushing into the corner, your eyes are observing the changes in the curvature as you wish, omitting the above means to achieve that. Same is the 'I out-track to set radius' concept for me, while out-tracking your first achievement is modulation in the sideways force ( we have no sensors to directly feel that road is pushing our tire's contact patches more or less sideways ) and then comes the changed track or radius ( which we first notice and accept for granted direct control ). Guys experienced in fighting their bikes on muddy off roads are more willing to buy the idea of 'something' in between their steering inputs and the resulting track change.

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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 04/11/2012 :  7:44 AM
quote:
The "we choose a lean angle" scenario doesn't make sense to me but I can understand why riders believe it. ...

If the prime idea is maintaining speed into the corner, we need certain amount of force to push us into this corner in order to pass it safely. When we lean we have these forces provided in our contact patches, the camber effect indeed. So it is reasonable to say : leaning provides cornering forces, they provide the cornering radius.

quote:
It's far more sensible that lean angle results as a function of the speed and curved path.

Lean angle is not determined/ set/ or result from speed and radius. Equilibrium lean angle can be approximately calculated or found by calculations from speed and radius. This calculation relies on the balance of two opposite rotation tendencies - If we follow a bike into right corner, gravity pulls it clockwise, cornering forces at tire's contact patches pulls it counter-clockwise.

quote:

The point was, he wasn't dialling up a lean angle to obtain a certain path. He manipulated his feet and arms and legs to conspire to get him onto a particular path. His body however, leaned as a result of the chosen path and speed. That to me should seal it for anyone who still thinks they lean or choose the lean angle to turn.



I see no problem here :)
The guy needs a tighter turn and knows he needs a stronger cornering force for that at certain speed. If he pushes with feet while not leaning over, he would only move up/down and faster/slower as a result. He has learned that he should lean to push the ground sideways out, and being pushed back by the ground sideways in. That's why he amends his lean balance by choosing path ( but indirectly by manipulating the sideways forces ) - leaning more he can push harder sideways and navigate a tighter track faster without falling down.
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shaddix
Ex-Member

Posted - 05/11/2012 :  8:51 AM
I did something interesting yesterday! I turned the bars ever so slightly to the left around 30mph and held them there fixed in position. The bike leaned to the right ever so slowly and continued to lean to the right to the point where I nearly fell. I had to turn the bars to the right to keep from dropping the bike.

My bars turning to the left was extremely slight, probably not more than 2 degrees, how come gyroscopic precession did not keep the bike upright as I was falling over? It couldn't have been enough front wheel turn to overwhelm that force could it?
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DataDan
Advanced Member
542 Posts
[Mentor]


Central Coast, CA
USA

Yamaha

FJR1300

Posted - 05/11/2012 :  8:30 PM
quote:
Originally posted by shaddix

I did something interesting yesterday! I turned the bars ever so slightly to the left around 30mph and held them there fixed in position. The bike leaned to the right ever so slowly and continued to lean to the right to the point where I nearly fell. I had to turn the bars to the right to keep from dropping the bike.

My bars turning to the left was extremely slight, probably not more than 2 degrees, how come gyroscopic precession did not keep the bike upright as I was falling over? It couldn't have been enough front wheel turn to overwhelm that force could it?

What is sometimes called "gyroscopic precession" is really a torque produced by changing angular momentum. When you turn the handlebars on a motorcycle, the change to front wheel angular momentum is in its direction. Steering a little bit to the left produces a torque that leans the bike a little bit to the right. But the lean torque is produced only while steer angle is changing. So holding that steer angle does not continue to produce lean torque since angular momentum is no longer changing.

But never mind that. The concept that explains your experiment is centrifugal force. In your car, if you make a sharp left turn, the dog in the back seat slides to the right. You, too, feel the force tossing you to the right, though the seat and belts keep you in place.

Same deal on a motorcycle. Steer to the left and the motorcycle tries to turn to the left. But in doing so, centrifugal force pushes the weight of the machine to the right. Since it is balanced on two tires in line with each other--rather than four forming a nice, stable rectangle--it leans very easily to the right in response to even a small steering input to the left. Continue to hold that steer angle, and centrifugal force continues to act on the mass of the motorcycle, leaning it further to the right.

In normal turning, what arrests the lean motion--so you don't lean over to the 90-degree limit imposed by the pavement--is the front wheel steering back to the right due to "trail". Because the point of tire contact is aft of the steering axis, leaning to the right tends to steer the front wheel to the right. That stops the bike from leaning further and coaxes it into a right turn.
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shaddix
Ex-Member

Posted - 05/15/2012 :  1:58 PM
quote:
Originally posted by DataDan

quote:
Originally posted by shaddix

I did something interesting yesterday! I turned the bars ever so slightly to the left around 30mph and held them there fixed in position. The bike leaned to the right ever so slowly and continued to lean to the right to the point where I nearly fell. I had to turn the bars to the right to keep from dropping the bike.

My bars turning to the left was extremely slight, probably not more than 2 degrees, how come gyroscopic precession did not keep the bike upright as I was falling over? It couldn't have been enough front wheel turn to overwhelm that force could it?

What is sometimes called "gyroscopic precession" is really a torque produced by changing angular momentum. When you turn the handlebars on a motorcycle, the change to front wheel angular momentum is in its direction. Steering a little bit to the left produces a torque that leans the bike a little bit to the right. But the lean torque is produced only while steer angle is changing. So holding that steer angle does not continue to produce lean torque since angular momentum is no longer changing.

But never mind that. The concept that explains your experiment is centrifugal force. In your car, if you make a sharp left turn, the dog in the back seat slides to the right. You, too, feel the force tossing you to the right, though the seat and belts keep you in place.

Same deal on a motorcycle. Steer to the left and the motorcycle tries to turn to the left. But in doing so, centrifugal force pushes the weight of the machine to the right. Since it is balanced on two tires in line with each other--rather than four forming a nice, stable rectangle--it leans very easily to the right in response to even a small steering input to the left. Continue to hold that steer angle, and centrifugal force continues to act on the mass of the motorcycle, leaning it further to the right.

In normal turning, what arrests the lean motion--so you don't lean over to the 90-degree limit imposed by the pavement--is the front wheel steering back to the right due to "trail". Because the point of tire contact is aft of the steering axis, leaning to the right tends to steer the front wheel to the right. That stops the bike from leaning further and coaxes it into a right turn.



Makes perfect sense thank you sir!
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