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 Physics and the theoretical
 Turning, center of gravity and camber thrust
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Lilley
Starting Member
7 Posts


Sydney, NSW
Australia

Suzuki

Posted - 07/16/2013 :  10:41 PM                       Like
While reading tip 48 again and letting it sit for a while, I found it couldn't get comfortable. As such I have some questions regarding the camber thrust mechanism and it's role in turning.

It seems rather inherent at first glance that rake and camber thrust both offer their efforts in a turn, particularly on a single track mechansim with two contact points, a pivot at one. Watching the downhill runs during tour de france made me wonder how does the concept of camber thrust translate to the elementally thin wheel? As I can understand it, for camber to provide thrust there must be camber inherent, and as such, width. Something road bike tyres have very little of is width. Further, the profile of such tyres is extreme, and internal air pressure of approximately 7 atm makes sure it stays that way. Yet pushbikes still turn with the same rules applied - speed, lean angle, radius of turn.

Take this one step further and imagine a scale bike that rolls on CD's. These have no camber at all, yet while leaning will still turn according to the same relationship, but without any apparent camber thrust.

So to summarise, how does camber thrust play its roll in turning in these examples of thinner tyres with little width or camber?

-----
A follow up question, it has generally been my impression that it is possible to lean a bike while travelling straight, as long as the resultant COG is still directly vertical. This is far easier to try on a pushbike than motor though it seems to contradict the principles of camber thrust as discussed in the article. A following of that logic suggests that a bike will turn when the resultant COG forms a lean despite the bike being vertical (rigid wheels are assumed) and generating no camber thrust. Someone suggested to me that camber thrust only has a role if the lean of the bike was initiated by steering input, but that seems to be a rather self defeating argument as wheels have no knowledge of steering input. So how does camber thrust fit within these situations? I suspect what I am misunderstanding is something to do with slip angle.

I really need to reread and reread Tony Foale's article on camber thrust, but in the meantime I hope people here can clear this up for me.

scottrnelson
Advanced Member
6890 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/17/2013 :  9:15 AM
quote:
Originally posted by Lilley

So to summarise, how does camber thrust play its roll in turning in these examples of thinner tyres with little width or camber?
It is my own personal opinion that camber thrust plays a negligible part in helping turn a motorcycle. I seriously doubt that you could measure it's effect.
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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/17/2013 :  9:40 AM Follow poster on Twitter  Join poster on Facebook as Friend  
What, exactly, is the mystery? A thin (narrow) tire provides essentially no camber thrust while a wider tire provides somewhat more (though not significant) camber thrust.

Camber thrust simply makes turning smoother than would a flat surfaced tire.
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The Meromorph
Male Moderator
834 Posts
[Mentor]


White House, TN
USA

BMW

R1100RT

Posted - 07/17/2013 :  11:39 AM
I respectfully disagree...
Camber thrust is usually a minor factor at higher speeds. On road cycles, I suspect is still a factor to the same extent despite the thinner tires. The high pressures in cycle tires do indeed minimize distortion, but that actually preserves the camber thrust, by preserving the 'conic section' of the tire when leaned over.
On a motorcycle the effect is quite dramatic at low speed with some tires, and very small with others.
Basically the profile of the tread determines the 'size' of the camber thrust. Tires with a rounded profile progressively increase the camber thrust with more lean angle. Tires with a less rounded profile (nearly flat tread, and a sharp angle where the sidewall starts, don't give much camber thrust at all.
Way back in the '60s, on my first bike, I put new tires on it, changing from AVONs to DUNLOPs front and rear. The AVONs were a squared off flat profile, the DUNLOPs were a half circle profile on the tread.
The difference was dramatic at slower speeds, the bike turned much more readily.
Also, counter weighting on slow turns is much more effective and 'reassuring' on round profile tires than on flat profile tires.

People who have flat profile tires generally don't give camber thrust much credit as a factor even in slow turns (and they are right!).
People who have round profile tires, may not know what camber thrust is, but generally are more willing to 'counterweight' on slow turns, because 'it works' (and they are right!) the camber thrust actually helps significantly.

The forces involved in turning a motorcycle, are multiple, interacting, and varying with both design and dynamic factors.
I don't know of a complete model of the forces involved, but I do know that a lot of smart and knowledgeable people can and have discussed it for many hours, without resolving their differences...
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scottrnelson
Advanced Member
6890 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/17/2013 :  3:22 PM
quote:
Originally posted by The Meromorph

The difference was dramatic at slower speeds, the bike turned much more readily.
Also, counter weighting on slow turns is much more effective and 'reassuring' on round profile tires than on flat profile tires.
I'm trying to figure out how you determined that camber thrust was what made the difference.

The tube in the front tire of my XR650L loses a bit of pressure each day. If I go a week or two without topping it off, it gets down in the range that is good for riding on dirt, but not for pavement. This morning it was way down, probably at 15 pounds or so. The steering was heavy at lower speeds and noticeable at all speeds. Then I topped it back up to 32 psi or so and the steering was much lighter. But it felt roughly the same when leaned over into a turn.

Camber thrust isn't likely to have any effect at all when steering from a straight up position until you at least the the bike leaned over a bit (this, assuming that it has any effect at all, which I'm not convinced of).

How far do you have to lean a bike over for camber thrust to start working?
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The Meromorph
Male Moderator
834 Posts
[Mentor]


White House, TN
USA

BMW

R1100RT

Posted - 07/17/2013 :  4:14 PM
quote:
Originally posted by scottrnelson

quote:
Originally posted by The Meromorph

The difference was dramatic at slower speeds, the bike turned much more readily.
Also, counter weighting on slow turns is much more effective and 'reassuring' on round profile tires than on flat profile tires.
I'm trying to figure out how you determined that camber thrust was what made the difference.

The tube in the front tire of my XR650L loses a bit of pressure each day. If I go a week or two without topping it off, it gets down in the range that is good for riding on dirt, but not for pavement. This morning it was way down, probably at 15 pounds or so. The steering was heavy at lower speeds and noticeable at all speeds. Then I topped it back up to 32 psi or so and the steering was much lighter. But it felt roughly the same when leaned over into a turn.

Camber thrust isn't likely to have any effect at all when steering from a straight up position until you at least the the bike leaned over a bit (this, assuming that it has any effect at all, which I'm not convinced of).

How far do you have to lean a bike over for camber thrust to start working?


Yes, camber thrust doesn't have any effect at all until you are leaned over. (It's nothing to do with the camber of the road surface...)
How far you have to have the bike leaned over to feel the effect depends on:
1. The profile of your tires (both of them).
2. The speed you are going.
3. Wheelbase, rake and trail.
4. Proper (high enough) tire pressure (if you're distorting your profile, you will get less (maybe no) camber thrust...
5. Tread Pattern - any 'normal' road tire with a rounded profile can experience. Knobbies with widely separated tread blocks may have camber thrust negated by 'tread squirm'.

Have you never experienced being able to turn more sharply at low speed by counterweighting/counterleaning?
I don't discount your experience, or that of the many others who say they feel no such effect on their bikes.
Please, don't discount my experience, or that of the many others who frequently use the effect they experience for their own benefit.

Neither are 'right' or 'wrong', they are truthfully relating their own experience on their own machines.

Back on that first bike of mine, I didn't know then what camber thrust was, let alone how it worked. But the difference was so dramatic that I never again bought flattish profiled tires for any bike, and changed out flat profile tires immediately on any used bikes I bought with them on (with the same 'dramatic results').
I've only worked out the camber thrust explanation to explain why that works, in the last few years...

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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/17/2013 :  4:41 PM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
1. The profile of your tires (both of them).


What??? Since camber thrust imposes a modest torque to assist TURNING the direction the front wheel is pointed toward relative to the body of the bike, it requires a degree of freedom that enable that turning. The rear wheel is completely constrained from changing the direction it's pointing toward relative to the body of the bike. Camber thrust (from the rear tire) CANNOT result in TURNING the direction the rear wheel is pointing toward relative to the body of the bike.

The profile of the rear tire is rounded primarily to maximize its life.
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Lilley
Starting Member
7 Posts


Sydney, NSW
Australia

Suzuki

Posted - 07/17/2013 :  8:45 PM
quote:
Originally posted by James R. Davis

What, exactly, is the mystery? A thin (narrow) tire provides essentially no camber thrust while a wider tire provides somewhat more (though not significant) camber thrust.

Camber thrust simply makes turning smoother than would a flat surfaced tire.


Forgive me for not understanding. What causes turn in cases where camber thrust is not applicable? Is it simply orientation of the front wheel? Tony Foale suggests that camber thrust is the only mechanism/turning force that causes turning in a leaning tyre.

Also, why do you say the camber thrust provided is not significant for wide tyres?
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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/17/2013 :  10:07 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Perhaps we have started off on the wrong feet (yours and mine). If what you wanted to do was argue Tony Foale's positions, perhaps that is what you should have done.

I will discuss *my* position and *your* position, but Tony is not here to defend himself.

Now, did I misunderstand your argument which suggested that a motorcycle running on wheels like CDs provide no camber thrust, yet can turn, and now you suggest that camber thrust is the only mechanism that causes turning in a leaned motorcycle - those positions seem rather at odds to me.

What do *you* believe?

As a matter of interest, *I* believe that skiers can change direction as can ice skaters - neither employing camber thrust, both leaning in their turns. *I* also believe that if the front wheel is pointing in any direction other than straight ahead, however slight that angular difference is, so long as there is traction between the front tire and the roadway, the body of the motorcycle will TURN from its current path into that same direction. It's called 'steering'.

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scottrnelson
Advanced Member
6890 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/17/2013 :  10:08 PM
quote:
Originally posted by Lilley

Tony Foale suggests that camber thrust is the only mechanism/turning force that causes turning in a leaning tyre.
I'm not going to go analyze Tony Foale's writings from 27 years ago to figure out exactly where he's right and wrong, but looking at the pictures in his camber thrust article, I don't see any showing normal motorcycle steering in a corner. Drifting and sliding the rear wheel while cornering is certainly a different topic.

The turning force that makes a motorcycle turn is the same thing, for the most part, that makes a car turn. Pointing the front wheel in the direction that you're turning.

A few years ago we had a big discussion here about whether or not the front wheel actually had to turn to steer around a corner, and after logic failed to answer the question, I finally did a little bit of "redneck engineering" using a coat hanger, some duct tape, and a digital camera and proved that the front wheel actually turns a bit when you go around a corner at 30 mph and at 50 mph. It doesn't turn by much, but it most definitely turns to make you steer around the corner.

What keeps the bike balanced in that turn and the countersteering forces are a separate issue that belong in some other discussion.
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Lilley
Starting Member
7 Posts


Sydney, NSW
Australia

Suzuki

Posted - 07/17/2013 :  10:48 PM
quote:
Originally posted by James R. Davis

Perhaps we have started off on the wrong feet (yours and mine). If what you wanted to do was argue Tony Foale's positions, perhaps that is what you should have done.

Not at all, just presenting options and trying to find the answer is all.

quote:
Now, did I misunderstand your argument which suggested that a motorcycle running on wheels like CDs provide no camber thrust, yet can turn, and now you suggest that camber thrust is the only mechanism that causes turning in a leaned motorcycle - those positions seem rather at odds to me.

What do *you* believe?

Sorry to confuse. Firstly, I don't really have a "belief" per se because I don't really understand it well - I'm trying to find the solution. I can understand the concept of camber thrust, but I can't understand its application in all circumstances. In tip 48, you appear to suggest camber thrust is the primary turning force which acts at the wheels for a leaning motorcycle. "But there is another, more powerful, reason that the lean is translated into a turn - Camber Thrust... This camber thrust forces your wheel to turn in response to a lean. I mentioned Tony Foales article because he effectively says the same thing. The trouble I have is that if in some circumstances it is effectively negligible or non-existant, why in other circumstances is it significant?

quote:
*I* also believe that if the front wheel is pointing in any direction other than straight ahead, however slight that angular difference is, so long as there is traction between the front tire and the roadway, the body of the motorcycle will TURN from its current path into that same direction. It's called 'steering'.


So what you are saying is the same as scottrnelson that in a steady state turn, the wheel is inherently turned inwards. This helps, thanks.

If you don't mind my asking, for objects with one point of contact (single wheels, skiers etc), why does a lean cause turn? The fact that it does in obvious. Is it simply a combination of velocity and gravity? Ie. gravity is pulling the wheel/skier sideways and down but the skier's velocity resists this by creating centrifugal force and so tracks a radius? I hope that makes sense.

Edited by - Lilley on 07/17/2013 10:56 PM
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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/17/2013 :  11:09 PM Follow poster on Twitter  Join poster on Facebook as Friend  
In fact, a turn causes lean, not the other way around. That is, turning causes centrifugal force, which, in turn, causedsleaning.

Because a bike's front-end has a rake angle, leaning the bike, along with gravity, causes the front-end to turn modestly into the lean - but that is not what we are talking about.

In the case of skiers and ice skaters, their leaning does not cause turning. They change the direction their skies/skates are pointing to cause turning, which causes leaning (or falling), or they encounter a slope on the surface which their surfaces or edges follow.

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gymnast
Moderator
4267 Posts
[Mentor]


Meridian, Idaho
USA

Harley-Davidson

Sportster Sport

Posted - 07/17/2013 :  11:23 PM
Lilley, Are you familiar with the term "out-tracking" as it relates to initiating a turn using counter steering? Here is a brief article that explains the relationship.

http://en.wikipedia.org/wiki/Countersteering
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Lilley
Starting Member
7 Posts


Sydney, NSW
Australia

Suzuki

Posted - 07/17/2013 :  11:37 PM
Ok, at this point I think I need to continue reading quietly before making a fool of myself.
quote:
Originally posted by gymnast

Lilley, Are you familiar with the term "out-tracking" as it relates to initiating a turn using counter steering? Here is a brief article that explains the relationship.

http://en.wikipedia.org/wiki/Countersteering


Yeh mate, I'm familiar with counter-steering and how it happens.
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James R. Davis
Male Administrator
17292 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/17/2013 :  11:41 PM Follow poster on Twitter  Join poster on Facebook as Friend  
In fairness, what I said about skiers and ice skaters are simply beliefs - I have not studied either.
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gymnast
Moderator
4267 Posts
[Mentor]


Meridian, Idaho
USA

Harley-Davidson

Sportster Sport

Posted - 07/17/2013 :  11:50 PM
quote:
Originally posted by Lilley

Ok, at this point I think I need to continue reading quietly before making a fool of myself.
quote:
Originally posted by gymnast

Lilley, Are you familiar with the term "out-tracking" as it relates to initiating a turn using counter steering? Here is a brief article that explains the relationship.

http://en.wikipedia.org/wiki/Countersteering


Yeh mate, I'm familiar with counter-steering and how it happens.



Did you read the article "mate"?
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Lilley
Starting Member
7 Posts


Sydney, NSW
Australia

Suzuki

Posted - 07/18/2013 :  12:06 AM
quote:
Originally posted by gymnast
Did you read the article "mate"?


I read it.
I'm from Australia, don't sweat the "mate" it's just how we speak. Cheers.

Edited by - Lilley on 07/18/2013 12:17 AM
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JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 07/20/2013 :  8:50 PM
The following basic physics and quotes from the book "Motorcycle Dynamics" http://www.amazon.com/Motorcycle-Dy...p/1430308613 might clear things up. Forgive me if I state the obvious or repeat what you may already know.

For any object to follow a circular path, a so-called centripetal force (or acceleration) must be present. The centripetal acceleration must be equal to v^2/R, where v is the speed and R is the radius of the path. See http://hyperphysics.phy-astr.gsu.ed...e/cf.html#cf for diagrams.

This is a general law and it holds true for any object in a circular path - a skater, motorcyclist, rock on a string rotated above your head, a satellite in orbit,... The difference between these examples is the way this centripetal force is generated. Gravitational force accelerates the satellite towards the centre of the Earth. For a rock on a string it is the force in the string.

On a car, assuming the undercarriage geometry is such that the tires remain flat on the ground, this force is called cornering thrust and is generated by the slip angle. Note that on a car, the slip angle is always positive - the wheels are always slipping towards the outside of the cure.

Since a car is a two-track vehicle, it does not need to lean into the corner. A motorcycle, on the other hand, is a single-track vehicle, thus it needs to lean into the corner in order to remain stable. The lean angle, similarly to the centripetal force, is determined only by the speed and the radius of the turn (https://en.wikipedia.org/wiki/Bicyc...mics#Turning). This lean angle, coupled with the rounded profile of the tire, generates the camber thrust that is the topic of this thread.

The total lateral (centripetal) force for a motorcycle in a corner is a sum of these two forces, the camber thrust and the cornering thrust. This force depends on exactly three variables, the normal force (i.e., the load on the tire), the camber angle (the lean angle) and the slip angle. It also depends on parameters that describe characteristics of the tire. The relationship between these quantities and the lateral force is quite complicated and I will not write it down.

However, for easier understanding, under the assumption of small camber and slip angles, the relationship can be linearised and the lateral force can be written as F = (klambda*lambda + kphi*phi)*N, where lambda is the slip angle, klambda is the cornering stiffness coefficient, phi is the camber angle, kphi is the camber stiffness coefficient and N is the vertical force, i.e., the load on the tire. The proportion of each contribution depends on the tire parameters, such as the profile, internal construction details, the type of rubber compound used, air pressure, temperature,... all of which which influence both of the stiffness coefficients,...

Before quoting a few paragraphs from Cossalter, let me reiterate: Once the radius of the path and the desired speed along this path have been chosen, both the lateral force necessary to maintain this path and the lean angle necessary to keep the bike from toppling over will have been determined and fixed.

quote:
This means that [for a certain tire and for the camber angle] between 0 and 28o, the lateral force needed for the equilibrium is less than the thrust generate by camber alone. Since the lateral force generated must be exactly equal to that needed for the equilibrium, the diminuition of the lateral force is obtained through a negative sideslip angle. That is, the wheel presents a lateral velocity component towards the interior of a curve.

Figure 2-18 shows, for example, that in the case of camber angle 10o, the equality of the force generated with that needed is obtained when the slip angle is -0.3o. With a camber angle 28o the lateral slip is zero. For values of the camber angle greater than 28o the lateral force produced by camber alone is not sufficient for the equilibrium of the motorcycle and therefore the increase in the lateral force is obtained with the lateral slip of the tire (positive slip).

This behaviour is characteristic of motorcycle tires in which the lateral force generated is, up to determined roll angles, almost entirely due to the camber component. Since this component appears more rapidly with respect to the component due to slip, it plays a fundamental role in safety. The camber component appears more rapidly because it depends on the carcass deformation while the cornering component depends on the slip angle which needs some time to occur.

Now consider the graph in Figure 2-19, which refers to a different type of tire. In this case the camber thrust is always inferior to the lateral force needed for equilibrium. This means it is always necessary to have lateral slip in order to generate the additional lateral force required for equilibrium. The lateral forces, the way they are produced and their dependence on the camber angle and slip angle, play a fundamental role in motorcycle's under-steering or over-steering behaviour.

If the generation of front lateral force requires a slip angle larger than that needed for generation of rear lateral force, the motorcycle will tend, as the roll angle increases, to skid more with the front wheel. This behaviour causes the vehicle to under-steer. On the other hand, of the slip in the rear wheel is greater than that of the front wheel, the behaviour will be over-steering. Neutral behaviour occurs when the slip angles are equal. On the basis of these considerations, the tire's ideal behaviour occurs when the slip angle is zero, that is, when the lateral force necessary for equilibrium is produced by camber alone.



Some comments:

I believe the preceding paragraphs match The Meromorph's experience with square vs. round tires quite well.

The camber force can indeed be measured. Cossalter has constructed a device for measuring both camber and cornering thrusts, which can be seen in an on-line article http://www.dinamoto.it/dinamoto/8_o...atii_eng.htm at Cossalter's web site http://www.dinamoto.it/dinamoto/index_eng.html. Not incidentally, the article has good explanations of how both thrusts are generated.

The idea that turning causes leaning in skaters and skiers is wrong. The lean that precedes the turn is so instinctive and natural once you get the hang of it, that you might well think that you have stared turned without first leaning, but simple physics dictates that it would be impossible.

I do not claim to be skater or that I have studied skating physics, although I did some in-line skating (not ice skating) a few years ago. However two simple thought experiments prove quickly that merely changing the direction in which the skates point is not sufficient to make a turn - you need to lean to initiate the turn.

Imagine skating on a roadway and approaching a pavement kerb at an angle (not head-on). When the skate nearest to the kerb makes contact the kerb, it will forced parallel to the kerb. What will happen to the skater? If the turn-causes-lean theory would apply, the skater would magically lean away from the pavement and proceed in a direction parallel to the kerb. Of course this does not happen. Instead the skater's centre of mass will continue to cross the kerb in the direction of the pavement, the legs will be yanked from underneath the skater and he will fall down.

A similar situation would occur if the skater jumped slightly into the air and pointed his toes to one side before landing. The feet would start going in the direction the toes are pointing (assuming no slip of the wheels) and again the feet would be yanked from underneath him.

As for skiing, I consider myself an expert skier, having done it for 40 years for two to three weeks every winter. During this time I've skied on all sorts of skis, from 140cm almost totally flat hot-dog skis with perfectly parallel edges and raised tips at both ends, through classical long skis (218cm for my 180cm height) to current 165cm slalom carving skis and in all sorts of conditions, from "regular" piste snow, through 1m of powder snow, slush in the spring, to snow so hard it was almost like ice. In ALL instances, regardless of skis and conditions, the first thing you need to do is to lean into the turn to initiate it.

I have actually performed the "jump" thought experiment many times in real-life on skis and I can tell you from first hand experience that you do NOT magically lean into the direction that the skis point to. During my hot-dogging days, I had numerous falls when I tried to do a helicopter - jump over a mogul, turn 360o around the vertical axis, land and continue skiing.

If when landing the skis were not pointed in the direction the COG was travelling, there were problems. If the angle was acute enough, the situation was still salvageable - not by leaning towards the direction that the skis pointed to (as this is a physical impossibility), but letting the skis lean me into a turn in the opposite direction and then bringing around the skis into the turn - more or less what happens during the counter-steering maneuver If the angle was too obtuse, the rate of rotation around the horizontal axis was simply too high, recovery was not possible in the short time available and I would simply fall downslope, with the skis doing exactly what I described would happen with skates - yanking the feet from underneath me too quickly for me to be able to salvage the situation. If the turn-causes-lean were true, I would instead perform a very tight turn in the direction of the skis. This NEVER happened.
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ScooterCommuter
Male Junior Member
52 Posts


Saint Paul, MN
USA

Kymco

Xciting 500RI

Posted - 07/20/2013 :  10:26 PM
FWIW I think an analysis of the physics would reveal fundamental differences between the dynamics of a bike and those of skis or skates. In the first place, the dynamics of a bike are based on inline contact patches, only one of which (the front) can be moved out of line. Skis or skates are fundamentally side by side in their interaction with the underlying substrate. Secondly, a bike is powered in its maneuvering, skis and skates are low-friction that are effectively "coasting" the whole time. Skis and skates are much more dynamic in their geometry than a bike can ever be, and so the expertise of a skilled skier or skater is hugely different to that of a biker. As a really piss-poor cross-country skier I'm hardly qualified to express a definitive opinion but the differences in geometry and the degrees of freedom in that geometry, along with the differences in how the "operator" can influence that geometry and the variation between a powered and an unpowered situation might argue that there simply isnt an applicable analogy between the two.
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twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 07/25/2013 :  3:19 AM
quote:
Originally posted by JanK
...For any object to follow a circular path, a so-called centripetal force (or acceleration) must be present. The centripetal acceleration must be equal to v^2/R, where v is the speed and R is the radius of the path.....

Before quoting a few paragraphs from Cossalter, let me reiterate: Once the radius of the path and the desired speed along this path have been chosen, both the lateral force necessary to maintain this path and the lean angle necessary to keep the bike from toppling over will have been determined and fixed...


Jank, you really did a decent attempt to explain the complicated topic simpler, which indeed is not an easy task. Anyhow, let me ask you: You reiterate, once radius is established, both lateral force and lean have been determined and fixed. Peope can take this statetement as rider somehow magically choses radius which after establishes forces and lean. Such interpretation is misleading, radius does not establish forces nor the lean. As per the first part of your quote, forces are those which become in equilibrium, which dictates the lean angle and cornering radius as a result.
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The Meromorph
Male Moderator
834 Posts
[Mentor]


White House, TN
USA

BMW

R1100RT

Posted - 07/25/2013 :  8:42 AM
It's a relatively minor point, but the rear tire does exert a turning force under camber thrust. No-one is claiming that the rear wheel turns relative to the frame, but the frame turns (to some extent 'dragged round' by the steering head) and the camber thrust additively assists in this process.
The greater the contribution of camber thrust from the rear tire, the 'easier' the steering feels. Changing the rear tire from a square to a round profile causes the steering to feel 'quicker'.
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