Talk:Bicycle and motorcycle dynamics/Archive 2

Archive 1

Archive 2

Turning. Explanation?

Why does the article on turning suddenly start with "In order to turn, that is, change their direction of forward travel, bikes must lean to balance the relevant forces:..."

Because that is the interesting part which separates bikes from most other wheeled vehicles. Otherwise, the section on turning is merely "riders turn bikes simply by pointing the front wheel in the direction they want to go." -AndrewDressel (talk) 20:19, 14 October 2008 (UTC)

Shouldn't it explain how the bike takes a turn? The origin of centripetal force?

I suppose it could. -AndrewDressel (talk) 20:19, 14 October 2008 (UTC)

Now it does. -AndrewDressel (talk) 18:16, 15 October 2008 (UTC)

Why does friction-centripetal force- act in a direction perpendicular to the wheel plane?

It doesn't exactly. It acts radially from the instantaneous center of the turn in the plane of the ground. -AndrewDressel (talk) 20:19, 14 October 2008 (UTC)

And first of all, one needn't necessarily lean to initiate a turn-unless he relies completely on the gyro effect. He'd do it if he wants to stay on the bike during the turn. comments? —Preceding unsigned comment added by 59.96.13.188 (talk) 15:40, 14 October 2008 (UTC)

I'm not sure I understand what you mean by this last comment. Gyroscopic effect are irrelevent to the need to lean in order to maintain balance in a turn. -AndrewDressel (talk) 20:19, 14 October 2008 (UTC)

Yes. Thats what. In the reference 14 (Fajans, Joel (July 2000). "Steering in bicycles and motorcycles), the starting lines go like this "Centrifugal forces will throw your bike over on its side if you steer the handlebars in the direction of a desired turn without first leaning the bike into the turn". So leaning is done for balancing purpose and not exactly to "initiate" a turn. P.S: Anyway i have this doubt. Whats the reason for the friction to act in a radial direction? I get the oft-repeated answer "To balance the centrifugal forces" but can you explain it w.r.t an inertial observer? My guess is that the resultant of the (frictional) forces "on the wheels from the road surface" due to its rotation and translation lies approximately in a radial direction. Have i gone somewhere? —Preceding unsigned comment added by 59.92.25.109 (talk) 17:44, 17 October 2008 (UTC)

The question seems to be the cause of the centripetal force. When turning, both front and rear tires are pointed slightly inwards from the actual path they are moving in. This is called "slip angle", but it's normally due to the contact patches deforming, more than any actualy slippage at the contact patches. Slip angle, downforce (normally weight), and tire characteristics, result in an outwards force onto the pavement, and the pavement reacts with an inwards, centripetal force onto the tires and bicycle, resulting in circular motion. If you extend the axis from the front and rear wheel, they will cross at a point inwards of the bicycle. The point on the pavment directly above this point (it's below the pavement if the bicycle is leaned inwards) defines the circle that the tires are trying to get the bicycle to follow. The actual path has a slightly larger radius due to slip angle. Jeffareid (talk) 02:18, 18 October 2008 (UTC)

Regarding the other apparent question, slip angle determines the amount of force perpendicular to the direction of travel, and is independent of the lean angle. The side force from the pavement will accelerate the bicycle inwards with the rate of acceleration equal to the side force divided by the mass of the bicycle (and rider), regardless of the lean angle or rate of roll of the bicycle. The side force, being below the center of mass also generates a outwards rolling torque force on the bicycle, and unless the bicycle is leaned over so that the upwards force from the pavement (countering gravity) generates an equal countering torque, the bicycle will also experience angular acceleration along the roll axis. The main point here is that a bike will turn regardless of lean angle, but will fall over if the roll axis torques are not balanced. As a pratical example of an uncoordinated turn, sometimes it's easier to avoid a pothole by using positive steering inputs to guide the tires around the pot hole, even though the bicycle or motorcycle will be leaned the wrong way for a brief moment. Jeffareid (talk) 02:31, 18 October 2008 (UTC)

I don't believe slip angle is necessary for the generation of centripetal force. All that is necessary for centripetal force is a constraint that curves. This is usually supplied by a wheel, a skate, or a ski that is not oriented in the direction of straight line travel and that has non-zero friction with the ground. It can also be supplied by a track which would completely eliminate any possibility of side slip and slip angle. The only direction the wheel is free to move, either against flat ground or on a track, is perpendicular to its axis, or the track if the wheel isn't free to steer. Driving forces in any other direction, from the inertia of the vehicle for example, cause friction at the contact patch, or normal forces from the track, to produce reaction forces parallel to its axis. These cause centripetal acceleration and so are called centripetal forces. -AndrewDressel (talk) 13:47, 18 October 2008 (UTC)

Forces involve deformation at the contact surfaces. It's more noticable in the case of rubber tires because the contact patch is realtively flexible, compared to a steel blade on ice. Slip angle is how rubber tires generate side forces (I can find references that dispute camber thrust as a significant factor if that's an issue). Slip angle on tires in a turn increases with centripetal acceleration. It's similar to angle of attack on a wing having to increase in order to increase acceleration perpendicular to the direction of travel. (update A wing is not a great analogy, because for a wing force is a combination of speed and angle of attack because the air is a gas (or fluid), while tires are solid and slip angle is dependent mostly on force and not affected much by speed). Even in the case of a steel wheel in a track, the actual turning radius will be slightly larger than the track's radius because the surface of the track deforms slightly, and the differences between track and actual turning radius could be converted into the equivalent of slip angle (it would be a very tiny angle). Jeffareid (talk) 23:34, 19 October 2008 (UTC)

I get that in the real world, deformation occurs. I question the assertion, though, that deformation is necessary. I would argue that slip angle is a consequence of rubber deformation, not how tires generate side forces. Can you point to a good source for this? -AndrewDressel (talk) 02:24, 20 October 2008 (UTC)

Only necessary because forces are coincident with deformations; every real world object has a stress versus strain relationship. My generic version of the term "slip angle" is the angle between the geometric (zero deformation) path and the actual (non-zero deformation) path. This eliminates the issue of classic "slip angle" versus "camber thrust", since deformation occurs in both cases. Jeffareid (talk) 17:49, 12 April 2009 (UTC)

The only direction the wheel is free to move, either against flat ground or on a track, is perpendicular to its axis.

Why is this the case? Why cant the tire just slide on the ground when it's steered? The wheel cannot by itself provide the centripetal force as it's a part of the system-the automobile. Friction is the only external force that should provide it. ````Ganesh —Preceding unsigned comment added by 59.92.61.106 (talk) 15:40, 27 October 2008 (UTC)

So, you are answering your own question here, right? Friction between the tire and the ground provides the centripital force. I think it is a matter of point-of-view or semantics as to whether the wheel itself provides the centripetal force or friction does. From the point of view of the bike, the wheel provides the centripetal force since it is the only part of the bike in contact with the ground. -AndrewDressel (talk) 17:07, 27 October 2008 (UTC)

But my doubt is why is this friction acting in a radial direction? For eg, when u push a crate along a "plain" surface, the friction acts opposite to the direction of motion only right? Why should it be radial in case of tires? Or can non-rigidity alone explain this problem? ````Ganesh —Preceding unsigned comment added by 59.92.61.106 (talk) 15:40, 27 October 2008 (UTC)

Suppose that suddenly the front wheel is turned 45° with respect to the direction of travel. Then, as with a sliding block, you would suppose that the friction force vector is exactly opposite the direction of travel. As with any vector, this force vector can be decomposed into perpendicular components: one aligned with the direction the wheel is pointing, and the other aligned with the axis of the wheel. If we suppose that the wheel is free to rotate about its axis, then the component of the friction vector aligned with the direction of the wheel must go to zero (in the idealized case where there are no disipative forces, such as bearing friction or aerodynamic drag, slowing the rotation of the wheel) because any non-zero friction force would simply cause the wheel rotation rate to change and quickly become zero. The only significant non-zero component of the friction vector remaining is the one aligned with the axis of the wheel.

Of course, the reality of a non-rigid tire interacting with pavement is far more complex. The finite size of the contact patch, the camber angle of the wheel (especially in the case of bikes), carcass deformation, etc. all conspire to generate competing torques, a "camber thrust", a slip angle, rolling resistance, etc. However, I suggest that these realities are not necessary to explain the simple approximation that the front wheel rolls in the direction it is pointed, just as Coulomb friction ignores all the complexities of exactly how friction is actually implemented and is still useful for calculations. -AndrewDressel (talk) 17:07, 27 October 2008 (UTC)

Alright. So your explanation goes like this: "The total friction force is opposite to the instantaneous velocity vector but centripetal force turns out to be one of its components in the radial direction, the other getting killed in the process of changing the angular momentum of the wheel". ````Ganesh —Preceding unsigned comment added by 59.92.12.124 (talk) 04:54, 28 October 2008 (UTC)

No. I do not say that the total friction force is opposite to the instantaneous velocity vector. I say suppose it is, as with the sliding block example you suggest. Then I show how that is not the case for a wheel. -AndrewDressel (talk) 12:49, 28 October 2008 (UTC)

Whatever be the direction of the decomposed forces, Rigid body dynamics says the acceleration of the COM should be in the direction of the net unbalanced force on the system. In this case the net unbalanced force on the whole system- the bike- is longitudinal(i.e., friction alone). So should the direction of the COM's acceleration right?. ````Ganesh —Preceding unsigned comment added by 59.92.12.124 (talk) 04:54, 28 October 2008 (UTC)

Well, the net unbalanced force must not be longitudinal since the only acceleration is centripetal. And, since the only sufficient force available to act on the bike is friction between the wheels and the ground, that friction must also be centripetal. The decomposition of forces shows how the centripetal component of friction is the only componet that cause acceleration, unless the brakes are applied. -AndrewDressel (talk) 12:49, 28 October 2008 (UTC)

Ok. Consider a single wheel in "space". The wheel plane is turned in a particular angle to its COM's velocity vector. Now the base of the wheel suddenly encounters a "frictional region". According to your explanation it will take a turn which is obviously not the case right?. I'm still confused. :-( ````Ganesh —Preceding unsigned comment added by 59.92.12.124 (talk) 04:54, 28 October 2008 (UTC)

I don't know what you mean by in "space". Literally not touching the ground? Then, what kind of "frictional region" do you mean? Coming in contact with the ground? If so, then yes, it would take a turn. -AndrewDressel (talk) 12:49, 28 October 2008 (UTC)

I'm sorry but I really don't think this section quite passes muster yet. We have a graphic of a model that has a consistent torque applied and turns away from the direction of the applied torque. The text warbles on about all sorts of things and never really refers to the graphs, which are really all there is to it. Never mind all the drivel above in this section, the reality is that steer torque is a roll acceleration demand and that the rider applies a PID control to achieve a target roll angle using primarily steer torque. It is that simple. Yes, there are other control methods possible but they are utterly feeble compared to the authority that steer torque has. (OK, you can talk about how off-centre the roll acceleration demand is the applied torque minus the trim torque, which is why your machine settles to a steady lean angle with an open loop torque input as it reaches equilibrium with the trim torque if you want to pad it out a bit.) For me there is a great opportunity - to simplify all the nonsense - that hasn't been taken. The existence of the BMW C1 suggests that even when the upper body is fully constrained to the machine there is no difficulty in controlling the machine for users who aren't "weirded out" by it. Harty, D., "The Derivation of Roll Reluctance Rating", C553/010/98, IMechE, 1998 contains a mathematical model that uses a PID controller as described to measure motorcycle agility. It can steer a bike that is backing in on the rear brake, pushing the rear out under power, wheelying, stoppying - with no change in parameters. In 2004 I made a free-leaning steer-by-wire vehicle that just ran a no-frills PID controller to chase a target set by a video game steering wheel and it just worked. In 1995 John Lenkeit made a servo rider for a motorcycle that uses the same principle - you can find the paper on the SAE website. Same deal, PID on lean, negative gain, it just works. You've clearly put a lot of work in and the graphs from the matlab model are very clear - why not lean on them? Black bird blue (talk) 14:34, 5 December 2008 (UTC)

Sigh, sadly there's more. Gyro roll torques are overstated. I make the spin velocity of the front wheel about 11 x 2 x pi radians/sec (I'm presuming a typical motorcycle radius of 0.32 m). For the inertia (0.6 kg m^2) and the turn rate of 2 degrees/second (1/30th of a radian/sec) then I'm getting about 1.4 Nm. Moreover, 1 degree of slip angle will generate substantially more than 50N of lateral force under the weight of the front end of a motorcycle. (It must be a motorcycle, no bicycle would have a CG at 0.6m and a front wheel inertia of 0.6 kg m^2.) The value would be more like 250 N, giving 150 Nm of roll moment - the gyro torques are 1% of the total, not 10%. Black bird blue (talk) 14:51, 5 December 2008 (UTC)

Yup, the "Gryo roll torque" is overstated. I also get about 1.4 Nm. -AndrewDressel (talk) 15:50, 5 December 2008 (UTC)

What I should have said is "I agree with your arithmatic." -AndrewDressel (talk) 18:59, 3 June 2009 (UTC)

Let's see if I can sort this out. Cossalter, on page 297 of Motorcycle Dynamics, gives an expression for the gyroscopic moment due to turning a spinning front wheel: $M_{g}=I_{wf}\omega _{f}{\dot {\delta }}$

Then, on page 303, he begins a numerical example with $v=22{\text{ m/s}}$ and $I_{wf}=0.6{\text{ kgm}}^{2}$

He then concludes that the "gyroscopic roll moment is about 3.5 Nm."

He doesn't specify a wheel radius, but uses 0.3 m earlier in the book, as in examples 1 and 2 on page 12.

That would yield a spin rate of $\omega _{f}={\frac {v}{r}}={\frac {22{\text{ m/s}}}{0.3{\text{ m}}}}=73.33{\text{ rad/s}}$

Although he does not state it explicetly, this requires that he is using a steering rate of ${\dot {\delta }}={\frac {M_{g}}{I_{wf}\omega _{f}}}={\frac {3.5Nm}{0.6({\text{ kgm}}^{2})(73.33{\text{ rad/s}})}}=0.07955{\text{ rad/s}}=4.56{\text{ }}^{\circ }{\text{/s}}$

This number appears to come from a graph on page 304 that shows steering angle reaching -2° in about 0.5 seconds: about 4°/s. Perhaps he is using a peak value.

Finally, at the top of page 304, he simply states "The leftward lateral force reaches a maximum value of 50 N after about [0.1 seconds]. Its tilting moment with respect to the mass center (height = 0.6 m) is equal to about 30 Nm."

I'm afraid that without a different source, I'm going to have to let the current numbers stand. -AndrewDressel (talk) 18:55, 3 June 2009 (UTC)

Andrew are you still here? ~~Ganesh —Preceding unsigned comment added by 117.193.198.6 (talk) 02:35, 14 April 2009 (UTC)

Please post messages directed to AndrewDressel on his talk page. -AndrewDressel (talk) 03:14, 14 April 2009 (UTC)

Andrew first of all I'd like to thank you immensely for clearing my doubt on steering. My peanut brain wasnt able to understand how the centripetal force was in the radial direction. Your explanation clarified the matter "a little". Circular geometry of wheel is the reason i've now realised. Thats why a crate given an initial velocity will be brought to rest by friction (whatever be its orientation) whereas a wheel (with its plane at an angle to its velocity vector) will take a turn because a component of the friction will change its W and bring the V(along that direction) equal to R*W thereby vanishing the friction. We discussed it here : http://en.wikipedia.org/wiki/Talk:Bicycle_and_motorcycle_dynamics#Turning._Explanation.3F

Now my doubt:

The basis of your explanation is the decomposition of the initial longitudinal friction to two components. The one along the wheel plane will modify W and V(along the wheel plane) such that V=R*W. (same story. But don't yawn!) But what if the wheel's initial W is set such that the component of velocity along the wheel plane is equal to R*W. The work of the component of the friction has already been done. So will the net friction become zero? What would happen then?

117.193.196.97 (talk) 14:43, 15 April 2009 (UTC)Ganesh

Role of rear wheels in steering.

Rear wheels' rigid positioning parallel to the longitudinal axis of the vehicle somehow makes the vehicle (two as well as a four wheeler) go around a curve when steered. If the hub of the rear wheel is hinged to the frame making deflection about the vertical axis possible (Like the wheels in TV stands) then the vehicle would deviate from the straight line path only initially and continue along that direction. How does the positioning of rear wheels accomplish this? Sganesh 88 (talk) 14:03, 11 May 2009 (UTC)

Very crudely speaking, the front and rear wheels are both tangent to the curve the vehicle is making. Kinematically, it doesn't matter whether the front wheel turns or the rear, it is the angle they make to each other and distance between them that matters. Another way to look at it is that the point where axes through both axles intersect is the center of the turn. Any side-slip of the tires alters this somewhat, and in bikes, the lean also alters this as described in the article. -AndrewDressel (talk) 23:33, 18 May 2009 (UTC)

What i observed is a pivot effect of the rear wheel. If you give a side push to the cycle by completely deflecting the handle bars to one side-(either left or right), then the whole cycle will rotate around the vertical axis passing through the rear wheel's contact point with the ground. So this can be generalised and said that the circular motion of the cycle during a turn is caused by the combination of the cycle's rotation around this axis and the forward motion (longitudinal component of the cycle's velocity vector). Sganesh 88 (talk) 06:20, 8 June 2009 (UTC)

I believe both wheels must exhibit this pivot effect you describe, to varying degrees, as both would continue in a straight line unless constrained to follow a curved path by the other. -AndrewDressel (talk) 14:33, 8 June 2009 (UTC)

I don't think the front wheel's contact point with the ground will act as a pivot in any case. Of course both the wheels play a major role in steering (Angle between their axes precisely) but the pivot will always be the rear wheel's contact point with the ground. In the experiment i mentioned in my above post, instead of pushing the handle bars(after deflecting them totally to one side), even if you pull them(analogous to reversing), the ensuing rotation will be about the axis passing through the rear wheels' contact point(patch) only. I also think this rotation will explain why the rate of rotation of the entire vehicle w.r.t the turn centre is equal to the rate of rotation w.r.t its center. (like the tidal-locking of moon) —Preceding unsigned comment added by Sganesh 88 (talk • contribs) 05:06, 10 June 2009 (UTC)

On stability during steering.. edit done!

I edited the initial portion on the stability during steering which said "leaning is done to balance the centrifugal force with the gravitational force"s" of the lean". There is no such thing called gravitational force of a lean and the forces aren't balanced. Only their torques w.r.t contact patch. Sganesh 88 (talk) 06:18, 20 June 2009 (UTC)

reference link fails to work!

The third reference link "a b c d e f Klein, Richard E.; et al.. "Bicycle Science". http://www.losethetrainingwheels.org/default.aspx?Lev=2&ID=34. Retrieved on 2008-09-09." doesn't work.. Check it out. Sganesh 88 (talk) 06:08, 20 June 2009 (UTC)

Found an archived copy and linked it in the ref. DMacks (talk) 08:15, 16 July 2009 (UTC)

Spinning engine parts and stability

• Gaetano Cocco, on page 13 of Motorcycle Design and Technology, says, without calculation or reference,:

we can state that when a motor is provided with a large flywheel and is kept revved up, it is easier to keep the bike balanced in a stationary position (so-called "sure place").

• The new Harley-Davidson museum in Milwaukee, WI, has an exhibit which attempts to demonstrate the same thing.
• However, neither Cossalter nor Foale mention the phenomenon in their respective books.
• Also, McGill and Wilton, in Engineering Mechanics, An Introduction to Dynamics, provide an example of a spinning disk constrained by an arm to rotate about an axis perpendicular to its spin axis, similar to a Harley engine constrained to pivot about the line between the two wheel contact patches, and conclude that the spinning of the disk provides no resistence to the arm swinging about its axis.

Does anyone have a definitive reference either way? -AndrewDressel (talk) 23:03, 6 August 2009 (UTC)

Andrew. If you will be a bit patient, you will find the answer on the rewrite of Rigid Rotation, when finished. It has to do with precession. See Gyroscope precession and nutation on YouTube. - Petr (talk) 13:14, 26 March 2010 (UTC)

Possibly, but I doubt it. Of course it has to do with precession, but simple demonstrations of gyroscopes are why the myths that wheels acting as gyroscopes are what keeps a bike upright are so persistant. Of course, gyroscopic effects do play a role, but they do not simply resist tipping as is often stated. Finally, without a difinitive reference, we can't included it in this article. -AndrewDressel (talk) 21:10, 26 March 2010 (UTC)

What does Parking problem have to do with Bicycle and motorcycle dynamics?

n/t. --Dennis Bratland (talk) 14:31, 10 July 2011 (UTC)

They both represent nonholonomic systems. Perhaps User:Slawekb is just trying to get a few links to his new article so that it is not an orphan. Maybe this and the falling cat problem are too much of a stretch. -AndrewDressel (talk) 15:04, 10 July 2011 (UTC)

Way too much of a stretch. b.t.w. The problem is really easy to solve - I just put my wife's VW Tiguan into reverse and it does it for me. Brilliant. --Biker Biker (talk) 15:17, 10 July 2011 (UTC)

It's not a stretch. The parking problem is governed by the same nonholonomic system that governs the admissible path of a bike (provided you require that the paths be C¹, as older solutions did). So it's very much directly relevant to the subject of this article. Incidentally, it's also virtually the same system as the system that governs the falling cat problem, so there is no stretch there either. It's sort of irrelevant whether it's "easy" for humans to solve the parking problem. It's "easy" for humans to ride a bike without knowing anything about its dynamics, yet the subject of this article is the dynamics of bicycles and motorcycles. By the way, the parking problem was not considered to be an easy mathematical problem in control theory. Sławomir Biały (talk) 18:55, 10 July 2011 (UTC)

OK, fair enough, but only if you explain that relationship in one or preferably both articles. Unless it's obvious what the relationship is, just tagging an article on the end in the see also section only creates confusion rather than enlightening the reader. --Dennis Bratland (talk) 20:10, 10 July 2011 (UTC)

On the other hand, the relationship may simply not be sufficient to warrant a see-also link. The pendulum article lists other types of pendulums but not other harmonic oscillators. Readers should expect instead that the harmonic oscillator article may contain a list of examples. In this case, it is the nonholonomic system article that should contain the list of examples, which it already does. I don't see the need for each example to link to each other example.-AndrewDressel (talk) 06:36, 11 July 2011 (UTC)

A Simple Explanation Could Help Many

I read through the bicycle article as well as this article. It seemed to me that it would help a lot of people if the fundamentals of how a bike is stable and controllable were put into simple terms. Here's what I recently added to the bicycle article, and I think a similar edit would get a lot of mileage here in this article:

Bicycle Revision as of 20:20, 2 August 2011

I hope most people find this helpful. I'd appreciate hearing other people's views on how best to incorporate a simple-language explanation into this article. The rigorous math of trig & calculus certainly has its place, but math is a language that communicates meaning. It should be possible to translate the fundamental info into plain English so that anyone arriving at this article can take away a solid understanding, leaving the rigorous one to those who want to dive down to that level.--Tdadamemd (talk) 05:13, 5 August 2011 (UTC)

I appreciate your efforts to improve this and the bicycle articles, but I disagree that a spate of analogies to walking, walking on stilts, tripods, and tricycles are necessary. If there is some detail that is not already communicated in plain English, please point it out. -AndrewDressel (talk) 10:51, 5 August 2011 (UTC)

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Balance

Maybe this should be mentioned in the article--NNeilAlieNN ^{Talk to me} 22:28, 19 January 2012 (UTC)

Has been since 07:15, 24 December 2011‎, but without that nice url. Thanks. -AndrewDressel (talk) 22:46, 19 January 2012 (UTC)

Effect of Trail on Stability

The effect of trail on stability can be interpreted in the following two ways (and possibly more); one could be talking about a stability in a strictly dynamics sense in which the stability of the bicycle system can be estimated by traditional linear methods and verified with modern non-linear computational methods, or one could be talking about stability in the subjective, does-this-bike-feel-nice-to-ride sense. Although the two are undoubtedly related, there are differences. One sentence in the Trail section of this article reads, "The more trail a traditional bike has, the more stable it feels." This is patently untrue in the strict dynamics sense of the word 'stability.' If, however, this sentence is taken to suggest that a bike would feel more stable with more trail then it has limited truth. A bike with no trail at all would feel squirrelly because small translational inputs to the handlebars would create larger-than-expected rotations of the front wheel. A bike with too much trail has a sluggish, floppy feeling. So in either sense of the word 'stability' the sentence "The more trail a traditional bike has, the more stable it feels" is untrue. — Preceding unsigned comment added by Marshall2389 (talk • contribs) 07:45 UTC, 1 February 2012‎

Excellent. You clearly have some experience in this area. Now to your points. First, the sentence explicitly states "the more stable it feels", and dynamic self-stability is extensively covered elsewhere in the article, so the "strict dynamics sense of the word" should not apply. Second, I have not seen where the words that describe how a bike feels with too much trail are well defined, but "stable" is obviously used. We are left to rely only on what sources say. As Lennard Zinn is a well-respected author in the bicycle field, and I have not yet seen scientific papers discussing how a bike with too much trail feels, he is about as reliable a source we are going to find on this topic. Finally, I believe that the concept that too much stability is not a good thing is also established, so there is no need to avoid using the word "stable" when a bike feels too stable and therefore "sluggish". If you have additional sources that do assert that a bike with too much trail does not feel stable, or more to the point, that more trail does not make a bike feel more stable, please point them out. -AndrewDressel (talk) 08:10, 1 February 2012 (UTC)

Sorry, I don't have any experience with opinions on bicycle stability related to trail. My own experience with various trails is on various bikes, so there are many other contributing factors. -Marshall2389 (talk) 04:33, 3 February 2012 (UTC)

I meant your comments on the meaning of stability. -AndrewDressel (talk) 10:29, 5 February 2012 (UTC)

One thing though, the third sentence of the same paragraph says, "Bikes with too much trail can feel difficult to steer." This seems a bit contradictory. Ease of steering and perceived stability seem related. The sentences could both be accurate if high-trail bikes are 'easy' to ride in a straight line, thus feeling stable, but difficult to control through tight, quick maneuvers. This very well may be the case. -Marshall2389 (talk) 04:33, 3 February 2012 (UTC)

I've combined the two sentences with an although in an attempt to represent the idea better: more is good up to a point, then it takes too much force. -AndrewDressel (talk) 10:29, 5 February 2012 (UTC)

One more thing. The paragraph also makes some general statements about trail for various bikes and then the survey results below contradict those statements. Marshall2389 (talk) 04:33, 3 February 2012 (UTC)

Yes, I cannot find a source that reconciles the discrepancy. The best I could do is insert "these ranges are not hard and fast." It is not ideal. -AndrewDressel (talk) 10:29, 5 February 2012 (UTC)

Two-wheel-steering motorcycle

James G P Jones constructed a two-wheel-steering motorcycle for a Wolverhampton University engineering undergraduate's dissertation in 2011. The motorcycle had an incrementally adjustable constant 2-wheel-steering mechanism and was designed primarily to establish the optimum rear-to-front opposite sense steering ratio. The key findings of the dissertation were that a high and forward centre of gravity is desirable, and that a 'steering ratio' of around 0.6 gave the best compromise between improved manoeuvrability and reduced stability around a range of slalom courses. Contrary to Milton's findings, it was found that full "true circle" (or having a steering ratio of 1) cycle was possible to control successfully with the appropriate high and forward centre of gravity.

This looks interesting, but I would like to see more than just a self-published source. -AndrewDressel (talk) 22:08, 28 April 2012 (UTC)

Countersteer

I found the following sentence under the Countersteer section:

"This brief turn moves the wheels out from directly underneath the center of mass, causing a lean in the desired direction."

This statement incorrect on several counts. First of all, it does not agree with the reference cited at the end of the section.

Which ones? Could you provide a quotation of the relevant passage(s)? -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

Second, it does not agree with the Countersteering main article.

All that suggests is that one or the other is incorrect or incomplete, but does not identify which one. -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

Third, it does not correctly explain what actually happens.

It is true that when you steer, the contact patch of the front tire does move out from under the center of mass, or to put it more correctly a line drawn from the front to rear contact patches will no longer be directly under the cetner of mass. This would indeed cause the bike to begin to roll (fall to one side) like an inverted pendulum. However, this effect would cause the bike to roll *in the direction* of the turn, not away from it. This is because of mechanical trail. Consider a right turn. Because of the mechanical trail, when you turn the handlbars to the right, the contact patch (which is behind the steering axis) moves to the left of the centerline. Now the center of mass of the bike is to the right of the line between the front and rear contact patches. This would cause the bike to start to roll to the *right*, which is the direction of the turn.

This is only true of the bike is not rolling forward. -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

But the actual result of steering to the right is that the bike rolls to the *left*. The reason for this is because the dominating force is the lateral force generated at the front contact patch because the front tire is now at an angle to the direction of motion. This force creates a counterclockwise torque around the roll axis, causing the bike to roll to the left.

Yes, steering to the right does generate a roll torque, but that does not mean that it dominates. Cossalter points out that initially, the roll moment due to gyroscopic effects of turning the front wheel can dominate. I don't believe I've read yet a source that compares the roll moment due to friction of the tires vs the roll moment due to gravity on the center of mass. I imagine it depends on the parameters of the particular bike, the amount and duration of steering torque applied, and the actual lean angle achieved. Do you have such a source? -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

The inverted pendulum effect does counteract the countersteer, but the effect is very small and very slow (read the article about inverted pendulum motion).

This article: inverted pendulum? All it says is that the rate is inversely proportional to the height of the center of mass and proportional to sine of the angle. I don't see anything about the relative sizes of the moments due purely to the motion of the cart, in the absence of gravity, for example, vs due to gravity itself. Again, one is a function of the force on the cart, and the other is a function of the current lean angle. At 90 degrees, of course, the roll torque due to the force on the cart goes to zero and the roll torque due to gravity is at its maximum. -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

To be able to lean the bike by steering solely because of the inverted pendulum effect, you would need to do it under conditions that prevented the front tire from generating significant lateral forces. For example, the bike would have to be practically stopped, or you would have to be on an extremely slippery surface, or the front wheel would have to be locked and sliding. Jayward (talk) 16:43, 25 April 2009 (UTC)

This is an interesting point, though. Thanks for catching it. -AndrewDressel (talk) 21:20, 25 April 2009 (UTC)

I figure this should go under the counter steering section. The following sentence is not necessarily true:

In order for a bike to turn, that is, change its direction of forward travel, the front wheel must aim approximately in the desired direction, as with any front-wheel steered vehicle.

This is not the case, especially with regard to motorcycles. The picture in the aricle even depicts race motorcycles countersteering through the apex of a turn. -VWWV —Preceding undated comment added 19:00, 18 May 2012 (UTC).

I cannot measure the steer angle of the pictured motorcycles, but I don't believe I recall seeing a reliable source describing a steer angle counter to the direction of a steady-state turn other than in situations of significant over steering, such as on an unpaved racing oval. -AndrewDressel (talk) 19:43, 18 May 2012 (UTC)

I have the same concern, indeed this sentence is directly contradicted further down the article. We all know that it is possible for a bike to enter and maintain a turn without the front wheel ever turning into the direction of the turn (it may need to in order to straighten from the turn). I agree the photograph appears to illustrate this. "In order for a bike to turn, that is, change its direction of forward travel, the front wheel must aim approximately in the desired direction, as with any front-wheel steered vehicle." is therefore wrong. Rolo Tamasi (talk) 21:55, 30 August 2012 (UTC)

Vittore Cossalter

I'm not sure what the issue is with Vittore Cossalter's work. He's clearly one of the leading researchers in the field. --Dennis Bratland (talk) 15:36, 30 October 2012 (UTC)

A quick look at MrOllie's recent contributions, with an apparent focus on removing bad references, suggests that he may have considered the additions by a new, single-purpose account to be something on the order of link spam. Given than Cossalter's book is already cited over 30 times in the article, however, and Padua is already listed as a research center, I also do not see a problem with having a short paragraph in the recent history section. I agree that the shear number of articles in respected, peer-review publications, such as Vehicle System Dynamics, does indeed indicate importance. AndrewDressel (talk) 16:32, 30 October 2012 (UTC)

OK. It's good to do cleanup and remove redundancy, but in order to be comprehensive the article should tell the reader who the important players in the field are. --Dennis Bratland (talk) 17:19, 30 October 2012 (UTC)

Riding on a treadmill

The issue is whether riding on a treadmill must be limited to "straight ahead".

• On one hand, the single cited source does state in the abstract "from a theoretical point of view, bicycling straight ahead on a treadmill with constant belt velocity should be identical to bicycling on flat level ground with constant forward speed."
• On the other hand, the same source states, also in the abstract "measurements are recorded for the case in which the laterally perturbed bicycle coasts freely on the treadmill," (emphasis added) and, more importantly, states in the conclusion "Therefore we conclude that riding a bicycle on a treadmill with constant belt velocity is dynamically equivalent to riding a bicycle on flat level ground around the straight ahead direction with constant speed." (emphasis added) The equivalence is only proven by the response of the bike to deviations from straight ahead motion, and no mention of forward momentum is made.

Therefore, I conclude that this article should not include "straight ahead", or at minimum also include some modifier such as "around" or "approximately", in the statement about the equivalence of riding on treadmills and riding on flat, level ground. -AndrewDressel (talk) 12:27, 1 September 2014 (UTC)

It must be hard to test turns with leaning on a threadmill, and I reckon this is why they only conclude with the bike moving "around straight ahead". I think it's fine with the around in there, that's more accurate also. Cheers, Atlesn (talk) 15:59, 1 September 2014 (UTC)

I think the "approximately" modifier is needed. Assuming a wide enough treadmill, riding at a significant angle to the treadmill will create a destabilizing lateral force on the tires that does not happen on pavement.

Other than rolling resistance, riding on a treadmill shouldn't be substantially different from riding on training rollers. Rollers have a significant learning curve, and even experienced riders can expect to fall the first couple times they use them. Usually, it's due to "lateral perturbations" in the steering by the rider. I can't tell if the bicycle's response to steering input is different on the rollers than on pavement, but it might make an interesting project for a grad student somewhere. --Triskele Jim 01:33, 5 September 2014 (UTC)

The "about" modifier in the conclusion of the cited source could just as well be due to their use of bicycle equations of motion that have been linearized about the straight ahead and level orientation. Outside of that regime, they should expect that their model will less accurately predict behavior, even on stationary pavement. The authors don't say one way or the other.

Riding on rollers may be expected to be less like riding on a treadmill than riding on a treadmill is like riding on pavement, for several reasons:

• The contact patch shapes are the same on treadmills and pavement, but different, shorter and wider, and changing with yawing and steering on rollers. This can alter the magnitudes of the torques generated in the contact patches.
• The interaction between the front wheel and the front roller, and to a perhaps less important extent between the rear wheel and the rear rollers, is more complicated because the bike wheelbase changes as the front wheel steers, thus causing the bike to pitch about the rear axle differently than it would on a treadmill or pavement. Even just yawing will move the front wheel contact point with respect to top-dead-center on the front roller.
• Of course the narrow width of rollers, compared to most pavement and to the wide treadmill used in the cited source, means that riders need to exercise much tighter control on their heading, and this can cause over-controlling in inexperienced riders.

Graduate students have started looking into this. Papers can be found here and here.

The lateral force of the treadmill belt on the tires should be no more destabilizing than the lateral force of stationary pavement on the tires. Because of Galilean invariance, riding on a sufficiently large constant-speed treadmill should be identical to riding on the flat and level deck of a sufficiently large ship moving in a straight line at the same speed, which, in turn, is identical to riding on stationary pavement. The only difference should be the limited maneuvering room on any practical treadmill. -AndrewDressel (talk) 15:52, 5 September 2014 (UTC)

That ship metaphor was interesting. Let's say you're cycling in the opposite direction of the ship but at the same speed (relative to the ship), then you would have zero air resistance (ruling out any wind). So you're going "backwards" but standing still, and then you start making a 90 deg turn in one direction. Wouldn't the ship start pushing on the side of the wheel trying to accelerate the bike up to the same speed? And the bike doesn't have any forward speed, so does it have enough kinetic energy to make the turn? It would need to accelerate also to continue moving in the new direction. When I visualize this, I just see the bike stopping and falling over :-P Cheers, Atlesn (talk) 20:28, 5 September 2014 (UTC)

The forces from the ship's deck on the tires, which cause the bike to accelerate, are identical to the forces from the pavement on the tires, which also cause the bike to accelerate. In both cases, the acceleration is towards the center of the turn, and in both cases, the bike would tip over if it were not first leaned into the turn by countersteering. -AndrewDressel (talk) 21:52, 5 September 2014 (UTC)

As for the kinetic energy of the bike, that is calculated relative to a chosen inertial frame of reference and so can have the same value on the ship with constant velocity as it does on "stationary" pavement (which, of course, is actually rotating about the center of the earth at about 1040 mph at the equator, orbiting the sun at about 67,000 mph, orbiting the center of the Milky Way at abouat 490,000 mph, etc.). That the bike on the ship might have zero kinetic energy with respect to the shore means no more than a bike on pavement having zero kinetic energy with respect to a video camera mounted on a cart moving at the same speed as the bike. -AndrewDressel (talk) 21:52, 5 September 2014 (UTC)

Motorcycle cornering when out of the saddle

Please add some information on this topic: the page today deals with combined CoM only. Based upon race behaviour, transferring rider weight into the corner improves cornering speed of motorcycles, but why? Is the effect due to reduced CoM cornering radius, or more vertical disposition of the tyres? Does the latter have an effect prior to reaching the edge of the tread or does it only help once the edge of the contact patch is reached? The article states: "The actual lean angle between the frame and the vertical must increase with tire width and decrease with center of mass height. Bikes with fat tires and low center of mass must lean more than bikes with skinnier tires or higher centers of mass to negotiate the same turn at the same speed". Tyre width is supported by an equation, and I don't know if the explanation that the angle between the CoM and contact patch decreases more slowly with a fat tyre is correct, but intuitively I can visualise this. However the assertion that bikes with higher CoM lean less is supported by a reference to a book, without further information, whereas previously circular motion is invoked to state the lean angle independed of CoM. Accepting it does change, hanging out of the saddle would require greater aggregate lean angle, so why does it help? ThanksMike163 (talk) 19:06, 22 October 2014 (UTC)

There already is one other little detail provided in the article in the section on rider control inputs

On heavy bikes, such as motorcycles, rider lean mostly alters the ground clearance requirements in a turn, improves the view of the road, and improves the bike system dynamics in a very low-frequency passive manner.

with a reference to a 2007 paper by Sharp, in which he examines the effect of rider torso lean by simulating a lane change maneuver, with and without torso lean input, with an optimal controller. He writes:

It is apparent that a motorcycle rider’s movements relative to the machine contribute to improving the dynamics in a largely passive manner. Such movements can often be observed to occur, in racing for example, even before the application of significant steering torque, in preparation for a maneuver and not as a part of it.

He concludes:

The optimal steering torque control is hardly affected by the inclusion of lean torque as a control or by the weighting attached to it. Neither is the tracking performance much influenced by the lean torque. It is clear in the results that lean torque control is quite a minor issue, mainly influencing the rolling motions of the rider’s upper body relative to the motorcycle. It appears that riders’ body motions are relatively prosaic, contributing to improving the view of the road and improving the system dynamics in a very low-frequency passive manner.

Meanwhile, Cossalter writes, on page 108 his Motorcycle Dynamics

If the rider leans his torso towards the interior of the turn and at the same time rotates his leg so as to nearly touch the ground with his knee, he manages to reduce the roll angle of the motorcycle plane.

When racing, the riders mover their entire bodies to the interior of the turn, both to reduce the roll angle of the motorcycle and to better the control the vehicle on the the turn. The displacement of the body toward the interior and in particular, the rotation of the leg cause an aerodynamic yawing moment that facilitates entering and rounding the turn.

I'll work on adding that last detail to the article. -AndrewDressel (talk) 14:56, 15 November 2014 (UTC)

A critical purpose of hanging off a motorcycle in a corner is to drag the knee, or sometimes foot, on the ground, for the purpose of feeling the pavement, gauging the distance to the ground, and giving the rider a feeling of stability. See [1][2]. --Dennis Bratland (talk) 17:09, 15 November 2014 (UTC)

Noted. -AndrewDressel (talk) 17:48, 15 November 2014 (UTC)

Thanks Andrew and Dennis. Some of this chimes with experience - certainly touchdown gives an idea of lean angle, and subjectively greater stability (I will have to review the links on 'feeling' of stability :)). Also preferable to sit out on approach as I find hanging off does destabilize esp. a light bike. I'll wait till it stops raining (UK) before assessing 'view of the road'. I have not read the paper yet, but so far no direct effect on cornering speed, which does not tally with today's racers. My observation is that where the bike is power limited e.g.: old 125cc class, aerodynamic drag dominates, but otherwise hanging off dominates. Perhaps where ground clearance is mentioned the issue is actually reduced contact patch on the rear. Mike163 (talk) 22:48, 19 November 2014 (UTC)

Brianhe (talk · contribs) pointed out another article, by David Hough, explaining how hanging off became common in racing after the 1970s due to improvements in motorcycle tires, allowing wider tires with a less rounded shape. There are various benefits to wider tires, such as handling road irregularities better, less heat, and longer tire life, plus a more oval contact patch, but it meant the contact ring moved outwards more for the same lean angle, making it harder to hold a line in a turn. This made it beneficial to move the rider's center of gravity into the turn, allowing less lean for a given line and speed. Hough also discusses the benefits of hanging off for a street rider, not just on the racetrack as in the above to links.

Elsewhere, I think in Proficient Motorcycling, Hough writes about the benefits of a motorcycle rider hanging off the outside of the turn, which you often see in low speed tight turns, like in Gymkhana.

We should expand coverage of both tire width and hanging off, and how it affects handling. Note also that ultra-wide tires are all the rage now on bicycles, and we should give some explanation of how that affects handling too. --Dennis Bratland (talk) 19:59, 23 November 2014 (UTC)

Hanging off on the outside during low-speed cornering is recommended in MSF course materials. They call it "counterweighting" in the BRC handbook, p. 9. It may also be in Hough. — Brianhe (talk) 21:54, 23 November 2014 (UTC)

Part on braking is vague or incorrect

A quote from the article: The rear brake of an upright bicycle can only produce about 0.1 g (1 m/s2) deceleration at best (...) All bikes with only rear braking are subject to this limitation This simply cannot be true. From a simple calculation a bicycle riding at 15 km/h, or 4.2 m/s, would then need at least 4.2 seconds to come to a full stop. This is quite hard to believe as I observe bikers with only rear wheel brakes stopping within 2-3 seconds from those kinds of speeds on a daily basis (living in the Netherlands). In fact, I would say that a bike would be rather uncontrollable and unsafe should it could not achieve a 1.5 to 2 m/s2 deceleration. Can anyone verify this 'fact' in the quoted source? I think the author might have misread the statement there, as I can imagine it being about the additional deceleration the rear break gives when combined with a front brake. — Preceding unsigned comment added by 145.94.78.63 (talk) 20:58, 19 June 2011 (UTC)

In the third edition of the reference Bicycling Science (quoted was the second edition) there is an example for a typical racing bike with max. 0.256 g at the rear wheel and 0.56 g at the front wheel. I worked out a slightly more utilitarian bike to have slightly higher values (0.3 g rear, 0.65 g front), so the article is wrong at least in theory (may be correct in practice). I have started expanding on the longitudinal braking dynamics, but recent reverts by Dennis Bratland showed me, that for this "good article" rated article, careful work and sourcing is required, so I'll have to get it all together first. One interesting result gleaned so far, is that the maximum possible deceleration is about the maximum coefficient of friction maintainable at the front wheel before lifting the rear wheel, times g. Theosch (talk) 16:29, 2 April 2015 (UTC)

Sorry I deleted too much. I've added it back with specific reference to page 245 of Wilson's Bicycling Science, 3rd edition. I also deleted the false statements about cruiser and chopper motorcycles having braking comparable to a recumbent. They have a much higher cg. A Feet forwards motorcycle (these are very, very rare prototypes) might qualify, but not cruisers and certainly not choppers. --Dennis Bratland (talk) 17:51, 2 April 2015 (UTC)

Thank you! Have now pointed "rear braking" to this reference as well (even if it is on the next page) and corrected the value accordingly. Theosch (talk) 10:00, 4 April 2015 (UTC)

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