[syhprum]

Geezer

When you cycle without your hands on the handlebars what you do is push the saddle from side to side with your bottom which moves the cycle from the vertical position which in turn rotates the steering due to the castor angle.
Try riding just standing on the pedals no hands you will find it near impossible.

[lyner]

lightarrow
I think you are ignoring the couple which acts to pull the bike upright when it is following a curve and leaning in. It cannot be ignored at speed (see speed skating - no wheels).  The castor angle causes the bike to lean in the appropriate direction, producing the same required couple.
This requires movement - as does the gyroscopic action.

Edit - added skating thing

[RD]

RD, are you up for a small experiment? LeeE has something in mind involving a bike with no steering and a steam catapult.
 I think the idea is to launch a volunteer off the deck of an aircraft carrier or something

No thanks. We need a stuntman like Evel Knieval* to prove a bike does not need a rider ...

 [ Invalid Attachment ]

http://www.firebox.com/product/1430/Evel-Knievel-Stunt-Set?itc=30&src_t=nwt&src_id=119

[* the plastic toy version]

Trials bikers and BMX riders can balance on two wheels without going forward by rapidly shifting their body, so the gyroscopic effect of the wheels is not necessary for balance, but the gyroscopic force does exist  particularly at high speed.

[lyner]

The trucks on skateboards must have a similar self-righting effect but the skill of actually staying upright is so great that the effect is probably not noticed by anyone who can actually ride on one. (There can hardly be any significant gyroscopic effect)

[lyner]

RD, your last comment made me realise - the effect which keeps you up once the bike starts to tilt and then turn due to the castor action is also, effectively, a gyroscopic / angular momentum vector effect. Will not changing the angular momentum vector by turning from a straight path involve a couple at right angles, which will tend to rotate the bike 'upwards'? I am not referring to the wheels - I refer to the curved path taken by the bike and the associated angular momentum.

[Karen W.]

Hey besides all the reasons that have been mentioned in the two pages here I would simply like to add my observation.

Is it quite possible that the smaller wheels cover less ground per rotation, and thus, take more effort to reach a required distance due to size when larger wheels actually cover more ground surface and distance per rotation then smaller wheels, and would that not make them more efficient simply because of distance covered per rotation...?

[lyner]

Karen
We did mention the increased frictional losses from small wheels, earlier. There is no, inherent, difference between the energy needed for large or small wheels - 'just' the frictional effects. If you use the right, lossless, gearing, there is no difference in the work needed to be done on the pedals for large or small wheels to, say, go up a given hill.

[Geezer (OP)]

So far, as Turveyd pointed out, the main advantage of large wheels (apart from the obvious machismo impact of course  []) seems to be that they really do reduce the likelyhood of "wedging" the front wheel on rocks and causing an A over T while riding off-road.

[Bored chemist]

I have a recollection that someone with nothing better to do produced a bike with a contra rotating gyroscope to cancel out the gyro effect of the wheels- it didn't make much difference.

Don't know what to say, because I really have difficulties to believe it.

Seeing is believing.
http://www.rainbowtrainers.com/default.aspx?Lev=2&ID=34
"Zero-Gyroscopic Bike I is a clever and yet simple experiment that dispels once and for all the centuries old conventional wisdom that a bike stays upright primarily due to the gyroscopic action of the two rotating tires. "

With wheels put in that way, you certainly don't have exactly zero gyroscopic effect.

Why not, or, at least, why isn't the remaining gyro effect so small as to be unnoticable?

[Geezer (OP)]

Geezer, I still can't understand why all that you wrote is impossible to do when the bicycle is stationary. First answer me this question: which is the *only* difference between stationary and moving bicycle? Don't tell me that it is the bicycle speed or linear momentum because you can always take the frame of reference where the bicycle is not moving and nothing must change, for what we are considering here.

OK - Let me try again.
Same conditions as before - constant speed, no pedalling etc. (I'm changing my story slightly!) Let's say the rider transfers weight slightly to the right. This weight transfer exerts a turning moment in the steering because of the castor angle, and the steering turns slightly right. The tires follow a path to the right, but the combined center of mass of the rider and bicycle tends to continue in a straight line - (think inverted pendulum). Because the path of the tires moved to the right relative to the center of mass of the rider/bicycle, the turning moment in the steering now reverses and the path of the tires now swings to the left, etc. etc.

Now, when the ensemble is stationary and the rider transfers weight to one side or the other it has no effect on the points where the tires contact the road relative to the rider/cycle center of mass. To stay upright, he must rapidly transfer weigh to the other side. He gets no assistance from the bicycle to stay upright. In the stationary case, the cyclist has lost the ability to alter his contact point with the road surface.

Here's a possible experiment. Set the bicycle and rider on a platform that can move left and right relative to the rider. Lock the bicycle's wheels. Drive the platform with a servo that is controlled by the attitude of the rider. When he leans right, the platform moves right. When he leans left, the platform moves left.

I suspect that, after a little training, the rider will be able to keep the bicycle upright indefinitely.

Should be "as easy as riding a bike!"
  

[Karen W.]

Karen
We did mention the increased frictional losses from small wheels, earlier. There is no, inherent, difference between the energy needed for large or small wheels - 'just' the frictional effects. If you use the right, lossless, gearing, there is no difference in the work needed to be done on the pedals for large or small wheels to, say, go up a given hill.

Don't you have to do more rotations with the pedal to cover the same area, and therefore expend more energy in doing so, at least for getting from point A to point B?

(Increased friction losses) am I mistaken, that means there are definitely losses of energy from pedaling moreas you stated, correct.. So how does pedaling more not require more energy, as you would have to increase the rate at which you made each rotation in order to keep up with a bigger wheeled bike...?
Am I missing something?

[lightarrow]

Why not, or, at least, why isn't the remaining gyro effect so small as to be unnoticable?

Because the axis of rotation don't coincide.

[lightarrow]

Geezer, I still can't understand why all that you wrote is impossible to do when the bicycle is stationary. First answer me this question: which is the *only* difference between stationary and moving bicycle? Don't tell me that it is the bicycle speed or linear momentum because you can always take the frame of reference where the bicycle is not moving and nothing must change, for what we are considering here.

OK - Let me try again.
Same conditions as before - constant speed, no pedalling etc. (I'm changing my story slightly!) Let's say the rider transfers weight slightly to the right. This weight transfer exerts a turning moment in the steering because of the castor angle,

I have to admit that I don't see it (my problem...) but if this effect is not _also_ gyroscopic (but of course I claim it is), then you must have it even when the bicycle is stationary.

 and the steering turns slightly right. The tires follow a path to the right, but the combined center of mass of the rider and bicycle tends to continue in a straight line - (think inverted pendulum).

I reminded you in my previous post that you cannot invoke linear momentum, since the same effect must be present in a frame of reference where the bicycle is stationary.

[Geezer (OP)]

Geezer, I still can't understand why all that you wrote is impossible to do when the bicycle is stationary. First answer me this question: which is the *only* difference between stationary and moving bicycle? Don't tell me that it is the bicycle speed or linear momentum because you can always take the frame of reference where the bicycle is not moving and nothing must change, for what we are considering here.

OK - Let me try again.
Same conditions as before - constant speed, no pedalling etc. (I'm changing my story slightly!) Let's say the rider transfers weight slightly to the right. This weight transfer exerts a turning moment in the steering because of the castor angle,

I have to admit that I don't see it (my problem...) but if this effect is not _also_ gyroscopic (but of course I claim it is), then you must have it even when the bicycle is stationary.

 and the steering turns slightly right. The tires follow a path to the right, but the combined center of mass of the rider and bicycle tends to continue in a straight line - (think inverted pendulum).

I reminded you in my previous post that you cannot invoke linear momentum, since the same effect must be present in a frame of reference where the bicycle is stationary.

Lightarrow - I gave you an example of the effect when the frame of reference IS stationary and there is no linear momentum. Did you not understand it?

[syhprum]

Karen

Why you get more friction losses with small wheels is because with a shorter contact area with the road they follow the minor hills and dales of the road with more squishing of the tyre.
On a dead smooth road there would be no difference but we don't have them.
I am amazed how much correspondence this trivial subject has provoked I thought the gyroscope myth had been laid to rest long ago.

[Geezer (OP)]

Like all good myths, it's not quite dead yet.

[Bored chemist]

Why not, or, at least, why isn't the remaining gyro effect so small as to be unnoticable?

Because the axis of rotation don't coincide.

Both are fixed to a rigid frame so (I think) the only effect of the gyro effect is that you put some forces on the frame.
Angular momentum is a vector. The direction is parallel to the axis. The sum of the two momenta of the two wheels is zero.

[lyner]

Karen
There is theory and there is practice and the two may give different answers but you need to identify which is which.
You would need to use different (higher) gearing, natch, to go at the same speed but, in the end, the speed of the periphery of any size wheel needs to be the same as the road speed. Gears will just multiply the actual speed your feet move and achieve the road speed you want. I am ignoring any effect of friction in the gearing system, which may be a bit worse (in practice) for a higher gear. Not very thorough, I know but right in principle. The actual power delivered to the road wheels is force times speed and that will be the same, whatever the wheel size. The actual losses are very low in modern chains and sprockets. It would be details of the tyres etc that would make all the difference to the friction losses, through contact with the road and internally in the tyres, I'm pretty sure.

[lyner]

To LeeE and Geezer: if you remove unessential things from the physical model of the problem, that is air friction ecc, the only difference between a moving and a not-moving bicycle is the fact wheels spin, in the first case. So if you want to find a cause of the different equilibrium, you have to look for here.

There is a major difference because of the interaction with the ground. Stationary, the bike tips and there is no inherent restoring force.
Moving, there will be a force against the direction of motion because the wheel has turned itself (castor action, as I keep repeating). This force produces a couple because it does not act through the cm of the bike and rider. The couple will tend to return the bike upright.

The temporary equilibrium which a trick cyclist (and a wire walker) achieves is a very different matter and must rely on using skill to change the moment of inertia and twisting body / frame. With experience, all cyclists get to have a bit of skill with this but it isn't necessary on a moving bike.
How can this not be relevant?

SO  - the cyclist doesn't make it happen, the gyroscopic effect can be reduced as much as you like and the bike still stays upright so what else is there but the castor effect and friction with the road? (It wouldn't work with a bike on ice even with massive wheels - would it?)

[Geezer (OP)]

To LeeE and Geezer: if you remove unessential things from the physical model of the problem, that is air friction ecc, the only difference between a moving and a not-moving bicycle is the fact wheels spin, in the first case. So if you want to find a cause of the different equilibrium, you have to look for here.

There is a major difference because of the interaction with the ground. Stationary, the bike tips and there is no inherent restoring force.
Moving, there will be a force against the direction of motion because the wheel has turned itself (castor action, as I keep repeating). This force produces a couple because it does not act through the cm of the bike and rider. The couple will tend to return the bike upright.

The temporary equilibrium which a trick cyclist (and a wire walker) achieves is a very different matter and must rely on using skill to change the moment of inertia and twisting body / frame. With experience, all cyclists get to have a bit of skill with this but it isn't necessary on a moving bike.
How can this not be relevant?

SO  - the cyclist doesn't make it happen, the gyroscopic effect can be reduced as much as you like and the bike still stays upright so what else is there but the castor effect and friction with the road? (It wouldn't work with a bike on ice even with massive wheels - would it?)

SC - Please refer to my latest post to Lightarrow which does include the importance of castor action. Castor action might also be described in the context of negative feedback which is, intrinsically, a stabilizing mechanism. Further, you might note that I have described a rather simple experiment that might dispel the confusion regarding bicycle stability once and for all.