[lyner]

Yes, Geezer, I agree but two points need amplifying.
1. I don't think enough has been said regarding the mistaken assertion that changing the frame of reference gives you a stationary bike with no outside influence. The fact is that, if you take the bike as the reference, you have the road coming towards it and it can  produce a force on the bike. Either way the force is the same.
2. Gyroscopic action is not a resilience effect - like a spring. It is the equivalent of a mass(a reactive effective) , merely slowing the bike's fall but doing nothing to actually return the bike upwards. A simple toy gyroscope never goes back up again - it just sags as it loses angular momentum to the surroundings.
But syphrum's comment  is right. This has run its course, and has run the same course several times. I think it demonstrates how a problem with so many degrees of freedom (particularly involving angular momentum) is beyond intuition. 'Opinions' start to take over from Mechanics.
It's still magic how you can, so easily, ride with no hands and no concentration. Hands in pockets and whistling are essential, too.

[Geezer (OP)]

SC - I like LeeE's observation about "being in a controlled state of falling". It seems that the combination of castor action (with motion) and the the rider's ability to transfer weight, provide a near perfect mechanism for the rider to "prevent falling". As you point out, about the only way you can "get surprised" is if you encounter a low friction surface like ice or gravel (I think I still have some of the scars on my knees, but not enough to prevent me wearing my kilt  [])

It's remarkable when you consider how old the invention is (the bike that is).

[lightarrow]

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?

No, I didn't, sorry, however now I'm beginning to understand. Thank you for your patience...  [:I]

[lightarrow]

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

That's right. Now (finally!) I understand what is "castor action" [:I]....

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.

...but I don't understand this. Let's say the the wheel has turned right, then shouldn't the couple be such to rotate the system to the right? Or do you mean that not only the horizontal component of the friction on the front wheel increased, but the vertical too? I can't see how  the couple is oriented.

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?)

Good question.

[Geezer (OP)]

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?

No, I didn't, sorry, however now I'm beginning to understand. Thank you for your patience...  [:I]

No problem Lightarrow. Here's another experiment we could try (if we had lots of money!)

Construct a large "rolling road" (like a treadmill*, but larger).
Get a bicycle with the smallest possible wheels and a "volunteer" cyclist (a little brother would be good if you have one.)
Run rolling road at 20km/hour.
Get volunteer to pedal bicycle on rolling road at 20km/hour.
Supply refreshing drinks to volunteer.

I think the volunteer will be able to stay balanced, even though he has no linear momentum. However, he can easily adjust the point of contact with the rolling road relative to his center of mass while the road is rolling. When the road stops rolling he loses that ability and falls over. So he should wear a helmet  []

* treadmill - Ornaments frequently purchased by Americans to decorate their houses.

[Geezer (OP)]

Here's a link to a nearby waterpark. It's possible (for some people anyway) to surf and stay upright, even without forward motion, and apparently, no gyroscopic action at all.

http://www.silvermt.com/Waterpark/default.aspx?page=WP-Surf-Club

[lyner]

Lightarrow
The axis is through the two points of contact with the ground. The force, I guess, is the centripetal force, due tu the curve followed by the mass. This is in the appropriate sense to pull you upright.

[lyner]

Ahh. I looked at your post again, lightarrow. The bike does deviate from it's course so there wll be a moment, initially, about a vertical axis.

[lightarrow]

Here's a link to a nearby waterpark. It's possible (for some people anyway) to surf and stay upright, even without forward motion, and apparently, no gyroscopic action at all.

http://www.silvermt.com/Waterpark/default.aspx?page=WP-Surf-Club

It's a rather different situation, isnt'it?

[syhprum]

Staying upright requires similar skills to balancing on a slack wire, as there is virtually no friction between the board and the water with training the bodies automatic balancing reflex learns to keep the board beneath ones center of gravity.
Rather like balancing a broom on ones hand (robots can easily be programed to do this).

[Geezer (OP)]

I love the broom balancing analogy. Wait a minute, I think I used that about twenty posts ago.