Originally Posted by ExMachina
Linear momentum is a minor factor in deciding bicycle stability, so the bike and rider need not translate over ANY path to be stable. As others here have explained quite well, what the bike DOES require is the ability of the wheel to drive under itself (via its own rotation and the motion of the steering mechanism) in order to self correct the bike's perpetual state of falling. The rollers allow this action to occur, albeit to a diminished degree (because the rollers are laterally fixed and the front wheel of the bike is slaved to the rear wheel).
You can further verify the minimal contribution the angular momentum of a bicycle wheel makes by comparing the amount of effort you must exert to change a spinning wheel's plane of motion at 3 revs per second (equiv to riding at 15mph on a 700x23 wheel) to the amount of effort you must exert to stabilize a friend who is balancing themselves on a bike. The latter requires magnitudes more effort, especially if they lean off of their center of gravity by even a single degree (to make this even more accurate, try balancing them by only holding onto only the front axle!!!).
..yep... i see this now... so you are talking about the force that's generated by the tire turning on the pavement, while spinning? you can actually propel your bicycle forward with this force, just by 'waggling' the front wheel... fairly fast too. you'll have to excuse me i didn't go through all the post... is there some indication of just what those figures are at a given speed? i.e. at a given speed how much contribution from the angular momentum of the wheel vs. 'wheel driven under it's self'.. but one thing i think is a little off with your example is that i doubt when riding in a straight line that anyone would lean themself and bike more than .5 degrees or so (or at least their center of gravity)... if you did that on rollers you would likely fall off... at those sort of angles only a few fingers are required to keep your freind upright