why do we lose balance in a bike when at a standstill?

krab

which is mostly correct when learning to ride a bicycle, since when learning, you are initially at a very slow speed. But to proclaim this and use it to sell training videos... people try to make money off anything these days. Myself and my kids learned very efficiently, with no videos (and no training wheels).

rcgldr

Someone else posted a link to the article. I don't think anyone was suggesting buying a video. Somehow we've managed to learn how to ride bikes without any videos for many years. Part of my point is that bicycles self-correct once they're going anything faster than a slow walk.

mathlete

Can someone explain what exactly the gyroscopic action is?

rcgldr

When a torque force is applied to a rotating mass, the reaction is along an axis perpendicular to the torque force.

Using a helicopter as example, there's a cyclic control that changed the pitch of the blades as they travel around in a circle. If the cyclic is tilted forwards, creating a pitch down torque force, the helicopter instead responds with a roll towards the rearward rotating blades. If the cyclic is tilted sideways towards the forward rotating blades, the helicopter pitches down.

To keep the pilot from getting confused, the cyclic control is shifted 90 degrees to compensate for this, so the pilot can just push forwards to pitch downwards.

Chi Meson

If you have ever ridden a bike on rollers, you will find out how very little help "conservation of angular momentum" will be for balance.

As opposed to stationary trainers, rollers are thin barrels that the wheels sit on. The rear wheel spins normally, which turns the rear roller, which is attached to a long band which turns the front roller which turns the front wheel. There is just as much angular momentum as when on the road. The wheels are not kept in place "side-to side" and consequently, a novice bicyclist will fall repeatedly, no matter how fast the wheels are spinning.

The reason why rollers are more difficult than actually riding, is you just can't lean! Leaning will turn the front wheel which will send you 10 inches to the side and off you go!

So angular momentum does connect the lean with the turn of the front wheel, but it does nearly nothing to keep us "gyroscopically" upright.
(Basically, everything Krab has said is Bang on.)

Tullebukk

Another question: let's say we only have a bicycle wheel, a dinner plate or another circular object. If we roll this object pretty slow down the floor, it will very well hold balance. What is the main reason for this? The gyroscopic effect?

rcgldr

It doesn't hold balance for very long. Gyroscopic forces will reduce the rate of lean, but a wheel will lean over until camber thrust counteracts the leaning torque due to gravity pulling downwards at the center of mass, and the upwards force at the contact patch, or until the wheel slips and falls.

I don't know if I mentioned this before, but I can balance a 10 speed for about 10 to 30 seconds without moving. This is because steering the front tire translates the contact patch sideways. Steer left, and the contact patch move right if not resisted. With the front tire on the ground, the contact patch doesn't slide much, so steering left moves the front end left. It's enough movement to balance a bike, but it's difficult. Velodrome racers can remain still for very long times, as this is part of the tactic used to try and get the other racer(s) ahead for those racers that want to start from behind (draft).

boit

Let me throw in a controversial twist to this discussion. Someone pointed out that gyroscopic effects plays no part in bicycle's balancing and threw in the counter rotating added wheels argument to back it up. Someone disagreed with him and said if anything the gyroscopic stability will increase. The first guy insisted that angular momentum is a vector hence this cancels out blah blah blah (they lost me somewhere. I agree with whoever say gyroscopic stability will improve, intuitively though cause my phyisic background is wanting. Now to the twist and this is helped by a chap who said motion in itself yield stability and called to meind ice skaters among other wheel-less locomotion.
Mine is simple. Riding a bike a structured perpetual falling to a particular direction much like a satellite in orbit. Take a flag pole and stand it on a flat surface. It is highly unstable (an inveted pendulam-like) Now tip it to fall to one direction. It is extremly had now to make it fall to any other direction. Left to its own divices it will hit at an exact spot. Slowing a bicycle to a stop is analogous to standing the flag pole on its foot again. how about that for food for thought?

sophiecentaur

I can't find any comments in this thread, about the effect of castor action and the fact that a line through the top bracket (the steering axis) on all stable bikes, when produced, always meets the ground ahead of the tyre footprint. This will always cause the wheel to steer into a lean. The result, when traveling forward, will be to produce a force on the ground, 'inwards' and a corresponding moment to turn the bike upright. The faster the bike is travelling, the more this effect will be.

Whilst the gyroscopic effect may be significant on morotbikes, it will be very small on light wheeled bikes - particularly with small diameter wheels, whereas the castor effect only depends on the distance between footprint and the forward produced line of the steering axis on the ground. Folding and 'delivery' bikes have very small front wheels but are still stable. It even works on kids' scooters with plastic wheels of less than 10cm diameter where angular momentum is extremely small.
There is more to this than just one mechanism at work, I'm sure.

rcgldr

Both Krab and I mention this effect (trail), starting with post 21 in this thread.

A.T.

There is more indeed. Guys from the TU Delft build a "bike" that eliminates the gyroscopic effect and the castor effect.... and it still balances itself:

sophiecentaur

That link is good and very inventive in its thesis. (I wonder how many takes they did to get that bike to stay up so well) At least it puts to bed the gyro theory, which applies to very few cases and which, I think only applies a damping / reactive force rather than a restoring force. If the gyro action were actually to bring the bike upright, would not the force be downwards again - by the same precession argument- as the rotation would then be the other way? Once a bike has leant into the curve, after the (truly) restoring couple is there until it is actually upright again.
"Turning into a fall" is a very good way of putting things; both "trail" and castor action, will achieve this. Their model has this forward pointing rod, which achieves the same thing. BUT how many of the bicycles we see on the road are loaded that way?

They do not dismiss trail as a mechanism so they are not disagreeing with my contention that it is due to trail on 'real bikes'. They just achieve the 'leaning in' by a different mechanism.
Funny thing is that I tried a 'butcher's delivery' bike, once. That had a large load out over the small front wheel. It was a real mother to ride.

rcgldr

As explained in that video, the front wheel was weighted to produce the equivalent of trail effect, using weight distribution to cause the front end to "fall inwards" more than the bike. I would have liked to see a true "2 skate bicycle" being tested on a ice rink, to show gyroscopic forces are not required.

TU Delft also ran into a conflict between their math and their testing of an actual bicycle regarding capsize speed, I don't know if they've since resolved the issue. Link to link to article, showing image of bicycle:

Koo06intro.pdf

page 4 of this article includes a graph where the upper limit of the "stable" range is just below 8 m/s = 28.8 kph:

Koo06.pdf

link to treadmill video where 30 kph is described as "very stable", even though it's greater than the 28.8 kph end of the "stable" speed range from the graph in the artitcle linked to above. My guess is this is due to the fact that the tires are not infinitely thin disk, and when leaned, the fact that the contact patch is on the side of the tire results in a outwards torque that keeps the bike from falling inwards as predicted by the capsize speed shown in the graph.

treadmill measurments

sophiecentaur

Is there any more need to 'prove' that gyro forces are irrelevant?

A.T.

It doesn't. It just says that there is more to it, and that there are others ways to achieve self stability.

Sounds like you simply don't understand the "gyro theory" here. The precession is not supposed to bring the bike upright directly, it merely turns the front wheel into the direction in which the wheel falls over.



Here a more complete explanation of the different theories:



So will the gyro effect. But what is the difference between "trail" and "castor action" again?

On real bikes the gyro effect also plays a role.

rcgldr

From what I recall, although precession turns the front tire in the direction of lean, at some speeds, the precession effect produces insufficient turn in to correct the lean, so trail is needed as well. Trail alone without gyroscopic effect can be enough to correct a lean angle.

Trail is the distance from where the pivot axis intercepts the ground back to the contact point between wheel and ground. Trail is normally used to refer to what happens when you lean a castored wheel (the wheel turns in the direction of the lean). Normally castor effect refers to the tendency of a vertical castored wheel to pivot away from the direction of motion so it lines up the wheel with the direction of travel.

sophiecentaur

But, once the bike starts to lift and right itself, won't that change the couple on the wheel so it steers out? I am confused.