I Bicycle physics: Keeping balanced....

[Delta2]

So my main question is why we maintain balance easily when the bicycle has some horizontal velocity, but it isn't the same when bicycle is at rest.

The best answer I could find at the web is that when the bicycle wheels rotate, essentially they form gyroscopes and gyroscopes resist the change in their axis of rotation (they tend to keep their axis parallel with respect to the ground).

Is this the answer to my question or not?

 

[A.T.]

So my main question is why we maintain balance easily when the bicycle has some horizontal velocity, but it isn't the same when bicycle is at rest.

The best answer I could find at the web is that when the bicycle wheels rotate, essentially they form gyroscopes and gyroscopes resist the change in their axis of rotation (they tend to keep their axis parallel with respect to the ground).

Is this the answer to my question or not?

Not really. The gyroscopic precession plays a role in steering the front wheel into the lean, if you don't hold the handles. But it's not the whole story. See videos below.

But if you are steering, then it's you who is steering to straighten up the bicycle. And that doesn't work if the bicycle is not moving.

 

[Delta2]

So I see the real explanation is a bit more complex, video 2 mentions 3 factors that combined are what makes the bicycle stable

 

[wrobel]

I think it would be a good exercise to write down equations of coin's rolling without slipping on a horizontal table and study stability of its straight vertical motion in the linear approximation.

to begin with:

 

[PeroK]

So I see the real explanation is a bit more complex, video 2 mentions 3 factors that combined are what makes the bicycle stable

The gyroscope effect might be relevant to the bicycle itself, but must be fairly negligible if we compare the mass of the wheels to the rider.

 

[Delta2]

The gyroscope effect might be relevant to the bicycle itself, but must be fairly negligible if we compare the mass of the wheels to the rider.

According to video 2 the gyroscope effect alone can't explain the stability of the bike (in the absence of a rider).

 

[wrobel]

The gyroscope effect might be relevant to the bicycle itself, but must be fairly negligible if we compare the mass of the wheels to the rider.

is not the stability\instability dependent on the velocity?

 

[PeroK]

is not the stability\instability dependent on the velocity?

That's true. More precisely, the gyroscope effect must be negligible at low speeds.

 

[PeroK]

We could try an experiment.

1) Turn a bike upside down, so that it is resting on its handlebars and seat.

2) See how much effort it takes to tip the bike over.

3) Spin the wheels as fast as possible and repeat step 2).

 

[A.T.]

The gyroscope effect might be relevant to the bicycle itself, but must be fairly negligible if we compare the mass of the wheels to the rider.

The gyroscope effect is about steering the front wheel into the lean, not about holding up the entire bike.

 

[ergospherical]

$\boldsymbol{q} = (x,y,\varphi,\psi, \theta)$\begin{align*}

\dfrac{d}{dt} \left( \dfrac{\partial L}{\partial \dot{\boldsymbol{q}}} \right) - \dfrac{\partial L}{\partial \boldsymbol{q}} = \lambda \begin{pmatrix} 1 \\ 0 \\ 0 \\ R\cos{\varphi} \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ 0 \\ R\sin{\varphi} \\ 0 \end{pmatrix}

\end{align*}Too tired to evaluate the gradients.

 

[Lnewqban]

So my main question is why we maintain balance easily when the bicycle has some horizontal velocity, but it isn't the same when bicycle is at rest.

The best answer I could find at the web is that when the bicycle wheels rotate, essentially they form gyroscopes and gyroscopes resist the change in their axis of rotation (they tend to keep their axis parallel with respect to the ground).

Is this the answer to my question or not?

The gyroscopic effect resists any roll of the wheels, steering movements and from leaned to vertical as well: it is not inducing verticallity of the chassis.

Please, see:
http://ezramagazine.cornell.edu/SUMMER11/ResearchSpotlight.html

The rider must steer in the right direction in order to keep the moving bike from falling over.

 

[wrobel]

$\boldsymbol{q} = (x,y,\varphi,\psi, \theta)$\begin{align*}

\dfrac{d}{dt} \left( \dfrac{\partial L}{\partial \dot{\boldsymbol{q}}} \right) - \dfrac{\partial L}{\partial \boldsymbol{q}} = \lambda \begin{pmatrix} 1 \\ 0 \\ 0 \\ R\cos{\varphi} \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ 0 \\ R\sin{\varphi} \\ 0 \end{pmatrix}

\end{align*}Too tired to evaluate the gradients.

I would use D'Alambert-Lagrange
$$[L]_\theta:=\frac{d}{dt}\frac{\partial L}{\partial \dot\theta}-\frac{\partial L}{\partial \theta},$$
$$[L]_\theta\delta\theta+[L]_\psi\delta\psi+[L]_\varphi\delta\varphi+[L]_x\delta x+[L]_y\delta y=0$$

 

[wrobel]

$$\{(\delta\theta,\delta\psi,\delta\varphi,\delta x,\delta y)\}=span(a,b,c)$$
$$a=(1,0,0,0,0),\quad b=(0,0,1,0,0)$$
$$c=(0,1,0,-R\cos\varphi,-R\sin\varphi)$$

 

[Spinnor]

Hold a yard stick or meter stick vertical in your open hand. Now move your hand so as to try and keep the stick vertical, similar effect when you steer a bike? When you steer a bike you are shifting the bike under your center of mass?