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I am looking for a curve traced by a moving bicycle when its steering bar is fully rotated either clockwise or anti-clockwise.

How to model it mathematically?

Is the curve a circle?

My attempt is as follows:

Let $\vec{r}_1(t)$ and $\vec{r}_2(t)$ be the position vectors for the tangent points (between road and tires) on the rear and front tires, respectively. I know that $|\vec{r}_2(t)-\vec{r}_1(t)|$, $|\dot{\vec{r}}_1(t)|=|\dot{\vec{r}}_2(t)|$ and $\dot{\vec{r}}_1(t)\cdot\dot{\vec{r}}_2(t)$ are constants. $\dot{\vec{r}}_1(t)$ is in the direction of $\vec{r}_2(t)-\vec{r}_1(t)$.

Assuming the tires rolls without slipping then their linear velocity is the same.

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2 Answers 2

Let's simplify the situation. Let's assume ideal wheels and ground, that contact only at a point. Let's assume the front steering axis is perfectly vertical, and there is no caster - i.e. the contact point between the wheel and the ground is on the line of the steering axis. Also assume the bicycle remains perfectly upright.

If the front wheel is moving forward at a constant speed, and if the steering bar is rotated at a constant speed with respect to a grid ruled on the ground, then the front wheel traces a circle on the ground. If the steering bar is only turned a fraction of a full turn, then the wheel will only trace a fraction of a circle.

What the rear wheel does is trace a curve whose displacement from the original line of motion goes as the integral of the area traced by the front wheel.

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If by "fully rotated" you mean that the front wheel is turned 90 degrees, then there is no solution in which the wheels roll without slipping. The only time the no-slip condition is satisfied while the wheels are orthogonal is when the bike is stationary.

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