5. Torques Acting on the Steering Axis

The equilibrium position of the handle bar is given by the condition that the sum of all torques is zero. In this section we discuss intrinsic torques which act on the steering axis. External torques applied by the rider to the handle bar are not considered. In no-hands riding, the equilibrium position of the handle bar has to be such, that it causes equilibrium of the system bicycle-rider.

The dominating intrinsic torque results from the vector product of the trail and the force acting on the contact point K of the front tire:

F_g is the force component in the vertical direction caused by gravity. F_c is the counterforce to the centrifugal force. Only the component of T in the direction of the steering axis has to be considered.

Under normal riding conditions gyroscopic torques are about an order of magnitude smaller, but have to be included in numerical calculations.

5.1 Gravity induced torque

The normal (gravity induced) force acting on the tire under stationary conditions is (A/L)gM. L is the distance between the wheel hubs and A the horizontal distance rear wheel hub - center of mass. This force results in a torque T_g acting on the steering axis given by:

(σ = turn angle of handle bar, θ_f = lean angle of frame, θ_cm = lean angle of center of mass, Φ = steering axis angle, H = z-coordinate of center of mass). Δ_scalar is the scalar value of the trail vector (see Section 5). Please remember the sign convention: A trail corresponds to a negative Δ_scalar because th contact point of the front tire is behind the steering axis. Under most circumstances the dynamic part in the above equation can be neglected.

In linearized form and for stationary conditions the above equation reduces to

For small angles his torque acts to turn the handle bar into the direction of the turn and of the lean.

For a bicycle at v = 0, T_g is the only torque. Equilibrium of the handle bar requires T_g to be zero. There are two solutions to this equation. In the first solution the expression in parenthesis in the above equation is zero. This corresponds to the symmetric solution with θ_f = σ = 0. This solution is not stable. If a bicycle is pushed with the handle bar free to move, then the handle bar will turn to an angle of approximately 60⁰ for an upright bicycle. This new equilibrium is stable and corresponds to the second solution given by Δ_scalar = 0. For an upright bicycle this is reached at a turn angle (see Sect. 4)

From the above equation it is evident, that the prime parameter which determines σ_zero is the hub offset k. With k = 0, the equilibrium turn angle of the handle bar would be 90⁰. This is not only inconvenient for handling a bicycle a zero or very small speed, but also dangerous. At 90⁰ the bicycle will stall. A too high value of k would cause other problems. σ_zero is the angle at which the trail becomes zero. Thus only turn angles below this value exhibit a trail and lead to stable riding. A good compromise is a value of k which reduces the trail to about half its k = 0 value.

An analytical expression for Δ_scalar = 0 for the tilted bicycle exists, but is rather intransparent. For this reason only the first order expansion expression is given. The validity range of this expression is very limited, but it illustrates the general behavior.

σ_zero rapidly decreases if the bicycle is leaned into the handle bar turn. This leads to a corresponding reduction of the stability range of a tilted bicycle.

5.2 Torque due to the centrifugal force

For a bicycle in motion, additional torques exist. The dominant additional torque is the castoring torque. It results from the vector product of trail and the force F_c, the reaction to the centrifugal force. Under stationary conditions F_c is given by:

For the definition of the coordinate system, see Figs. 1 and 2 in section 2 "Geometry and Kinematics". α is the steering angle of the front wheel and α_cm the corresponding angle of the center of mass trajectory. L is the distance between the wheel hubs and A the horizontal distance rear wheel hub - center of mass. This force has to be split into two vectors, the first perpendicular to the front wheel and the second perpendicular to the back wheel. The front wheel component becomes

The component in the steering axis of the vector product of trail and F_c is the castoring torque T_c and becomes

σ is the turn angle of the handle bar. Below the linearized expression valid for small angles is shown

For small angles T_c has the opposite sign as the torque T_g. It acts to turn the handle bar back into the σ = 0 position.

Under stationary conditions, the front wheel exerts a gyroscopic torque on the fork. As pointed out in sec. 3. "Torques on the lean axis", This torque is irrelevant for the stabilization of the center of mass. However, for a steering axis angle different from 90⁰, it has a component in the steering axis which has to be taken into account. This has been neglected in many previous models, because only the 90⁰ situation was considered. In linearized form the stationary torque becomes

J_w is the moment of inertia of the front wheel. The torque has the same sign as T_c and also is proportional to the velocity v squared. The ratio of the two torques is thus independent of velocity. For small angles and representing the moment of inertia I₀ of the wheel as mr², the ratio is

For a typical bicycle with rider m/M = 1/100, L/A = 3, r/Δ_scalar = 6 and cos(Φ) =0.34 resulting in a ratio of 0.06. The conclusion is that the gyroscopic torque has to be included in numerical calculations, but that it can be omitted in qualitative considerations. In numerical calculations its contribution can be absorbed by multiplying T_c by a correction factor K with K = 1.06.

5.3 Handle bar equilibrium at moderate and high velocity.

Equilibrium under stationary conditions is reached, if the sum of T_n and T_c is zero. In linearized form this is fulfilled for:

where σ_h is the equilibrium turn angle of the handle bar. This equation has only a solution for v > v_crit, given by

For a typical bicycle, v_crit is in the range 6 - 8 kph.

It is also of interest to have a look at the behaviour in the nonlinear regime. This is shown in Fig. 7.

Figure 7. Equilibrium position of the handle bar as a function of frame lean angle at 10 kph. The dotted lines represent instable solutions, the solid lines stable solutions. The blue lines were calculated by omitting the gyroscopic torque of the front wheel, in the red lines it is included.

It is illustrative to look at the blue lines of Fig. 7 (without gyroscopic torques). The initially instable Δ_scalar = 0 solution crosses the initially stable T_g + Tc = 0 solution at a lean angle of about 18 deg. At this point the two solutions change stability. If the gyroscopic torque of the front wheel is included, the two solutions do not cross, but repel each other. The gyroscopic torque is only of importance in the crossing range.

5.4 Handle bar equilibrium at very small velocity.

Below v_crit, there are not stable small angle solutions for small lean angles. As was shown in the first part of this section, the equilibrium turn angle at v = 0 is given by the condition Δ_scalar = 0. In the following we investigate the handle bar equilibrium for v < v_crit. The major result is shown in Figure 8.

5.5 Gyroscopic torque due to dθ_f/dt

There exists also a dynamic gyroscopic torque proportional to v dθ_f/dt acting on the steering axis. In linear approximation its value is given by:

This torque acts to turn the handle bar towards the side of the increasing lean. It has to be compared with T_n, which has the same sign. As opposed to T_n which is proportional to the lean, this torque is proportional to the first derivative of the lean. In the second order differential equations for bicycle motion it acts thus qualitatively as a damping. It has no influence on bicycle equilibrium, but influences bicycle dynamics at high speed. This will be discussed in section 6. The ratio between this torque and the torque of the normal (gravity) force becomes for σ = 0:

The ratio is proportional to the velocity v and to the time derivative of θ_f. Obviously the ratio is maximum if not only v is high, but also the denominator is small. This is the case for riding in a nominally straight line. To make an example, we assume v = 40 kph, a nominally straight trajectory with a "noise" due to perturbations such as pedaling of σ = 1 deg, θ_f = 1 deg and dθ_f/dt = 5 deg/sec. This results in a ratio of approximately 0.7. At smaller speed and in turns (σ > 0), the ratio is smaller. The influence of this gyroscopic term on the control loop in no-hands riding will be discussed in a later section.

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