Motorcycle Gyroscopic Effects¶
Introduction¶
While it is technically possible for a bicycle or motorcycle to self-stabilize in the absense of gyroscopic effects from the spinning wheels (see Koojiman et al., 2011), eliminating gyroscopic effects from most 'normal' bicycle or motorcycle designs will have drastic effects on their transient behavior and stability, even for small light two-wheelers with small wheels. Below, see how the eigenvalues of the fourth order model of Lafayette College's Razor MX350 minibike change when gyroscopic effects are excluded.
as the figure shows, without the gyroscopic effects from the wheels (which are only around 12" in diameter including tires!), the minibike is never self-stable, because there is no speed at which all four of the eigenvalues' real parts are less than zero. With gyroscopic effects, the 4th-order Whipple model predicts that the Razor has a small window of self-stabilizing speeds between just over 3 m/s and just below 4 m/s.
Approach¶
This notebook will walk through where gyroscopic effects on motorcycle stability come from. The expressions for gyroscopic torque that we develop here will be incorporated into the derivation of the fourth order model of motorcycle dynamics by including them as "non-conservative torques" in Lagrange's equations. Especially if you have some background in physics, you might balk at the idea of treating gyroscopic reactions as "non-conservative," since they arise from the conservation of angular momentum, and in the absense of friction, an idealized gyroscope will certainly satisfy the first law of thermodynamics, even if the direction of angular velocities change.
However, in the fourth order Whipple-style model of motorcycle dynamics, the system's energetic boundary only includes motion in the bike's yaw, roll, and steer directions. The angular momentum of a motorcycle's front or rear wheel is actually a result of the bike's forward ($\hat{\imath}$) motion, which is assumed constant in the Whipple model. This means that the spin motion of the front and rear wheels is an "infinite well" of power, so the gyroscopic effects can technically cause energy to cross into or out of the system's boundary, acting potentially as a "source" or a "sink" of energy.
In the remaining sections, we will use the conservation of angular momentum to derive the reaction torques that must be present in order for changes in the wheels' directions to occur using the rotational form of Newton's Second Law for a rigid body:
\begin{equation} \left. \frac{d \vec{H}}{dt}\right|_g = \sum \vec{T}_g \end{equation}
where "g" denotes "about a body's center of mass. The idea here is that as the bike rolls, yaws, and steers, the rear and front frames of the bike exert torques on the wheels in order to motivate changes in the direction of the wheels' angular momenta. Once we know what torques must have been exerted on a wheel in order for roll, yaw, and/or steer motino to occur, we can infer that by Newton's 3rd law, those same torques are exerted by the wheel on the body it is attached to, but in the opposite direction. Collecting all of those torques will allow us to come up with expressions for the nonconservative gyroscopic reaction torques needed for the fourth order Whipple model derivation in the direction of each of its three generalized coordinates: $\tau_{\phi, GY},\tau_{\delta,GY}$, and $\tau_{\psi,GY}$.
Angular Momentum¶
The front or rear wheel of the motorcycle has nearly all of its angular momentum in the body-fixed (moving) $\hat{\jmath}$ direction due to the "spinning" of the wheel.
Under the assumption that the tires roll without slip, we can write:
\begin{equation} \begin{aligned} \vec{\omega}_{f} &=& \frac{U}{R_{rfw}}\hat{\jmath}_f \\ \vec{\omega}_{r} &=& \frac{U}{R_{rrw}}\hat{\jmath}_r \end{aligned} \end{equation}
Where $U$ is the bike's forward speed, and $R_{fw}$ and $R_{rrw}$ are the front and rear wheel radii, respectively. $\hat{\jmath}_f$ is the unit vector pointing in the front frame's local $y$-direction, and $\hat{\jmath}_r$ is the unit vector pointing in the rear frame's local $y$-direction, assuming ISO body-fixed vehicle coordinates where $x$ is forward, $y$ is left, and $z$ is up. These unit vectors move with the bike as it yaws, rolls, and steers.
If the front and rear wheel moments of inertia about the spin axes are $J_{yyf}$ and $J_{yyr}$ respectively, we can write the rear wheel's angular momentum as:
\begin{equation} \vec{H}_{r} = \frac{J_{yyr}}{R_{rrw}}U \hat{\jmath}_r \end{equation}
And we can write the front wheel's angular momentum as:
\begin{equation} \vec{H}_{f} = \frac{J_{yyf}}{R_{rfw}}U \hat{\jmath}_f \end{equation}
Rear Wheel Gyroscopic Reaction Torque¶
By Newton's second law, we know that the torque required to change the rear wheel's angular momentum is given by Eqn 1, with angular momentum given by Eqn 3. Then, the reaction of the rear wheel on the bike's rear frame will have the same magnitude, but act in the opposite direction. In other words, the gyroscopic reaction torque of the rear wheel on the rear frame can be written as:
\begin{equation} \vec{\tau}_{GY,r} = -\frac{d\vec{H}_r}{dt} \end{equation}
By the product rule of differentiation, we can write: \begin{equation} \require{cancel} \frac{d\vec{H}_r}{dt} = \cancelto{0}{\frac{d}{dt}\left(\frac{J_{yyr}}{R_{rrw}}U \right)\hat{\jmath}_r} + \frac{J_{yyr}}{R_{rrw}}U\frac{d}{dt}\hat{\jmath}_r \end{equation}
The first term in the product is zero if the bike's forward speed is constant, but the derivative of the $\hat{\jmath}_r$ term is not zero because the $\hat{\jmath}_r$ vector moves as the bike moves. The derivative of the $\hat{\jmath}_r$ vector can be written as the cross product of the vector's angular velocity with the vector itself:
\begin{equation} \frac{d}{dt}\hat{\jmath}_r = \vec{\omega}_{xyz,r} \times \hat{\jmath}_r \end{equation}
where $\vec{\omega}_{xyz,r}$ is the angular velocity of the coordinate system attached to the rear wheel. in our case, the rear wheel is attached to the rear frame of the bike, so the angular velocity of the rear wheel's coordinate frame is the same as the angular velocity of the bike's rear frame. Therefore,
\begin{equation} \vec{\omega}_{xyz,r} = \left(\dot{\phi}\hat{\imath}_r + \dot{\psi}\hat{k}_r\right) \end{equation}
where $\dot{\psi}$ is the bike's rear frame yaw rate and $\dot{\phi}$ is the bike's rear frame's roll rate. This means that we can express $\frac{d}{dt}\hat{\jmath}_r$ as:
\begin{equation} \begin{aligned} \frac{d}{dt}\hat{\jmath}_r &=& \left(\dot{\phi}\hat{\imath}_r + \dot{\psi}\hat{k}_r\right)\times \hat{\jmath}_r\\ &=& -\dot{\psi}\hat{\imath}_r + \dot{\phi}\hat{k}_r \end{aligned} \end{equation}
Now, from the kinematics notebook, we know that we can write $\dot{\psi} = \frac{U\sin\lambda}{b}\delta + \frac{c\sin\lambda}{b}\dot{\delta}$, where $\delta$ is the bike's steering angle, $b$ is the bike's wheelbase, and $\lambda$ is the inclination of the steer axis relative to the negative $\hat{\imath}_r$ direction. Making this substitution into Eqn 9, and substituting the result into Eqn 5 yields:
\begin{equation} \begin{aligned} \vec{\tau}_{GY,r} &=& \frac{J_{yyr}}{R_{rw}}U\left(\left(\frac{U\sin\lambda}{b}\delta + \frac{c\sin\lambda}{b}\dot{\delta}\right)\hat{\imath}_r - \dot{\phi}\hat{k}_r\right) \\ &=& \tau_{GY,\phi,r}\hat{\imath}_r + \tau_{GY,\psi,r}\hat{k}_r \end{aligned} \end{equation}
This is our final equation for the gyroscopic reaction torque of the rear wheel on the rear frame of the bike as the bike rolls and yaws.
Front Wheel Gyroscopic Reaction Torque¶
For the front wheel, we repeat the same process. By Newton's second law, we know that the torque required to change the front wheel's angular momentum is given by Eqn 1, with angular momentum given by Eqn 4. Then, the reaction of the front wheel on the bike's front frame will have the same magnitude, but act in the opposite direction. In other words, the gyroscopic reaction torque of the rear wheel on the rear frame can be written as:
\begin{equation} \vec{\tau}_{GY,f} = -\frac{d\vec{H}_f}{dt} \end{equation}
By the product rule of differentiation, we can write: \begin{equation} \require{cancel} \frac{d\vec{H}_f}{dt} = \cancelto{0}{\frac{d}{dt}\left(\frac{J_{yyf}}{R_{rfw}}U \right)\hat{\jmath}_f} + \frac{J_{yyf}}{R_{rfw}}U\frac{d}{dt}\hat{\jmath}_f \end{equation}
The first term in the product is zero if the bike's forward speed is constant, but the derivative of the $\hat{\jmath}_f$ term is not zero because the $\hat{\jmath}_f$ vector moves as the bike moves. The derivative of the $\hat{\jmath}_f$ vector can be written as the cross product of its angular velocity with itself:
\begin{equation} \frac{d}{dt}\hat{\jmath}_r = \vec{\omega}_{xyz,f} \times \hat{\jmath}_f \end{equation}
where $\vec{\omega}_{xyz,r}$ is the angular velocity of the coordinate system attached to the front wheel. In our case, the front wheel is attached to the front frame of the bike, which is then attached to the rear frame of the bike. The only difference between the front and rear angular velocities is that the front frame has an additional angular velocity in the $\hat{e}_\lambda$ direction (see the kinematics notebook). Therefore,
\begin{equation} \begin{aligned} \vec{\omega}_{xyz,f} &=& \left(\dot{\phi}\hat{\imath}_f + \dot{\psi}\hat{k}_f + \dot{\delta} \hat{e}_{\lambda} \right)\\ &=& \left( \dot{\phi} - \dot{\delta}\cos\lambda \right)\hat{\imath}_f + \left(\dot{\psi} + \dot{\delta}\sin \lambda \right)\hat{k}_f \end{aligned} \end{equation}
where $\delta$ is the bike's steering angle, $\hat{e}_\lambda = -\cos\lambda \hat{\imath}_f + \sin\lambda \hat{k}_r$ is the unit vector of the steer axis, $\dot{\psi}$ is the bike's rear frame yaw rate and $\dot{\phi}$ is the bike's rear frame's roll rate. This means that we can express $\frac{d}{dt}\hat{\jmath}_f$ as:
\begin{equation} \begin{aligned} \frac{d}{dt}\hat{\jmath}_f &=& \left(\left( \dot{\phi} - \dot{\delta}\cos\lambda \right)\hat{\imath}_f + \left(\dot{\psi} + \dot{\delta}\sin \lambda \right)\hat{k}_f\right)\times \hat{\jmath}_f\\ &=& - \left(\dot{\psi} + \dot{\delta}\sin \lambda \right)\hat{\imath}_f + \left( \dot{\phi} - \dot{\delta}\cos\lambda \right)\hat{k}_f \end{aligned} \end{equation}
Again, from the kinematics notebook we know that we can write $\dot{\psi} = \frac{U\sin\lambda}{b}\delta + \frac{c\sin\lambda}{b}\dot{\delta}$, where $b$ is the bike's wheelbase, and $\lambda$ is the inclination of the steer axis relative to the negative $\hat{\imath}_r$ direction. Making this substitution into Eqn 15, and substituting the result into Eqn 11 yields:
\begin{equation} \begin{aligned} \vec{\tau}_{GY,f} &=& \frac{J_{yyf}}{R_{fw}}U\left( \left(\dot{\delta}\left(\sin\lambda +\frac{c\sin\lambda}{b}\right)+\delta \frac{U\sin\lambda}{b}\right)\hat{\imath}_f - \left(\dot{\phi}-\dot{\delta}\cos\lambda\right)\hat{k}_f \right) \\ &=& \tau_{GY,\phi,f}\hat{\imath}_f + \tau_{GY,\psi,r}\hat{k}_f \end{aligned} \end{equation}
Combining Reaction Torques for Use in the LaGrangian Derivation¶
In the fourth order model notebook, we consider the gyroscopic reactions developed in the preceding two sections as "non-conservative" torques that appear in LaGrange's equations for the bike. This means that we need to find scalar reaction torques representing total gyroscopic reactions from the front and rear wheels about each of the bike's degrees of freedom in its 'generalized coordinates' $\begin{bmatrix} \phi & \delta & \psi \end{bmatrix}^T$.
The total gyroscopic reaction torques about each degree of freedom can be written:
\begin{equation} \begin{aligned} \tau_{GY,\phi} &=& \tau_{GY,\phi,f}&+\tau_{GY,\phi,r}\\ \tau_{GY,\delta} &=& \tau_{GY,\delta,f}&\\ \tau_{GY,\psi} &=& \tau_{GY,\psi,f}&+\tau_{GY,\psi,r}\\ \end{aligned} \end{equation}
The $\phi$ and $\psi$ torques are easy to compute. They are just the sums of the front and wheel torques in the $\hat{\imath}$ and $\hat{k}$ directions, respectively. Note that I have dropped the "f" and "r" subscripts-- the fourth order model is linearized about zero roll and steer angles, so these body-fixed vectors will be the same for the front and rear frames. Adding front and rear reactions and collecting terms, we can compute these two torques as:
\begin{equation} \begin{aligned} \tau_{GY,\phi} &=& \delta \frac{U^2\sin\lambda}{b}\left(\frac{J_{yyf}}{R_{fw}}+\frac{J_{yyr}}{R_{rw}} \right) + \dot{\delta}\left(U\frac{c\sin\lambda}{b}\left(\frac{J_{yyf}}{R_{fw}}+\frac{J_{yyr}}{R_{rw}} \right) + U\frac{J_{yyf}}{R_{fw}}\sin\lambda \right) \\ \tau_{GY,\psi} &=&\dot{\phi} \left(-U\left(\frac{J_{yyf}}{R_{fw}}+\frac{J_{yyr}}{R_{rw}} \right)\right) + \dot{\delta}U\frac{J_{yyf}}{R_{fw}}\cos \lambda\\ \end{aligned} \end{equation}
The $\delta$-direction, however, is less straightforward. Eqn 17 states that only the front wheel contributes to reaction torques in the $\delta$-direction. This is because only the front frame is able to move freely in the $\delta$-direction, so only the front wheel's reaction torques are capable of doing work about that axis. To find this torque, we can take the dot-product of $\vec{\tau_{GY,f}}$ with the $\hat{e}_\lambda$ unit vector to find the portion of the gyroscopic reaction torque about the steer axis. This yields:
\begin{equation} \begin{aligned} \tau_{GY,\delta} &= \vec{\tau}_{GY,f}\cdot\hat{e}_\lambda \\ &= \vec{\tau}_{GY,f}\cdot \left(-\cos\lambda \hat{\imath} + \sin\lambda \hat{k} \right) \\ &= -U\cos\lambda\frac{J_{yyf}}{R_{fw}} \left( \dot{\delta}\left(\sin\lambda +\frac{c\sin\lambda}{b}\right)+\delta\frac{U\sin\lambda}{b}\right) -U\sin\lambda\frac{J_{yyf}}{R_{fw}}\left( \dot{\phi}-\dot{\delta}\cos\lambda \right) \end{aligned} \end{equation}
A close look at Eqn 19's expanded form will show a $\pm \dot{\delta}U\frac{J_{yyf}}{R_{fw}}\sin\lambda\cos\lambda$ that cancel each other out, leaving:
\begin{equation} \tau_{GY,\delta} = -\dot{\delta} U\frac{J_{yyf}}{R_{fw}}\frac{c\sin\lambda\cos\lambda}{b} - \delta U\frac{J_{yyf}}{R_{fw}}\frac{U^2\sin\lambda\cos\lambda}{b} - \dot{\phi}U\sin\lambda\frac{J_{yyf}}{R_{fw}} \end{equation}
Eqns 18 and 20 represent the final equations we need to incorporate gyroscopic reaction torques into the fourth order model of bicycle/motorcycle dynamics.