The loop-the-loop is by now a standard stunt—a vehicle completes a vertical circle without losing contact with the track in a seemingly gravity-defying feat. One of the earliest mentions of the stunt is in the early 1900s when the American cyclist Conn Baker, stage name Diavolo (pictured above in action), travelled the country performing the feat on a track about 20 feet in diameter. His crowds, naturally, feared for his life.

The anxiety is understandable: most people grasp intuitively that if the speed is too low, the vehicle will come off the track at some point beyond where the track becomes vertical. But have you ever wondered about the maths behind it all? What, exactly, would happen to a dawdling Diavolo?

In order to understand how this trick works, we will need to understand something about the mechanics of the motion. In what follows we will make some modelling assumptions:

- That the track is circular.
- That there are no forces directed along the track, such as driving forces (ie Diavolo isn’t pedalling during the feat), or forces of resistance, such as friction.
- That the vehicle is point-like.

Under these simplifying assumptions we can make progress using Newton’s laws. These laws are all about the concept of force. His first law says that, in the absence of a net force, an object will carry on moving in the same direction and with the same speed (which might be zero) for ever. His second law $F=ma$, tells us that an object’s acceleration, $a$ (the rate of change of its speed and direction) is determined by the forces acting upon it. His third law introduces the concept of a *reaction force*. This force is present between any two objects in contact. For example, when you stand barefoot on the ground your weight (the force of gravity acting on your body) is exactly balanced by the reaction forces between your feet and the ground. Newton’s first law then says that you will stay in that state indefinitely—or until a new force is applied to your body.

In order for an object to follow a circle, there must be a net force directed toward the centre of that circle. This so-called centripetal force (which must be applied in order to keep the object in circular motion) is given at any time by $mv^2/r$, where $m$ is the mass of the object, $v$ is its current speed, and $r$ is the radius of the circle. In the loop-the-loop that force is supplied by two different mechanisms: the weight $mg$ of the object itself, which acts vertically downwards, and the reaction force $F_R$ between the object and the track, which always acts perpendicular to the track.

A *free body diagram* is a drawing used to visualise and analyse the forces on an object and how it responds to them. We represent the object by a point, and we represent the forces by arrows. When the object is climbing the track vertically, the free body diagram looks like this:

Its weight makes no contribution toward achieving the required $mv^2/r$. At this point in the motion the reaction force is supplying the entire centripetal force. By contrast, when the object is at the top of the circle, both the weight and the reaction force are aligned and they sum up to give $mv^2/r$. In fact, there is a critical speed $v_c$ at which the weight supplies the entire centripetal force: $mv_c^2/r=mg$, at this speed the reaction force is zero, ie the object is at the point of losing contact with the track.

The world record for the largest loop-the-loop completed by a car is held by Terry Grant. In 2015 he successfully completed the stunt on a track about three times as high as that of Diavolo. During a performance under wet conditions, he came very close to undershooting this critical speed and at the apex of the loop his wheels began to lose contact with the track.

## Coming off the track

Let’s imagine what would happen if Terry’s car, or Diavolo’s bicycle, had a speed less than $v_c$ and left the track. At this point I feel the need to say: *please don’t try this at home*. When the object leaves the track it is in free-fall, and will follow a parabolic trajectory. At some later time that trajectory will crash back into the track. You can imagine that this unfortunate event has happened to many daredevils over the years. Let’s see if we can find out at which point along the track this collision happens.

Let the track’s radius be $r$ and let the origin of our coordinate system be at its centre. If the object loses contact with the track at the angle $\theta$ (measured increasing anti-clockwise from the positive $x$-axis), then we can build the equations for $x(t)$ and $y(t)$ which describe the subsequent free-fall trajectory as a function of time $t$. We will set $t$ to be zero at the point contact is lost. Then $x(0)=r\cos\theta$ and $y(0)=r\sin\theta$.

As we learned above, for the object to be following a circle we must have a force of magnitude $mv^2/r$ pointing radially inwards. At the point of departure the reaction force is zero and the only force being applied to the object is its weight. We can think of the weight $mg$, which acts vertically downwards, as being the sum of two perpendicular components: a radial component of magnitude $mg\sin\theta$, and a tangential component of magnitude $mg\cos\theta$. The tangential component is responsible for slowing the object down as it climbs the track. The radial component must be equal to $mv^2/r$, or else we would not have circular motion. Solving for the speed in the equation $mg\sin\theta = mv^2/r$ we find $v=\sqrt{gr\sin\theta}$.

Which direction is the object moving in at the point of departure? When an object is moving along a track, it means that at any given time it is moving tangent to the track. The tangent to a circle is always at right angles to the radial direction. With some trigonometry, this allows us to promote the speed $v$ to a velocity: the object is moving with speed $v_x(0)=\mathrm{d}x/\mathrm{d}t=-v\sin\theta$ in the horizontal direction, and $v_y(0)=\mathrm{d}y/\mathrm{d}t=v\cos\theta$ in the vertical, at the time $t=0$.

Acceleration is the second derivative of position with respect to time. For an object in freefall, there is a constant force acting downwards on it: its weight, $mg$. This means that the vertical coordinate of our object, $y(t)$, should obey

\[\frac{\mathrm{d}^2y}{\mathrm{d}t^2}=a_y=-g,\]

while $\mathrm{d}^2x/\mathrm{d}t^2=0$ as there is no horizontal force applied.

Armed with these facts about the initial values of $x$, $y$, and their first and second derivatives, we can use the equations of a freefall trajectory:

\begin{align*}

x(t) &= x(0) + v_x(0) t,\\ y(t) &= y(0)+ v_y(0) t + \frac{a_yt^2}{2},

\end{align*}

giving us

\begin{equation}\label{para}\left.\begin{split} x(t) &= r\cos\theta – vt\sin\theta,\\ y(t) &= r\sin\theta + vt\cos\theta – \frac{gt^2}{2}. \end{split} \hspace{4mm} \right\} \hspace{-8mm} \tag{1} \end{equation}

These equations tell us that the object will be found at coordinates $(x(t),y(t))$ at time $t$ seconds after it leaves the track.

Going back to our original question of where and when our unfortunate (and purely theoretical) daredevil will land on the track again, we should try to solve the equation $x^2(t) + y^2(t) = r^2$ for $t$. This is just imposing the constraint that the point $(x(t),y(t))$ is at a distance $r$ from the origin, thereby intersecting the circular track once again. Something very important happens in this equation; it ends up having just two terms: one of order $t^4$ and one of order $t^3$. In fact, it reduces to

\begin{equation}\label{time} t^3\left(t-\frac{4v}{g}\cos\theta\right)=0,\tag{2} \end{equation}

and tells us immediately that at $(4v/g)\cos\theta$ seconds after leaving the track, the object crashes back into it. The simplicity is for a very good reason which we will explore below. Let’s find out *where* the crash has occurred.

Substituting the time of collision into $x(t)$ we find

$$ \begin{split} x(t) &= r\cos\theta – v \left(\frac{4v}{g}\cos\theta \right)\sin\theta\\ &= r\cos\theta – 4 r\cos\theta \sin^2\theta\\ &= r\left(4\cos^3\theta – 3 \cos\theta\right). \end{split} $$

The expression $4\cos^3\theta – 3 \cos\theta$ happens to be one side of the triple-angle identity for cosine: \begin{equation}\nonumber \cos3\theta\equiv 4\cos^3\theta – 3 \cos\theta. \end{equation}

We have found that at the collision point $x(t) = r \cos 3\theta$, which is the same as $r \cos(-3\theta)$. It must be that $y(t)$ is $\pm r \sin 3\theta$ since $x^2+y^2 = r^2$. In fact the sign turns out to be negative. We have found that the collision point is $x=r\cos(-3\theta)$, $y = r\sin(-3\theta)$. This is thrice the angle of departure, measured in the opposite direction.

This relationship between the angles of departure and collision is a purely geometric result: independent of the radius of the track and the strength of gravity. It is as true on the moon with a track the size of a mouse, as on Mars with a track 50 feet high.

One can also see the limiting case when the object just manages to stay on the track: this is when $\theta=\mathrm{\pi}/2$ and then $3\theta = 3\mathrm{\pi}/2$ (measured in the opposite direction) corresponds to the same point on the track, ie the top-most point. This is what Grant experienced at the top of his loop: his speedometer showed a much higher speed than he was actually travelling at—his wheels began to spin faster as they lost contact with the track!

Now that we know where and when to put a landing mattress (or, in Terry’s case, perhaps something more elaborate) to keep our daredevil safe, let’s think a bit more about the result that we have found. Returning to equation \eqref{time}, we might wonder why it is so simple. Why did we not end up with terms of order $t^0$, $t^1$, or

$t^2$? The answer is tied up with the theory of polynomial curves and with the physics of Newton’s laws.

The freefall trajectory in equation \eqref{para} is, in fact, a quadratic curve $y=ax^2+bx+c$. In order to see this we eliminate the time $t$ by rewriting the equation for $x(t)$ as follows:

\[ t= \frac{r\cos\theta-x}{v\sin\theta}, \]

and then replacing the instances of $t$ in the equation for $y(t)$ with this expression. The result is the following parabola:

\begin{equation}\label{yofx} y(x) = r\sin\theta – \cot\theta \,(x-r\cos\theta) – \frac{(x – r\cos\theta)^2}{(2r\sin^3\theta)}. \tag{3} \end{equation}

The curve of the track itself, $x^2+y^2=r^2$, is also quadratic in $x$. To find intersection points we substitute $y$ from the parabola into this equation and end up with a polynomial of degree four which can be factorised into the form $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=0$. There will, therefore, be four solutions, but they do not need to be distinct (or even real, for that matter). In fact, asking for the curves to be tangent is the same thing as requiring two of $\alpha,\beta,\gamma,\delta$ to be equal.

To see this consider a curve which intersects another curve at two nearby points $x=\alpha-\varepsilon$ and $x=\alpha+\varepsilon$. In the following picture, we have chosen the second curve to be the $x$-axis. In the limit as we take $\varepsilon \to 0$, these two points merge into one and so the curves are tangent at the point of contact.

We saw that $y(t)$ in equation \eqref{para} has the property that two derivatives in $t$ gives us a constant: $\mathrm{d}^2y/\mathrm{d}t^2=-g$. The curve $y(x)$ of the parabola in equation \eqref{yofx}, being a quadratic in $x$, also has a constant second derivative $\mathrm{d}^2y/\mathrm{d}x^2$. That’s no accident—the derivatives in $x$ are determined by the derivatives in time. When the object is at the point of losing contact with the track, the only force acting on it is its weight—exactly as is the case for the freefall trajectory. This means that not only are the first derivatives matched between track and trajectory (as they are tangent), but the *second* derivatives are as well. We could have just asked for the parabola which is tangent to the circle at the angle $\theta$, and also matches the *second* derivative of the track there as well.

Returning to the equation $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=0$: asking for matching *second* derivatives is the same thing as requiring *three* of $\alpha,\beta,\gamma,\delta$ to be the same. This mechanism is the reason for the simplicity in equation \eqref{time}, even if the time $t$ is playing the role of $x$.

## When conic sections hug circles

The parabolic freefall trajectory is an example of a conic section—the curves produced when a plane intersects a cone. These curves come in different varieties: circles, ellipses, parabolas, and hyperbolae.

They can all be defined by a common construction. Let $F$ be a point and $L$ a line not through $F$. Then let the locus of points $P$ have the property that the ratio of distances $e=\overline{PF}/\overline{PL}$ is a constant, where $\overline{PL}$ is measured perpendicular to $L$. The curve consisting of the points $P$ is a conic section, and the value of the ratio $e$ determines whether it is a circle ($e=0$), ellipse ($e<1$), parabola ($e=1$), or hyperbola ($e>1$).

Since we have studied the case of the parabola, let’s move on to other conic sections. Whereas the equation of the parabola (compare with equation \eqref{yofx}) has no $y^2$ term, a completely general ellipse or hyperbola, whose line $L$ is parallel with one of the coordinate axes, has the form

$$ A(x-a)^2 + B(y-b)^2 = 1, $$

where $a$, $b$, $A$, and $B$ are constants and $A$ and $B$ are not both negative. In this section we will seek to constrain these constants so that the resulting conic section has the following three properties:

- intersects the circle $x^2+y^2=r^2$ at the point $C=(r\cos\theta,r\sin\theta)$,
- matches the first derivative of the circle at $C$,
- matches the second derivative there as well.

Imposing the first condition amounts to requiring:

\begin{equation}\label{point} A(r\cos\theta-a)^2 + B(r\sin\theta-b)^2 = 1. \tag{4} \end{equation}

To make this more concrete, here is a picture for $r=1$, $\theta=\mathrm{\pi}/4$, with some randomly chosen values of $A$, $B$, $a$, and $b$ which satisfy equation \eqref{point}:

Now we would like to impose the second condition. What is the first derivative of the circle at $C$?

\begin{align*} \frac{\mathrm{d}}{\mathrm{d} x}(x^2+y^2)&=2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ \implies \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y}=-\cot\theta. \end{align*}

To find the first derivative of the conic section at $C$, we can employ the same strategy:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left[A(x-a)^2 + B(y-b)^2 \right] =0\\\implies \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{A(x-a)}{B(y-b)}, \end{align*}

and so our condition is:

\begin{equation}\label{frst} -\frac{A(r\cos\theta-a)}{B(r\sin\theta-b)} = -\cot\theta. \tag{5} \end{equation}

Here is our previous picture, but now with this second condition imposed on $A$, $B$, $a$, and $b$:

Finally we would like to impose the matching of the second derivatives. For the circle, we may proceed as follows: \begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= \frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{x}{y}\right) \\ &=-\frac{1}{y}+\frac{x y’}{y^2}\\ &= -\frac{y^2+x^2}{y^3} = -\frac{1}{r\sin^3\theta}. \end{align*} The same strategy is applied to the conic section: $$ \begin{split} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= \frac{\mathrm{d}}{\mathrm{d}x}\left[-\frac{A}{B}\frac{(x-a)}{(y-b)}\right]\\ & =-\frac{A}{B(y-b)}+\frac{A(x-a) y’}{B(y-b)^2}\\ &= -\frac{A}{B^2} \frac{B(y-b)^2+A(x-a)^2}{(y-b)^3}\\ &=-\frac{A}{B^2(y-b)^3} \\ &=-\frac{A}{B^2(r\sin\theta-b)^3}. \end{split} $$

We therefore require \begin{equation}\label{sec} -\frac{A}{B^2(r\sin\theta-b)^3} = -\frac{1}{r\sin^3\theta}, \tag{6} \end{equation}

and our picture now looks like the following:

Equations \eqref{point}, \eqref{frst} and \eqref{sec} constitute a system of three equations constraining the four constants $a$, $b$, $A$, and $B$. The solution can be expressed as

\begin{align*} a &= \frac{\lambda-1}{\lambda}r\cos^3\theta,\quad b = (1-\lambda)r\sin^3\theta,\\ A &= \frac{\lambda^2}{r^2\left[(\lambda-1)\sin^2\theta + 1 \right]^3},\\ B &= \frac{\lambda}{r^2\left[(\lambda-1)\sin^2\theta + 1 \right]^3}, \end{align*}

where $\lambda$ is a free parameter not equal to zero. We therefore have a family of conic sections, parameterised by $\lambda$, given by

$$\lambda^2 \left[ x – \frac{\lambda-1}{\lambda}r\cos^3\theta\right]^2 + \lambda\left[y \,\, – (1-\lambda)r\sin^3\theta\right]^2 =r^2\left[(\lambda-1)\sin^2\theta + 1 \right]^3, $$

which intersect the circle $x^2+y^2=r^2$ at $C$ and also match the first and second derivatives of the circle at $C$. It is difficult to give a direct geometric interpretation for the parameter $\lambda$, but it controls the tilt of the plane (or, equivalently, the ratio

$e$). When $\lambda=1$ the tilt is zero: we have the circle itself. If we decrease $\lambda$ the tilt grows rapidly, stretching the circle out into an ellipse elongated along the $x$-axis and becoming a (horizontal) parabola in the limit as $\lambda\to 0$, then hyperbolae for $\lambda < 0$. If we increase $\lambda$ from $\lambda=1$, the circle gradually stretches out along the $y$-axis instead. Here, the tilt of the plane grows more and more slowly as $\lambda$ is increased, and a (vertical) parabola is only reached in the limit as $\lambda\to\infty$. Specifically, if one carefully opens all the brackets above, and keeps only those terms with the highest power of $\lambda$ (which turns out to be $2$), one recovers $y(x)$ from equation \eqref{yofx}. So, even though we excluded the parabolas in our treatment, they are actually included here through a limiting procedure.

For any value of $\lambda\neq 0$ or $1$, these curves also intersect the circle at a single further point: $(r\cos(-3\theta),r\sin(-3\theta))$. This result is a special case of a more general theorem which says that if a conic section (aligned parallel with the coordinate axes) intersects the unit circle in four places, the values of $\theta$ at those places add to give zero (or a multiple of $2\mathrm{\pi}$). Here, we have engineered a triple point of contact at $C$, and so thrice $\theta$ is equal to minus the angle at the other intersection point.

Finally, we note that under any general linear transformation: $x\to ax+by$, $y\to cx+dy$, the circular track is mapped (in general) to an elliptical one, while the freefall trajectory is mapped to another conic section. The fact that the two curves have matching first and second derivatives at the point of departure is unaltered by the transformation, and so there will be some transformed version of the result for elliptical tracks. For example, starting with a circular track and a parabolic freefall trajectory, if we apply a stretch parallel to the $y$-axis: $y\to ky$, we produce an elliptical track with a parabolic free-fall trajectory; the departure angle is then $\tan^{-1}(k\tan\theta)$ and the collision angle is $\tan^{-1}(k\tan(-3\theta))$.