Hyperloop: Height difference

To begin, let's parametrize the possible shape of the Hyperloop tunnel. We're given that the tunnel has to be given by some cubic, which we'll denote $f(x)$ . Since it has to be horizontal at 0 m and 1000 m, we know that $f'(0) = 0$ and $f'(1000) = 0$ , so we know $f'(x) = Cx(x-1000)$ for some positive constant $C$ . Integrating (and assuming without loss of generality that $f(0) = 0$ ), we find that $f(x)$ has the form $f(x) = C(x^3 - 1500x^2)$ .

Now, in order for the pod not to crash into the ceiling, the tunnel must not curve down too fast. Formally, this is equivalent to the following: at any point on the trajectory of Hyperloop pod, if we removed the tunnel and let the pod freely fall due to gravity, the free-fall trajectory must lie strictly below the trajectory of the tunnel (at least in the neighborhood of the point). This translates into the mathematical statement that the second-derivative of the "free-fall trajectory" from a point on the curve must be strictly less than the second-derivative of the curve at that point.

What does the "free-fall trajectory" look like? If the pod has a horizontal velocity of $v_x$ to the right and a downward velocity of $v_y$ , then as a function of $t$ , we have that

$x = v_x t$ $y = - \frac{1}{2}gt^2 - v_y t$

Using $t = \frac{x}{v_x}$ to parametrize in terms of $x$ , we have that:

$y = -\frac{g}{2v_x^2}x^2 - \frac{v_y}{v_x}t$

The second derivative of this curve at this point is equal to $-\frac{g}{2v_x^2}$ . On the other hand, the second derivative of $f(x)$ is given by:

$f''(x) = 6C(x - 500)$

so we must have

$-\frac{g}{2v_x^2} \leq 6C(x - 500)$

Now, to find $v_x^2$ , note that

$v^2 = v_x^2 + v_y^2 = v_x^2 \left(1 + \left(\frac{dy}{dx}\right)^2\right) = v_x^2(1 + 9C^2x^2(x-1000)^2)$

so in particular,

$v_x^2 = \frac{v^2}{1+9C^2x^2(x-1000)^2}$

Finishing this off, we have that

$-\frac{g}{2v^2}(1 + 9C^2x^2(x-1000)^2) \leq 6C(x - 500)$

or equivalently,

$\frac{g}{2v^2} \geq \frac{6C(500-x)}{1 + 9C^2x^2(x-1000)^2}$

To find the maximum possible height of this tunnel, we want to find the maximum value of $C$ such that this holds for all $x \in [0, 1000]$ . But note that the RHS is decreasing in $x$ (at least until $x \geq 500$ , in which case the RHS is negative and the inequality automatically holds). So it suffices that

$\frac{g}{2v^2} \geq 3000 C$

and therefore the maximum value of $C$ is

$C = \frac{g}{6000v^2}$

Substituting this back into $f$ and computing $f(1000)$ , we find that $f(1000) = 18.15$ , as desired.

We know that the Hyperloop follows some kind of cubic path from the beginning to the end of the drop. We also know that its horizontal velocity is $v_0=$ 300 m/s at the top and at the bottom. At any point along the path it must be true that the velocity vector maintains $v_0$ as its magnitude, i.e. $v_0^2 = \dot{x}^2 + \dot{y}^2$ .

We also want to keep the pod from crashing into the ceiling of the Hyperloop's tunnel. For this to be true, the local projection of the pod into projectile motion must result in a steeper acceleration than does the path of the Hyperloop tunnel. Instantaneous projectile motion implies a constant increase in horizontal distance $t = x/\dot{x}$ . The vertical acceleration is then

$\begin{aligned}\frac{d^2y}{dx^2} &= \frac{d^2}{dx^2}\left[-\dot{y} t - \frac12g t^2\right] \\ &= \frac{d^2}{dx^2}\left[-\frac{\dot{y}\dot{x}}{x} - \frac12g\frac{x^2}{\dot{x}^2}\right] \\ &= -\frac{g}{\dot{x}^2} \end{aligned}$

The Hyperloop path is given by some function of $x$ , $y = f(x)$ . We know that $f(0) = 0$ , $\displaystyle\frac{df(0)}{dx} = v_0$ and $\displaystyle\frac{df(x_f)}{dx} = v_0$ .

The cubic function that satisfies these boundary conditions must take the form given by $\displaystyle f(x) = \gamma x^2\left(x-\frac32x_f\right)$ with $\gamma$ constant.

As we argued above, the vertical acceleration with position of the Hyperloop tunnel must be less than that of projectile motion, for the duration of the trajectory.

The vertical acceleration implied by $f$ is $\displaystyle\frac{d^2f}{dx^2} = 6\gamma x- 3\gamma x_f$ , thus we must have

$\displaystyle \frac{-g}{\dot{x}^2} \leq 6\gamma x-3\gamma x_f$ .

$6\gamma x - 3\gamma x_f$ is a minimum at the same point that $-g/\dot{x}^2$ is a maximum, i.e. $x=0$ where $\dot{x}=v_0$ , therefore, the maximum value for $\gamma$ is given by $g/v_0^2 = -3\gamma x_f$ or $\gamma=g/3v_0^2x_f$ .

The vertical drop from the beginning to the end of the descent is then given by $\begin{aligned} f(x_f) &= \frac{g}{3v_0^2x_f}x_f^2\left(x_f-\frac32x_f\right) \\ &= -\frac{g}{6v_0^2}x_f^2 \end{aligned}$

It is curious that this is equal to $\displaystyle\frac13\left(\frac12gt^2\right)$ , one third the descent for the Hyperloop-less pod sent sailing off the edge of a cliff with the same initial velocity. Is these a more general reason why the result must hold?

In order for the pod to remain at the bottom of the tunnel, its trajectory due to gravity must be lower than the trajectory of the cubic spline.

Assume that the cubic spline is $y_{s}=\frac{-hx^{3}}{10^{9}} +h$ (since $y=h$ when $x=0$ and $y=0$ when $x=1000$ )

Consider the pod at ( $x_{1};y_{1}$ ):

The trajectory due to gravity $y_{g}=y_{1} -300\sin\theta t - \frac{1}{2}gt^{2}$ ( $\tan\theta$ is the gradient of the tangent of the cubic spline at ( $x_{1};y_{1}$ ))

Since $x = 300\cos\theta t \rightarrow y_{g}=y_{1} -\tan\theta - \frac{1}{2}g\times \frac{x^{2}}{300^{2}\cos^{2}\theta}$

In order for the trajectory of $y_{g}$ to be lower than $y_{s} \Rightarrow |\frac{d^{2}y_{s}}{dx^{2}}| < |\frac{d^{2}y_{g}}{dx^{2}}|$ with all value of $x$ $\Rightarrow \frac{6hx}{10^{9}} < \frac{g}{300^{2}}(1+\frac{9h^{2}x^{4}}{10^{18}})$ with all value of $x$

Substitute the boundary values ( $x=0$ and $x=1000$ ) $\Rightarrow h=18.15 m$