In my very lengthy story in the near future, there are some great 'Modern Marvels'. The next one I'm going to introduce might not be feasible, but if it is, I'd like to attach some numbers to it.

The Secretary for Urban & Regional Development with the Federal Highway Administration have secured the funds to build a straight tunnel from New York City to Los Angeles, through the Earth. It is a near vacuum. Therefore the modernized high speed rail cars simply "fall" into the tunnel, accelerating continuously for the first half of the tunnel, then slowing down to almost a perfect stop at the other end. The only friction is that of the rolling coefficient, and perhaps what little air remains.

Simply board your pressurized rail car, and go through the airlock.


(1) What is the deepest the tunnel goes? Is it in the crust still?

(2) What is the length of the tunnel? On the great arc, they're about 4,000km apart, but the tunnel will be shorter.

Bonus: Any ideas how fast a 'falling' train can get going? Should I add support from the train?

High level figures and speculation fine, of course.

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    $\begingroup$ Why would anyone build a system that sticks you halfway between NYC and LA? $\endgroup$ – Oldcat Apr 28 '15 at 17:25
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    $\begingroup$ If you want to get into the realistic engineering complexities of a project like this, look at the research going into the Hyperloop project. Both Elon Musk's "Hyperloop Alpha document" and the subsequent research by Hyperloop Transportation Technologies (the company working to build a working system out of it) are freely available, as the people in charge believe in open-source engineering. Maintaining a vacuum tunnel would be a huge mess in and of itself. Nature abhors a vacuum, and the more air you pump out, the harder it is to keep it out! $\endgroup$ – Mason Wheeler Apr 28 '15 at 18:19
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    $\begingroup$ Note also that you can get from New York to Los Angeles using no fuel with 19th century technology - the sailing ship - and get to admire the view while doing so. $\endgroup$ – jamesqf Apr 28 '15 at 18:53
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    $\begingroup$ You will have to use fuel to overcome the losses from the various types of friction. $\endgroup$ – Samuel Apr 28 '15 at 20:39
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    $\begingroup$ If the depth of the tunnel is not feasible just make a stop in Kansas City so you have a tunnel from there to LA and another one from there to NY. If you need more stops to further reduce the depth (or you just want to stop in cities that would actually be worth visiting) NY->Chicago->Denver->LA might be a good approach. $\endgroup$ – Roger Apr 28 '15 at 22:09

Curved Tunnels

It turns out that a longer path can actually be faster than the straight-line path. The fastest possible path in a uniformly dense Earth is a hypocycloid. We can define this path parametrically as follows:

$$ x = (R-r)\cos\theta + r\cos\left(\frac{r-R}{r}\theta\right) \\ y = (R-r)\sin\theta + r\sin\left(\frac{r-R}{r}\theta\right) $$

$R$ is the radius of the planet, and $r$ is the 'radius' of a single loop of the cycloid. $\theta$ goes from $0$ to $2\pi r/R$ over a single loop. For example, the case $R=1,\ r=0.1$ looks like this:

enter image description here

Geometric Properties

We can integrate to find the length $S$ of a loop:

$$ S = \int |d\vec{s}| = \int_0^{2\pi r/R} \sqrt{\left(\frac{\partial x}{\partial\theta}\right)^2+\left(\frac{\partial y}{\partial\theta}\right)^2}d\theta \\ = \int_0^{2\pi r/R} 2(R-r)\sqrt{\sin^2\left(\frac{R}{2r}\theta\right)}d\theta \\ = 8r(1-\frac r R) $$

The endpoint has coordinates:

$$ x = R\cos\left(2\pi\frac r R\right) \qquad y = R\sin\left(2\pi\frac r R\right) $$

This means that the distance across the surface $d$ is:

$$ d = R\times 2\pi\frac r R=2\pi r $$

Equations of Motion

First, we calculate the distance $\rho$ from the center of the planet:

$$ \rho^2 = x^2+y^2 \\ \rho^2 = 2r^2-2rR+R^2+2r(R-r)\cos\left(\frac R r\theta\right) \\ \rho^2 = R^2-2r(R-r)\left(1-\cos\left(\frac R r\theta\right)\right) $$

Then we calculate the velocity $v$ as a function of the rate of change of $\theta$:

$$ v^2 = \dot x^2 + \dot y^2 \\ v^2 = 2(R-r)^2\left(1-\cos\left(\frac Rr\theta\right)\right)\dot\theta^2 $$

We can calculate the total energy of the train (per mass) as:

$$ \frac{v^2}2 + \frac g{2R}\rho^2 $$

Where $g$ is the surface gravity of the planet. Plugging in our above expressions, adding the initial conditions $\theta=0,\ \dot\theta=0$, and solving for $\dot\theta$ gives us:

$$ \dot\theta = \sqrt{\frac{gr}{R(R-r)}} $$

Therefore, the period of motion over one loop is:

$$ T = \frac{\Delta\theta}{\dot\theta} = 2\pi\frac rR\sqrt{\frac{R(R-r)}{gr}} = 2\pi\sqrt{\frac{(R-r)r}{gR}} $$

For the Earth, this results in a curve like this:

enter image description here

You can see that the time reaches 42 minutes and 14 seconds for the limiting case of traveling to the other side of the Earth (where the cycloid degenerates to a straight line). We can divide the distance by this time to obtain an equivalent speed; that is, how fast you would have to go around the surface to make the same time:

enter image description here

In the through-the-Earth case, the speed reaches $\sqrt{gR}$, $7.9~\text{km}/\text{s}$. This happens to be the same as orbital speed, meaning that a surface-skimming satellite makes it to the other side at the same time as a straight-line gravity train.

In the case of a New York to Los Angeles trip ($d=3914~\text{km}$):

  • The track length is $4500~\text{km}$
  • The maximum depth is $1250~\text{km}$
  • The equivalent surface speed is $2.6~\text{km}/\text{s}$
  • The maximum speed is $4.7~\text{km}/\text{s}$
  • The peak acceleration is $1.8~g$
    • Acceleration is zero (free-fall) at both ends, and peaks in the center of the track. Peak value is $2~g$ for a very short track, and decreases linearly with length to $0~g$ (free-fall) for a through-the-Earth track.
  • The ratio $r/R$ is about $0.098$, meaning that the picture of the hypocycloid in the beginning is a pretty close approximation to the shape of this track.

Practical Problems

  • Such a tunnel is impossible to build, since no known material could withstand the heat and pressure in the mantle of the planet.
  • Any air friction or friction with the tunnel will slow the train down, and it will not be able to reach the other side without propulsion.
  • $\begingroup$ Tiny nit-pick, but I think in your first diagram you have r = 0.2. $\endgroup$ – user6511 Apr 29 '15 at 2:13
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    $\begingroup$ @user6511 Nope, the radius is 0.1. This means that the diameter, or maximum depth of the cycloid is 0.2. $\endgroup$ – 2012rcampion Apr 29 '15 at 2:16
  • $\begingroup$ I see now. The wording confused me. Cheers! $\endgroup$ – user6511 Apr 29 '15 at 2:21
  • $\begingroup$ @user6511 The usage of r comes from the fact that one of the ways to generate a hypocycloid is by rolling a circle with radius r inside a circle of radius R and tracing a point on the edge of the small circle. $\endgroup$ – 2012rcampion Apr 29 '15 at 2:24
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    $\begingroup$ @DougMcClean I guess what I meant to say was peak perceived acceleration. For a straight hole, you're in free-fall the whole way and experience 0 g. By this metric, standing stationary on the ground is 1 g. $\endgroup$ – 2012rcampion Apr 29 '15 at 14:58

Assuming an Earth radius of 6378 km, an arc of 4000 km would represent about 36 degrees of arc, so the direct line between the points (i.e. the length of the tunnel) would be about 3935 km. Interestingly, if you assume Earth is uniform density (not actually true, but not a horrible approximation for this), the travel time for the train is the same between any two points on earth, about 42 minutes. The top speed depends on the points, and would take a bit of high school physics to work out (if someone wants to add it in the comments, I'll incorporate it into the answer.)

Edit: I neglected to add that with the calculations above, the tunnel would extend about 300 km below the surface, which is about 10 times the thickness of the Earth's crust. Also to clarify, this is very infeasible with current engineering techniques. The depths, the attendant pressures, the friction, and other problems mean we seem to be quite a ways away from achieving this.

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    $\begingroup$ Your calculations are correct, the chord length is 3935km. $\endgroup$ – Samuel Apr 28 '15 at 17:43
  • $\begingroup$ Thanks for the check, @Samuel. I'll remove the qualification on my calculations that I'd left in in case I flubbed the basic geometry. $\endgroup$ – user3499545 Apr 28 '15 at 17:45
  • $\begingroup$ The tube would have to be evacuated, the car would essentially be a space capsule, and the car would ride on a magnetic cushioning system. There would be some friction but it would be very minute. The car could be accelerated and decelerated using a sort of LINEAC system that would require power but no propellant (the Earth is the propellant). $\endgroup$ – Jim2B Apr 29 '15 at 2:56
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    $\begingroup$ Jim2B, I agree that the friction is the most tractable problem here, although the vacuum part is pretty difficult. I'd be more concerned about the extreme pressure and temperature situation and the extreme difficulty in building a tunnel rather than the difficulty of running a train through it. $\endgroup$ – user3499545 Apr 29 '15 at 3:02

I came up with this same idea about 20 years ago as a thought experiment, and did calculations to arrive at the 42-minute travel time as well!

But I want to point out that the tunnel does not need to be a straight line, especially since it would travel deeper than the earth's crust. You can have a 1000-foot drop (on a non-terrifying angle for the passengers) to get up to speed, and then travel "level" for most of the distance before ramping up at the end.

And to avoid the rolling coefficient, I recommend maglev.

  • $\begingroup$ The fastest path will be a hypocycloid, which starts and ends with a vertical drop. $\endgroup$ – 2012rcampion Apr 28 '15 at 18:40
  • $\begingroup$ @2012rcampion That's a lot faster than having the painfully gradual acceleration that would come with a straight path, and would probably be faster than my suggestion too. But passengers wouldn't be comfortable with a vertical drop! That's an amusement part ride! But your suggestion is very smart; you should add it as an answer on its own! $\endgroup$ – BrettFromLA Apr 28 '15 at 18:49
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    $\begingroup$ I'm going to once I get home and can work up some graphics to work along with it. A side note, the vertical drop worked fine for Total Recall. $\endgroup$ – 2012rcampion Apr 28 '15 at 18:52

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