Light at the end of a long tunnel (calculating the horizon)

So I had a ride in the cab of a diesel train recently. The route took us through the Remutaka Tunnel which is relatively long at 8.9 km (5.5 miles). Of note was the fact that you can’t see the end of the tunnel from the start, the reason being that there is a gradual climb to the middle of the tunnel and then a descent out to the other side. I jokingly told my daughter that you can’t see the light at the end of the tunnel because of the curvature of the Earth. Then after a moments thought I realised that with the length of the tunnel you maybe couldn’t see the end even if it was flat*.

So what say the collective genius of the SDMB? If the Remutaka Tunnel had no gradient, would the end of the tunnel be visible?

Some details, the height of the tunnel is 5.18 metres (17 feet)
I’m not sure what our eye height would’ve been, we were in a loco similar to this one:

so maybe 9 - 10 feet off the ground.

*Flat defined as how a body of water would sit, i.e., it follows the curve of the Earth rather than being exactly flat.

The distance to the horizon will be roughly proportional to the square root of your height, e.g. if you were 3 metres off the ground then it would appear to be a distance of about 6.2 km away from your vantage point

The dip is only about 3 minutes of arc, so it is marginal in terms of being able to see something if it were flat and 5 metres high

A nice 2-dimensional geometry/trigonometry problem. You can lay this out on paper and work out the math easily enough, and some approximations help.

Draw a line from the center of the earth to one tunnel entrance, and draw an other line from the center of the earth to the midpoint of the tunnel. The distance between those two points along the curvature of the earth is exactly 2.75 miles, but the true straightline distance is extremely close to that, so we’ll just say that it’s the same, 2.75 miles. Now we’ve got a triangle (shortest leg, 2.75 miles), and the angle where two legs meet at the midpoint of the tunnel is so close to 90 degrees that we will just say it’s exactly that, 90 degrees. So now we can figure out the angle between those first two long lines we drew, since we know the distance to the center of the earth is 4000 miles:

angle = arcsin(2.75/4000)

The angle is 0.03939 degrees.

Now take that line from the center of the earth to the midpoint of the tunnel, and draw an exactly perpendicular line over to the tunnel entrance. Extend the very first 4000-mile line we drew until it meets this new 2.75-mile line at some height X above the earth’s surface. What is the height, X, of that point above the earth’s surface? That’s where your eyeball needs to be in order to just see over the hump at the middle of the tunnel to a point equally high at the other end of the tunnel, assuming the floor of the tunnel is liquid flat.

cos(0.03939) = 4000/(4000+X)

X = 4000(1-cos(0.03939)/cos(0.03939)

X = 0.0009453 miles

X = 5 feet

If your eyeballs are five feet above the floor at the entrance of a 5.5-mile long tunnel, you’ll just be able to gaze into the eyes of your doppleganger at the far end. You won’t be able to see your doppleganger’s feet, but you’ll have a clear view of the sky above his head. Your eyes were 9-10 feet off the ground, so a perfectly “flat” and straight tunnel would have afforded an unobstructed view of the far end, assuming no optical refraction due to vertical air density gradients.

If you are truly enamored with and simply must see “the light at the end of the tunnel”, you could construct your tunnel in a manner that compensates for the curvature of the earth.

Or you could line your tunnel with mirrors. That should work for short distances such as this. Not sure how long the tunnel would need to be to negate the affect.

Not sure what arrangement of mirrors you’re suggesting here, but using a really long chain of mirrors isn’t the good idea it seems at first – each mirror scatters or absorbs some of the light. Even with 90% reflection, after ten bounces you’re down to 35% of your initial light. How much gets through a tunnel after a lot of bounces depends upon how the mirrors are arranged and how the light bounces off. If you want to just get the light from one end to the other you might direct a glimmer through.

But the idea that Egyptians could light the interior of tombs using multiply reflected sunlight really doesn’t work in practice.

I was thinking along the lines of a theoretical mirrored coating so that there would not be breaks where individual mirrors met. More of a thought scenario, like a hollow fiber optic in order to get to the theoretical 90% efficiency. And yes, not at all practical or even doable.

The Stanford Linear Acclerator, which zips particles at high speeds through a much narrower tube, was built in such a way. It is 20 cm “lower” in the middle to make it straight enough for particles to pass through.

20 inches, actually.

20 cm was the value used for the prototype: the Stanford Not Quite Linear Accelerator.

We would need to account for the height of the tunnel entrance/exit as well seeing as you would see light even if only the top of the exit was visible.

Thanks for that!

Like xkcd, there’s a Mythbusters for that.

.MythBusters Episode 169: Let There Be Light

Aziz, light!

Not only that, there’s a chapter in my latest book, Sandbows and Black Lights, devoted to that. Also an article in the next issue of Optics and Photonics News

A chain of mirrors to bring sunlight deep into a tomb far underground is impractical. But if you’ve got a tunnel that’s almost straight enough to get light already, then you might just need one or two bounces to finish the job. And if you’re just going for ambient lighting, not shining a spotlight onto a golden idol or something, then those one or two bounces don’t even need to be very precise.

Light tubes are a thing in modern construction.

Much better, thank you Aziz.

They are, indeed, but they’re limited to the distances allowed by scattering and absorption loss. And they’re a helluva lot less lossy than ancient Egyptian mirrors.(I mention them in the book)

Sorry; let’s work it out more carefully and account for everything. At least geometrically, assuming there are no extreme density gradients in the tunnel air.

Taking the geometric horizon distance, in km, to be 3.57√h, where h is the height of your eyes in meters… you said the length of the tunnel was 8.9 km, so if you stand at the entrance 3 m above the ground, your horizon is 6.18 km away. That leaves about 2.7 km, but that’s OK, you can still see almost to the ground: (2.7/3.57)2 = 0.6 m. If your eyes are 1.7 m high, the horizon is only 4.65 km off, and you will be able to see objects more than about 1.4 m off the ground, so still no problem seeing sky if the entrance is over 5 m high.

Conversely, if you work it out from the top of the exit (you said 5.18 m), the end of the tunnel will still be visible from anywhere over 5 cm off the ground.