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# Tunnel - How Far?

barry1010 | 21:39 Tue 13th Aug 2024 | How it Works

The tunnel exits above ground in bright daylight.  It is flat and straight

What is the maximum length that I could see the light at the end of the tunnel from the start?

I have good eyesight if that matters. There is nothing in the tunnel causing snn obstruction

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we need the diameter.

sorry if the tunnel is really straight then infinity but I suspect you mean on earth thus following the curvature thus we need the diameter.

Question Author

8ft high, 8ft wide, arched

on earth so not actually straight?

from 8ft the horizon is 3.5 miles.

Even if the tunnel is straight there will be a distance beyond which the image of the entrance falls below the resolution of the retina.

...assuming its on a great circle circumference.

21:50 he's talking about some light not a book.

The eye sees an image of the tunnel entrance which gets smaller as the distance increases. At some point the image will be so small it drops between the rods/cones on the retina so it will not be visible.

Question Author

Ah! The height of the tunnel is 8ft.

I'd be standing at 6'6

we are talking about 3.5 miles on earth.

baz just put the height of your eyes in here:

http://www.ringbell.co.uk/info/hdist.htm

As mentioned, the resolution capabilities of the human eye also need to be considered. Are you able to see a fairly bright object (the end of the tunnel) which is eight feet across, at a distance of three miles?

well I can see andromeda! he's talking about "the light at the end of the tunnel" not watching the two ronnies.

The question is "SEE". The light will reach the eye but the eye won't see it.

Question Author

I am thinking that unlike the horizon, which is huge, the light at the end of the tunnel will be small because it is far away.  When I have been through long tunnels the distant opening looks tiny when I first see it, growing larger the closer I get.  Could there be a point where it is too small to register?

barry - that's exactly the point I'm trying to make.

Question Author

So there is no straightforward answer?

OK, “apparent” size, then.

As an object gets further away its apparent size decreases with the square of the distance. Assuming the tunnel appears to be 2.5 metres (roughly 8 feet) in diameter at a distance of ten metres, at a distance of 5,000 metres (roughly three miles or the distance of the "horizon") it will appear to be 2.5/(5000/10)²

2.5/(5000/10)² = 2.5/500² = 2.5/250,000

If my arithmetic is correct at this early hour I make the apparent size of the “light at the end of the tunnel” at 5km 0.00001 metres or 0.01 millimetres.  That is one hundredth of a millimetre. I’ve no idea what the limits of resolution for the human eye are but I imagine it is far higher than that. A quick Google suggests it is about 0.1 mm, which seems reasonable.

So a further quick "backwards" calculation suggests the apparent size of the tunnel’s end would shrink to 0.1mm at about 1.5km (1.57km I think is near enough). So that would seem to be the distance at where the light at the end of the tunnel is no longer visible to the naked eye.

So the curvature of the Earth is not a consideration as the tunnel end becomes invisible long before the “horizon” sees it disappear.

Thanks NJ for taking the trouble to do the calculations.

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