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# Distance to horizon problem

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## Distance to horizon problem

(OP)
Any help/hint on the following would be appreciated.

Can we see (with naked eyes) a 150 ft tall, 2 ft wide pole from a 40 ft tower separated by 15 miles on a relatively flat terrain? I also wonder if "distance to horizon" approach would be applicable. Please suggest.

Thanks.

### RE: Distance to horizon problem

I used simple circle geometry, earth radius of 4000 miles.
I get that it would be visible, for  a separation of up to about 23 miles. As I recall there is curvature of light  involved which I did not consider but should be negligible.

I don't believe the "standard" naked eye is that good to see a 2' wide pole 15 miles away though, but maybe under the right light and contrast..

### RE: Distance to horizon problem

(OP)
Thanks CarlB, for the input. I did get the same separation distance. However, the question of seeing a 2 ft wide object from 15 miles away remains unanswered, at least for now.

### RE: Distance to horizon problem

A cargo ship at 25 miles is quite easy from an 80 m height.  Not the waterline though.  A motor boat at 3 miles is visible, but isn't easy all the time.  At 10 miles, its not much more than a few "specs" in width.  I think if you could see a 2 ft wide object at 5 miles, you'd qualify for eagle scout.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

### RE: Distance to horizon problem

The question, as posed, is not sufficiently detailed to derive an answer.  What do you mean by "see?"

If you mean detect a difference between the object and its background, then, under the right conditions, yes, you can "see" the tower.

If you mean whether you can tell that it's a 2-ft tower, then the answer is no.

2ft/15mi ~ 5 arcseconds.  All stars at night subtend less than 5 arcseconds, and we can "see" them, even at from 20 ly away.

### RE: Distance to horizon problem

I don't think you could see the width of a star except for the Sun.  You see the light rays spreading out from the source, the diameter of which is quite a bit larger than the diameter of the source by the time they reach here.  Wouldn't you agree?

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

### RE: Distance to horizon problem

Just look at the sky at night.  Almost all the stars that are visible in the night sky are at least as large as the Sun.  From a strict angular subtense perspective, they are almost all smaller in width than a 2ft tower at 5 miles.  The sun, if moved to Alpha Centauri's location, would subtend about 7 milli-arcseconds.  Now, of course, the atmosphere diffraction limit is about 2 arcseconds, so the objects aren't as small as they should be.  But, that's still smaller than the 5 arcseconds that the tower would subtend.

Nonetheless, you can "see" them, i.e., tell the difference between them and their background, which was the question I posed about what is meant by "see."  So, if the tower was adequately contrasted with its background, you could "see" it.  There would be no detail, nor any indication of actual width, but again, that's not necessarily required to achieve "seeing"

Note that this is fundamentally different that "resolving" 2ft at 15 mi.  That would require you to tell the difference between 2ft objects spaced 2ft apart at 15 mi, which is impossible with the naked eye.

As we know, detailed understanding of the requirements spells the difference between success and failure, so the OP needs to define his problem more precisely.

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