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Thread: The Summit

  1. #21
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    Default Line of reasoning

    HikerBob- many thanks for the (huge) topo map.

    Maybe the story about the WTC and Mt Washington was that if the earth were flat there would be no obstacle along the line connecting them. I'm not sure if that's a cool tidbit or not. It sounds cool. And it looks cool on the topo map. But this kind of situation would exist with many other sites I imagine. Taking away the curvature of the earth greatly compromises the geometry of the situation - kind of like saying I could lift a Hummer if gravity were way less...

    Patrad, I'm afraid your geometrical prowess way exceeds mine. But it sounds like you had fun pondering this question.

  2. #22
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    Default Here's the graph that I referred to in my previous post:

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    Veni, Vidi, Velcro!

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    Using the following formula for drop in Earth's surface as a function of distance due to the curvature of Earth:
    drop = [1/cos(distance*360/circumference) - 1] * radius

    where:
    earth's mean radius = 3958.26 miles
    equatorial circumference = 24901.55 miles
    distance of Mt.Washington to WTC = 282.13 miles

    We find that over a distance of 282.13 miles, the surface drops out approximately 10.0507 miles.

    So the WTC would need to be on top of a mountain nearly 9 miles high for an unimpeded "line-of-sight".

  4. #24
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    Default Beating a dead horse

    mk10: Though the issue was settled much earlier in this thread it is still interesting to see your calculations.

    In the end, this whole thread distills down to the mildly interesting fact that, were it not for the curvature of the earth, there would have been an unimpeded line between the top of the WTC and the top of Mt Washington. A reflection of how topographically muted Southern NH/Mass/Ct is.

  5. #25
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    Quote Originally Posted by John_Calif
    mk10: Though the issue was settled much earlier in this thread it is still interesting to see your calculations.
    I was merely showing how to calculate the drop in the earth surface due to curvature and simply used figures which seemed to be within the context of this thread. We all know that the original issue has been settled.

    Quote Originally Posted by John_Calif
    In the end, this whole thread distills down to the mildly interesting fact that, were it not for the curvature of the earth, there would have been an unimpeded line between the top of the WTC and the top of Mt Washington. A reflection of how topographically muted Southern NH/Mass/Ct is.
    Yes, this thread evolved from a discussion of line of sight between Mt.Wash/WTC and applied theory to a discussion purely hypothetical in nature--and although interesting in its own right, a hypothetical finding is absolutely meaningless in any practical sense.

    But to all of us who enjoy meaningless hypotheticals...

    Ignoring the earth's curvature, oblate spheroid properties, optical refraction, MTF, plus various other physical laws, the longest distance between any two points would probably be: a clear sight line from Ellef Ringnes Island in Canada, through the Bering Strait, all the way down the Pacific south to Antarctica, ending at the top of Vinson Massif--a distance of 10,866.08 miles. And yes, boring a hole and looking at these points through the center of the "real" earth is shorter (7,903.03 miles) than the distance in this "flat" fantasy world.

    Mark

  6. #26
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    Default ....yet another way of solving the problem

    As much as I love to figure out problems like this to 3 decimal places, there is a simple way to get a reality check on the answer to this problem that uses only simple math.

    Think of the radius of the earth (~3960miles) as the base (B) of a right triangle with the right angle formed where the radius hits the circumference. The hypotenuse (C) will be another line radiating from the center equal in length to the radius (3960 miles) plus the height of Mt. Washington (6288/5280=1.1909 miles). The third leg (A) is formed by the tangent and goes between the end points of ‘B’ and ‘C’, with the right angle at the end of ‘B’. Using the formula for right triangles, A squared +B squared=C squared, plug in 3960 for B, 3961.1909 for C and solve for A. Doing it quickly I get ~97 miles which is the distance you could see from the summit of Mount Washington to the horizon.

    Repeat this procedure using 3960.259 for C which is the radius +WTC and calculate this new line to the horizon to be ~45 miles. Adding the two parts together gives ~142 miles which is a fair estimate of the maximum distance Mount Washington and the WTC could be separated to have theoretical line-of-sight between them. Stephen Hawking might want to include the bending of light as it passes close to a large mass and optical engineers might want to include refraction of light similar to what creates a mirage, but this is a simple answer.

    These figures are close to the calculations presented by Yankee and John_Calif so I assume this is a good simple way to get an estimate.

  7. #27
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    Quote Originally Posted by Arthur Dent
    Think of the radius of the earth (~3960miles) as the base (B) of a right triangle with the right angle formed where the radius hits the circumference. The hypotenuse (C) will be another line radiating from the center equal in length to the radius (3960 miles) plus the height of Mt. Washington (6288/5280=1.1909 miles). The third leg (A) is formed by the tangent and goes between the end points of ‘B’ and ‘C’, with the right angle at the end of ‘B’. Using the formula for right triangles, A squared +B squared=C squared, plug in 3960 for B, 3961.1909 for C and solve for A. Doing it quickly I get ~97 miles which is the distance you could see from the summit of Mount Washington to the horizon.
    Not quite correct, the tangent (A) isn't exactly the same as the distance (D) along the surface (see attached diagram). But if the angle between B and C relative to the lengths of those lines is small (as is the case in our example) then the error introduced is insignificant. So, although not entirely accurate, this method is close enough for our purposes. Now, seeing that we've done all these rudimentary calculations and we have a better idea of what is realistic and what is not...

    I was on top of Mt Algonquin last week and tried to locate Mt Washington (133.55 miles at a bearing of 85°) and although the skies were fairly clear, an optimal view was still not to be had (photo attached with arrow drawn where MW would have been). So what is the furthest peak anyone has been able to see (aided or unaided) from atop the Rockpile?
    Attached Images Attached Images

  8. #28
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    Whoever thought there'd be so many different answers when adding 2+2.
    Bill
    Next up: Vermont City Marathon: May, 2011
    EasternLight

  9. #29
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    mk 10-"Not quite correct, the tangent (A) isn't exactly the same as the distance (D) along the surface (see attached diagram). But if the angle between B and C relative to the lengths of those lines is small (as is the case in our example) then the error introduced is insignificant."
    Ah, you are re-wording what I actually said. First I said "this is a good simple way to get an estimate" and I did specify that the answer would be an approximation of the line of sight, I never mentioned surface distance 'D' because that isn't where my line-of-sight travels. The scale in your drawing, although done for illustrative purposes, is quite distorted because the ratio of the radius to the summit elevation is about 4000:1.

    If we were to be truly technical about this we would actually have to traverse the route along the surface. What we would find is that the surface isn't flat and so the true distance would not be the arc but some distance greater because of all the ups and downs. My point was simply, that the estimate could be done simply, without making this needlessly complex, and I tried to stress this point in my post.

    As to what can be seen from the summit, I can tell you this past weekend it was nothing. On a clear day(or night) the land marks used might be Percy Peaks-35 miles N, Portland Light-65 miles E, and if I recall, Whiteface W, is about 135 miles.

  10. #30
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    Mount Marcy is 130 miles away and under normal conditions it cannot be seen from Mount Washington. Not because the air isn't clear enough, but because it lies below the horizon behind the Green Mountains.

    Under the right conditions Mount Marcy can loom above the horizon due to temperature inversions and atmospheric distortions. And with a set of binoculars it can be seen from Mount Washington.

    If it wasn't for the atmosphere the sun would rise approximately 7 minutes later than we see it rise. It would also set 7 minutes earlier.
    Bill
    Next up: Vermont City Marathon: May, 2011
    EasternLight

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