Comments on: This should be basic geometry...
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Comments on Ask MetaFilter post This should be basic geometry...Thu, 13 Aug 2020 20:17:02 -0800Thu, 13 Aug 2020 20:58:03 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: This should be basic geometry...
http://ask.metafilter.com/347461/This-should-be-basic-geometry
I want my camera to have an unobstructed view of the night sky down to about 10° above the horizon. There is a large thunderstorm several counties away which my weather app tells me has cloud tops up to 60,000 feet / 18,250 meters. My camera is 5 feet / 1.5 meters off the ground. Assume that I am on a level plain with no other obstructions. How far away do I need to be from the storm so that its top is less than 10° above my camera's horizon? What would be the equation for this?post:ask.metafilter.com,2020:site.347461Thu, 13 Aug 2020 20:17:02 -0800theorygeometrymathmathsequationhorizoncloudphotographyastrophotographyresolvedtrigonometryBy: true
http://ask.metafilter.com/347461/This-should-be-basic-geometry#4973317
If you're comfortable ignoring the curvature of the earth over this distance, I think it's just going to be <a href="https://www.mathsisfun.com/algebra/sohcahtoa.html">sohcahtoa</a>. You know the angle (10 degrees) and the opposite side (60,000 feet). You want the adjacent side. tan(10d) = OP/A, so A = OP/tan(10d), so it's 60,000 / .1736, which is about 364000 feet or about 65 miles.<br>
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So about 65 miles if the earth were flat. Curvature of the earth and height of the observer will both lower that number, so you can probably treat it as an upper bound, if my back of the envelope math is right.comment:ask.metafilter.com,2020:site.347461-4973317Thu, 13 Aug 2020 20:58:03 -0800trueBy: teraflop
http://ask.metafilter.com/347461/This-should-be-basic-geometry#4973322
Taking the curvature of the earth into account is a bit more work, but not that much.<br>
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You can draw a triangle using three points: your location (A), the top of the clouds (B), and the center of the earth (C). You know two sides of the triangle: AC is the earth's radius (6,378 km) plus 1.5 meters, and BC is the earth's radius plus 18,250 meters. And you know one angle: ∠BAC is 90 degrees (from the center of the earth to the horizon) plus 10 degrees (from the horizon to the cloud tops).<br>
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First we find the angle ∠ABC using the rule of sines: sin ∠ABC = sin ∠BAC × AC ÷ BC = 0.98200, so ∠ABC = 79.11°.<br>
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Then, since the angles in the triangle add up to 180°, we can find ∠ACB = 180° - ∠BAC - ∠ABC = 0.88°, or 0.0155 radians.<br>
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Finally, an arc of 0.0155 radians on the earth's surface corresponds to a circular distance of 0.0155 × 6378km = 99 km, or 61 miles.comment:ask.metafilter.com,2020:site.347461-4973322Thu, 13 Aug 2020 21:49:26 -0800teraflopBy: SemiSalt
http://ask.metafilter.com/347461/This-should-be-basic-geometry#4973379
Given the total uncertainty in the numbers you have, I think you can use the fact that for small angles, sin(a) approximately = a. To put this another way, <br>
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height of clouds / distance of clouds = angle.<br>
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or <br>
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distance of clouds = height of clouds / angle.<br>
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The complicating factor is units. The formula is for the angle in radians. You think of the cloud height is feet or meters, the distance in miles or kilometers. You need some conversion factors. Using distances in miles, you would get:<br>
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angle in degrees = 57.3 * (height of clouds in feet / 5280) / distance in miles<br>
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distance of clouds in miles = (height of clouds in feet / 5280) / (angle in degrees / 57.3)<br>
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For the example:<br>
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distance of clouds in miles = (60000/5280) / (10/57.3)<br>
= 11.36 / .1745 = 65.1<br>
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So the clouds would have to be 65 miles or farther away for the angle to be under 10 degrees.comment:ask.metafilter.com,2020:site.347461-4973379Fri, 14 Aug 2020 06:51:31 -0800SemiSalt