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"The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc." - Wikipedia

The meridian is when the sun is directly facing earth and is at exactly 90° overhead at some point along the longitudinal line at a given location.

My question is, if the sun is 93 million miles away and travels overhead at 90° between the Tropic of Capricorn and the Tropic of Cancer, how is it that there are Meridian solar elevation angles reported from relatively nearby cities on Earth of less than 89.99°?

Let's take for instance Calgary, Canada as of today May, 04 2016 -- estimated ground distance from Calgary to where the Sun is at 90° on Calgary's longitudinal line is ~3000 miles. The solar elevation angle (aka "altitude") is reported by the timeanddate website to be 55°! - TimeandDate

Reversing the angle, using right triangle math:

The base angle, our solar elevation angle, of a right angle triangle is:
    "solar elevation angle"= arctangent (h/a)
    89.998151749049 = arctangent (93000000/3000)

So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

That makes sense, since if something is 93 million miles away from us and you think of it as a triangle, then the distance between any place on earth and the location on earth where the sun is at exactly 90° overhead -- you should expect the other angle of the triangle to be at near 90°, connected by a relatively tiny sliver of distance in a much elongated triangle to the center point of the distant sun.

If we go the other direction and leave out the presumption that the sun is 93 million miles away. We take h as unknown, we know the reported solar elevation angle of 55°, we can again use right triangle math:

    Let a=3000, solar elevation angle=55°
    h = a * tangent ("solar elevation angle")
    4,284.4440202263 = 3000 * tangent(55°)

So according to my calculations, based on the advertised solar elevation angle of 55° at Calgary, Canada on May, 04 2016 at Meridian, at a ground distance of 3000 miles from where the Sun is at exactly 90° overhead --  the Sun is actually only 4,284.4440202263 miles high!

** Update **: The math would work out for a sphere Earth with a radius of 4000 miles, using the exact solar position and adding corrective angles due to a presumption of a spherical curvature:

    Calgary, Canada: 51°03′N 114°04′W
    Location of Sun at Calgary Meridian: 16°23′N 114°04′W
    Distance = 2395.386 miles
    Radius of Earth=4000 miles
    Arc distance = 4790.772 miles
    Width of arc = 4509.52059 miles
    Angle to center of the Earth=68.622754110849°
    68.622754110849°/2=34.31137705542450°

This math utilized calculations that may have had a bit less acurracy but to make it work you need to add a bandaid padding of approximately 34.31137705542450° to Calgary,Canada's advertised angles. If you do you arrive at 55°+34.31137705542450° = 89.31137705542450° which is close to what I would expect to see. If, we are truly living on a Flat Earth, then the calculations would be telling us that the Sun height is only 4,284.4440202263 miles high and we can use simple trignometry.
Arc Calculator
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If the math is difficult for you, you can check with an online triangle calculator such as: Ke!san

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  • $\begingroup$ This looks like an artefact of floating point numbers , but I have to check the calculations myself first. $\endgroup$ – SE - stop firing the good guys May 4 '16 at 15:28
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    $\begingroup$ Have you taken into account the curvature of the earth and how that affects your angles and triangle shape? $\endgroup$ – Tanenthor May 4 '16 at 15:32
  • $\begingroup$ Including mild curvature such as that of the Earth curvature formula should be negligible. It would change the base distance by a bit -- if you slide the base from 3000 to 5000 or 1000 you will still land on 89.99...0° for the expected solar elevation angle when presuming a height of 93 million miles. $\endgroup$ – A.Danischewski May 4 '16 at 15:46
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[[image31]]

Your confusion may stem from the way we sometimes draw the Sun (not to scale) as though it's close to the Earth.

In reality, as you note, the lines of vision to the Sun are nearly parallel:

[[image32]]

Note that the angles (measured from the red line, the horizon, to the arrow) are still different, but for entirely different reasons.

I've actually stolen @Andy's answer (and @Tanenthor's comment), but I'm hoping the diagram makes it easier to understand.

Re the parallel lines, the 11810x60 image below may help explain: even though the lines eventually converge (on an object about 10 times the distance to the moon, still nowhere near the distance to the Sun), they are nearly parallel when leaving Earth.

[[image33]]

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  • $\begingroup$ Your first picture is correct, not the second -- so "this is wrong" is correct in in describing it's own text --- the first picture depicts the actual solar elevation angles. To calculate a solar elevation angle (aka "altitude") you trace a ray from a given location up through the center of the Sun disc and the angle made with the horizon line is the solar elevation angle. The solar elevation angles at different locations on Earth should converge on the center of the Sun disc as you depicted in the first picture. $\endgroup$ – A.Danischewski May 5 '16 at 0:49
  • $\begingroup$ You are correct, but the sun is very very far away, so the lines would be nearly parallel. The first image shows the sun as though it were only a few thousand miles away, so the angles are seriously exaggerated. $\endgroup$ – barrycarter May 5 '16 at 0:53
  • $\begingroup$ Yes, that's true -- they should be nearly perpendicular (89.99...0°). So in the second picture, the rays should be converging slightly toward the center of the far distant sun disc as you have shown happening in the first picture. $\endgroup$ – A.Danischewski May 5 '16 at 1:13
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    $\begingroup$ Correct, but the slightness is so small, I couldn't accurately depict it in the second picture. For measuring the angle to the horizon, parallel lines are a better approximation. $\endgroup$ – barrycarter May 5 '16 at 1:21
  • $\begingroup$ Thank you @barrycarter, this is exactly what I was trying to get at, but it was late and I couldn't make a graphic. $\endgroup$ – Tanenthor May 5 '16 at 5:22
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So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

This is the wrong way to do it. A point 3000 miles south will be pointing in a very different direction, because the earth is round and only about 8000 miles in diameter.

For example:

Earth is around 25,000 miles in circumference. A point 3000 miles away is 0.12 times the circumference away. (3000/25000.) The circumference is equivalent to 360 degrees (assuming the Earth is a perfect circle, which is almost true for its polar section.) So 0.12 times 360 degrees is a difference of forty-something degrees. This probably accounts for your difference (very roughly).

Additional comment about roughness of these calculations: in fact the Earth is not a perfect sphere. Drawing a great circle along the meridian, the Earth will not appear perfectly round because the equator bulges slightly (and the poles are flattened). It's only something like 30-40 miles difference, but bear in mind this has a slight effect when calculating angles by dividing the circumference, as I did above. (Still better than assuming a straight line though, which is what I think the question is doing.)

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  • $\begingroup$ If you would descend around a curve the angle will get closer to 90° not further away. Another way to think about it is to view the approximation as straight, without any curve, then you have a simple right triangle. Now to include minor corrections for a curve you can simply draw the height down as far as the curve brings you -- because it's a straight line. Then we are back to simple right triangle math. $\endgroup$ – A.Danischewski May 4 '16 at 16:21
  • $\begingroup$ So let's use our example if your curve brings you down 1100 miles (and the Earth curvature formula is near that) -- with a solar elevation of 55° and base distance of 3000 miles. At 93,000,000 miles the expected solar elevation angle is 89.99815174904°, now if we include our correction for curvature the drop of 1100 miles we get a height of 93,001,100 miles and now an expected solar elevation angle of 89.998151770909°. As you can see it is now closer to 90°. $\endgroup$ – A.Danischewski May 4 '16 at 16:32
  • $\begingroup$ To clarify - somewhere in the Tropics, where the Sun is directly overhead, the horizon will be at 90 degrees to the sun, by definition. At (say) Calgary, Canada a person standing there will have a different horizon. At what angle will that horizons be? (Draw a circle and think about it.) $\endgroup$ – Andy May 4 '16 at 16:35
  • $\begingroup$ Also read this page which covers a guy who worked all this out a long time before we did... Eratosthenes and shadows in wells $\endgroup$ – Andy May 4 '16 at 16:44
  • $\begingroup$ Nice try but it still doesn't work out. Take my example yet again, a solar elevation angle of 55° from Calgary, Canada at 3000 miles from where the Sun is directly overhead. Presuming that the local horizon is curved, if we go along the same longitudinal line as that of Calgary, Canada into the Southern hemisphere the same distance away from where the Sun is at 90° we should have the same exact solar elevation angle of 55°. $\endgroup$ – A.Danischewski May 4 '16 at 17:10

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