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I'm a pilot, not a GIS expert. What I need is a formula or a web site that I can supply the variables to answer my question.

I need to know the angle down to the horizon from different flight altitudes. This is for a specific flight over the ocean so terrain is not a factor.

Knowing the angle to .1 degree will be sufficient accuracy. Knowing the angle for every 2 thousand feet from 25,000 feet to 41,000 feet will cover my needs.

Thank you,

Mike Wendt.

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There is a right triangle: the plane is at one vertex (A), the center of the earth is at another (O), and the most distant visible point on the horizon is the third (B), where the right angle occurs. alt text

That point on the horizon is about 6,378,140 meters = 20.9362 million feet from the center of the earth (the earth's radius)--that's one leg--and you are between 25,000 and 41,000 feet feet further from the center--that's the hypotenuse. A little trigonometry does the rest. Specifically, let R be the earth's radius (in feet) and h be your altitude. Then the angle from the horizontal down to the horizon (alpha) equals

Angle = ArcCos(R / R+h).

Note that this is purely a geometric solution; it is not the line of sight angle! (The earth's atmosphere refracts the light rays.)

For R = 20.9362 million feet and heights in 1000's of feet between 25000 and 41000 I obtain the following angles (in degrees) with this formula:

2.8, 2.85, 2.91, 2.96, 3.01, 3.07, 3.12, 3.17, 3.21, 3.26, 3.31, 3.36, 3.4, 3.45, 3.49, 3.54, 3.58

You could just linearly interpolate within this interval if you prefer, using a formula like

Angle = 1.5924 + 0.048892 (h / 1000)

for heights h in feet. The result will typically be good to 0.01 degree (except at the extremes of 25,000 and 41,000 feet, where it's off almost 0.02 degrees). E.g., with h = 33,293 feet, the angle should be around 1.5924 + 0.048892 * (33.293) = 3.22 degrees. (The correct value is 3.23 degrees.)

For all heights less than 300 miles, an acceptably accurate approximation (i.e., to 0.05 degree or better) is to compute

Angle = Sqrt(1 - (R / (R + h))^2).

This is in radians; convert it to degrees by multiplying by 180/pi = 57.296.

The ellipsoidal flattening of the earth won't make much difference. Because the flattening is only about 1/300, that ought to introduce only around 0.01 degree of error or so in these results.

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Part 1. Thank you whuber. I will explain more about what I need to accomplish. I’m working a charter flight that wants to see a ‘double sunrise’ in flight. The plan is to give a view of the sunrise on one side of the aircraft then drop altitude while making a 180 degree turn so that the passengers on the other side see a second sunrise. Since the apparent angular size of the sun is about .5 degree, I need to raise my horizon by descending something more than .5 degree, while making the 180 degree turn. – Mike in Guam. Dec 29 '10 at 0:05
Part 2. I need to descent more than .5 degree to adjust for the continued rising of the sun due to the earth’s rotation. The earth rotates 1 degree in 4 minutes. The 180 degree turn will take a little less than 2 minutes. So, I really need to descend at least 1 full degree. With the numbers you provide, descending from 41,000 feet to 25,000 feet only gives me .62 degrees. An additional problem is that much descent needs about 3 minutes, an additional .75 degrees of earth rotation. – Mike in Guam. Dec 29 '10 at 0:06
Part 3. My 737-800 has a ceiling of 41,000 feet and in this area, I can descend to 3,000 feet unrestricted. Is that enough? I can plan about 5,000 feet per minute descent. I’ve heard of double sunrise flights being successful. But your math is saying it may not be possible. Thanks, Mike. – Mike in Guam. Dec 29 '10 at 0:06
The earth radius is approx 20.9 million ft! Not 32.8 million. – user6684 Apr 2 '12 at 11:34
Good catch, seb! I have no idea how 32.8 million crept in, because it is so obviously wrong. I have recomputed everything in this reply and edited it to reflect the correct value. Unfortunately for @Mike (but fortunately for me), it does not change his situation: his 0.62 degrees has increased to 0.78 degrees but it's still not enough for success. – whuber Apr 2 '12 at 16:00

This is really more of a comment to @whuber's answer. (We can't put images in comments.)

Atmospheric refraction seems to be a significant factor.

enter image description here


I wonder if the equations in this NASA publication, "Method for the Calculation of Spacecraft Umbra and Penumbra Shadow Terminator Points", could be adapted for this.

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No, the shadow cone calculations are based on the size of the light source (ie the Sun), the size of the shadowing body (the Earth) and the distance between them. This is shown on pages 3 and 4 of the document you linked, showing how Umbral and Penumbral Cone geometries are defined and calculated. – Corey Apr 22 at 6:21

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