I'm a pilot, not a GIS expert. I need a formula or a website to 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 the terrain is not a factor.

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

4 Answers 4


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.

  • 1
    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. Dec 29, 2010 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. Dec 29, 2010 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. Dec 29, 2010 at 0:06
  • 1
    The earth radius is approx 20.9 million ft! Not 32.8 million.
    – user6684
    Apr 2, 2012 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, 2012 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.

  • 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, 2016 at 6:21

enter image description here

You should use this formula, which will include atmospheric refraction.

d - the final horizon dip value (in degrees) at the given altitude
h - the cruising altitude
R - the main radius of Earth (flattening neglected)
k - the standard refraction coefficient - 0.13
180/Pi - conversion between radians and degrees, as the basic formula, throws results in radians.

If you are flying at an altitude of 10km your dip of the horizon is around 3 degrees.


Actually, the horizon stays at the eye level of the viewer, Regardless of height, due to the fact that...curvature does not exist.

The only way the horizon can be at eye level of the viewer, regardless of altitude, is on a planular surface...or flat.

Earth is flat, Antarctica is a wall of ice containing the water of Earth and the firmament attaches there.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.