# Atmospheric Refraction vs Height

1. Apr 28, 2007

### natski

I just wanted to ask a quick question that maybe someone knows off the top fo their head. How does the refraction angle of the Sun at the sunset vary as a function of the observer's height. i.e. if you are standing on a tower looking over the ocean, is the refraction deviation of the Sun at sunset significantly different to that when you are on the beach?

Thanks

2. Apr 29, 2007

### Astronuc

Staff Emeritus
I don't think that 100 ft (33 m) or 500 ft (152 m) would make much of a difference.

Perhaps the more appropriate question is "How does the refraction angle of the Sun at the sunset vary as a function of the observer's altitude?" The higher one goes, the lower the density of the atmosphere.

Perhaps one can consider two concentric circles, and consider the chord which is tangent to the circumference of the inner circle, which passes through the outer circle to a point at a long distance. Assume the space between the circles is filled with a gas and the density is a function of 'r' from the center, or perhaps more appropriately, R1+h, where R1 is the radius of the inner circle and h is the height from the circumference of the inner circle measured toward the outer circle.

3. Apr 29, 2007

### pervect

Staff Emeritus
4. Apr 29, 2007

### Staff: Mentor

Well, the refraction caused by the elevation of the object sighted is significant enough to factor into celestial navigation calculations, but the height of the eye's effect on that angle is not, at least not for the height of any ship.

5. Apr 30, 2007

### natski

For those interested, I have managed to solve this problem by calculating a table of data points from a simulation program I wrote.

I have made the assumption that we are always looking at a sea level horizon where the Sun is setting. I used a program I have been writing for some time on atmospheric refraction to simulate an observer at different heights. I used the US Standard Atmosphere (1976) and Edlen's semi-empirical refracivity equation. I have attached the document as an Excel file so enjoy!

Interestingly, I have found it impossible to get an accurate fit of the data points. For the 700nm wavelnegth refraction angle, I tried using an exponential fit in Mathematica but couldn't really get close. A polynomial fit is a waste of time too. It seems to have the form:

theta=32*{2-Exp[-f(h)]}

If anyone can get a good fit, please do let me know as it would really help me out! I have also attached some of attempted fits.

File size:
20.5 KB
Views:
124
File size:
3.4 KB
Views:
77