# A method to predict visibility of the international space station

1. Nov 13, 2009

### Esran

Does anyone know of a good method for calculating the visibility of the international space station from an arbitrary point on earth's surface? I've always been curious how NASA makes all those predictions about fly-overs and such. Any links to web sites explaining the procedure in as much detail as possible would be greatly appreciated.

And no, this isn't homework. It's just a project I'm working on by myself to strengthen my skills and satisfy my curiosity.

The orbital path of the ISS does not appear to be exactly circular, but instead slightly elliptical (since its elevation supposedly varies between 278 km and 460 km). The eccentricity is so slight though, I think I'm safe in supposing that the orbit is circular. Assuming an altitude of 369 km, I also think I know how to calculate the speed of the ISS, and given the speed of the ISS with a presumed ideal circular orbit, I am pretty sure I can figure out its altitude.

Of course, I am neglecting atmospheric drag here, and other such details.

My question is this: is the ISS in a constant orbit? By that, I mean is its orbit wobbling all over the place or does it trace a fairly immobile ring around the earth? If the orbit is constant, I can envision calculating fly-overs for certain points on the earth’s surface by comparing the ISS orbital period to the earth’s rotation. It’s basically a really complex horse-race problem: if horse A passes point 1 every t seconds, and horse B passes point 1 every r seconds, find the period of their instantaneous synchronization at point 1.

Thus, I plan to proceed by calculating the great circle of the ISS orbit, rotating the earth until Austin (where I live) is immediately below it, then drawing a vertical line upward from the surface. The point where this line intersects the ISS orbit is what I’m interested in. I can obtain the frequency which Austin passes immediately below this point (using the earth’s rotation), compare it to the frequency which the ISS passes through this point, and calculate an equation to predict alignments. Although the concept seems easy to me, I can see that in practice this will probably take awhile. But given the angle of the ISS orbit to the north pole, it shouldn’t be too terribly bad. Unfortunately, I do not have this angle.

The part I see major difficulties with involves the calculation of visibility of the ISS from the earth’s surface. In my head, I can imagine drawing a line up to the ISS from a point on the earth’s surface. I see the line moving to follow the space station, almost like a radar dial, until the line bumps into the curved surface of the earth and the satellite is no longer visible. However, I do not have the slightest idea how to model this mathematically. I would assume the problem is similar to predicting when and where ships come over the horizon, but again I have no knowledge of how to approach that issue either.

2. Nov 13, 2009

### mgb_phys

I think the main problem is predicting when the ISS will be illuminated by the sun at the right angle to make it bright enough to see, rather than predicting when it will be overhead

The NASA Jpass and JTrack java applets have the orbital elements (which do change0 and info about how they are calulcated http://science.nasa.gov/realtime/Jpass/25/JPass.asp [Broken]

Last edited by a moderator: May 4, 2017
3. Nov 13, 2009

### Esran

Well, first I just want to predict the times that it would be visible from Austin, assuming that whenever it is in a position to be seen, that it can indeed be seen. I'll get to the other things later.

4. Nov 13, 2009

### Bob S

5. Nov 13, 2009

### Esran

That is good, but it is on too basic of a level. I want to come up with formulas to do the calculations myself, not just obtain data from computer programs.

6. Nov 13, 2009

### hamster143

If you assume that it's elevation is constant, it is a straightforward mapping from the distance between the observer and the space station, to the angle above or below the horizon. Draw a cross-section of the Earth that passes through center of the Earth, the observer, and the space station. The rest is trigonometry.

The space station is above the horizon if

cos(d/R) > R/(R+H)

where R is radius of the Earth, d is the distance between the observer and the point directly underneath the space station, H is the altitude of the space station.

Realistically, it has to be closer than that, because, unless you're in Kansas or on a boat in the ocean, you don't really _see_ the horizon. That can be calculated too.

I have to agree with mgb_phys, it's not enough to know if it's above the horizon, there's the issue of determining if it's visible. The station must be sunlit (or else you won't see it), but the sun must still be below the horizon for the observer (or else the sky will be too bright). Also, if the station is close to the line from the observer to the sun, it will be appearing in bright sky, and it will be lit from behind. NASA SkyWatch applet requires a minimum solar separation of 45 degrees.

7. Nov 13, 2009

### Count Iblis

8. Nov 13, 2009

### Count Iblis

9. Nov 13, 2009

### mgb_phys

http://www.amsat.org is a good source of maths, algorithms and orbit data

Your best chance of seeing the ISS is probably dawn/dusk, you need it to be above the horizon (duh) but it sun light, since it's orbit isn't very high (350km) compared to the radius of the earth a bit of simple trig shows that there isn't very much of an orbit where it's visible.

10. Nov 13, 2009

### Esran

hamster143, I know how to do the part you suggested. However, that is not my main concern.

Given a point on the earth's surface, say Austin, and where the ISS is at one particular time in its orbit, I would like to predicted at what times and for how long the ISS will be visible from Austin, assuming the ISS orbit remains constant.

I have fiddled around with drawing a tangent plane to our point on the sphere of earth, and using differential equations to describe how it changes as the earth spins, looking where it intersects the great circle of the ISS orbit, etc...

It's really, really complicated. And I'm lost.

11. Nov 13, 2009