Satellite Beam Mapping

[Huevos]

Can you reveal how you've calculated that ? I'd like to play a bit with the formula.

Yes, it's just simple trigonometry. Pick a location for the beam centre. work out the distance from that location to your location (or to the location of the pixel on the map). Work out distance of beam centre to satellite. Work out distance of your location to satellite. That gives the lengths of the 3 sides of the triangle. From that you can work out the angles.

If you want a more in depth answer (or to see the maths) we can start another thread so we don't drag this one off track.

Yes, it's just simple trigonometry. Pick a location for the beam centre. work out the distance from that location to your location (or to the location of the pixel on the map). Work out distance of beam centre to satellite. Work out distance of your location to satellite. That gives the lengths of the 3 sides of the triangle. From that you can work out the angles.

If you want a more in depth answer (or to see the maths) we can start another thread so we don't drag this one off track.

That's not giving you the distance the beam moved over the earth when the satellite changes its attitude i think.
Anyways to be precise you'd need to use spherical trigonometry, which i have completely removed from my mind. When using an approximation with normal trigonometry one would get the minimal distance only. Ie using 0.15 degrees on the satellite end, using the distance from me to the satellite as the adjacent and hypotenuse of the triangle. that would give me the opposite (ie the resulting distance of the pointing error on earth) as d * tan 0.15 with d=
38932. That gives me about: 101km. Which is near enough the expected value to make it plausible.
But you are right, probably the wrong thread. Let's look at transponders again. ;)

That's not giving you the distance the beam moved over the earth when the satellite changes its attitude i think.
Anyways to be precise you'd need to use spherical trigonometry, which i have completely removed from my mind. When using an approximation with normal trigonometry one would get the minimal distance only. Ie using 0.15 degrees on the satellite end, using the distance from me to the satellite as the adjacent and hypotenuse of the triangle. that would give me the opposite (ie the resulting distance of the pointing error on earth) as d * tan 0.15 with d=
38932. That gives me about: 101km. Which is near enough the expected value to make it plausible.
But you are right, probably the wrong thread. Let's look at transponders again. ;)

Like I said you can't give this a numeric value. For locations where the satellite is near the horizon the distance is huge. And I don't see why you are trying to complicate things talking about spheres. Triangles have 3 straight sides. If you know the length of each side you've got the angles too. No spheres involved.

Give me a location and I'll draw the 0.15 outline.

Like I said you can't give this a numeric value. For locations where the satellite is near the horizon the distance is huge. And I don't see why you are trying to complicate things talking about spheres. Triangles have 3 straight sides. If you know the length of each side you've got the angles too. No spheres involved.

Give me a location and I'll draw the 0.15 outline.

Because the earth is a sphere and we measure distances on the surface of the earth. Spheres involved. And i can give it a numarical value. i know one angle (0.15 degrees attitude change) and one side of the triangle, ie the distance from me to the satellite. And i want to have the distance the beam moves on earth when the satellite rotates by 0.15 degrees around the z axis, ie the axis through the solar paddels. I think you are talking about a completely different problem.

Because the earth is a sphere and we measure distances on the surface of the earth. Spheres involved. And i can give it a numarical value. i know one angle (0.15 degrees attitude change) and one side of the triangle, ie the distance from me to the satellite. And i want to have the distance the beam moves on earth when the satellite rotates by 0.15 degrees around the z axis, ie the axis through the solar paddels. I think you are talking about a completely different problem.

We are just measuring the distance between two points in space. That the media between those two points might be solid rock is irrelevant to the distance between the two.

As for how far the beam moves on the planet's surface is based on the location on the planet and the elevation of the satellite. And it is not circular and varies from one location to another. Look at the maps attached. Both are the satellite swinging the beam in a 0.15º perfect circle, one over Ankara, and the other over Dublin (with the satellite geostationary at 28.2E). Note that even though the satellite is moved in a perfect circle what is seen on the ground is far from circular. And certainly not a fixed distance.

Because the earth is a sphere and we measure distances on the surface of the earth. Spheres involved. And i can give it a numarical value. i know one angle (0.15 degrees attitude change) and one side of the triangle, ie the distance from me to the satellite. And i want to have the distance the beam moves on earth when the satellite rotates by 0.15 degrees around the z axis, ie the axis through the solar paddels. I think you are talking about a completely different problem.

We are just measuring the distance between two points in space. That the media between those two points might be solid rock is irrelevant to the distance between the two.

 

[cazhh]

for now i'm only interested in my local area. It's here where i do the measurements and it's here where i know how the beam looks like. i only talk about a rotation of the satellite around its z axis.
And i'm for now only interested in how many km of movement _on_ the surface of the earth that tiny rotation causes. To give you that distance you need to take into account that the earth is a sphere. And probably even more. Everything else would just give you an approximation.
Having said that, the approximation seems to be good enough to suggest that the difference in signal i saw over the past year here in my location could be caused by the typical pointing error of a satellite.
It would still be interesting if SES uses antenna tracking to reduce the pointing error. But we will probably not find out by looking at the measurements.

 

[Huevos]

That motion does not draw a circle and hence there is no linear distance. The only place on the planet where that is possible is directly below the satellite. If you give me your location I'll superimpose a 0.15º footprint on the map for that location.