[AusNOG] NBN Satellite Latency

Andrew McN andrew at mcnaughty.com
Sun Feb 5 23:10:34 EST 2017

On 03/02/17 17:38, Greg McLennan wrote:
> My typical pings for a geostationary global link I manage(melb to asia)
> from rf modem to rf modem(comtech) ~530ms. But expect to see slightly
> longer ping timed if say the bird is low on the horizon(e.g you at
> around 18 degrees elevation to see the satellite) for the ground station
> antenna compared to being located on the equator and pointing straight
> up.. You can do the physics math!!
> Cheers Greg

On 03/02/17 17:44, Ross Wheeler wrote:
> Yeah, but as Geoff hinted - it's not even that simple.
> Geostationary orbit is basically 26,200 miles from the earths centre
> (22,230 odd from sea level) - so 22,230/186282 = 119.3ms
> BUT - that's if the bird were directly overhead...
> If the sat is at say, 45 degrees elevation (and at the same latitude,
> which it probably isn't), then that path just became 31,438 miles =
> 169ms each way. The closer to the horizon, the longer the path and the
> higher the travel time...  and the further from the equator you are, the
> greater the distance too.

This is a substantial over-estimation of the difference in distance when
away from the equator.  Flat earth thinking even, which would be close
enough if the geostationary orbit height were small relative to the
radius of the earth, but it's not.

By my calculation, the distance to the satellite at the 45th parallel
should only be about 6% worse than at the equator.  Even at the pole
it's only 19% worse.

Define a coordinate system, centred on the earth, with the satellite on
the x axis, and the poles aligned with the y axis, and the scale is
kilometers.  r = radius of the earth (6371 km).  g = geostationary orbit
height from sea level (35786)km.  I'll ignore the fact that the earth is
not quite spherical, which amounts to less than a 1% error.

For the 45th parallel:
ground station (a) is at [ 1/sqrt(2)*r, 1/sqrt(2)*r ] = [ 4505, 4505 ].
satellite (b) is at  [ g + r, 0] = [ 42157, 0 ]
path length between the two is sqrt( |a-b| ) = 37921 (km).

If people are finding that where the satellite is in the sky matters
much to their ping time, then it seems that they should look to
explanations other than the travel time at the speed of light.

Andrew McNaughton

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