Why are Starlink "Racetrack" Flares [Mostly] Reported from Planes?

jarlrmai

Senior Member
If your graph was accurate, the flare ought to be reddish, from the sunlight passing through a lot of atmosphere.

Does the angle at the center of the Earth correspond to the actual positions of observer and identified satellites?

What are you basing your 90⁰ sighting angle on? (perfect observer line vs. radius)

I think it is possible that the normal vector of whatever surface is reflecting the sun on the satellite may not be aligned with the center of the Earth, i.e. it might be tilted slightly north or south. Can we determine this from the data?
We've seen from the videos of the pilots that the flares scintillate as we expect.
 

Mick West

Administrator
Staff member
If your graph was accurate, the flare ought to be reddish, from the sunlight passing through a lot of atmosphere.
I think if it it was reddish, then the sunlight would be being attenuated to the degree that it would not be visible to the naked eye. Consider that you can see the shape of the sun at sunset, which normally takes a very powerful ND filter. Starlinks in direct white sunlight are would probably not be visible with such a filter.

Red flares would happen though, just super dim. Possibly visible in NV.
 

Mendel

Senior Member.
Yes it does, however this is a 2d model that is taken from a slice of a 3d Globe. The actual angle is a function of both the lat and long of the observer and directly below (nadir) of the satellite. Haven't quite worked out the formula for that yet.
Article:
The haversine formula determines the great-circle distance between two points on a sphere given their longitudes and latitudes.

assuming you have lat/long for the satellite
 

flarkey

Senior Member.
Staff member
assuming you have lat/long for the satellite
Yes you can use the haversine furniture in specific scenarios when you have the lat long of the observer and the satellite. One you've calculatied the straight line surface distance you can compare this to the earth's circumference, which will produce a fraction. Taking this fraction of 360° then gives you the angle between the two radii from the centre of the earth.

In my post above I was referring more to modelling it in 3d (although admittedly I did state 'formula').
 
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