Astrofest Rocket "plume" IMAGE

[DavidTrap (David)]

Originally Posted by gary

A quick back-of-the-envelope calculation and it all adds up.
At UT 12:46 that would correspond to a LAST of 20:36 at Linville.
RA: 16:43 Dec -12 would give an Az of 274 degrees and an Alt of 36 degrees.
That corresponds to an HA of 3:42. So that would put it on the meridian at +96:36E
longitude at that time. So as you mention, indeed it puts it over the Indian Ocean
west of the WA coast and referring to the projected ground track, it looks highly
plausible. Not clear yet whether it is the resultant plume from the earlier burn
obeying the First Law or exhaust from separation.

Fascinated with this maths (and the images and their RA/Dec overlays too).

I can keep up with this until the fourth line where you mention HA? What is that. How do you then calculate the longitude from that?

High school maths is but a distant memory...

ta in advance,

DT

[gary]

Originally Posted by DavidTrap

I can keep up with this until the fourth line where you mention HA? What is that. How do you then calculate the longitude from that?

Hi David,

No problem.

HA is the Hour Angle, which is expressed as a time.
So for the object, its HA = LAST - RA
i.e HA = 20:26 - 16:43 = 03:42
(Apologies, in the prior post I made a typo for the LAST, as 20:36 rather than
the correct time of 20:26 - but the result was correct).

We know that there are 24 hours in 360 degrees.
Thus 1 hour = 15 degrees.

Thus 03:42 converted to degrees is 3.71 * 15 = 55.66
The Longitude at Linville is 152.274E.
Thus 152.274 - 55.66 = 96.61E
which is equivalent to 96:36E

Best Regards

Gary

[DavidTrap (David)]

Ok, I think I've got it,

You've identified the RA where the object was visible in the sky, and then calculated the longitude on the planet where that RA would have been on the meridian at that particular time?

I may have my terminology slightly out, but am I on the right track?

Thanks again

DT

[gary]

Originally Posted by DavidTrap

Ok, I think I've got it,

You've identified the RA where the object was visible in the sky, and then calculated the longitude on the planet where that RA would have been on the meridian at that particular time?

I may have my terminology slightly out, but am I on the right track?

Thanks again

DT

Hi David,

You have got it!

And your terminology is correct.

Now armed with an accurately set chronometer (a digital wrist watch will do)
and some star charts with RA marked on them, one could confidently set
sail and know what longitude one is at by noting the time a star crosses the
local meridian.

Best Regards

Gary

[DavidTrap (David)]

Excellent

ta
DT

[AstroTourist (Terry)]

Hi Gary,
I note your comment ... "back of the envelope"
But I wonder what the inaccuracy in the calculation is due to the object not being on the celestial sphere.
Granted the altitude is high, close to geo synchronous at 23,000 odd km. But there must be some parallax??
Rgds,
Terry

[gary]

Originally Posted by AstroTourist

Hi Gary,
I note your comment ... "back of the envelope"
But I wonder what the inaccuracy in the calculation is due to the object not being on the celestial sphere.
Granted the altitude is high, close to geo synchronous at 23,000 odd km. But there must be some parallax??
Rgds,
Terry

Hi Terry,

Indeed there is and it is not necessarily insignificant.

Normally to compute these things we would be transforming geocentric
to/from topocentric and be also taking into account the effects of Earth flattening
and so on.

Even for the Earth's biggest satellite, namely the Moon, as you are aware,
parallax plays a significant part in determining its apparent RA/Dec for a given
topocentric location.

For example, the geosynchronous communications satellites used here in Australia,
like the Optus C1, will have about a 15 minute difference in apparent RA between
a dish pointing here in Sydney and a dish pointing in Perth. These are all
parked up in the equatorial band.

Particularly for any object in a low Earth orbit, like the ISS, topocentric
transformations play a dominant factor in the apparent position.

So with regards the object observed at Duckadang, the one thing we certainly
don't know is its real altitude. All bets are certainly off if it
transpires it was a couple of pranksters with a helium balloon covered in Alfoil
taken from the kitchen and then suspended by string to float above the western
bunkhouse.

So at the risk of mixing metaphors, the back of the envelope calculation
will get us in the ball park if we assume it was the Atlas/AEHF and that
it was on its way to the high geostationary orbit. If we knew the elevation above
sea level of it at that moment in time from trajectory data, we could get a larger
sheet of paper and refine the calculation some more.

Once again, great image!

[DavidTrap (David)]

I was thinking about parallax error, but assumed that your estimate would have at least been in the ballpark for something heading for geosynchronous orbit.

DT