With Venus looking so "in your face" lately - and at a neck friendly angle - I decided to do an azimuth shot on it.
My baseline is an "x" in my drive to a floodlight illuminating the sign at the local church (300m)
It has an RTK derived geodetic azimuth of 214*34'05" (I know..static...I didn't feel like explaining it all to the vicar)
Here's what I got from W&G's "Adjust" software:
Star
Venus March 23, 2015
Observer's Astronomic Position:
Latitude = 48°23'40.0"
Longitude = 89°16'25.0"
StopWatch Start Time, UTC: 0:00:00.0
DUT correction: 0.0sec
GHA of Star at 0h UT : 145°55'30.00"
GHA of Star at 24h UT : 145°45'30.00"
Declination of Star at 0h UT : 14°07'42.00"
Declination of Star at 24h UT : 14°34'24.00"
Pointing 1
Time 0:25:52.0
Hor. Angle 44°34'07"
Declination 14°08'10.78"
L H A 63°06'54.2"
Azimuth to star 259°09'13"
Az of Line 214°35'06"
Average Astronomic Azimuth of Line = 214°35'05.5"
Deviation from the mean = 0.01"
For me - 1 minute- I'm impressed!
Of course I also wanted to crunch it by hand using W&G's "Elementary Surveying" as my guide
which gives me:
Z = 0.8919159946/(0.167232585-0.338126098) = -79*09'12"
and into this table,
where:
Venus is west of north, hence t = LHA.
Z is a negative number therefore < 0
LHA is between 0 and 180,
therefore my azimuth should be 360 + Z according to the table, but that gives:
280*50'47"
The only way I can get 259*09'12" is by 180 - Z which isn't even an option
So what am I missing?
I did Polaris last year ( a real pain in the neck in Thunder Bay) and Arcturus, and ran into the same issue.
Would appreciate a star shot guru taking a look - thanks!
> Here's what I got from W&G's "Adjust" software:
> Star
>
> Venus March 23, 2015
>
> Observer's Astronomic Position:
> Latitude = 48°23'40.0"
> Longitude = 89°16'25.0"
> 7
> StopWatch Start Time, UTC: 0:00:00.0
> DUT correction: 0.0sec
>
> GHA of Star at 0h UT : 145°55'30.00"
> GHA of Star at 24h UT : 145°45'30.00"
>
> Declination of Star at 0h UT : 14°07'42.00"
> Declination of Star at 24h UT : 14°34'24.00"
I don't have W&G's software, and haven't run the numbers through MICA yet, but are you sure that WEST Longitude is supposed to be entered as a positive number in that software?
Just a thought. I might have to try it myself, if it ever warms up around here.:-(
I don't have an answer for you on that (Chuck Ghilani might).
I use MICA to do astro comps, and I got 259°09'15.1" for the topocentric azimuth. The difference of 2" is possibly due to the fact that your computation is at the center of the earth and mine is at the surface. This does not matter for star shots, but definitely matters for the moon. I suspect it does matter at a smaller level for venus or mars.
I got the formulas for topocentric reduction out of the printed Astronomical Almanac. But, the online edition does not have that section. I do address it in my paper about azimuths.
Venus! Star shot help (stop watch vs. time?)
Is the observation not dependent on accurate time like a star shot is?
Venus! Star shot help (stop watch vs. time?)
He recorded the accurate time of observation.
Stop watch started at 00:00:00 (midnight UTC). His stop watch time is added to that to get the UTC time of Obs. He also noted that DUT1 is 0.0s
Venus! Star shot help (stop watch vs. time?)
Ah! thanks for the clarification.
The problem may be in the calculation of the Local Hour Angle LHA.
Venus! Star shot help (stop watch vs. time?)
Yes in "Adjust" you can either put in the actual time of observation or the time you started your stopwatch and the elapsed time
are you sure that WEST Longitude is supposed to be entered as a positive number in that software?
Good point - I'll look into that
> I don't have an answer for you on that (Chuck Ghilani might).
I'd love to talk to those fella's
The difference of 2"
Wow I can live with 2" !
Colin, where did you get the correction DUT= 0.0 seconds?
Looks to me that its -0.56107 seconds for the date you give.
JOHN NOLTON
Tombstone, AZ
USA
> Venus is west of north, hence t = LHA.
If LHA > 180deg, t= 360deg - LHA, t is east and positive (+)
If LHA < 180deg, t= LHA , t is west and negative (-)
For your example, t= -LHA.
Hi John, as I am just beginning to dabble in this I haven't really wrapped my head around that issue yet.
You could explain it to me if you want:-)
My main issue was why when I hand calc'd the numbers did I get a different answer to what the software did and I think I figured it out - the meridian angle t, in my case should've been negative.
> > Venus is west of north, hence t = LHA.
>
> If LHA > 180deg, t= 360deg - LHA, t is east and positive (+)
>
> If LHA < 180deg, t= LHA , t is west and negative (-)
>
> For your example, t= -LHA.
Yes! Thankyou - I found an old textbook that explained it better.
Unless I missed it, the Wolf and Ghilani text doesn't really clarify that t might be negative, although to be fair it does keep telling you to draw a diagram!
I recommend "A Manual on Astronomic and Grid Azimuth" by R.B. Buckner. It's available from Larry P, a frequent poster on the forum.
