Go Back   IceInSpace > Beginners Start Here > Beginners Talk

Reply
 
Thread Tools Rate Thread
  #1  
Old 07-07-2021, 06:12 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Planets location in the night sky

Hi, newbie to telescopes, just bought my first (off IIS) and have a couple of questions. Here is the first.



The first is how to know when and where a planet is visible. Venus (I'm pretty sure it's Venus) is visible here in Surfers Paradise at sunset for a few hours.



I wrote a little program to produce the planetary data (altitude, azimuth, declination & right ascension). According, with azimuth being 227 deg. at 6pm, I would expect Venus to be 133 deg from north, therefore in the south-western sky. However it is basically due west, so about 90 deg off magnetic north.


What am I doing wrong? (Please confine yourselves just to this question, please )

Last edited by daz; 08-07-2021 at 05:25 PM.
Reply With Quote
  #2  
Old 07-07-2021, 06:40 PM
Startrek (Martin)
Registered User

Startrek is offline
 
Join Date: Dec 2017
Location: Sydney and South Coast NSW
Posts: 5,990
Got a laptop ?
Do yourself a big favour and download Stellarium, it’s a free planetarium which provides a map of the night sky 24/7 in real time and can be set up to show all celestial objects including bright stars , star clusters , planets , moon , galaxies , emission nebula , planetary nebula and so on ....
You can roll time forward via its clock and see what’s coming up each night , night after night , month after month
After using it for a while it teaches you how the objects traverse across the night sky plus identify most common objects in the night sky. Leaves the old cardboard planetarium wheel behind.
I’ve attached a doc I put together last year on how to set up the basics in Stellarium
Hope the above answers your enquiry
Cheers
Martin
Attached Files
File Type: pdf Stellarium Basic Set Up and Use.pdf (29.2 KB, 101 views)
Reply With Quote
  #3  
Old 07-07-2021, 07:07 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Hi, thanks for the response, Martin, am aware of Stellarium, however would prefer to learn the basics. I've also just realised I was using local time in my program instead of utc. Which makes it even more puzzling as Venus should be at 62 deg. azimuth.


As you can see by the diagram, Venus should not be due west at 62 deg.
Attached Thumbnails
Click for full-size image (Altitude & Azimuth.jpg)
15.9 KB67 views
Reply With Quote
  #4  
Old 07-07-2021, 07:31 PM
Startrek (Martin)
Registered User

Startrek is offline
 
Join Date: Dec 2017
Location: Sydney and South Coast NSW
Posts: 5,990
Quote:
Originally Posted by daz View Post
Hi, thanks for the response, Martin, am aware of Stellarium, however would prefer to learn the basics. I've also just realised I was using local time in my program instead of utc. Which makes it even more puzzling as Venus should be at 62 deg. azimuth.


As you can see by the diagram, Venus should not be due west at 62 deg.
Stellarium teaches you the basics , gives you Alt and Az co ordinates every second of the night on every object
Also provides a simulated real time view of the night sky overhead on every celestial object
You can zoom in on objects see what their physical appearance and colour looks like and use ocular plug ins to advise what eye piece to use and or FOV for Astrophotography
And so much more ....
If your into the science of astronomy , Stellarium provides that as well
It will tell you exactly where Venus is suppose to on any night during planetary season
Cheers
Reply With Quote
  #5  
Old 07-07-2021, 07:40 PM
DeWynter's Avatar
DeWynter (ILYA)
Registered User

DeWynter is offline
 
Join Date: Aug 2018
Location: Sydney, NSW, Australia
Posts: 161
Hi Darren,

What programming language and libraries are you using? I'd suggest to use Python and the Skyfield library. After that calculating everything is a piece of cake and you can easily do timezone conversions. I'm using Skyfield all the time and it's fantastic.

Speaking about Venus - at 6:30pm in Sydney it should be at ~300 deg (West-North-West). Or alternatively -60 off North, but azimuth goes from North through East (90) to South (180) and then to West (270), so saying "due west at 62 deg" technically it's not correct.
Reply With Quote
  #6  
Old 07-07-2021, 08:08 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Hi, Ilya is it? Thanks for the reply. Using PyEphem, I've just found that in the southern hemisphere azimuth is calculated differently, plus I think True North is used instead of Magnetic North.

Off the net:
"The difference between true and magnetic north varies by location and over time. In most populated parts of the world, it currently ranges from 30 degrees west (south-eastern tip of Africa) to 26 degrees east (southern tip of New Zealand)."
Also not sure if PyEphem gives the azimuth angle or true azimuth. However,
360 - 62- 30 for the difference in north puts it about 270 deg., which is where I'm seeing it.
Still not sure why I haven't spotted Mars, as Venus & Mars are almost conjunct.
I'll take a look at Skyfield, I thought it wasn't being maintained, is there a Python 3 version?
Reply With Quote
  #7  
Old 07-07-2021, 08:09 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Martin, thanks for the info, I'll give it a whirl sometime. Cheers.
Reply With Quote
  #8  
Old 07-07-2021, 08:21 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Actually, that calculation should be:
180 + 62 + 30


"Azimuth angle is measured from 0o to 180o either westwards or eastwards from either north or south. If the observer is in the northern hemisphere, the azimuth is measured from north and if in the southern hemisphere, it is measured from south."
Reply With Quote
  #9  
Old 07-07-2021, 09:55 PM
gjr80 (Gary)
Registered User

gjr80 is offline
 
Join Date: Aug 2017
Location: Brisbane
Posts: 34
We use pyephem in some open source weather software I contribute to and find it very accurate. I am not aware of azimuth needing to be interpreted differently according to hemisphere, as far as I was aware it is always returned as degrees (or radians) East of North and is true not magnetic. I think you will find magnetic declination/variation is closer to 11.3 degrees for Surfers Paradise.

I just did a quick interactive pyephem compute of Venus viewed from Surfers Paradise at 1800 (local) today (7 July) and it gave the following:

Code:
Python 3.7.3 (default, Jan 22 2021, 20:04:44) 
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import ephem
>>> o=ephem.Observer()
>>> o.lat='-27.9985'
>>> o.lon='153.4227'
>>> o.elevation = 0
>>> o.date = '2021/7/7 08:00:00'
>>> v=ephem.Venus()
>>> v.compute(o)
>>> print("azimuth=%s altitude=%s" % (v.az, v.alt))
 azimuth=299:29:57.5 altitude=12:47:00.0
Which is pretty close to SkySafari which shows 299 deg 29' 55.1" and 12 deg 47' 19.6". Seems pretty close to me!


Gary
Reply With Quote
  #10  
Old 07-07-2021, 10:22 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Thanks, Gary, I'll have another look.
Reply With Quote
  #11  
Old 08-07-2021, 05:21 AM
DeWynter's Avatar
DeWynter (ILYA)
Registered User

DeWynter is offline
 
Join Date: Aug 2018
Location: Sydney, NSW, Australia
Posts: 161
Ahmm... With Skyfield azimuth is always azimuth. I just recently wrote an app which does alt/az calcs and tested with different locations from Southern and Northern hemisphere and it gives the correct numbers (verified with SkySafari/Stellarium). Yet we have real azimuth for planets between ~0-90 and ~270-360 (due North), while up North they've got ~90-270 (due South). So it's not from 0 till 180. It's the whole 360. Statement "Azimuth angle is measured from 0o to 180o either westwards or eastwards from either north or south." is not entirely correct. And North is true North. You still need to apply a correction for magnetic north later as it's different in different parts of the world if you want a precise location.

The Skyfield is maintained and developed. The developer just fixed some issues with the code caused by new JPL ephemerides I've found recently.

I'm using it with the latest Python 3.9 on my macOS and Windows 10. No issues at all.

Regards,
Ilya
Reply With Quote
  #12  
Old 08-07-2021, 09:48 AM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Gary, have confirmed via https://www.magnetic-declination.com/# that 11.3 deg is correct for Gold Coast.


The problem seems to be with my code, which produces different results depending on whether the values for lat & lon are enclosed in quotes or not, and the manner in which the variables are declared:


ob.lon, ob.lat, ob.elevation = -28.00274, 153.42999, 0


produces 286 deg


ob.lon, ob.lat, ob.elevation = '-27.9985', '153.4227', 0


produces 62 deg


ob.lon, ob.lat, ob.elevation = -27.9985, 153.4227, 0


produces 286 deg


ob.lat='-27.9985'
ob.lon='153.4227'


produces 299 deg


So, 299 - 11.3 gives me an angle to magnetic north of 287.7 deg or 72.3 deg., which seems about right based on observation. I'll pull out a compass to confirm if the skies clear.


Thanks for your help everyone!
Reply With Quote
  #13  
Old 08-07-2021, 11:07 AM
gjr80 (Gary)
Registered User

gjr80 is offline
 
Join Date: Aug 2017
Location: Brisbane
Posts: 34
The problem is the lat/long data you are feeding to your Observer object. Pyephem works internally using radians not degrees. When you pass lat/long as strings pyephem will attempt to parse the strings as degrees (hence the pyephem examples use lat long as decimal degrees in quotes) and store the results as radians. From memory there are a number of different formats that are supported. When you 'print' an Observer object (or otherwise display the lat and long properties of an Observer object) you are by default presented with lat/long in degrees. If you provide lat/long as numerics pyephem interprets the values as radians, so you need to convert the degrees to radians yourself first. So considering your examples, the first would need to be something like:

Code:
import math
 ob.lon, ob.lat, ob.elevation = math.radians(-28.00274), math.radians(153.42999), 0
the second is fine as is:

Code:
ob.lon, ob.lat, ob.elevation = '-27.9985', '153.4227', 0
The third and fourth would need to use math.radians() just like the first.

Try it and 'print' the observer details of each using print(ob) and you you should see the same lat/long for each:

Code:
<ephem.Observer date='2021/7/8 01:40:27' epoch='2000/1/1 12:00:00' lon='-27:59:54.6' lat='153:25:21.7' elevation=0.0m horizon=0:00:00.0 temp=15.0C pressure=1010.0mBar>
Gary
Reply With Quote
  #14  
Old 08-07-2021, 11:39 AM
Nikolas's Avatar
Nikolas (Nik)
Dazed and confused

Nikolas is offline
 
Join Date: May 2012
Location: Melbourne
Posts: 3,254
Quote:
Originally Posted by daz View Post


What am I doing wrong? (Please confine yourselves just to this question, please )
Use google and stellarium on your laptop
Reply With Quote
  #15  
Old 08-07-2021, 01:07 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Ha, ha, typo, thanks Gary. Looks like the skies are clear, so should get a sighter tonight.


Found my compass and lined it up to where Venus appeared over the top of of the neighbours, seems to be an angle of about 90 deg to magnetic north using 11.3 deg magnetic declination. Will check it later.
Reply With Quote
  #16  
Old 08-07-2021, 01:24 PM
daz (Darren)
Registered User

daz is offline
 
Join Date: Jun 2021
Location: Surfers Paradise, Australia
Posts: 37
Correction.



Found my compass and lined it up to where Venus appeared over the top of of the neighbours, seems to be an angle of about 90 deg to magnetic north.



Will check it later, hopefully, although the cloud cover has returned, so looking unlikely.


So, as it stands, 299 - 11 = 288 = 72 deg, I seem to be missing about 18 deg using a magnetic declination of 11.3 deg. (if indeed the 11.3 should be subtracted).


Ha ha.


This all started with me wanting to find a few stars to enter into the goto mount, so I thought I'd write a simple program to tell me what was where.


Then I saw Venus, but it wasn't where I thought it would be. Still, I'm closer now than I was.

Last edited by daz; 08-07-2021 at 05:23 PM.
Reply With Quote
Reply

Bookmarks

Thread Tools
Rate This Thread
Rate This Thread:

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT +10. The time is now 08:21 PM.

Powered by vBulletin Version 3.8.7 | Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Advertisement
Bintel
Advertisement
Testar
Advertisement