Monday, June 23, 2025

Orientating an image - a better procedure

Previously, I described a way of orientating an image so that the edges of the frame are parallel to the North-South and East-West directions. In a lot of situations this perhaps does not need to be that accurate but in the case of binary stars where you are trying to determine the position angle (PA) of one star relative to another, then it does become a bit more of an issue. For example, I had been taking some measurements of the separation and PA of the binary star Xi Bootis, and I wondered if my method for orientating the image could be improved. After all, I was using Photoshop and a chart to determine the angle that a line between two field stars made with the horizontal side of the frame. This was potentially inaccurate because I had to place a cursor on where I thought the centre of the field stars were on the chart.

Let me now revisit the problem. Below is the image I took of the binary star after I had rotated it to what I thought was the correct orientation:-

 

As well as the two components of the binary star Xi Boo A and B you can faintly see two field stars in the image. To show this better and get an idea of the problem have a look at the diagram below:-

The field stars have been marked as 1 and 2 and there is a dashed line between them. I have also indicated a dashed line that travels due north from 1 until it meets the line drawn due east from 2. In the situation where the angular distance between field stars 1 and 2 is small (less than a degree or thereabouts) then the right angle triangle approximately lies on a flat plane. Let the angular separation between the two field stars be S. This forms the hypotenuse of the right angled triangle. The vertical dashed line forms the adjacent side to the angle θ marked. This side represents a separation which is the difference in the declinations of the two stars. Let this be δ² - δ¹. It follows from basic trigonometry that the angle θ is to a good approximation θ = arccos (δ² - δ¹ / S).

Now this is all very well but we have to determine the equatorial coordinates of the field stars. If you view the page on the BAA website where observations of Xi Boo have been made you will see that, not only is there a link to In-The-Sky.org, but also a link to the SIMBAD astronomical database for this object. This is a very cool page! If you look near the top of the page there is a 'submit query' button and on the same line just before it there is a field where you can adjust the radius of the search. Changing this radius to 12 arc minutes gives a wider view of the locality round Xi Boo and on the left an image with circled objects. Each of these circled objects (in different colours) relates to a line in the table to the left. If you hover over an object in the image it highlights it in the table (very neat).

This enables us to determine what our field stars are. It turns out that star 1 has a designation of V* EO Boo and star 2 has a designation of BD+19 2872 (they have V magnitudes of 8.43 and 9.06, respectively). Furthermore the table lists the epoch 2000 coordinates of these two stars as RA 14 51 42.2211380760, Dec. +19 05 21.217940052 and RA 14 51 38.0435867520, Dec. +19 10 23.589683616, respectively. I am amazed that they are quoting the accuracy of the positions to 9 decimal places in seconds of arc! Using my routine to calculate the separation of these two stars I find that S = 0.085586862 degrees. Further, δ² - δ¹ can be determined from the stored variables of the program and this corresponds to Z-X. I find that δ² - δ¹ = 0.083992154. Hence  θ = arccos (δ² - δ¹ / S) = 11.07775584 degrees.

If I now compare this value of θ with some measurements of this angle from my image above using Photoshop I get a value of 11.3 degrees, as best as I can determine, so I think I did a pretty good job in orientating my image!

All text and images © Duncan Hale-Sutton 2025
 

Friday, June 20, 2025

Z Ursae Majoris beginning to fade

Just over a week ago, one day after full moon on the 12th June 2025, we had some more clear weather which enabled me to get another view of the variable star Z Ursae Majoris. The twilight at about 22:00 UT (23:00 BST) is now very bright but Z remains pretty much near its peak and so determination of its magnitude is not too much of a problem. Using 10x50 binoculars my estimate was:-

Z UMa, 22:23 UT, chart 217.02, A(1)V(1)B, mag. 6.8

The star is now beginning to fade. At the present time (20th June 2025) it is about 7th magnitude. It looks like the peak in brightness is going to be sharp as I predicted but it may yet brighten again or slow down its fade. We shall see!

All text and images © Duncan Hale-Sutton 2025 

Sunday, June 8, 2025

Z Ursae Majoris on the 30th May 2025

The recent weather has been far too cloudy for me to carry out any observations but, just over a week ago, on the 30th May I was able to make an observation of Z Uma. Even then the weather wasn't that clear as I had to observe through thin cloud. At that time the moon was 3 days after new and twilight was an issue as it always is in the middle of summer. My estimate was as follows:-

Z UMa, 22:20 UT, chart 217.02, A(3)V(7)B, mag. 6.6

From this observation it looks like Z is beginning to fade after its recent maximum. Looking at other data from the BAA, this now does seem to be the case.

All text and images © Duncan Hale-Sutton 2025 

Sunday, June 1, 2025

Using a program to calculate the angular separation of Mizar and Alcor

In my last post I described a program that could be used to calculate the angular separation of two celestial objects using their equatorial coordinates. I now want to revisit my calculation of the separation of the double stars Mizar and Alcor. I previously gave the coordinates of Mizar A as RA 13h 23m 55.543s Dec. +54 deg. 55' 31.30" and Alcor as RA 13h 25m 13.538s Dec. +54 deg. 59' 16.65". I previously found that their separation to 4 decimal places was 0.1968 degrees. Now using my own program I find that inputting the same coordinates I get the separation as 0.196820915 degrees which to four decimal places is the same value as before.

Unfortunately, as often happens on the web, the tool that I used previously to calculate the separation of these two stars is no longer accessible and so I can't compare the values to higher decimal places.

However I have found some other online calculators. For example, inputting these coordinates in  Clear Sky Tonight I obtain the separation as 11 minutes and 48.56 seconds which corresponds to 0.196822222 in decimal degrees and this differs from mine at the 6th decimal place. Another online calculator is this one by the Russian-Turkish 1.5m telescope. If I enter the coordinates here I get a separation of 0.196837 degrees which differs from mine at the 5th decimal place.

To be honest, this is more than I expected, so rounding obviously is an issue. However, 1 arc second is 0.00028 degrees so we are talking about a tenth of an arc second error or so at most.

All text and images © Duncan Hale-Sutton 2025

 

A program to calculate the angular separation between two stars

I have previously written a program to calculate binary star orbits using the algorithm given in Practical Astronomy with your Calculator by Peter Duffet-Smith. I wrote the code for a programmable calculator (a Casio fx-4500P) and it makes predicting the positions of binary stars fairly straight forward. Recently I had been thinking it would be useful to have some code that could calculate the angular separation of any two stars (or any two celestial objects) as this is something that is always cropping up. For example, in May last year I took a photograph of the double star Mizar and Alcor and I used their angular separation to determine the resolution of my camera (in arc seconds per pixel) when it was at the prime focus of my Orion OMC-140 telescope.

So now I have used another algorithm in Peter's book to write some code for the Casio to do this (see section (31) on page 52). It is based on knowing the equatorial coordinates of the two objects (in other words their Right Ascension and Declination):-

"STAR 1"
Pause 3
A"H": B"M": C"S"
W = A + B/60 + C/3600
E"D": F"M": G"S"
X = E + F/60 + G/3600
D"SIGN +/-1"
D = -1 → X = XD ∆
"STAR 2"
Pause 3
H"H": I"M": J"S"
Y = H + I/60 + J/3600
L"D": M"M": N"S"
Z = L + M/60 + N/3600
K"SIGN (+/-1)"
K = -1 → Z = ZK ∆
U = 15(W - Y) 
S = cosˉ¹ (sin X x sin Z + cos X x cos Z x cos U) ▲
S > 0.1 → Goto 1 ∆
"RECALC"
Pause 3
T =  √((cos X x U)² + (X - Z)²) ▲
Lbl 1
 
Before running this program it is best to set the mode to degrees and to clear the memories. The Right Ascension coordinates of the two objects are entered in the hours, minutes, seconds format and for the Declination in degrees, minutes, seconds. The angular separation of the two objects, S, is given in decimal degrees.
 
Further execution of the program is possible for small angles (less than 0.1 of a degree). This is because there may be a limitation in the accuracy of the calculator to return an exact result due to its precision. T gives an alternative calculation of the angular separation which may be more accurate. 

All text and images © Duncan Hale-Sutton 2025