CH Cygni 25th November 2022

We again had some clear weather last Friday the 25th November. The moon was 2 days past new and not a problem. I went out to observe the variable star CH Cygni. At 18.29 UT I discerned that CH was a bit brighter than star A (=6.5mag.) on BAA chart 089.04 but much fainter than star L (=5.7 mag.). My estimate was L(3)V(1)A which made it magnitude 6.3.

I think this result was right and that CH was the brighter side of star A, however, observations on the 23rd and 24th of November put this star at magnitude 6.7 and 6.9 which is much fainter. I am not worried by this because I am pretty sure of what I saw.

All text and images © Duncan Hale-Sutton 2022

TX and AH Dra 23rd November 2022

It looked like we were going to get a good dark sky last Wednesday, the 23rd November, but it didn't last. It was new moon and I went out early evening to observe the variable stars TX and AH Draconis. I planned to get estimates for the brightness of these two stars, go indoors and then come out again later to look at some other stars but it clouded up.

I was using the new version of the BAA star chart (106.04) and at 18.20 UT I found that TX was brighter than star K (=7.0 mag.) but fainter than star N (=7.7 mag.) but not by much. My estimate was K(2)V(1)N which put it at magnitude 7.5.

Going on to AH at 18.28 UT I found that it was about midway in brightness between star 6 (=7.8 mag.) and star 8 (=8.4 mag.) making it 6(1)V(1)8 or magnitude 8.1.

All text and images © Duncan Hale-Sutton 2022

Derivation of the obscuration percentage during a partial eclipse

After the partial solar eclipse on the 25th October I came up with a formula for how to calculate the percentage area of the sun that is covered by the moon (the percentage obscuration). I now want to show how I derived this formula.

I have assumed that the sun and the moon have the same apparent diameter during an eclipse. This won't necessarily be true but will be a reasonable approximation. So in the above diagram we will assume the moon is the circle to the left and the sun to the right. Where the two circles intersect represents the area where the moon eclipses the sun. If we can calculate the area of the sun to the left of the chord AB then the area of the intersection of the two circles is just twice this by symmetry.

Firstly, as ADC is a right-angled triangle we have that AD/AC is sin t. Let AC be the radius r and let AD be x. So (x/r) = sin t. If the chord length AB is l then l = 2x.

We can see that the area of the sun to the left of the chord AB is the area of the sector of the circle AEBC (the piece of cake) minus the area of the isosceles triangle ABC. The ratio of the angle 2t to the angle 2pi (pi being the constant 3.14...) is the same ratio as the area of the sector to the area of the circle (t being measured in radians). Thus the area of the sector is tr^2 (where r^2 represents r squared).

The area of the triangle ABC is half the base AB times the height DC. Half AB is x and DC is rcos t, so the area of the triangle is xrcos t. So the area of the sun to the left of the chord AB is tr^2 - xrcos t and the area of the intersection of the two circles is 2(tr^2 - xrcos t).

We can now calculate the obscuration fraction as the ratio of this intersection area to the area of the sun. We can see this is (2/pi)(t - (x/r)cos t). But (x/r) = sin t, so the obscuration fraction is (2/pi)(t - sin t cos t). But sin t cos t = (1/2)sin 2t. So the obscuration fraction becomes (1/pi)(2t - sin 2t). If we let 2t = w, then the obscuration fraction is (1/pi)(w - sin w) or as a percentage (100/pi)(w - sin w) as we stated before. Note that to find w we use that w = 2t = 2arcsin (x/r). But x = l/2 and r = d/2 where d is the diameter of the sun. So w = 2arcsin (l/d). Thus measurements of the chord l and diameter d give the angle w and hence the obscuration.

There is another measure of the depth of the partial eclipse and that is the magnitude. This is the fraction of the sun's diameter that is occulted by the moon. In the above diagram you can see that the portion of the diameter of the sun that is covered by the moon is twice the distance ED. Now ED is EC - DC which is r - rcos t = r(1 - cos t). So twice ED is 2r(1 - cos t) and so the magnitude is just (1 - cos t) which in terms of w is (1 - cos (w/2)). By convention the magnitude is usually quoted at maximum eclipse.

 

I now want to consider the more complicated obscuration calculation when the moon and the sun are not the same apparent diameter on the sky. The above figure shows the exaggerated situation. We again assume that the disc on the left is the moon and that on the right is the sun. It can be seen that the obscured area of the sun AEBF consists of two 'lens' shaped areas either side of the chord AB. We have already calculated the first AEBD. This is (r^2/2)(w - sin w) with w = 2arcsin (l/d). By a similar argument the lens shaped area AFBD is given by (R^2/2)(W - sin W) where W = 2arcsin(l/D) (D=2R). So the total area eclipsed is (1/2)(r^2(w - sin w) + R^2(W - sin W)) and as a percentage of the area of the sun this is (50/pi)(w - sin w + (D/d)^2(W - sin W)). Here W = 2arcsin((d/D)(l/d)).

All text and images © Duncan Hale-Sutton 2022

TX and AH Dra, CH Cyg, Z and RY UMa on the 30th October 2022

We had some more clear weather here on Sunday the 30th October. The moon was nearly at first quarter but it would set at 20:04 UT. The sky transparency wasn't great but I decided to go ahead with my observations anyway. The constellation of Draco was in the west but still high enough in the sky to make estimates of the semiregular (Srb) variables TX and AH Dra.

Beginning with TX on chart 106.03 with my 7x50 binoculars I could make out star P at visual magnitude 8.4. TX was fainter than star K (=7.0 mag.) and star N (=7.7 mag.) but brighter than P. At 21:16 UT I saw that TX was much closer to N in brightness than star P, so I estimated it at 1 point from N and 3 points from P (i.e. N(1)V(3)P). This put it at magnitude 7.9 to 1 decimal place.

I then moved on to AH Dra which is south of TX. At 21:40 AH was fainter than star 1 (=7.1 mag.) but brighter than star 8 (=8.4 mag). In fact it was probably middle way between these two stars and this was confirmed by comparison with star 6 (=7.8 mag.). So my estimate was 1(1)V(1)8 which made AH visual magnitude 7.8 to 1 d.p.

Whilst I have been writing up this observing session I have noticed that there is a new chart for TX and AH Draconis. Chart 106.03 has now been redrawn and is now 106.04. The only difference that I could spot was that the star labelled 1 is now magnitude 7.0 rather than 7.1. I will use the new chart in my next session.

Cygnus was also reasonably well placed so my next target was the ZAND+SR variable star CH Cyg. This was easy to estimate as its brightness indistinguishable from star A and was thus magnitude 6.5.

Finally, the constellation of Ursa Major is beginning to rise again about 11pm having swung round under the pole, so I could have a go at Z and RY UMa. At 22:34 UT Z UMa was brighter than star D (=7.9 mag.) but was equal to B meaning that my estimate was magnitude 7.3. At 22:43 UT RY UMa was fainter than star 1 (=6.7 mag.) but was brighter than star 4 (=7.7 mag.). In fact its brightness was indistinguishable from star 2 which made it magnitude 7.4.

All text and images © Duncan Hale-Sutton 2022

Partial Solar Eclipse 25th October 2022

We had a partial eclipse of the sun on Tuesday 25th October. From our location near Norwich first contact with the moon was to begin at 10:07, maximum coverage was to occur at 11:00 and last contact at 11:55 (all times BST). I had a flu jab appointment at 11am and so I didn't have much time to set anything up. In the end I decided to use my 8x24 binoculars to project the image (twice!) onto a sheet of white paper. The first image I took was at 10:39 and the last at 10:47. The best of them was this image at 10:45 (15 mins before maximum):-

I quite like the fact that this looks like two eyes looking down. Looking at this image I think that the "bite" taken out of the sun is much less than 25% of the total area of the disk. To see this imagine dividing the sun into four equal quarters of cake. The bite can be seen to roughly sit in one of those quarters but it by no means covers the whole quarter. However, I do reckon that the bite covers more than half of that 25% (i.e. more than 12.5%). So perhaps 15 or 16%. I must try and calculate it. London was predicted to be 15.2% at max.

I have now (18th November 2022) done some work on calculating the obscuration (the percentage area of the sun that is covered by the moon as it is called). Have a look at this diagram below:-

The shaded area represents the area of the sun that is eclipsed by the moon. I have assumed that the moon and the sun appear to be the same angular diameter on the sky. I have found that the percentage obscuration is given by 100 (w - sin w) / pi where the angle w is in radians. To find w you need two measurements; the length of the chord PP' (which we can call l) and the diameter of the sun (which we call d). Then w = 2 arcsin (l/d). 

In the above picture of the eclipse I chose the left-hand projected image which was more well defined and circular and measured l to be 209 pixels and d to be 311 pixels. This gave the obscuration to be 15.2%. But I think this is a coincidence that it is close to the London value!

All text and images © Duncan Hale-Sutton 2022

AC Her 18th October 2022

On Tuesday the 18th October we had clear patch in the early evening when I thought I could get some variable star observing done. The moon was at last quarter the day before and it wasn't due to rise until about 10:52 UT. At 19:03 UT the skies were clear and there was virtually no wind. I was hoping to observe more than one star but by 19:30 UT it had clouded up!

However, I did get a look at AC Herculis. Looking at BAAVSS chart 048.04 at 19:15 UT with my 7x50 binoculars I could see star E which has a visual magnitude of 8.2 so my limiting magnitude was fainter than this. At 19:20 UT AC was brighter than this star and also star D (at magnitude 7.4). However, it was fainter than star C (mag. 6.9) but not by much. So at 19:26 UT my estimate for AC Her was that it was 1 point from C in brightness and 2 points from D, namely C(1)V(2)D. This put it at magnitude 7.1. This was in good agreement with other observers from the BAA.

All text and images © Duncan Hale-Sutton 2022

The Sun on the 25th September

Over a week ago I had another go at projecting the image of the sun using my Celestron NexStar 102SLT. I used much the same set up as before, projecting the image onto white paper. Here is the image taken at 10:02 UT (using a Samsung Galaxy A21s phone camera):-

North is at the top and East is to the left. I have added the Active Region  numbers as they have been officially allocated. I would have preferred to have determined the East-West drift line myself but there wasn't time before cloud obscured the view. Instead I compared my image to that of the Solar Dynamics Observatory taken a few hours later at 13:45 UT (Courtesy of NASA/SDO and the AIA, EVE, and HMI science teams):-

 

To get how much my image had to be rotated by compared to the SDO image I examined both images in photoshop. By measuring the angle between the vertical and a line drawn between the same two sunspots on both images I was able to determine the difference between them. I came up with differences of 44.1, 44.6, 44.7 and 44.9 degrees, giving an average of 44.6 which is how much my image needed to be rotated by to get North at the top.

My image compares quite well with the SDO image. There was one active region to the far right of the SDO image that didn't show up in my image because it contained sunspots that were so small.

I have been wanting to try and quantify the activity of the sun and one way to do this is to calculate the Relative Sunspot Number or R number. The method for this is described in this BAA guide. It can be seen on both images that sunspots appear in groups (or active regions). If the number of groups seen is G and the total number of sunspots in all groups is S then R is 10*G+S.

A couple of things to be aware of is that a group is counted as separate from another group if its centre is more than 10 degrees away from the other group's center. However, it isn't that easy to quantify how much 10 degrees is on a 2D image of a 3D sphere! We can get an idea of how much it is by drawing on the sun's image a diameter from East to West and then a radius that at an angle of is 10 degrees to this line. The arc along the sun's edge subtended by this is then a guide to how far 10 degrees is on the surface.

Another point is that sunspots have a dark umbra and a lighter penumbra and sometimes several spots are linked by a contiguous penumbra. In counting sunspots it is the darker umbrae that should be counted rather than a "spot" linked by penumbrae.

Given all this I think from my image AR3110 has 4 spots, AR3107 has 6 spots, AR3105 has 5 spots and AR3108 has 2 spots. That is 4 groups and 4+6+5+2=17 spots. So R=10*4+17=57.

I did find this confusing because clearly a better image like the SDO one has much more detail and so many more spots and a few more groups can be counted. However, I have learned that R is a relative number that reflects the observer's estimate. It is therefore important for observers to use the same methods/equipment) each time.

All text and images © Duncan Hale-Sutton 2022 except for the SDO image.