Mobile phone picture of the Moon

Later on Monday evening (the 2nd January) I thought I would have a go at photographing the moon using my mobile camera (a Samsung Galaxy A21s) and my Orion OMC-140 Maksutov-Cassegrain. I have heard that it is possible to get quite good pictures by holding the camera lens up to the eyepiece of a telescope and taking a picture 'afocally'. The difficulty in doing this by hand is getting the telescopes image of the moon lined up with the camera of the phone and then holding it steady enough so that a picture can be taken. However, after some experimentation I did get one shot that I thought was pretty good:-

Here I used a Celestron 25mm eyepiece and the camera settings were 1/50s at f/2, ISO40. This was how I saw the image in the eyepiece. I was using a star diagonal, so even though North is at the top East is to the right and West to the left (which is the opposite way to which we usually see the moon with the naked eye). I was really impressed with the results and may try and do more work with my phone camera, perhaps purchasing a holder that can attach to the eyepiece to keep everything steady. If you 'click' on the image you can get an enlarged view which will show more of the detail of the 'seas', mountain ranges and, of course, craters.

All text and images © Duncan Hale-Sutton 2023

A low apparition of Venus on the 2nd January 2023

On Monday this week we had some nice clear conditions in the late afternoon and so I thought I would go out and see if I could find Venus and Mercury. In fact, it would have been virtually impossible to see Mercury as it was lost in the glare of the setting sun. You can use the BAA computing section's applet to find the position of the planets (and other objects) at any time. I did find Venus and so I got this image at 16:39 UT:-

This was about 34 minutes after sunset and Venus was about 4.4 degrees above the horizon (at this time Mercury was just 2.2 degrees above that level). This is looking almost SW. At the present time Venus is emerging from conjunction with the sun and will reach its greatest elongation east on June 4th. Its apparent diameter is quite a small 10.4 arc seconds and the phase is about 96%.

All text and images © Duncan Hale-Sutton 2023

Geminid Meteor Shower 13th December 2022

Two weeks ago we had the Geminid meteor shower which peaked over the night of the 13th/14th December. The moon on this date was waning gibbous and about 3 days from last quarter. It rose about 20:44 UT and so would interfere with observations after this time. Nevertheless, this is now the best meteor shower of the year and so I decided to see if I could capture any events on my camera.

I am glad to say that I was quite successful in spite of the moon. In all I caught 6 meteors on camera in the period 22:00 UT (approximately) to 23:17 UT. I had a few problems to begin with and it took me a while to settle on a successful method. After 22:18 I had the correct date and time (UT to the nearest second) recorded in my camera (prior to this I still had the year as 2021! and the time was out by a few minutes). At 22:27 I settled on taking successive 15 second exposures at ISO1600. I was using a D90 camera with a 18-105mm lens set at 18mm. This gave me a field of view of about 66x46 degrees. I made a note of the frame numbers where I had seen a meteor which was in or near the part of the sky where the camera was pointing.

I noted that I had seen 10 meteors in all. Initially I was pointing the camera south and then later I pointed it east. The first meteor I detected was this one at 22:18 UT.

This was a 30s rather than a 15s exposure. If you click on the image you will see an enlarged version. The meteor crosses the boundary between Orion and Taurus near pi 1 Orionis. The Hyades, Mars and the Pleiades can be seen clearly above the short trail of the meteor.

The second meteor that was recorded was at 22:42.

This one is more difficult to see but it is found just above the cloud to the right of Orion. This meteor lies entirely in the constellation of Eridanus near 47 Eridani.

The third Geminid was the best of the night and recorded at 22:45 UT.

This was a bright meteor and as bright as Mars which was nearby and magnitude -1.8. It also had a similar yellowish orange colour. The trail starts in Taurus (passing 5 Tauri) and ends in Cetus crossing a little bit of Aries. A couple of things to note: 1) there is a bit of ghostly train left behind the trail where the meteor has passed and 2) the trail is wiggly in structure near the start. This latter point may have been caused by camera shake as the meteor appeared very shortly after I pressed the camera shutter. However, Mars is just as bright and there is no camera shake noticeable in its image.

The third meteor I detected was at 22:54 UT.

This was a noticeable meteor and can be seen just above the figure of Orion the hunter. The trail of this meteor starts in Orion but then crosses into part of Taurus (it ends between 131 and 133 Tauri).

The next meteor to be captured was at 23:08 UT. By this time I had moved the camera to look eastwards.

This one is very faint and difficult to see. The short trail appears in the constellation of Cepheus not far from the triangle of stars made by delta, epsilon and zeta.

The final capture was at 23:12 UT.

This meteor is in virtually the same part of the sky as the previous one but, perhaps, a bit brighter. It passes between 18 and 19 Cephei.

All text and images © Duncan Hale-Sutton 2022

RW Cephei 12th and 24th December 2022

I decided to have a go at observing a different variable star this month and that is RW Cephei. The star is located near the triangle of stars made by zeta, delta and epsilon Cephei (the "foot" at the base of the constellation's main rhombus of stars). RW is a red hypergiant whose apparent brightness ranges between 6.0 and 7.6 over a period of about 346 days. It is classified as a semi-regular SRd type. Earlier this month this star has become the focus of attention because it may be going through a period of exceptional dimming as did Betelgeuse in Orion in January 2020.

I had a look at this star on Monday the 12th December 2022 in the early evening. At this time the moon was 4 days past full and not due to rise until 19:32 UT. At 18:30 the sky was clear and dark and astronomical twilight had just ended.

At 18:32 UT RW was fainter than star E (=7.3 mag.) on chart 312.02. At 18:50 UT I thought it was marginally brighter than star H (=7.8 mag.) and my estimate was E(2)V(1)H or magnitude 7.6 (to 1 d.p.).

I had another go at this star a few days ago on Christmas Eve (24th December). Then it was just one day past new moon and again the sky was clear and dark but there may have been a slight mist.

At 18:45 UT RW was again fainter than star E (=7.3 mag.) but this time it was closer brightness to this star than star H (=7.8 mag.). My estimate at 18:55 UT was that it was E(1)V(2)H or magnitude 7.5. So not much different to 12 days ago.

I don't have much to compare my results to as there hasn't been any further data added to the BAA VSS database by other observers since the 10th December, though one BAA VSS member did say that he had measured it to be visual magnitude 7.7 on the 21st December (via baavss-alert).

All text and images © Duncan Hale-Sutton 2022

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