Tagged: Sun
- This topic has 28 replies, 9 voices, and was last updated 1 year, 11 months ago by Steve Holmes.
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16 October 2022 at 11:19 am #613062Nick JamesParticipant
I’m planning to livestream the eclipse from Chelmsford weather permitting using a Megrez 72 refractor and an IP camera. I’ve been checking the system out this morning and have a livestream running here: https://youtu.be/a33mI9szL9k. It should be up for the next few hours. There are a few small spot groups on the Sun at the moment but the big dark blobs are dust spots. I need to clean these off before the big day.
16 October 2022 at 12:22 pm #613063Grant PrivettParticipantSeems to be working. Looks like thin cloud presently.
There are a couple of small dust patches – as you said – but nothing thats going to detract from a good chunk of the Sun being obscured…
22 October 2022 at 8:51 am #613216Mr Giovanni Di GiovanniParticipantGreetings everyone, I am looking for a few amateur colleagues who can collaborate in observing the phenomenon. More precisely, to establish the timing of the contacts and the maximum phase, together with the contact angle (clockwise from the north). For arrangements cross19@libero.it.
Thank you and greetings to all.
Giovanni (Italy)23 October 2022 at 8:29 am #613221Mr Giovanni Di GiovanniParticipantDear Nick
I saw the solar disk on the youtu.be page. I download images every 2-4 minutes. I have a few things to ask you:
1) Does the image come in real time? What is the phase shift?
2) The disc image would be fully suitable for my study of the phenomenon. The only small problem is that the disc is too big, it is not entirely included in the picture.
3) If the disc remains stationary within the frame, it is not easy to establish the North-South direction. To do this, it would be necessary to bring (for example every 15-20 minutes) the disc close to the east side of the frame and NOT chase it until it reaches the west side. In this way, the alignment of the various images will make it easy to establish the North-South line on the painting itself. Can you make these breaks?This is the procedure I will follow with the camera (nikon) on my telescope:
– Entire disc in the frame;
– Periodically interrupt the tracking and take 3 or 4 images until the solar disc starts to leave the frame.
Here (L’Aquila near Rome the weather should be permissive)Thank you and good observation
Translated with http://www.DeepL.com/Translator (free version)
23 October 2022 at 11:14 am #613223Nick JamesParticipantMy livestream on Tuesday morning will be here:
The forecast for Tuesday here is light cloud all morning so it may not be a very interesting livestream! At least it is not likely to be raining…
23 October 2022 at 1:57 pm #613224Mr Giovanni Di GiovanniParticipantSkies with high but thin clouds are also forecast for my locality, I hope well. However, for my study of the phenomenon it is only necessary to have the cusps of the solar sickle visible.
25 October 2022 at 9:43 am #613252Nick JamesParticipantClear here at the moment and the livestream is running. Under 30 mins to go.
https://www.youtube.com/watch?v=1nUMDx8PiF4
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25 October 2022 at 11:32 am #613255Bill BartonParticipant10:00UT, projected image H25 eyepiece, 63mm Aperture OG, 840mm f/l.
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25 October 2022 at 11:49 am #613258Paul G. AbelParticipantBeen a nice sunny morning here in Leicester. Observed the eclipse with the students from Leicester University Astro-soc who had organised the event. Made some drawings with my 40mm PST including one showing the lunar limb close to a sunspot.
25 October 2022 at 12:22 pm #613259SheridanParticipantSuccess from near Milton Keynes
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Sheridan Williams
25 October 2022 at 1:32 pm #613262Mr Giovanni Di GiovanniParticipantDear friends, I present to you my photographs of today’s solar eclipse. I have just one question for you: How much would you have bet that the cloud would not move for the duration of the phenomenon? Is this a statistically significant phenomenon? It went wrong for me, very wrong. Best regards to you all.
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25 October 2022 at 1:40 pm #613265Mr Giovanni Di GiovanniParticipantHi Nick, I will do a thorough processing and study of the images I extracted from your You Tube film. Thanks to you I was able to follow the whole phenomenon. You will hear from me soon. Thanks again. Regards.
26 October 2022 at 7:59 am #613278Nick JamesParticipantI’ve put a speeded up version of my eclipse video here:
https://www.nickdjames.com/Eclipses/20221025/eclipse_20221025_ndj.mp4
The entire eclipse in 35 seconds.
Someone on Youtube yesterday said that watching a partial eclipse was like watching paint dry. As a public service this video is short enough that it should not stress the attention span of most social media users.
26 October 2022 at 12:53 pm #613280Grant PrivettParticipantYou probably need to add dancers, a 120bpm backing track, some dry ice, lasers, a conspiracy theory and someone being humiliated to keep social media fans tuned in.
18 November 2022 at 4:45 pm #613695Duncan Hale-SuttonParticipantI am going to stick my neck out here but I wondered if observers actually calculate the obscuration of the sun by the moon from the pictures that they take of the partial eclipse or does this have no value because the earth/moon system dynamics is so well known? I think that the obscuration percentage is given by the simple formula 100(w – sin w)/pi if you assume that the sun and moon appear to be the same anglular size. If you draw a chord between the points where the moon’s shadow touches the sun’s edge (the chord PP’ in the attached diagram) the angle w (in radians) is the angle subtended by P and P’ from the centre of the sun’s image. If you have an image of the partial eclipse you can measure (in pixels) the diameter of the sun’s disc and the length of the chord l and then w = 2 arcsin (l/d).
For example, for my image of the partial eclipse taken at 09:45 UT on the 25th from 52 degrees 44 minutes north and 1 degree 28 minutes east, l was 209 pixels, d was 311 pixels and so the obscuration was about 15% at this time (this was about 15 minutes from maximum coverage).
Even if the calculation has no scientific value, it may still be interesting to calculate and compare with prediction.
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21 November 2022 at 10:02 pm #613808Alex PrattParticipantHi Duncan,
The dynamics of solar eclipses are very well known, although that’s no reason why you shouldn’t measure the obscuration for yourself.
The Observers’ Challenge and the BAA Handbook listed the obscuration percentages for various locations, whereas the article in the October Journal gave the magnitude values (fraction of diameter obscured by the Moon). Some casual observers were put off by the small obscuration percentages, not realising that the magnitude values signified impressive partial phases. I prefer that ‘eclipse magnitudes’ are quoted in eclipse work.
Some teams observe total solar eclipses to measure the Sun’s radius, aiming to determine any evidence of long-term change in its diameter, e.g.
‘Estimating the Eclipse Solar Radius from Flash Spectrum Videos’ in
https://www.iota-es.de/JOA/joa2022_2.pdf
(17 MB download)
Cheers,
Alex.
23 November 2022 at 2:21 pm #613872Duncan Hale-SuttonParticipantHi Alex. Thanks very much for replying to my query. All very interesting.
I must admit that I understood eclipse obscuration (or percentage coverage) better than I did magnitude as you do tend to see predictions for these more in articles in a run up to a partial eclipse. I did look at the BAA handbook and the observer’s challenge beforehand. One thing I could point out as a relative newcomer to all this is that I didn’t understand what the blue line predictions were in the handbook as there didn’t seem to be a key (that I could see, unless perhaps, someone could point me to where it is)! I see now the numbers on the blue lines are probably magnitudes rather than maximum obscuration (I have just learnt something new in that magnitudes are quoted for maximum coverage). On the next page, though, the numbers are given as max obscuration (so even the handbook can’t decide what to display!). However, I do agree with you that magnitudes may be better because they are linear rather than depending on area. I must say that quite a lot of the other numbers quoted on page 13 of the 2022 handbook don’t mean much to me.
I think that what trying to measure the percentage obscuration did for me was to really make me understand what these terms meant and the bit of maths required to calculate the area of intersection of two identical circles was enlightening (no one has yet told me whether what I quoted was right or wrong yet!). Clearly it is more complicated if the sun and the moon are not the same apparent diameter but the same rules apply.
I guess what I am trying to say, perhaps, is that even though the mechanics of the earth moon system is very well known and calculating magnitudes or obscurations from our observations is not going to improve that knowledge, for those of us on the learning curve it is still very instructive!
Thanks for the link to the article. I will have a read of this at some point.
Duncan.
- This reply was modified 2 years ago by Duncan Hale-Sutton.
24 November 2022 at 12:33 am #613910Steve HolmesParticipantHi Duncan,
Inspired by your remarks, I derived a series of formulae to calculate the obscuration from the chord length but “did it the hard way” i.e. by using just basic trigonometry, in order to get a completely different method from yours. Comparing my result with yours, for the chord length and sun diameter you quote, I found that the two answers are precisely the same. I think we may therefore conclude that your equation is correct (and very much simpler than my deliberately straightforward but multi-step derivation!).
I then applied the formulae to the observations I had made. I was fortunate to be able to take an image within seconds of maximum eclipse at my location, and as it was captured with a camera using a “hyper-zoom” lens the image was much larger than your projected image, enabling me to get a greater precision in the lengths involved. I found that the chord length was 846px and the Sun’s diameter was 1200px, resulting in an obscuration of 17.98%. I also calculated the magnitude using the same lengths, which was 0.2908. These numbers should be compared to the predicted values of 17.96% and 0.2910 given by the eclipse circumstances predictor written by the well-known “computer” Xavier Jubier. Pretty good result, I thought!
And now addressing myself to Nick James, concerning his message about the eclipse video above:-
Hi Nick!
I was quite lucky here in north Suffolk as although I had patchy cloud cover I was able to capture images between the clouds about every 5 mins, and from them assemble not a video as such but rather a “video animation”. This was after quite a lot of processing to get the images consistent of course! It shows the entire eclipse in just 25secs, as per the attachment.
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24 November 2022 at 12:43 am #613912Steve HolmesParticipantHi Nick,
And here’s another animation I constructed showing that, although it appears to “roll” in an arc across the Sun in the previous animation, the Moon’s track across the Sun is indeed a straight line. Made by not removing previous frames when adding a new one until the straight line position is reached, and then successively removing early frames until last contact.
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24 November 2022 at 11:23 am #613916Steve HolmesParticipantAnd hello to Duncan again!
I didn’t have my BAA Handbook at hand when typing my previous replies last last night but now I have I can perhaps shed some light on the eclipse diagram to which you refer.
Diagrams such as these are taken from the NASA eclipses website rather than being generated by the BAA, and there is an explanatory page about them at https://eclipse.gsfc.nasa.gov/SEplot/SEplotkey.html which might help you. It is, unfortunately, worded in very technical terms but the last part should help you understand the various numbers and times around the map. But basically the blue lines are lines of equal eclipse magnitude and the green lines join places where maximum eclipse for that place occurs at the same time. A larger and clearer version of the magnitude/time plot for the UK appears on page 273 of the current version of the Journal, where the text clearly says that the numbers are magnitudes rather than obscurations.
Hope that helps!
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