Tagged: Sun
 This topic has 28 replies, 9 voices, and was last updated 1 year, 5 months ago by Steve Holmes.

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24 November 2022 at 4:29 pm #613930Duncan HaleSuttonParticipant
Hi Steve.
Wow, that’s great. It is nice to see that you could get a measurement for the magnitude and the obscuration which so nicely agrees with the prediction. Very satisfying. As regards my formula I am pleased that you could verify it. I too slogged through some trig but I spotted this simplified form of it at the end. I must work out what roughly the errors would be on these values given that you could measure things to within a pixel or two. I haven’t actually checked the assumption that the moon and the sun were approximately the same apparent diameter for this particular eclipse. Oh, actually, on that diagram of the eclipse in the handbook it has the apparent semidiameters of the sun and moon at greatest eclipse (S.D.=…). Taking these to be the numbers we need it would imply that the moon was slightly smaller than the sun?
Thanks also for explaining about the eclipse diagrams in the handbook. Yes, that link you gave to that NASA page does definitely help! I will keep a note of it.
I also liked your two videos of the eclipse, especially the second one which shows that the path of the eclipse is a straight line. Very clever.
Duncan.
24 November 2022 at 11:01 pm #613931Steve HolmesParticipantHi Duncan,
Happy to hear you liked my video animations. Given the “rather variable” (!) conditions I experienced during the eclipse; the need to take several shots on different exposures each time to combat the variable cloud cover, as well as experimenting with two different density filters for the same reason, and the problem of taking shots at the required 5 minute interval if clouds decided to intrude meant that the individual frames needed A LOT of processing to get them consistent enough to be assembled into the animations. I’m thus glad to know that my efforts have been appreciated! The second one is particularly intriguing, as you say – I haven’t seen anything similar on the Internet so I was rather pleased it came out so well.
I also found equations of the form you derived on the Internet but wanted to explicitly avoid that approach so I could get my results via an entirely different route and thus be confident in the comparison between the two methods. Doing it “the hard way” also enabled me to sort of work backwards from the answer and derive an again multistep process for converting from magnitude to obscuration, which I’m sure will prove useful.
In terms of the measurement and calculation errors which might be expected, the fact that I was able to use a much larger image gave me advantages in both precision and accuracy. More precise as I was working at the 1 in 1000 level whereas you were at 1 in 250, and potentially more accurate as, because of the greater precision, I could determine my measurements at a smaller “step length”. That is, in my case a 1 pixel step hardly moved the “marquee” lines I was using to measure distances, and so the lines could be more closely aligned with the Sun, whereas your 1 pixel change will have moved the measurement point by four times as much so you might not have been able to line things up exactly. I still had the problem of estimating the correct points to measure from of course, given that my image was not sharp (because of the limitations of the homemade filters I was using), but even so I believe that my distances were probably correct to +/ 1 pixel. Translated to the difference this would make to the final result, as compared to the nominal 1200 by 846 figures, the maximum difference (in the cases 1201 / 845 and 1199 / 847) was just less than 0.5% which is still pretty acceptable.
The same website I used to determine the theoretical magnitude and obscuration also gave the ratio of the diameters at my maximum eclipse, which was 0.99253. The assumption of equal sizes is thus highly reasonable, involving potential errors of only the same order of magnitude as those involved in taking the measurements and probably less as the radii are used as part of more complex equations rather than as direct parameters. The SD values on the NASA chart are of course those relating to the point of absolute greatest magnitude (over eastern Siberia) so, while indicative of the situation, will not be applicable to the UK as the Moon will have moved along its orbit during the eclipse.
Steve
28 November 2022 at 6:41 pm #613970Duncan HaleSuttonParticipantHi again Steve,
I completely agree with you with what you said about the errors. The bit I meant to calculate was the error on the estimate of the obscuration and magnitude. If we take your +/ 1 pixel example, the largest value of the angle subtended by the chord w is derived from the ratio 847/1199 which corresponds to w=1.568860462 radians. The smallest angle is derived from the ratio 845/1201 which corresponds to w=1.56084678 radians. So the upper and lower limits to the obscuration percentage are 18.11% and 17.85% respectively, which roughly means that your observed value is 17.98 +/ 0.13 (compared to the predicted value of 17.96%). With respect to the magnitude the upper and lower limits are 0.2922 and 0.2894 respectively, which means that your observed value is 0.2908 +/ 0.0014 (compared to the predicted value of 0.2910). So it seems we would have to work a bit harder in accuracy before we could start to determine if the observed value really differs from the prediction (and this is what would be of interest).
Duncan.
 This reply was modified 1 year, 6 months ago by Duncan HaleSutton.
 This reply was modified 1 year, 6 months ago by Duncan HaleSutton.
29 November 2022 at 12:44 am #613981Steve HolmesParticipantOddly, although before I made this reply the front page of the website said the latest reply was from Duncan, and the total number of replies before this one was given as 23 with the last made by Duncan 5 hrs 57 mins ago (at time of typing), the number of actual posts was shown as only 22 with nothing visible from Duncan. An aborted reply, or a glitch somewhere?
29 November 2022 at 9:20 pm #614014Duncan HaleSuttonParticipantHi Steve, there was a problem with the system. I created a post and then made a couple of small edits and then it disappeared! I was hoping it could be resurrected but I think it has gone into the ether.
The short summary of what I said was this. I agree about the errors. Taking your results, the ratio 847/1199 gives an upper limit for the angle w and the ratio 845/1201 gives a lower limit. Taking these to be limits of the error we get for your measurement of the obscuration and magnitude 17.98 +/ 0.13 and 0.2908 +/ 0.0014 respectively. My conclusion is that we would have to work a bit harder to get these errors down if we were going to see any deviation from the predicted values of these quantities (which would be interesting).
Duncan.
30 November 2022 at 12:58 am #614015Steve HolmesParticipantHi Duncan,
Interesting! I experienced exactly the same problem when I tried to edit a reply. Like yours, it just vanished. Maybe something to report to “the powers that be”?
Anyway – I agree with your “error bars”. Probably the only practical way to improve the accuracy and precision of the observations would be to increase the resolution of the image (mine was 1920×1920 pixels, cropped from 3264×2448 to centre it) and to use a better filter so the image was sharper. Mine was just a homemade item, constructed by inserting a small piece of Mylar sheet – formerly a lens from a pair of eclipse specs! – into the top of a closed cardboard cylinder sized to fit over my camera barrel. Cheap, but not exactly high quality! Then of course one would have to redo the calculations to take account of the possibly differing apparent sizes of Sun & Moon. I’ve seen the required analysis on the Internet, but the result is by no means as simple as your “oneliner” formula! One must also consider the accuracy of the predictions themselves – do they assume a spherical Earth and Moon, for example, or use a “true geoid” and reliable limb profiles? At the level we are considering, such things become important.
I shall certainly be repeating these calculations the next time round though, on 29th March 2025!
Steve
5 December 2022 at 6:34 pm #614312Duncan HaleSuttonParticipantHi Steve,
the problem with my missing post was kindly sorted (it had been marked as spam!) and it now appears in the thread above.
I have had a look at the situation where the sun and the moon are different apparent sizes and I don’t think the situation is too bad. You have to take into consideration that the two “lens” shaped areas either side of the chord PP’ in my diagram are not the same size, but the formulae for their areas have similar forms. I think that the percentage obscuration in this instance is given by (50/pi)(w – sin w + f^2(W – sin W)). As before w is the angle subtended (in radians) by the chord of length l at the centre of the sun. However, now there is also an angle W subtended by the chord at the centre of the moon. You can see that the w – sin w part is the same as before, but now there is an additional part that comes from the area of the “lens” to the right of the chord which is governed by the shape of the moon. The factor f is the ratio of the moon’s diameter to that of the sun’s. Now as before w = 2 arcsin(l/d) where l is the length of the chord and d is the diameter of the sun. But we also have that W = 2 arcsin ((1/f)(l/d)). I hope this make sense.
I now calculate that if f=0.99253 (assuming that the moon was smaller than the sun at the time of your measurement) then the measured obscuration is now 18.08 +/ 0.13. So it does make a small difference. I guess that this formula works up until W or w is less than or equal to pi. I haven’t figured out what happens otherwise!
Duncan
 This reply was modified 1 year, 5 months ago by Duncan HaleSutton.
5 December 2022 at 6:46 pm #614314Duncan HaleSuttonParticipantSorry Steve, my forum post is missing again as soon as I made a small edit to it! Here it is (again):
Hi Steve,
the problem with my original missing post was kindly sorted (it had been marked as spam!) and it now appears in the thread above.
I have had a look at the situation where the sun and the moon are different apparent sizes and I don’t think the situation is too bad. You have to take into consideration that the two “lens” shaped areas either side of the chord PP’ in my diagram are not the same size, but the formulae for their areas have similar forms. I think that the percentage obscuration in this instance is given by (50/pi)(w – sin w + f^2(W – sin W)). As before w is the angle subtended (in radians) by the chord of length l at the centre of the sun. However, now there is also an angle W subtended by the chord at the centre of the moon. You can see that the w – sin w part is the same as before, but now there is an additional part that comes from the area of the “lens” to the right of the chord which is governed by the shape of the moon. The factor f is the ratio of the moon’s diameter to that of the sun’s. Now as before w = 2 arcsin(l/d) where l is the length of the chord and d is the diameter of the sun. But we also have that W = 2 arcsin ((1/f)(l/d)). I hope this make sense.
I now calculate that if f=0.99253 (assuming that the moon was smaller than the sun at the time of your measurement) then the measured obscuration is now 18.08 +/ 0.13. So it does make a small difference. I guess that this formula works up until W or w is less than or equal to pi. I haven’t figured out what happens otherwise!
Duncan
8 December 2022 at 11:38 pm #614489Steve HolmesParticipantWell, after the saga of the disappearing edits and the successful outcome of my “Mars occultation” project, plus all the other diversions life throws at one, where were we? Ah yes – the problem of different diameters for Sun and Moon.
On which topic, I can but say “Yes, you are quite right”. I have to confess that I committed the cardinal error of looking for the answer before fully analysing the problem! The analyses of the general problem on the Internet all seem to take the centretocentre distance as the variable parameter rather than the chord distance, which does indeed result in some rather involved calculations – sometimes involving integration between limits! Conversely, using the chord distance and the wsin(w) formula makes the extension to unequal diameters almost trivial, and I agree with your answer.
Using this principle in my spreadsheet, I also agree that the result for f=0.99253 is 18.08%. Interestingly, if I use the values 1201 and 845 for the Sun diameter and chord length respectively (instead of 1200 and 846) together with f=0.99253, the result is 17.955%, or a magnitude of 0.29093, compared to the online answer of 17.960% and 0.29098, which is exceedingly close and tends to confirm my estimate of “accurate to 1 pixel”. I am thus still happy with my result as it agrees with the online answer to within a variation of 1 pixel whether the online calculator used the f factor or not.
As to the operative limit of the formula, do you not mean “until w or W is GREATER THAN pi”? Fortunately, if f=1 this can never happen as when the angle is pi the event is an exact total eclipse and the formula gives the correct answer of 100% and 1.00 for the obscuration and magnitude. If f<1 the formula can indeed fail as we are then heading for an annular eclipse, for which the chord length actually decreases at some point and then becomes undefined during the annular phase of the eclipse. If f>1 the formula seems to give sensible answers until the chord length is equal to the Sun’s diameter i.e. the case which would be an exact total eclipse if f=1, after which it will give wrong answers becuase, as in the case of an annular eclipse, the chord length will decrease as the Moon moves further over the Sun’s disc. The formula will thus give reducing values of obscuration whereas it is of course increasing. Further analysis required, I think!
Steve

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