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Post by Ron Walker on Oct 21, 2022 10:35:36 GMT -7
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Post by Ron Walker on Oct 21, 2022 10:36:05 GMT -7
Posted by: Ron Walker Mar 20 2021, 10:40 AM This is great stuff Scott (as my eyes glaze over) I might not understand all of what you said (the older I get, the less I understand) but it all logically falls into place. There was a time that I actually understood this stuff.
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Post by Ron Walker on Oct 21, 2022 10:36:24 GMT -7
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Post by Ron Walker on Oct 21, 2022 10:37:09 GMT -7
Posted by: Ron Walker Mar 20 2021, 10:47 AM Scott, fear not, all of these unrelated posts will be moved to a new thread called, "Posts that have nothing to do with the thread they were posted in".
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Post by Ron Walker on Oct 21, 2022 10:37:33 GMT -7
Posted by: moonmagic Mar 20 2021, 12:00 PM QUOTE(Scott T @ Mar 15 2021, 06:26 AM) * Notes on co-ordinate systems
Before converting our star positions and star plate centres to latitude and longitude a couple of helpful notes about co-ordinate systems:
1. Cartesian co-ordinates are simply distances along three perpendicular axes ( x,y and z) from a starting point (origin) and can be expressed in any suitable unit – inches, mm or just plain numbers. Cartesian co-ordinates are not particularly convenient for talking about positions on the surface of a small sphere, because any change in position on the surface, results in a change in the value of at least two of the axes.
2. Right Ascension and Declination. RA and Dec are traditionally used to define the position of stars on the celestial sphere they are nearly the celestial equivalents of latitude and longitude. Right Ascension divides the celestial sphere into 24 divisions (hours) around its equator (each hour is subdivided into 60 minutes and each minute subdivided into 60 seconds). Declination divides the elevation between the equator and the pole into +90 degrees for the north pole and -90 degrees for the south pole. Each degree can be subdivided into minutes and seconds. There are plenty of far better, more thorough explanations out there on the internet. An important complication to be aware of; right ascension and declination can be expressed in ‘full, longhand’ form or in decimal form. Remember not to mix them up.
If you need to convert longhand to decimal form:
Decimal Right Ascension = hours + (minutes/60) + (seconds/3600)
For example, if the right ascension is 2 hours, 30 minutes and 45 seconds, then the decimal form is:
2 + (30/60) + (45/3600) = 2.5125
The same sort of principal applies to converting degrees from longhand to decimal:
Decimal Degrees = degrees + (minutes/60) + (seconds/3600)
3. Longitude and latitude. These are traditionally used to define positions on the surface of the earth but can be applied to any sphere. There are some very clear descriptions are out there on the internet. Longitude divides the distance around the equator into 360 degrees (which as always can be subdivided into minutes and seconds). Latitude divides the arc from the equator to the north pole into 90 degrees. The south pole is at -90 degrees.
The same complication applies i.e. longitude and latitude can be expressed in degrees, minutes and seconds or in decimal form.
One further complication exists, just as it is possible to express any angle in degrees or radians, latitude and longitude can be expressed in degrees or radians. I tend to naturally think about angles in degrees, but we are going to use radians for three very good reasons:
a) The ready-made formulae for calculating distances between points on a sphere are expressed in radians b ) The ready-made formulae for converting cartesian co-ordinates to latitude and longitude are expressed in radians c) Trigonometric functions in Excel work in radians. If you want to work in degrees you have to add in an extra conversion.
I think in degrees but I like radians. Alien mathematicians would use radians (they are universal) whereas they would have to work out why we divide circles up into 360 degrees.
Converting the star catalogue Right Ascension and Declination into Latitude and Longitude
To convert the decimal form of Right Ascension and Declination into longitude and longitude in radians use the formula:
Longitude in radians = (Right Ascension ÷24)x2π
To convert declination (decimal form) into latitude in radians use the formula:
Latitude in radians = Declination * (π÷180) While RA runs all the way around the sphere continuously from 0 to 24 hours, longitude divides the circle into +180 degrees heading east and -180 degrees in a westerly direction rather than counting from 0 to 360.
It is easy to apply a correction to the longitude using a spreadsheet or code using the ‘if’ function. If the longitude is greater than 180 degrees (or π radians), you simply subtract 360 (or 2π in radians).
The syntax for Excel is =IF (logical test, value if true, value if false). =IF(longitude<180,longitude, longitude-360) I will edit this bit in so that the story is in chronological order.
I too must re-read posts several times on this, but I do note your logical format. I got my latest laugh as you expressed pros and cons of latitude/longitude vs. degrees of R.A. & declination against cartesian points finally to radians with the line "ALIEN mathematicians would use radians." Indeed! mm
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Post by Ron Walker on Oct 21, 2022 10:37:58 GMT -7
Posted by: Ron Walker Mar 20 2021, 01:46 PM QUOTE(moonmagic @ Mar 20 2021, 12:00 PM) * I too must re-read posts several times on this, but I do note your logical format. I got my latest laugh as you expressed pros and cons of latitude/longitude vs. degrees of R.A. & declination against cartesian points finally to radians with the line "ALIEN mathematicians would use radians." Indeed! mm What did he say? ?
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Post by Ron Walker on Oct 21, 2022 10:38:54 GMT -7
Posted by: Scott T Mar 22 2021, 12:17 AM QUOTE(Ron Walker @ Mar 20 2021, 05:47 PM) * Scott, fear not, all of these unrelated posts will be moved to a new thread called, "Posts that have nothing to do with the thread they were posted in".
No worries Ron
I have been trying to get the hang of gnuplot to start to visualise the coverage of the plates. A couple of initial plots attached - these are based on about 8000 stars to mag 6.5. All just plotted as dots with no account for magnitude.
Serendipity has struck a second time - I was hoping not to split Orion over more than one plate because I think alignment becomes far more critical if really easily recognisable asterisms are split. It turns out it is in one piece on plate 5.
The second pic has 12 plates.
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Post by Ron Walker on Oct 21, 2022 10:39:17 GMT -7
Posted by: Ron Walker Mar 22 2021, 10:41 AM QUOTE(Scott T @ Mar 22 2021, 12:17 AM) * No worries Ron smile.gif
I have been trying to get the hang of gnuplot to start to visualise the coverage of the plates. A couple of initial plots attached - these are based on about 8000 stars to mag 6.5. All just plotted as dots with no account for magnitude.
Serendipity has struck a second time - I was hoping not to split Orion over more than one plate because I think alignment becomes far more critical if really easily recognisable asterisms are split. It turns out it is in one piece on plate 5.
The second pic has 12 plates.
I do remember that! When I had just the northern star ball in the house, all of Orion was projected. As I think now, every other section is projected below the celestial equator from the north and the opposite ones from the south ball. There is one section that is replaced by the base of the lamp that is totally projected from the south and the reverse is true 180 degrees away on the south ball. This all comes together like a jigsaw puzzle on the dome. Perhaps this is why they say any given projector is only good for such a small variant in dome diameters (though I am not at all convinced of that).
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Post by Ron Walker on Oct 21, 2022 10:39:39 GMT -7
Posted by: Scott T Mar 27 2021, 06:31 AM A couple more plots just before we move on to the next bit - the complete celestial sphere grouped into 32 faces using ~8000 stars.
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Post by Ron Walker on Oct 21, 2022 10:39:58 GMT -7
Posted by: Ron Walker Mar 27 2021, 10:38 AM Can't wait till this all plays out. How many stars do you plan for your final project?
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Post by Ron Walker on Oct 21, 2022 10:40:26 GMT -7
Posted by: Scott T Mar 27 2021, 11:19 PM Hi Ron - that is a really good question. The truth is I have not decided yet. The star catalogue I have (and the capacity of Excel) means I could go up to 50,000 but at the moment I am probably going to go around the 9000 mark for no other reason than it would have a similar number to some of the classic big Zeiss projectors.
I am a bit torn - On the one hand I don't want to swamp the constellations with too many background stars and on the other I don't want to invest huge amounts of time and not have a really pretty sky.
An idea is emerging to design it so that the star plates are reasonably easy to swap out - I could start with relatively few stars to get it up and running then invest the time into making very elaborate star plates.
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Post by Ron Walker on Oct 21, 2022 10:40:49 GMT -7
Posted by: Ron Walker Mar 28 2021, 01:27 PM QUOTE(Scott T @ Mar 27 2021, 11:19 PM) * Hi Ron - that is a really good question. The truth is I have not decided yet. The star catalogue I have (and the capacity of Excel) means I could go up to 50,000 but at the moment I am probably going to go around the 9000 mark for no other reason than it would have a similar number to some of the classic big Zeiss projectors.
I am a bit torn - On the one hand I don't want to swamp the constellations with too many background stars and on the other I don't want to invest huge amounts of time and not have a really pretty sky.
An idea is emerging to design it so that the star plates are reasonably easy to swap out - I could start with relatively few stars to get it up and running then invest the time into making very elaborate star plates.
I would go for the larger amount of stars. You would be surprised how easy it is to hide the very faint stars with just a little overall dome light. You can do either the normal white dome light or a blue dome light. As in nature it doesn't take a lot so it doesn't look like you have a lot of light on to wash out the fainter stars. I find that as I go into night and then come from night I can easily control all the the visibility of the stare except first magnitude or brighter with just the amount of room light I add. So now the first star I see tonight is Serious and then the first magnitude stars come out. By this time the star projectors are at full brilliance and the viewability of the fainter stars is controlled by the amount of light pollution I add with either the room lights, the moon blue light pollution, or my light dome from the nearby cities.
Also I would not worry about swamping the constellations as when you get to 5th and 6th magnitude stars they are really faint and the 6th ones are almost invisible if you look directly at them. The constellations will stand out because the are made up of brighter stars anyway and will still stand out with all of the faint ones around.
I would consider projecting the Milky Way with a different set of projectors and not put the stars on the main star plates. It is very effective to be able to control the brightness in relation to the star background. I am torn between the two types of projections, the cloudy type of Milky Way which to me is the more accurate, and then the billions of individual stars which look more like a time exposure of the Milky Way rather then a naked eye view. If I was limited to only one type I think I would stay with the cloud type as it looks more natural. If you ever devise plates with individual stars I would like to get a set as it would be nice to have the availability of both types. Either way, do keep a separate projector system so you can always adjust the lever with regard to the star field.
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Post by Ron Walker on Oct 21, 2022 10:41:12 GMT -7
Posted by: Scott T Apr 5 2021, 12:13 AM Thanks Ron for the condenser information on the other thread - really good stuff.
You may have noticed a pause in the posts - I have hit a simple but time consuming problem.
I have settled on 19000 stars at the moment -about 600 stars per plate. The spreadsheet sorts the stars to the correct plates in an instant. However, there are around 5000 stars duplicated on more than one star plate (in the overlapping zones of the pates).
I do not want to keep the duplicate stars because if they don't align perfectly (very likely) there will be a higher density of stars in the overlapping regions and if they do align perfectly the stars will be double brightness.
While Excel has a 'remove duplicate' function, it works by retaining the first occurrence and deleting the rest. This would create some large star plates which retain all the stars and some clipped plates where the duplicates are all missing. I want a fairly even spread so that the projections of two star plates merge seamlessly.
The only way I can think of to do this is to simply go down the list and randomly delete duplicate stars, sometimes from one plate, sometimes from another. This is pretty easy it just takes a long time. There are 19000 rows to pick through.
I am pretty sure I will think of a better way to do it when I finally get through them.
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Post by Ron Walker on Oct 21, 2022 10:41:33 GMT -7
Posted by: Scott T Apr 5 2021, 12:25 AM Sharing the spreadsheet.
I am perfectly happy to send anyone a copy of the first spreadsheet who wants it - I have a small version which only contains about 300 stars for illustration with a file size of 72kB. It contains all the necessary sorting formulae so you can paste in as many extra stars as you want.
I will share the full 19000 version too when it is ready - it is already quite a big file 11MB.
Just send me a message - at the moment it looks like I will need to e-mail it to you unless there are any clever suggestions on how to share files.
I notice that you can upload non-photographic files to this board but it does not support xls files. When the star plates are finalised I might be able to upload each individual plate as a txt files.
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Post by Ron Walker on Oct 21, 2022 10:42:00 GMT -7
Posted by: Ron Walker Apr 5 2021, 12:04 PM QUOTE(Scott T @ Apr 5 2021, 12:13 AM) * Thanks Ron for the condenser information on the other thread - really good stuff.
You may have noticed a pause in the posts - I have hit a simple but time consuming problem.
I have settled on 19000 stars at the moment -about 600 stars per plate. The spreadsheet sorts the stars to the correct plates in an instant. However, there are around 5000 stars duplicated on more than one star plate (in the overlapping zones of the pates).
I do not want to keep the duplicate stars because if they don't align perfectly (very likely) there will be a higher density of stars in the overlapping regions and if they do align perfectly the stars will be double brightness.
While Excel has a 'remove duplicate' function, it works by retaining the first occurrence and deleting the rest. This would create some large star plates which retain all the stars and some clipped plates where the duplicates are all missing. I want a fairly even spread so that the projections of two star plates merge seamlessly.
The only way I can think of to do this is to simply go down the list and randomly delete duplicate stars, sometimes from one plate, sometimes from another. This is pretty easy it just takes a long time. There are 19000 rows to pick through.
I am pretty sure I will think of a better way to do it when I finally get through them.
I can see an accuracy dilemma raising its ugly head right now. I assume we are talking about 19K stars in the general individual star population. As I said before, one can run a small amount of ambient light to wash out the dimmer stars but that destroys the jet black background that makes an optical/mechanical machine so good. This might not be much of a problem as the reflected light from the Milky Way might just do that job by itself. Thinking this through, one might wonder is it worth projecting more then 6K stars or even less. The Morrison projector only projected 3,800 stars and many consider it was one of the best star fields ever projected. What this all boils down to (in this fevered mind anyway) is how to project 19K stars but only allow approximately 6K to be seen by the average viewer without optical aid. That is, if a viewer comes to a show with birding or opera glasses (ones that can focus down to the dome screen) and with that aid be able to see those very faint stars.
I have read on more then one occasion that people love the bright skies of the Zeiss IX and how well it competes with the video projected at the same time, but then a lot of those same people say the Zeiss VI projected the most accurate skies. It would seem that to project and be seen with a video projection, the stars must be overly bright so the representation of a clear night sky is actually inaccurate. To be really accurate, faint stars should take as many as ten minutes to show up as the viewer becomes accustom to the dark night. This would never work in a now typical rush rush program that only runs twenty minutes or so, but would still work, and work well, in a more expanded hour show.
The problem comes, it would seem, when we try and project all of these various magnitudes of stars where the faint ones don't get brighter because of refraction from such a small star hole or project such a large star that it would be confused with other Suns or Moons in our sky. Since the laws of optics doesn't change, it appears that an optical projector can handle about five stops of magnitude before problems arise at either end of the spectrum. The typical answer has been to add optics for the brightest of stars since there are fewer of them and we could easily add them for the first two levels of magnitude and then let the star projectors take over to magnitude eight. I would imagine the Korsakov projector built for Boston had a spectacular sky as it projected three sets of stars from as many starballs. 1st and 2nd from individual projectors mounted on one ball, 3rd and 4th on another, and finally 5th and 6th on a third. They used the same sized holes but changed the wattage of the lamps. One wonders why they didn't project 1st and 2nd from individual projectors mounted on one ball, 3rd, 4th, 5th, and 6th, from the second, and then 7th, 8th, 9th, and 10th from the third. What a sky...or perhaps the reasoned that no one would be able to see anything projected from the third starball without optical aid, and how many people would bring opera glasses or for that matter, ever care enough to bring them.
So then how do we project these fainter stars and should they be seen even thought they can not be seen in real life. How many stars is enough? Is a Spitz Jr. with only 400 projected stars enough. It certainly can teach locations of the major constellations in their proper positions even in a small room with ceiling and walls. Actually more accurately then many of the newer projectors reproducing many more stars then any of us will ever see. These projectors project a flat "star chart" onto the ceiling of a room. Useful as any but as inaccurate as most. No, to be accurate these very faint stars should not be seen by the average viewer without optical aid. The only way without multiple projectors is to use some kind of neutral density filtering on those dimmer stars which is a whole new thread. Perhaps thr Morrison was right all along...3800 stars are enough.
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