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johannes susato
Joined: 04 Jan 2009 Posts: 1394

Posted: Wed Mar 03, 2010 11:17 pm Post subject: 


Hi Mark,
thank you very much for your detailed answer to my question of possible errors in Ptolemy's star positions.
Please don't forget to start the thread "Paranatellonta . . . " some day. 

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Mark Moderator
Joined: 30 Sep 2005 Posts: 4540 Location: Edinburgh, Scotland

Posted: Thu Mar 04, 2010 1:16 am Post subject: 


Quote:  Hi Mark,
thank you very much for your detailed answer to my question of possible errors in Ptolemy's star positions.
Please don't forget to start the thread "Paranatellonta . . . " some day. 
Indeed. Although Margherita is really the authority on the traditional basis of the Paranatellonta here.
I just think there is quite a lot of confusion out there which is leading many people to assume Brady's methods are identical with the traditional Paranatellonta. _________________ ‘’As thou conversest with the heavens, so instruct and inform thy minde according to the image of Divinity…’’ William Lilly 

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johannes susato
Joined: 04 Jan 2009 Posts: 1394

Posted: Thu Mar 04, 2010 1:58 am Post subject: 


Not wanting to doubt the authority of margherita or any other member of this forum  only it was your idea, Mark, to compare Brady's and the real traditional technique of paranatellonta. 

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Clelia Romano
Joined: 31 Mar 2008 Posts: 353 Location: São Paulo

Posted: Tue Jun 29, 2010 6:42 pm Post subject: 


Hi Eddy:
I know this is now an old thread but I was reading exactly on the same subject that motivated it: the Brady´s book. Mark caughts my attention into this thread and then I saw your reference on the coordinate´s conversor and it really drew my eye.
I had one doubt, though: when we convert the equatorial coordinate to the ecliptic one, we give the RA and the declination,okay? Well,I did it, using Spica , that is more or less at 23º of Libra
The RA was 13º 32’ and the declination 10º52’
The program gave me (in ecliptic coordinates) a longitude of 8° 6'
What this means? The difference between Equator and ecliptic coordinates? Initially I thought that it was referring to ecliptic longitudes departing from 0º of Aries, but it was impossible, since Spica can´t be in Aries. At least not now!
I agree with Mark´s concerns, and I was trying to figure out if the result using Equator coordinates would give a so big variety comparing to Ecliptic coordinates. If the difference was not be so big it would be useful to give a larger orb to cover both systems.
Thanks in advance for your attention!
Clelia _________________ http://www.astrologiahumana.com 

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Mark Moderator
Joined: 30 Sep 2005 Posts: 4540 Location: Edinburgh, Scotland

Posted: Tue Jun 29, 2010 7:19 pm Post subject: 


Hello Clelia,
Hopefully Eddy will reply to your post. However, in the meantime I just wanted to state that I recently used that coordinate calculator/convertor myself to plot the ecliptical position and latitude of a comet. I was initially very confused as I had calculated the Comet's postion was around 26º Taurus near to the fixed star Algol. However, I got a figure nowhere near that. The reason was that I had failed to convert the equatorial coordinates to hours and minutes first. Instead I was dealing with decimal coordinates. You need to do that first before you can enter the star or comet position into the coordinate calculator. Could that be at the root of your problem? Plus the result you get from the calculator doesn't give you the sign just the postition along the zodiac with the beginning obviously 00 as Aries. Hence my comet position in Taurus was listed as simply 56º.
If its any use to anyone here is a calculation method I got from the web. Although Eddy had already given me the basic formula to convert between the two systems of equatorial coordinates.
Right Ascension is traditionally measured in hours, but degrees are a common alternative that is being seen more frequently in modern times. A complete circle of 360° is equal to 24:00 hrs of right ascension.
The fact that a sidereal day is equal to 23:56 hours of solar time is irrelevant to this discussion. A sidereal day is equal to 24:00 hrs of sidereal time.
Now to solve your RA problem. I'll assume you are using a handheld calculator, and will provide the simplest algorithm for that case.
Lets use an example of 66.918277°. Divide that by 15. The result is 4.461218467. The truncated integer value of 4 is the number of hours of RA.
Multiply the fraction 0.461218467 by 60. The result is 27.673108. The truncated integer value of 27 is the number of minutes of RA.
Multiply the fraction 0.673108 by 60. The result is 40.386, which is the number of seconds of RA.
Therefore, 66.918277° of RA equals 4:27:40.386 hrs of RA.
Remember to distinguish between northern and southern declinations by preceding the numerals with (N or S) or (+ or ).
Eddy mentioned another important point in a PM which I am sure he will not mind me sharing here:
Quote:  Note that the astronomical ephemerides are set for a so called epoch http://en.wikipedia.org/wiki/Epoch_(astronomy) in the table you linked to you can see 'RA (2000)'. The positions are set for a date (usually 50 years apart) so astronomers don't have to make a new starmap every year. So what you do when you converted to longitude/latitude is converting to 2010 as if it were sidereal to tropical chart. For years after 2000 you add the number of years divided by the rate of precession which is about 72 years per degree (or rather 71.583). In the example it would be about (for the middle of the year 2010) 10.5/72 = ca. 0.15° = 0°9'. 
Hope this helps,
regards
Mark _________________ ‘’As thou conversest with the heavens, so instruct and inform thy minde according to the image of Divinity…’’ William Lilly 

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Clelia Romano
Joined: 31 Mar 2008 Posts: 353 Location: São Paulo

Posted: Tue Jun 29, 2010 8:10 pm Post subject: 


Hi Mark!
I printed your explanation, only to see that I wrote it wrong in the message : Janus gave me the RA in hours and not in degrees.
BUT your hint was important since that what I have to do is to transform the hours in degrees, since the coordinates Eddy sent use degrees instead of hours.
Now I´ll do exactly the opposite: at the moment you said to divide I´ll multiply.
I´ll try it and i´ll let you know!
Thank you a lot!
Clélia _________________ http://www.astrologiahumana.com 

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Clelia Romano
Joined: 31 Mar 2008 Posts: 353 Location: São Paulo

Posted: Wed Jun 30, 2010 2:59 am Post subject: 


Hi Mark and All:
I found a handy calculator online to transform the Right Ascension given in hours in decimal time: http://www.springfrog.com/converter/decimaltime.htm
All you have to do after that is to transform the result obtained in hours.
So, as I supposed that the star was at 13hs 32’ I filled the provided space with this number.
Then I pressed the button "convert hours minutes and second in decimal time" and I got: 13.533055555555556
I divided the number by 15 and I got 0,90220370.
Now I multiplied the above result by 60= 5413222200
Simplifying, I think the result( if I did it right) is 0hs 54’
I don´t know if I did it right, as I said, but if I did, how to convert this last number into a position on the ecliptic?
Sorry to bother you!
Clélia _________________ http://www.astrologiahumana.com 

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Altair
Joined: 27 Jun 2012 Posts: 1

Posted: Thu Jun 28, 2012 5:28 am Post subject: 


I am so glad I found this thread as the longitude method V parans has bothered me for some time. I have been working with the fixed stars for three years and initially I was a big fan of Brady's because there aren't many modern interpretations of the fixed stars out there.
However I have never gelled with using parans. Mainly because if I want to do a chart about, say the current new moon, I would have to do a chart for every major city on earth for it to be meaningful. The late Diana Rosenberg uses "starsets" which look to be an interesting way of working, but her book is not out in the UK yet. If anyone has read it I'd be interested to know if she goes into the history of projected ecliptic/celestial equatorial degrees at all.
Can anyone answer these questions for me as there is so little information on the use of fixed stars on the web.
1) Is there an english translation of Ulagh Begs work anywhere? I have googled and just found the very beautiful arabic text. Is it purely an astronomical work or does it go into astrology and meaning of the fixed stars?
2) What is the difference between using ecliptic and equatorial measurements for the fixed stars in minutes? Are they really the same?
My pet peeve is that Brady is really seen as the authority on fixed stars and she is very vocal about her condemnation of Ptolemy's PED method. But if Ptolemy's tables have really have been found to be inaccurate why on earth did she put tables at the back of her own book (which recommends the use of parans) of Ptolemy's positions rather than the updated and more accurate measurements of Ulagh Beg? It almost seems to be a deliberate red herring and attempt to fudge all understanding of the longitude method.
Then we have Brady referencing "Anonymous of 379" who recommends only using stars near to the ecliptic for longtitude method and parans for those far. If that was the case then why do all the modern day fixed stars writers (Robson, Ebertins etc) include the meaning of stars like Vega, Phecda, Phact, Polaris which are of high latitude? It doesn't make sense. They do not mention using parans at all.
I don't mean to bash Brady as I have a great respect for her work. But she is such a authority in this field that I think she has put a lot of people off using Fixed stars due to the arguments over which method to use. Even her paran system is different from Solar Fires which means you would have to buy her software "Starlight" to use it.....
Constellation of words is a wonderful resource, but even they have written on some of the stars in Ursa Minor (quoting the Ebertins) that they may not have a strong effect because they are too far off the ecliptic. I'm inclined to favour the stars closer to the ecliptic because of all this confusion 

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