# Lagna calculations - help req with hour angle Classic List Threaded 20 messages Open this post in threaded view
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## Lagna calculations - help req with hour angle

 Indian calendrics has a concept called "lagna". The period during which each sidereal segment of the ecliptic (http://samvit.org/calendar/astro/side.htm) is in contact with the eastern horizon is called the "lagna" of that zodiac sign. For example, the instant the first point of Aries touches the eastern horizon starts the lagna of Aries. Similarly for the other segments too. Now I was searching for a method to calculate the point in time when each lagna starts (which calculation is very much needed in Indian calendrics) and have developed the following: These are adapted from Astronomical Algorithms by Meeus. The local hour angle corresponding to the rise (or set) of a celestial body is given by: cos H = - tan phi tan delta where: H is the local hour angle phi is the observer's latitude delta is the declination of the celestial body Now from the local hour angle, how do I calculate the instant in UT of the lagna's start? /*--------rest can be skipped by those not interested in lagna-------------*/ The value delta for the first points of each sidereal segment can be found by the formula: sin delta = sin epsilon sin lambda where lambda is the ecliptic longitude of the first point of the required sidereal segment (which is the sidereal longitude plus the ayanamsha [http://en.wikipedia.org/wiki/Ayanamsha]) The formula above is adapted from the coordinate conversion formula by substituting beta = 0, since all points in question are *on* the ecliptic. The ecliptic tropical longitude of the sidereal segments may be found by the formula: ayanamsha = eclipticlongitude(ra(spica2000),dec(spica2000)) + precessionfrom2000(now) - 180 tropiclong = (raashi-no - 1) * 30 + ayanamsha -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 Op 6-dec-2005, om 12:42 heeft Shriramana Sharma het volgende geschreven: > Indian calendrics has a concept called "lagna". The period during   > which each > sidereal segment of the ecliptic (http://samvit.org/calendar/astro/  > side.htm) > is in contact with the eastern horizon is called the "lagna" of   > that zodiac > sign. For example, the instant the first point of Aries touches the   > eastern > horizon starts the lagna of Aries. Similarly for the other segments   > too. In occidental astrology, that would be called the ascendant, which is   measured from the aequinox. > Now I was searching for a method to calculate the point in time   > when each > lagna starts (which calculation is very much needed in Indian   > calendrics) and > have developed the following: These are adapted from Astronomical   > Algorithms > by Meeus. > > The local hour angle corresponding to the rise (or set) of a   > celestial body is > given by: > > cos H = - tan phi tan delta > > where: > H is the local hour angle > phi is the observer's latitude > delta is the declination of the celestial body > > Now from the local hour angle, how do I calculate the instant in UT   > of the > lagna's start? simple: Sidereal Time = Hour Angle of an object + Right Ascension of that   object.  So you need to also compute the right ascension (as measured   from the true or mean aequinox of the date) of that particular point   on the ecliptic.  From the date and the S.T. you can compute the local mean solar time. Subtract from that the geographic longitude East of Greenwich (15   deg. == 1 hour) to obtain UT. Also see the final section, "Ecliptic and Horizon", of the chapter on   Transformation of Coordinates in Meeus' Astronomical Algorithms   (form. 12.9 in the first edition).
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## Re: Lagna calculations - help req with hour angle

 Tuesday, 06 December 2005 22:26 samaye, Tom Peters alekhiit: > > Now from the local hour angle, how do I calculate the instant in UT > > of the lagna's start? > simple: Very good, so it is correct what I have assumed that the said hour angle calculated by Meeus' formula is the hour angle of the object which is rising or setting? > Sidereal Time = Hour Angle of an object + Right Ascension of that > object. >  From the date and the S.T. you can compute the local mean solar time. > Subtract from that the geographic longitude East of Greenwich (15 > deg. == 1 hour) to obtain UT. Well I only have the formulae for GST, I guess I will convert the local sidereal time calculated as above to GST by subtracting my longitude east of Greenwich (15 deg -> 1 hour). And from that, I must find UT, and after that I come back to Indian Standard Time by adding 05:30. Now to find the UT corresponding to the GST, how do I do that? I think I can calculate the GST at 0000 UT for that day and then subtract it from the GST of the ascendant-start, then add 24 hours if it is negative. Right? (If yes, please say yes.) > Also see the final section, "Ecliptic and Horizon", of the chapter on > Transformation of Coordinates in Meeus' Astronomical Algorithms > (form. 12.9 in the first edition). It has been shifted to chap 14, "Parallactic Angle and Three Other Topics" in the second edition. The only thing is, I am thinking whether I have to do a binary search with this formula: tan lambda = - cos theta / (sin epsilon tan phi + cos epsilon sin theta) [lambda = eclip long of horiz-eclip intersection; theta = local sidereal time; epsilon = obliquity of eclip; phi = local latitude] since I need lambda as input and theta as output. Or is there some other way? I don't think I can rearrange this formula using ordinary trigonometry and expect a simplification, since obviously this involved spherical trigonometry which I have little real experience with. So, any suggestions? -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Tom Peters-2 Please peruse the following and point out errors if any, and give suggestions for improvement ----------------------------------------------------------------------------- /*    Program to calculate the start of a lagna, when the first point of a    raashi coincides with the eastern horizon */ #import /*    the above library contains the following functions which are used in this pseudo-program    greenwich-sidereal-time-at-0000-UT (juliancenturies-since-J2000.0)    julian-date (time)    precessionfrom2000 (time) */ /* all angles are in degrees all celestial quantities are those of the raashi first point all terrestrial quantities are those of the observer lambda = celestial longitude of raashi first point phi = terrestrial latitude of observer L               = terrestrial longitude of observer (east is positive) alpha = right ascension of raashi first point delta = declination of raashi first point E epsilon = obliquity of ecliptic H = local hour angle of raashi first point at start of lagna theta = local sidereal time at start of lagna theta0          = Greenwich sidereal time at start of lagna bigtheta0 = Greenwich sidereal time at 0000 UT on that day T = Julian centuries elapsed since J2000.0 at 0000 UT of the required day JD = Julian Date of 0000 UT of the required day raashinum = the number of the raashi (Aries = 1) for which the start of lagna is to be found lagnaadiUT      = the required time of the start of lagna in UT lagnaadi = the required time of the start of lagna in Indian Standard Time */ JD = julian-date (today, 00:00 UT) // remember that the current date at Greenwich is to be taken T = (JD - 2 451 545.0) / 36525 const A ra2000spica     = 13 h 25 m 11.5793 s const D dec2000spica    = -11 h 09 m 40.759 s const L lambda2000spica = arctan ((sin A cos E + tan D sin E) / cos A) const ayanamsha2000 = lambda2000spica - 180 ayanamsha = ayanamsha2000 + precessionfrom2000(now) lambda = (raashinum - 1) * 30 + ayanamsha alpha = arctan (sin lambda cos epsilon / cos lambda) delta = arcsin (sin epsilon sin lambda) H = arccos (tan phi tan delta) + 180 theta = H + alpha theta0 = theta - L bigtheta0 = greenwich-sidereal-time-at-0000-UT (T) lagnaadiUT = mod ((theta0 - bigtheta0) / 360) / 15 lagnaadi = lagnaadiUT + 05:30 // do NOT do a modulus by 24; if it is 18:30 or later, a new day has started in Indian Standard Time -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Shriramana Sharma Op 7-dec-2005, om 4:30 heeft Shriramana Sharma het volgende geschreven: > Very good, so it is correct what I have assumed that the said hour   > angle > calculated by Meeus' formula is the hour angle of the object which   > is rising > or setting? Yes; if you want to make a distinction between rise and set then you   should use sin's and cos's in the numerator and denominator instead   of a tan function. >> Sidereal Time = Hour Angle of an object + Right Ascension of that >> object. >>  From the date and the S.T. you can compute the local mean solar   >> time. >> Subtract from that the geographic longitude East of Greenwich (15 >> deg. == 1 hour) to obtain UT. > > Well I only have the formulae for GST, I guess I will convert the   > local > sidereal time calculated as above to GST by subtracting my   > longitude east of > Greenwich (15 deg -> 1 hour). And from that, I must find UT, and   > after that I > come back to Indian Standard Time by adding 05:30. That is a detour: you asked for the UT time of a lagna, but I suspect   that you are really interested in the local time or zone time of a   rise.  The formula for "sidereal time at Greenwich at 0h UT"  is also   valid for " sidereal time at X at 0h local mean solar time at X". Then subtract that from the sidereal time of rise that you found; then divide the difference by by 1.0027379... to go from sidereal   interval to mean solar interval; then add whatever difference there is between your local time and   zone time (or GMT, if you really want that). > Now to find the UT corresponding to the GST, how do I do that? I   > think I can > calculate the GST at 0000 UT for that day and then subtract it from   > the GST > of the ascendant-start, then add 24 hours if it is negative. Right?   > (If yes, > please say yes.) Not quite, there are 1.0027379... sidereal days in a mean solar day,   so you have to apply a factor to the time passed since 0h as I   explained above. > It has been shifted to chap 14, "Parallactic Angle and Three Other   > Topics" in > the second edition. The only thing is, I am thinking whether I have   > to do a > binary search with this formula: > > tan lambda = - cos theta / (sin epsilon tan phi + cos epsilon sin   > theta) > > [lambda = eclip long of horiz-eclip intersection; theta = local   > sidereal time; > epsilon = obliquity of eclip; phi = local latitude] > > since I need lambda as input and theta as output. Or is there some   > other way? > > I don't think I can rearrange this formula using ordinary   > trigonometry and > expect a simplification, since obviously this involved spherical   > trigonometry > which I have little real experience with. So, any suggestions? Yes it is tricky, since theta appears in both the numerator and the   denominator.  No doubt there is an inverse formula but I don't know   how to derive it either.  From symmetry I suspect that you can   interchange lambda and theta, but I am not sure.  You could do a   binary search, which always converges to a value with known accuracy   in a limited number of steps.  Alternatively iterate: make an initial guess, see how much the calculated lambda differs from the desired one, subtract that from theta, compute again.
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## Re: Lagna calculations - help req with hour angle

 Thursday, 08 December 2005 02:21 samaye, Tom Peters alekhiit: > Yes; if you want to make a distinction between rise and set then you > should use sin's and cos's in the numerator and denominator instead > of a tan function. I don't get it - what does it matter? I am going to do an arccos operation, which return a value between 0 and 180. If it were an arctan function then splitting the argument of the arctan into numerator and denominator might have the purpose of identifying the quadrant. For an arccos operation, the second value must be obtained by just changing the sign of the output of the arccos, since cos(-x) = cos(x). Or subtract the output from 360 to get a meaningful hour angle. Don't you agree? > That is a detour: you asked for the UT time of a lagna, but I suspect > that you are really interested in the local time or zone time of a > rise.  The formula for "sidereal time at Greenwich at 0h UT"  is also > valid for " sidereal time at X at 0h local mean solar time at X". OK great. So local mean solar time for Chennai would instance be UTC + 05:21:12, assuming Chennai longitude = +80.3 E. So I just need to add 00:08:48 to get Indian Standard Time. Thanks... -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Tom Peters-2 On Dec 7, 2005, at 15:51, Tom Peters wrote: >> It has been shifted to chap 14, "Parallactic Angle and Three Other >> Topics" in >> the second edition. The only thing is, I am thinking whether I have >> to do a >> binary search with this formula: >> >> tan lambda = - cos theta / (sin epsilon tan phi + cos epsilon sin >> theta) >> >> [lambda = eclip long of horiz-eclip intersection; theta = local >> sidereal time; >> epsilon = obliquity of eclip; phi = local latitude] >> >> since I need lambda as input and theta as output. Or is there some >> other way? >> >> I don't think I can rearrange this formula using ordinary >> trigonometry and >> expect a simplification, since obviously this involved spherical >> trigonometry >> which I have little real experience with. So, any suggestions? > > Yes it is tricky, since theta appears in both the numerator and the > denominator.  No doubt there is an inverse formula but I don't know > how to derive it either.  From symmetry I suspect that you can > interchange lambda and theta, but I am not sure.  You could do a > binary search, which always converges to a value with known accuracy > in a limited number of steps.  Alternatively iterate: make an initial > guess, see how much the calculated lambda differs from the desired > one, subtract that from theta, compute again. BROMBERG says: Is this not the correct inverse of the above expression? Tan[theta] = -Cot[lambda] Sec[epsilon] - Tan[epsilon] Tan[phi] -- Irv Bromberg, Toronto, Canada
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Shriramana Sharma Shriramana Sharma wrote: > The only thing is, I am thinking whether I have to do > a binary search with this formula: > > tan lambda = - cos theta / (sin epsilon tan phi + cos epsilon > sin theta) > > [lambda = eclip long of horiz-eclip intersection; theta = > local sidereal time; epsilon = obliquity of eclip; phi = > local latitude] > > since I need lambda as input and theta as output. Or is there > some other way? > > I don't think I can rearrange this formula using ordinary > trigonometry and expect a simplification, since obviously > this involved spherical trigonometry which I have little real > experience with. So, any suggestions? No spherical trig is needed - you can rearrange this to   A sin theta + B cos theta = C with A = cos epsilon tan lambda      B = 1      C = - sin epsilon tan phi tan lambda As A sin theta + B cos theta = D sin (theta + X), with D*D = A*A + B*B and tan X = B/A, theta follows from   theta = asin(C/D) - X or  theta = pi - asin(C/D) - X ======================================================= * Robert H. van Gent                                  * * E-mail: [hidden email]                     * * Homepage: http://www.phys.uu.nl/~vgent/homepage.htm * =======================================================
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Irv Bromberg Thursday, 08 December 2005 08:33 samaye Irv Bromberg alekhiit: > >> tan lambda = - cos theta / (sin epsilon tan phi + cos epsilon sin > >> theta) > > Is this not the correct inverse of the above expression? > > Tan[theta] = -Cot[lambda] Sec[epsilon] - Tan[epsilon] Tan[phi] No! See: tan lambda = -cos theta / (sin epsilon tan phi + cos epsilon sin theta) => tan lambda = -1 / (sin epsilon tan phi sec theta + cos epsilon tan theta) => sin epsilon tan phi sec theta + cos epsilon tan theta = -cot lambda => tan theta = (-cot lambda - sin epsilon tan phi sec theta) / cos epsilon => tan theta = -cot lambda sec epsilon - tan epsilon tan phi sec theta which is of no use at all since it has a function involving theta on the right hand side too. You forgot the sec theta on the RHS and hence you thought it was the solution. It is *not* possible to isolate the functions containing theta here to one side of the equation. What Tom and I are musing is that there must be some separate solution derived from the same spherical trigonometry that was used to derive the original tan lambda expression. -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Gent van R.H. Thursday, 08 December 2005 14:13 samaye, Gent van R.H. alekhiit: > As A sin theta + B cos theta = D sin (theta + X), with D*D = A*A + B*B and > tan X = B/A, Great! I didn't know this. Does this identity have a special name or something? I would like to search the net for its derivation and proof... (Not that I don't believe you, but I would like to know how this equation works.) >   theta = asin(C/D) - X >  theta = pi - asin(C/D) - X How do I know which of these two times is for the rise of longitude point lambda and which for the set? -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 This stems directly from the formula for sin(x+y), if you substitute A = D*cos(y) and B = D*sin(y). >From: Shriramana Sharma <[hidden email]> >Reply-To: East Carolina University Calendar discussion List               ><[hidden email]> >To: [hidden email] >Subject: Re: Lagna calculations - help req with hour angle >Date: Thu, 8 Dec 2005 19:56:46 +0530 > >Thursday, 08 December 2005 14:13 samaye, Gent van R.H. alekhiit: > > > As A sin theta + B cos theta = D sin (theta + X), with D*D = A*A + B*B >and > > tan X = B/A, > >Great! I didn't know this. Does this identity have a special name or >something? I would like to search the net for its derivation and proof... >(Not that I don't believe you, but I would like to know how this equation >works.) > > >   theta = asin(C/D) - X > >  theta = pi - asin(C/D) - X > >How do I know which of these two times is for the rise of longitude point >lambda and which for the set? > >-- > >My website : http://samvit.org>My OS      : SUSE Linux 10.0 with KDE 3.5 _________________________________________________________________ Don't just search. Find. Check out the new MSN Search! http://search.msn.click-url.com/go/onm00200636ave/direct/01/
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Tom Peters-2 Thursday, 08 December 2005 02:21 samaye, Tom Peters alekhiit: > The formula for "sidereal time at Greenwich at 0h UT"  is also   > valid for " sidereal time at X at 0h local mean solar time at X". Hmm - I wonder. Let theta be the sidereal time at Greenwich at 0000 UTC. You state that theta is also the sidereal time at say: Allahabad (Indian Standard Meridian) at 0000 UTC+0530 or 0530 UTC. I think I can disprove this. I take the special case theta = 0 for this. If I can disprove your statement for theta = 0 then your statement as such stands disproved. Now if the sidereal time at Greenwich is zero, it means that the vernal equinox lies exactly above the Greenwich meridian. The vernal equinox point (neglecting precession for the present discussion) takes 23 h 56 m (as an approximation) to come back to the Greenwich meridian. This means that the vernal equinox covers 360 on the earth's surface in 23 h 56 m. It also means that to cover the 82.5 degrees from the Greenwich meridian to the Indian Standard meridian, the vernal equinox will only take 23 h 56 m * 82.5 / 360 = 5 h 29 m 5 s. So, if the vernal equinox was above the Greenwich meridian at 0000 UTC, causing the sidereal time there to be zero, it (the vernal equinox) will reach the Indian Standard meridian already at 052905 UTC, and by 0530 UTC, which is 0000 UTC+0530 Indian Standard Time, it will have *crossed* the Allahabad meridian, thereby causing the sidereal time at Allahabad to be *greater* than zero. Therefore it is not true that the same formula that gives us the Greenwich sidereal time for 0000 UTC for a particular date gives us also the sidereal time for any other location for 0000 local mean time of the same date. Any faults in my reasoning? -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 In reply to this post by Shriramana Sharma Shriramana Sharma wrote: >> As A sin theta + B cos theta = D sin (theta + X), with D*D >> = A*A + B*B >> and tan X = B/A, > > Great! I didn't know this. Does this identity have a special > name or something? I would like to search the net for its > derivation and proof... > (Not that I don't believe you, but I would like to know how > this equation > works.) http://www.education2000.com/demo/demo/botchtml/sinplcos.htm> >   theta = asin(C/D) - X > >  theta = pi - asin(C/D) - X > > How do I know which of these two times is for the rise of > longitude point lambda and which for the set? If you can relate your theta to hour angle (H) then it is simple. When H < 0 (or between 180 & 360 degrees), then it rises. When H > 0 (or between 0 a& 180 degrees), then it sets. Once you have figured out the correct hour angle, then convert to sidereal time and finally to local (zone) time. ======================================================= * Robert H. van Gent                                  * * E-mail: [hidden email]                     * * Homepage: http://www.phys.uu.nl/~vgent/homepage.htm * =======================================================
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## Re: Lagna calculations - help req with hour angle

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## Re: Lagna calculations - help req with hour angle

 Offlist. > Uhm, not quite: we need to apply the factor 1.0027... I tried to process this, but I did not quite digest it, and still had the gut feeling that something was not right. I forward the following text right from Meeus' words: ----------------------------------------------------------------------------- <  i.e., the formula which is valid for sidereal time at Greenwich <  at 0000 UTC is also valid for the sidereal time at any given <  location X at 0000 local mean solar time at X? I replied "yes", but now I see that was incorrect. The correct answer is "no". This can be seen from the following example. Suppose that the location X is exactly at longitude 75 degrees East. This corresponds to exactly 5 hours with respect to Greenwich. Suppose we want to find the sidereal time at X on a certain date D, and that on this date the sidereal time at Greenwich at 0h UT is Ts. Question: Find the sidereal time at X at 5h (exactly), local time. 5h local time at X corresponds to 0h local time at Greenwich; thus this is 0h UT. At this instant, the sidereal time at Greenwich is Ts, hence at X the sidereal time is Ts + exactly 5 hours. But what is the sidereal time at Greenwich at 5h UT? Because it is Ts at 0h UT, add 5h 00 minutes 49.3 seconds, which is the sidereal equivalent of 5h 00m 00s mean solar time. Hence, the sidereal time at Greenwich at 5h UT is Ts + 5h 00m 49.3 seconds. This is numerically different from Ts + 5h 00m 00s, so your friend's statement is incorrect. Regards. Jean Meeus -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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## Re: Lagna calculations - help req with hour angle

 Op 13-dec-2005, om 18:01 heeft Shriramana Sharma het volgende   geschreven: > Offlist. > >> Uhm, not quite: we need to apply the factor 1.0027... > > I tried to process this, but I did not quite digest it, and still   > had the gut > feeling that something was not right. I forward the following text   > right from > Meeus' words: > > ---------------------------------------------------------------------- > ------- > <  i.e., the formula which is valid for sidereal time at Greenwich > <  at 0000 UTC is also valid for the sidereal time at any given > <  location X at 0000 local mean solar time at X? > > I replied "yes", but now I see that was incorrect. The correct answer > is "no". This can be seen from the following example. > > Suppose that the location X is exactly at longitude 75 degrees East. > This corresponds to exactly 5 hours with respect to Greenwich. > Suppose we want to find the sidereal time at X on a certain date D,   > and > that on this date the sidereal time at Greenwich at 0h UT is Ts. > Question: Find the sidereal time at X at 5h (exactly), local time. > > 5h local time at X corresponds to 0h local time at Greenwich; thus   > this > is 0h UT. At this instant, the sidereal time at Greenwich is Ts,   > hence at > X the sidereal time is Ts + exactly 5 hours. > > But what is the sidereal time at Greenwich at 5h UT? Because it is   > Ts at > 0h UT, add 5h 00 minutes 49.3 seconds, which is the sidereal   > equivalent > of 5h 00m 00s mean solar time. Hence, the sidereal time at   > Greenwich at > 5h UT is Ts + 5h 00m 49.3 seconds. > > This is numerically different from Ts + 5h 00m 00s, so your friend's > statement is incorrect. As I stated in my response of 10 Dec., the formula as listed in AA   indeed is valid for 0h UT at Greenwich only, as specified.  But it is   not difficult to translate it to another longitude, so you don't need   to go back and forth to Greenwich and UT to compute the local   sidereal time.  My example is one way of doing it. Meeus in his text above compares the sidereal time in India at 5h   local mean solar time (0h UT) with the sidereal time in Greenwich at   5h UT.  Indeed they are different (by a factor of 1.0027... over the   interval).  But you are interested in the local sidereal time at 0h   local mean solar time at your place, and for that you only need to   change the constant value but not the rate.  That was my point. To be fully exact, you need to take the higher order terms into   account.  If you keep measuring T in UT from J2000.0, then there is   no change, but that destroys the benefit of having a local   expression.  Alternatively you might change T such that it refers to   your local meridian, but then you get cross-terms that have to be   added to the constant and linear term.  Essentially you correct for   the fact that at your new epoch (0h local time) the ratio of the mean   sidereal (actually tropical) day and the mean solar day, was sligthly   different from what it were a few hours later at 0h UT in Greenwich.   That is a minute correction though.
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## Re: Lagna calculations - help req with hour angle

 Wednesday, 14 December 2005 00:09 samaye, Tom Peters alekhiit: > interval).  But you are interested in the local sidereal time at 0h > local mean solar time at your place, and for that you only need to > change the constant value but not the rate.  That was my point. > So the constant term   > for time zone GMT+5.5h is 99.7419172 deg.   You mean the constant expression instead 100.460... in formula: bigtheta0 = 100.460... + 36000.770...T + 0.000...T^2 - T^3 / 38710000 ? > Add 360.98564736629 (=   > 360*1.00273790935) deg. for every fraction of a day after 1 Jan 2000   > 0h zone time. I do not comprehend. Do you not mean "add 360.9856... **per day** for every fraction of a day"? And *to what* are you asking me to add this fraction? > account.  If you keep measuring T in UT from J2000.0, then there is > no change, but that destroys the benefit of having a local > expression.   In essence, what you are saying is that I can create, from the formula 12.3 given above, a similar formula for my local meridian? > the fact that at your new epoch (0h local time) the ratio of the mean > sidereal (actually tropical) day and the mean solar day, was sligthly > different from what it were a few hours later at 0h UT in Greenwich. I thought you said that: "the ratio between mean solar day and mean sidereal day is fixed for all the Earth and does not depend on the meridian."? I had a new thought. Meeus also gives the formula for the Greenwich Sidereal Time at any given instant in time, as 12.4 (second edition). It goes smalltheta0 = 280.460... + 360.9856...(JD - J2000) + ... Now for any given location, calculate UT for 0 h local time and the JD at from that, and use the above formula to calculate the sidereal time at Greenwich, and then add the longitude east of Greenwich (15 deg -> 1 hour) to obtain the local sidereal time at 0 h local time. This seems a more straightforward solution. What do you say? -- My website : http://samvit.orgMy OS      : SUSE Linux 10.0 with KDE 3.5
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