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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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). |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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 <meeus's astronomical algorithms.h> /* 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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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. |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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 <http://www.sym454.org/> |
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 * ======================================================= |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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/ |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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 * ======================================================= |
In reply to this post by Shriramana Sharma
Op 9-dec-2005, om 6:05 heeft Shriramana Sharma het volgende geschreven:
> 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. Yes, the sidereal (actually tropical) rotation rate of the Earth w.r.t. the aequinox. Not w.r.t. the mean Sun which moves ahead of the aequinox and therefore makes the mean solar day 4 min. longer. > 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. Uhm, not quite: we need to apply the factor 1.0027... But only to correct the constant term. The relation between the local mean solar time and local mean sidereal time has the same evolution throughout the year at any meridian, because the ratio between mean solar day and mean sidereal day is fixed for all the Earth and does not depend on the meridian. Also mind that Allahabad is ahead of Greenwich: the aequinox crosses the meridian there 5h29m (SI time) before it does so at Greenwich. From formula 11.4 of Meeus AA (1st edition) I compute that the S.T. at Greenwich at 0h UT at 1 Jan-2000 (JD 2451544.5) was 99.9677947 deg.: this is the hour angle of the vernal aequinox point w.r.t. the meridian at Greenwich at that moment. Then at 82.5 deg. East of Greenwich the hour angle at that time was exactly 82.5 degrees more, because the circumference of the globe is 360 degrees. The local solar time at that meridian was already 5h30 min. The hour angle at 0h local time then was -5h30m * 1.00273790935.. * 15 deg. = -82.7258775 less, so net -0.2258775 deg. less. So the constant term for time zone GMT+5.5h is 99.7419172 deg. Add 360.98564736629 (= 360*1.00273790935) deg. for every fraction of a day after 1 Jan 2000 0h zone time. |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
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. |
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.org My OS : SUSE Linux 10.0 with KDE 3.5 |
Op 14-dec-2005, om 3:23 heeft Shriramana Sharma het volgende geschreven:
> 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 > > ? No, the theta(0) of formula 11.4 (in my 1st edition) like I said, that gives the hour angle at any instant. The constant term there is 280.46... deg., but that is for the epoch J2000.0 so at 12h UT: so subtract 180.4928... deg. for 0h UT: this gives 99.9677947.. deg. like I wrote. Then correct as I explained to refer to the meridian and mean solar time of 82.5 deg. East of Greenwich. >> 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? add 360.9... per day, but that need not be integer days but may be fractional days. This is in contrast with formula 11.2 and 11.3 (in my 1st edition) which are valid only for 0h UT every day: so you can not just substitute any value for T there, it must correspond to 0h UT. In formula 11.4 you can subsitute any value for JD or T. > 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? Yes, but I was working from formula 11.4 which I think may be more useful. >> 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."? It does not depend on the meridian but it does depend on the epoch. The expressions have quadratic and cubic terms so the rate has a secular variation. If you move your meridian to -82.5 deg., then you will also want to move your epoch to 0h local time there, which differs 5.5h from the Greenwich epoch. Then to be very exact, the T**2 and T**3 terms need to be evaluated. Just to pick all nits. > 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? Partly: you can get rid of Greenwich and UT altogether, like I tried to explain in my message of 10 December using that same formula. Give me the precise meridian you want to use and I will show you. |
Thursday, 15 December 2005 00:43 samaye, Tom Peters alekhiit:
> you can get rid of Greenwich and UT altogether, like I tried > to explain in my message of 10 December using that same formula. > Give me the precise meridian you want to use and I will show you. The situation is that I want to calculate the lagna start times for the specific location for which the user inputs the lat and lon, take for instance Chennai at approx 13 N 80 E, but return those times to the user in IST which is exactly UTC+0530. -- My website : http://samvit.org My OS : SUSE Linux 10.0 with KDE 3.5 |
Op 16-dec-2005, om 13:08 heeft Shriramana Sharma het volgende
geschreven: > Thursday, 15 December 2005 00:43 samaye, Tom Peters alekhiit: > >> you can get rid of Greenwich and UT altogether, like I tried >> to explain in my message of 10 December using that same formula. >> Give me the precise meridian you want to use and I will show you. > > The situation is that I want to calculate the lagna start times for > the > specific location for which the user inputs the lat and lon, take for > instance Chennai at approx 13 N 80 E, but return those times to the > user in > IST which is exactly UTC+0530. So we come back where we started. The spherical trigonometry formulae will give you the right ascensions, and then the hour angles, of these points when they are on the horizon. You can then also compute the local sidereal time (which is the hour angle of the point with ecl.long. = right ascension = 0). For a given geometry that is independent of the geographic longitude (only depends on the latitude and sidereal time). From that you can compute the local mean solar time, which is related to the local mean sidereal time in a cyclic fashion with a period of 1 year: so it depends on the time of year but essentially not on the geographic longitude. The local time finally can be reduced to UT and zone time. By working the other way round, i.e. starting with UT and computing the sidereal time at Greenwich at 0h UT, you take a detour. |
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