Calculating Local Mean Sidereal Time(LMST), Meeus.

I thought I'd try to write some Python code for Subject via chapter 7 of
Meeus, fourth edition. His method is for Greenwich. It seems like one
could use his method for mean sidereal time, and then apply my longitude
(121.02) west divided by 15.0 to that. One way or another I'm not
getting the result I would expect. Is Meeus the way to go, or is there
something else? My code does work for his two examples which leads to
the mean.

W. eWatson wrote:
I thought I'd try to write some Python code for Subject via chapter 7 of
Meeus, fourth edition. His method is for Greenwich. It seems like one
could use his method for mean sidereal time, and then apply my longitude
(121.02) west divided by 15.0 to that. One way or another I'm not
getting the result I would expect. Is Meeus the way to go, or is there
something else? My code does work for his two examples which leads to
the mean.

In my opinion, Meeus is a good way to go.

In what way do the results you're getting differ
from what you expect? Maybe you could provide
an example.

On 6/23/2010 7:28 AM, Greg Neill wrote:
W. eWatson wrote:
I thought I'd try to write some Python code for Subject via chapter 7 of
Meeus, fourth edition. His method is for Greenwich. It seems like one
could use his method for mean sidereal time, and then apply my longitude
(121.02) west divided by 15.0 to that. One way or another I'm not
getting the result I would expect. Is Meeus the way to go, or is there
something else? My code does work for his two examples which leads to
the mean.

In my opinion, Meeus is a good way to go.

In what way do the results you're getting differ
from what you expect? Maybe you could provide
an example.


As a further check, in addition to the book's examples, I selected May
23, 2010 at 21:00 hours here on the west coast. To find the LST, I used
The Sky 6, and found the LST as 12:02:27 with a UT of 04:00:00. My
longitude is 121.02 west.

Here are my results. The first is for the book problem and the second
for my local time.

***Prime Meridian
-----MEAN
Here is JD: 2443825.5
T: 0.788651608487
DATE/TIME 1978/11/13 04:34:00
int(sidereal_Zulu_revs): 79
Zero hours solution
G Time: 3.45036951176 degrees
Mean ST at Zulu: 3:27:01.3
sidereal_Zulu: 3:27:01.3
Zulu_hrs: 4.56666666667
Zulu_adj 4.57916977987
Zero hour solution: 3.45036951176 agrees with book
Any time solution: 4.57916977987 agrees with book
Local solution: -0.0384607083689 subtract out hrs to to 121
Local in hh:mm:ss: 0 -2 -18.4585501282

***Nevada City******************* 1 My location calcs
-----MEAN
Here is JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8
sidereal_Zulu: 16:01:59.8
Zulu_hrs: 21.0
Zulu_adj 21.057496068
Zero hour solution: 16.0332795258
Any time solution: 21.057496068
Local solution: 29.0227755938
Local in hh:mm:ss: 29 1 21.9921376568

Adding Zero and Any gives: 37.090775593800004, or 13.09077...

Without using 121.02, 13 5 26.7921376568

None of this agrees with The Sky 6.

Note too that the JD I use is not the same as what TS6 produced. TS6
produced 2455340.667. To me this indicates it using some other method.

Another approach to this is given at the USNO site as
http://www.usno.navy.mil/USNO/astronomical-applications/astronomical-information-center/approx-sider-time/?searchterm=sidereal.

W. eWatson wrote:

As a further check, in addition to the book's examples, I selected May
23, 2010 at 21:00 hours here on the west coast. To find the LST, I used
The Sky 6, and found the LST as 12:02:27 with a UT of 04:00:00. My
longitude is 121.02 west.

So, let's see if I understand what you're giving here.

Longitude: 121.02 degrees west
Date: May 23, 2010
Local civil time: 21:00
UT: 04:00 (21:00 + 07:00 Standard Time to UT Conversion)

[snip Meeus example]

***Nevada City******************* 1 My location calcs
-----MEAN
Here is JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8

Okay. That's the mean sidereal time at Greenwich for
0:00 UT on the given date. That is,

GMST0= 16:01:59.8

sidereal_Zulu: 16:01:59.8
Zulu_hrs: 21.0

Shouldn't that be 04:00? You want to convert your
local time to local time at Greenwich (timezone
adjustment) then use that to adjust the sidereal time
at Greenwich to account for it not being 0:00 UT.

So, 04:00:00 x (day/sidereal_day) = 4.01095164 hrs
= 04:00:39.43 in H:M:S format

note that "day/sidereal_day" is the constant 1.002737908

Adding this sidereal time offset to GMST0 yields

GMST = GMST0 + 04:00:39.43
= 16:01:59.8 + 04:00:39.43 = 20:02:39.23

That's the mean sidereal time for Greenwich at the
instant you're interested.

Now, you're located at 121.02 degrees west longitude.
In terms of hours in the sidereal day it's lagging
Greenwich by:

(121.02/360)*24hr = 8.06800000 hr

= 08:04:04.80 in H:M:S notation

which is your OFFSET from GST due to longitude in
siereal hours.

That gives a local sidereal time of:

LMST = GMST - OFFSET

= 20:02:39.23 - 08:04:04.80

= 11:58:34.43

Which is pretty close to your 12:02:27 value.


Zulu_adj 21.057496068
Zero hour solution: 16.0332795258
Any time solution: 21.057496068
Local solution: 29.0227755938
Local in hh:mm:ss: 29 1 21.9921376568

Adding Zero and Any gives: 37.090775593800004, or 13.09077...

Without using 121.02, 13 5 26.7921376568

None of this agrees with The Sky 6.

W. eWatson wrote:
Note too that the JD I use is not the same as what TS6 produced. TS6
produced 2455340.667. To me this indicates it using some other method.

Another approach to this is given at the USNO site as

http://www.usno.navy.mil/USNO/astron...nomical-inform
ation-center/approx-sider-time/?searchterm=sidereal.

The JD given by The Sky 6 is probably the Julian day
for the given instant at Greenwich.

The Julian day you start with for your calculations is
for the given date at 0.0 UT at Greenwich. (You should
adjust the date accordingly if the day has already
'ticked over' at Greenwich. This will happen when the
number of hours to midnight (your local time) is less
than the number of hours of offset for your timezone)

W. eWatson wrote:
Note too that the JD I use is not the same as what TS6 produced. TS6
produced 2455340.667. To me this indicates it using some other method.

Another approach to this is given at the USNO site as

http://www.usno.navy.mil/USNO/astron...nomical-inform
ation-center/approx-sider-time/?searchterm=sidereal.

You should be able to get lots of test cases using this site:

http://www.jgiesen.de/SiderealTimeClock/index.html

On 6/23/2010 1:10 PM, Greg Neill wrote:
W. eWatson wrote:

As a further check, in addition to the book's examples, I selected May
23, 2010 at 21:00 hours here on the west coast. To find the LST, I used
The Sky 6, and found the LST as 12:02:27 with a UT of 04:00:00. My
longitude is 121.02 west.

So, let's see if I understand what you're giving here.

Longitude: 121.02 degrees west
Date: May 23, 2010
Local civil time: 21:00
UT: 04:00 (21:00 + 07:00 Standard Time to UT Conversion)
Yes, these are all correct.

[snip Meeus example]

***Nevada City******************* 1 My location calcs
-----MEAN
Here is JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8

Okay. That's the mean sidereal time at Greenwich for
0:00 UT on the given date. That is,

GMST0= 16:01:59.8
Yes, correct.
In fact, all below here looks correct. I made some adjustments in the
program, and it very closely agrees with all of your calcs.

I looked at my longitude and found that it was 121 deg 2 min and 32 sec.
(Nevada City, CA 39°15'06.9" N 121°02'31.9" W) I used that in the calcs
and got the final result as 16:01:59.8, which is a bit strange, since
the long changed a bit, and this is pretty much the same.
Nevada City, CA 39°15'06.9" N 121°02'31.9" W

I'm looking at TS6 and it shows the LST as 12:02:27, so this differs by
a bit over 4 hours, so there's something missing in all this.

Where I made a blunder in the code is that I mixed the Meeus examples
with my example, and managed to get a few pieces of code turned around.
I should have worked with two separate programs.

Here's what I pumped out for this example:
***Nevada City******************* 1
Longitude: 121.042222222
Hours from local: 8.06948148148
-----MEAN
Here is JD: 2455339.5
Here is UT: 4.0
Ephem JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8
sidereal_Zulu: 16:01:59.8
Zulu_hrs: 4.0
Zulu_adj 4.010951632
Zero hour solution: 16.0332795258
Any time solution: 4.010951632
Local solution: 11.9747496763
Local in hh:mm:ss: 11 58 29.0988347234


sidereal_Zulu: 16:01:59.8
Zulu_hrs: 21.0

Shouldn't that be 04:00? You want to convert your
local time to local time at Greenwich (timezone
adjustment) then use that to adjust the sidereal time
at Greenwich to account for it not being 0:00 UT.

So, 04:00:00 x (day/sidereal_day) = 4.01095164 hrs
= 04:00:39.43 in H:M:S format

note that "day/sidereal_day" is the constant 1.002737908

Adding this sidereal time offset to GMST0 yields

GMST = GMST0 + 04:00:39.43
= 16:01:59.8 + 04:00:39.43 = 20:02:39.23

That's the mean sidereal time for Greenwich at the
instant you're interested.

Now, you're located at 121.02 degrees west longitude.
In terms of hours in the sidereal day it's lagging
Greenwich by:

(121.02/360)*24hr = 8.06800000 hr

= 08:04:04.80 in H:M:S notation

which is your OFFSET from GST due to longitude in
sdiereal hours.

That gives a local sidereal time of:

LMST = GMST - OFFSET

= 20:02:39.23 - 08:04:04.80

= 11:58:34.43

Which is pretty close to your 12:02:27 value.


Zulu_adj 21.057496068
Zero hour solution: 16.0332795258
Any time solution: 21.057496068
Local solution: 29.0227755938
Local in hh:mm:ss: 29 1 21.9921376568

Adding Zero and Any gives: 37.090775593800004, or 13.09077...

Without using 121.02, 13 5 26.7921376568

None of this agrees with The Sky 6.

W. eWatson wrote:
On 6/23/2010 1:10 PM, Greg Neill wrote:
W. eWatson wrote:

As a further check, in addition to the book's examples, I selected May
23, 2010 at 21:00 hours here on the west coast. To find the LST, I used
The Sky 6, and found the LST as 12:02:27 with a UT of 04:00:00. My
longitude is 121.02 west.

So, let's see if I understand what you're giving here.

Longitude: 121.02 degrees west
Date: May 23, 2010
Local civil time: 21:00
UT: 04:00 (21:00 + 07:00 Standard Time to UT Conversion)
Yes, these are all correct.

[snip Meeus example]

***Nevada City******************* 1 My location calcs
-----MEAN
Here is JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8

Okay. That's the mean sidereal time at Greenwich for
0:00 UT on the given date. That is,

GMST0= 16:01:59.8
Yes, correct.
In fact, all below here looks correct. I made some adjustments in the
program, and it very closely agrees with all of your calcs.

I looked at my longitude and found that it was 121 deg 2 min and 32 sec.
(Nevada City, CA 39°15'06.9" N 121°02'31.9" W) I used that in the calcs
and got the final result as 16:01:59.8, which is a bit strange, since
the long changed a bit, and this is pretty much the same.
Nevada City, CA 39°15'06.9" N 121°02'31.9" W

The civil time is blocked into time zones. The GMST0
value depends only on the Julian day (and not fractions
thereof), so the result should be independant of the
longitude. Even in cases where the date at Greenwich
has advanced a day over the local date, the fractional
rotation will be very close indeed to the position one
sidereal day previous.


I'm looking at TS6 and it shows the LST as 12:02:27, so this differs by
a bit over 4 hours, so there's something missing in all this.

Well, the 16:01:59.8 value corresponds to the sidereal
time at zero hours UT for the date in question. The
offset due to the Earth rotation corresponding to the
current time of day (the current UT) has not yet been
applied. Once that is done you'll have the local mean
sidereal time for Greenwich.


Where I made a blunder in the code is that I mixed the Meeus examples
with my example, and managed to get a few pieces of code turned around.
I should have worked with two separate programs.

Here's what I pumped out for this example:
***Nevada City******************* 1
Longitude: 121.042222222
Hours from local: 8.06948148148
-----MEAN
Here is JD: 2455339.5
Here is UT: 4.0
Ephem JD: 2455339.5
T: 1.10388774812
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0332795258 degrees
Mean ST at Zulu: 16:01:59.8
sidereal_Zulu: 16:01:59.8
Zulu_hrs: 4.0
Zulu_adj 4.010951632
Zero hour solution: 16.0332795258

That's GMST0

Any time solution: 4.010951632

That's the offset (in sidereal hours) due to the
current UT at Greenwich.

Local solution: 11.9747496763

That's the GMST0 with the UT time offset and
the longitude offset applied: the local sidereal
time (in sidereal hours) at your location.

Local in hh:mm:ss: 11 58 29.0988347234

Well, that's what I got, too.

You might be interested in checking out Meeus'
book, Astronomical Algorithms, which has updated
methods for finding the sidereal time (new Julian
epoch and an additional (cubic) term in the
expression for Theta_0. He also shows how to
correct for nutation when determining the apparent
sidereal time at Greenwich, where high accuracy is
required.

And the solution is!
***Nevada City******************* 1
Longitude: 121.042222222
Hours from local: 8.06948148148
Time zone from local: 7.0
-----MEAN
Ephem JD: 2455340.5 --- The problem
Here is local time: 21.0
Here is UT: 4.0
T: 1.10391512663
DATE/TIME 2010/05/23 21:00:00
int(sidereal_Zulu_revs): 110
Zero hours solution
G Time: 16.0989893497 degrees
Mean ST at Zulu: 16:05:56.4
sidereal_Zulu: 16:05:56.4
Zulu_hrs: 4.0
Zulu_adj 4.010951632
Zero hour solution: 16.0989893497
Any time solution: 4.010951632
Local solution: 12.0404595002 (GMST0)
Local in hh:mm:ss: 12 2 25.6542008239 -- Solution, 1.35 sec off
-----APPARENT
correction (seconds): -0.206526739992
APP 12.0405168688
app result: 12 2 25.8607275639

It looks like the JD was the culprit. The date should have been bumped
up by one day. There's still a 1.35 sec or so difference.

Note I computed apparent sidereal from Meeus, fourth edition. Is there a
fifth edition? Maybe some constant has changed for the 2000 epoch? The
fourth edition was printed in 1988.

Thanks for sticking in there.