|
Hello Daves and others:
This has all been very interesting and
eductional! Yesterday morning at 9:00, when the sunlight first hit the
sundial, I carefully measured the distance of the center of nodus shadow
from the equinox line on my Flandrau Heliochronometer. This point was 3
millimeters south of the line and 27 cm) west of the meridian (high noon).
I measured again at 4:05 pm and the center of the shadow was directly over the
equinox line 31 cm east of the meridian!
This was all very gratifying because it showed
that the nodus was correctly designed, constructed and installed. I also
saw with my own eyes that it is true that the equinox line on a sundial
does not exactly show the path of the sun during the day because the declination
changes rapidly throughout the day. The shadow actually traveled from a point
south of due west towards a point slightly to the north of due east. (I
calculate that this equals about .6 of a degree off of due west to
east).
This small deviation is probably too
small to be seen on small dials, but was quite noticible on my
heliochronometer.
The important lesson I learned was that to check
the functioning of the nodus, you should observe its shadow when the
solar declination equals zero, not when it is equinox. If this moment
occurs at night, you're out of luck. Because if you check it the next day,
the shadow will already be north of the equinox line.
Thanks you all for your help with the
numbers.
Can't wait to check it each time a cusp rolls around!
John
John L. Carmichael Jr.
Sundial Sculptures
925 E. Foothills
Dr.
Tucson Arizona 85718
USA
----- Original Message -----
Sent: Saturday, September 22, 2001 8:32
PM
Subject: RE: equinox solar
declination
Dear John,
Here are the
correct figures for 7hours west of Greenwich (my mistake earlier setting it to
6hrs)
I calculated what you need (according to
Meeus' truncated VSOP87 theory):
|
Time (Tucson, AZ) |
Declination
(Degrees:Minutes:Seconds) |
|
|
|
|
2001/09/22
08:50:00.0000 [+7 gives UT]
|
0:07:03.0740 |
|
2001/09/22
08:55:00.0000 [+7 gives UT]
|
0:06:58.2075 |
|
2001/09/22
09:00:00.0000 [+7 gives UT]
|
0:06:53.3410 |
|
2001/09/22
09:05:00.0000 [+7 gives UT]
|
0:06:48.4745 |
|
2001/09/22
09:10:00.0000 [+7 gives UT]
|
0:06:43.6080 |
|
|
|
| September Equinox |
2001/09/22
16:04:30.6638 [+7 gives UT]
|
0:00:00.1298 |
| September Declination Zero |
2001/09/22
16:04:38.5840 [+7 gives UT]
|
0:00:00.0000 |
By way of example: the time of the September Equinox may be found using
the Find Annual Feature function The Find Annual Feature function
takes two parameters. The first is any date in the year that is to be
searched and the second is the feature that we are searching for. The
Find Annual Feature function returns a Modified Julian Day.
sdxFindAnnFeatx(sdMJDx(2001,1,1,0,0,0,6),sdFeatEquinoxSep())
= "52174.9614660165<MJD>
52174.9614660165</MJD><TimeZone>6</TimeZone>"
Repeating the example and using the Time
To Text function gives us the calendar day for Tucson, AZ
sdxTime2Text(sdxfindannfeatx(sdmjdx(2001,1,1,0,0,0,6),sdfeatequinoxsep()))
= " 2001/09/22
17:04:30.6638 [+6 gives UT]"
Regards
David Pratten
Author - The Sun API
Hi all:
I'm getting this message out a little late, but
tomorrow on the equinox, I'd like to check the solar declination of
the nodus's shadow. (Of the Flandrau heliochronometer). I want to
see if the center of the shadow is centered exactly on the 0 degree line at
the exact moment of equinox.
My astronomical calendar says that the equinox
occurs here (in Tucson AZ) at exactly 4:05 pm. But since the
declination is changing so quickly this time of year, I'm wondering if the
shadow will be exactly on the 0 dec. line in the morning too.
Does anybody know what the solar
declination is at around 9:00 my time? (6 hours and 55
minutes before the actual equinox). I wonder if I'll be able
to detect this small amount?
Thanks
John
John L. Carmichael Jr.
Sundial
Sculptures
925 E. Foothills Dr.
Tucson Arizona
85718
USA
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