|Sep17-12, 08:31 PM||#1|
Estimating Local Sidereal Time
Pretty elementary question here, but I'm new to astronomy and I'm having trouble estimating LST. Online sources give me the exact formula, but this doesn't really help me on the fly. My location is 97.7-ish degrees W of Greenwich, which I know means I can subtract roughly 5.5 hours from Sidereal time at Greenwich.
I know that at the vernal equinox in march is when the ecliptic cross the equator, solar noon at Greenwich (lat. 0) = RA of 0, and the autumnal equinox in September is at an RA of 12h, so the change between sidereal time and solar time is about 4 minutes per day or about a degree. I can kind of picture how this is happening qualitatively, but I'm kind of lost on roughly quantifying it for a particular day/time/location.
My professor will say something like "Ok, it's September 17th, and it's 8:15pm here, which means it's 1:15am in Greenwich, so roughly what is our LST?"
Obviously he's just looking for a rough estimate within a few minutes or so of the actual LST, so how do I go about doing this in my head?
|Sep17-12, 09:52 PM||#2|
at noon in your local solar time, the sun is due south. and i think it's the same for solar and sidereal at some equinox, just can remember which.
60 minutes per 15 degrees of longitude is something i understand. dunno about 4 degrees per day for sidereal. i thought that adjustment would be more like 1 degree longitude per day which would be about 4 minutes. (or more precisely 360/365.25 degrees per day.)
is that where 360 degrees per rotation comes from? that we have approximately 360 days per year and the Earth (in it's non-circular orbit) moves in one day about 1 degree around the Sun (with respect to distant stars)? i always thought it was something about being divisible by so many factors; 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 60, 75, 90, 120, 180.
|Sep17-12, 10:52 PM||#3|
|Sep21-12, 04:53 PM||#4|
Estimating Local Sidereal Time
You need Julian Date and Days Since 1-1-2000 to calculate LST correctly:
JD = 367*yr - floor(7/4*(yr+floor((mo+9)/12))) + floor(275*mo/9) + dy + (hr -tz -dst+mn/60)/24 + 1721013.5
where tz - timezone
and dst =1 when Daylight Savings Time is on dst=0 otherwise.
d = JD - 2451545.0
LST = 280.46061837 + 360.985647366*d - LongitudeEastPositive
|Sep27-12, 10:18 PM||#5|
I knew the correct algorithm for calculating LST precisely, but didn't know how to do a "back of the napkin" or mental calculation for it. Just figured it out, so I figured I'd post it on the off chance someone looks up this thread needing it:
For Greenwich, their local celestial meridian is at 0h RA is at noon on the Spring Equinox (March 21). This also means it's at 12h RA at noon on the fall equinox. For ease of calculation you can also assume (although it's approx. 2 minutes off) that at midnight on these dates LST is approx. 12h and 0h, respectively.
Now, the sidereal day is about 23hrs, 56 minutes per solar 24hr day. This means that at solar midnight (or noon) your local sidereal time will be about 4 minutes later than it was on the previous day, at the same civil time.
You can use this (offset) * (number of days) to get an approximate Greenwich sidereal time, then offset for your latitude.
Right now, it's approx. 10pm (22:00) local civil time (5 hrs earlier than GT due to daylight savings). This means it's about 3:00 in Greenwich (universal time).
Today is Sept 27. (fall equinox was Sept. 23rd).
This means that at civil midnight in Greenwich today:
==> LST was about 0:00 + 4*(4 days) = 0:16
==> LST in Greenwich right now is approximately 0:16 + 3:00 = 3:16
==> LST in Austin right now (97.7 W or about 6.5 hrs W) = 3:16 - 6:30 = 20:44
Now, I made some assumptions here, and according to an online LST calcular, I'm about 9 minutes off. But again, I was just looking for the approximate LST. This estimation is generally good enough for the purpose of figuring out which objects will be observable of the course of the night, etc.
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