Estimating Local Sidereal Time

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In summary, the conversation discusses the process of estimating Local Sidereal Time (LST) in astronomy. The formula for calculating LST is mentioned, but the individual is looking for a rough estimate that can be done mentally. They explain how to do this by using the sidereal day and offsetting for latitude. The example given shows how to estimate LST for a specific date and location.
  • #1
bossman27
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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?

Thanks!
 
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  • #2
bossman27 said:
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.

97.7/15 is more like 6.5 hours.

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.

Greenwich is 0 degrees longitude.

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. don't know 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.
 
  • #3
rbj said:
97.7/15 is more like 6.5 hours.

Greenwich is 0 degrees longitude.

Right. The adjustment is 6.5 hours, I just wrote 5 on accident because during daylight savings our clocks are 5 instead of 6 hours earlier than Greenwhich, but obviously that doesn't affect longitude. Latitude was a misstatement on my part.

rbj said:
60 minutes per 15 degrees of longitude is something i understand. don't know 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.)

I believe I did say it's about 4 minutes or one degree.

rbj said:
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.

Interesting question! I've always thought that as well, but Wikipedia says that both of those are possible reasons for using 360 degrees, although apparently no one knows the exact origin.http://en.wikipedia.org/wiki/Degree_(angle)#History
 
  • #4
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
 
  • #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.

An example:
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.
 
Last edited:

1. What is Local Sidereal Time?

Local Sidereal Time (LST) is a measure of the Earth's rotation and is used to determine the position of stars in the sky at a specific location and time. It is based on the Earth's rotation relative to the fixed stars, rather than relative to the Sun like solar time.

2. How is Local Sidereal Time calculated?

LST can be calculated using the formula LST = GST + Longitude / 15, where GST is Greenwich Sidereal Time and Longitude is the observer's longitude. Alternatively, it can be calculated using astronomical software or by referencing an LST table.

3. Why is Local Sidereal Time important for astronomers?

Astronomers use LST to determine the position of stars in the sky and to coordinate observations with other astronomers globally. It is also necessary for celestial navigation and for calibrating telescopes and other instruments.

4. How does Local Sidereal Time differ from Universal Time?

Local Sidereal Time is based on the Earth's rotation relative to the fixed stars, while Universal Time is based on the Earth's rotation relative to the Sun. Therefore, LST and Universal Time can differ by up to a few minutes, depending on the observer's location.

5. Can Local Sidereal Time be converted to other time systems?

Yes, LST can be converted to other time systems, such as Coordinated Universal Time (UTC) or Local Mean Time (LMT), by adding or subtracting the observer's longitude or using conversion formulas. However, this conversion may introduce slight errors due to the Earth's irregular rotation.

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