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The J2000 Epoch and RA

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#1 N5SE

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Posted 06 June 2020 - 05:08 PM

Hi all,

 

At the moment I'm trying to put together a little spreadsheet calculator. The idea is to roughly indicate the position of an object based on a viewing time and date.

 

To do this I need to work out how RA coordinates are is calculated from the J2000 epoch.

 

From what I understand RA is calculated from an equinox at a given point in time. Throughout the months of the year the object gains time behind the Sun and ends up in the same position behind Sun 12 months later.

 

What I need to know is from which equinox (if it is an equinox) is the RA from J2000 epoch calculated.

 

Is it the northern hemisphere vernal equinox (roughly March 21 2000)?

 

Or is it not an equinox and January 1st 2000 12:00 Terrestrial Time?

 

Slightly confused, if anyone could shed some light on this it would be appreciated.

 

-Sal.

 



#2 Paul Skee

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Posted 06 June 2020 - 05:21 PM

This isn't bad:

https://en.wikipedia...Right_ascension



#3 ButterFly

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Posted 06 June 2020 - 06:49 PM

The idea is to roughly indicate the position of an object based on a viewing time and date.

 

 

When you say "position of an object based on a viewing time and date", do you mean the object's current RA and Dec, or something else?

 

Based on your description, you may be confusing precessing J2000.0 to the present epoch with the much simpler calculation of "hour angle" (and then presumably onto Alt/Az).

 

If it really is precession you are looking for, consult the math at: Precessing positions from B1950 to J2000, and view the javascript source code of the following calculator: Astronomical Coordinate Calculator version 0.31.

 

BTW: you can run the above source code on pretty much any browser, even on a phone.


Edited by ButterFly, 06 June 2020 - 07:05 PM.


#4 catalogman

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Posted 06 June 2020 - 08:13 PM

All of the usual libraries do the calculation but the OP specifically asked for a spreadsheet.

 

To compute the change of mean place at equinox and epoch over 20 years, one method is to compute
the annual rates and then multiply by 20.

 

As an example, take the mean place B1980.0 for Arcturus and bring it up to J2000.0.

 

From the Astronomical Almanac, the B1980.0 positions are

 

RA0 = 14 14 44.9, DEC0 = +19 17 09

 

Then

 

RA0 = 15*(14 + 14/60 + 44.9/3600) = 213.6870 deg
DEC0 = 19 + 17/60 + 9/3600 = +19.2858 deg

 

Compute the times

 

t1 = 1980.0
t2 = 2000.0
dt = t2 - t1 = 20
t = (t1 + t2)/2 = 1990.0
T = (t - 1900.0)/100 = 0.90

 

the angles

 

e0 = 23.452294 - 0.0130125*T = 23.464005 deg

psi' = (50.3708 + 0.0050*T)/3600 = 50.3753/3600 deg

l' = (0.1247 - 0.0188*T)/3600 = 0.10778/3600 deg

 

and the annual precession rates

 

m = psi' * cos e0 - l' = 0.012806 deg

n = psi' * sin e0 = 0.0055717 deg

 

Apply precession over dt = 20 years to get

 

RA' = RA0 +(m + n*sin RA0 * tan DEC0)*dt = 213.9215 deg

DEC' = DEC0 + (n*cos RA0)*dt = +19.19308

 

This is the position at equinox 2000.0.

 

To apply proper motion, get the annual rates from SIMBAD

 

d(muRA) = -1093.39 mas/yr
d(muDEC) = -2000.06 mas/yr

 

Over dt = 20 yrs,

 

d(RA) = d(muRA)/3.6E6 * dt = -0.006074 deg
d(DEC) = d(muDEC)/3.6E6 * dt = -0.011111 deg

 

so the position corrected for proper motion is

 

RA = RA' + d(RA) = 213.9154 deg = 14.26103 h = 14 15 39.7
DEC = DEC' + d(DEC) = +19.18197 deg = +19 10 55

 

The value in SIMBAD rounds to RA = 14 15 39.7, DEC = +19 10 57.

This is the mean place at equinox and epoch 2000.0.

 

-- catalogman


Edited by catalogman, 06 June 2020 - 08:16 PM.


#5 Tony Flanders

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Posted 07 June 2020 - 06:08 AM

What I need to know is from which equinox (if it is an equinox) is the RA from J2000 epoch calculated.
 
Is it the northern hemisphere vernal equinox (roughly March 21 2000)?
 
Or is it not an equinox and January 1st 2000 12:00 Terrestrial Time?


J2000.0 is 12:00:00h, January 1, 2000 TT. Julian days run from noon to noon rather than midnight to midnight, which is handy for people interested in viewing the night sky.

As is so often the case, the Wikipedia article on the subject is excellent.



#6 Waddensky

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Posted 20 June 2020 - 07:59 AM

Don't confuse epoch and equinox, an epoch is a moment in time and the equinox defines the zero point of the coordinate system. The catalogues based on the Hipparcos data for example, list coordinates for epoch 1991.25 and equinox 2000.0. The zero point of J2000.0 is the vernal equinox (intersection between the equatorial plane and the ecliptic plane) on 1 January 2000, 12:00 TT (JD 2451545.0).

 

If you want to calculate the correct coordinates for another epoch in the same coordinate system, only apply corrections for proper motion of the star. If you want to express the coordinates in the epoch and equinox of the date, first correct for proper motion, then reduce the coordinates for precession (because the zero point has also changed as the Earth's axis has shifted). Be careful, most precession formulae (like the examples in Astronomical Algorithms by Jean Meeus) are valid only for a few centuries around J2000.0. More rigorous expressions, valid for long time intervals, can be found here.


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#7 Andrekp

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Posted 20 June 2020 - 10:29 AM

J2000.0 is 12:00:00h, January 1, 2000 TT. Julian days run from noon to noon rather than midnight to midnight, which is handy for people interested in viewing the night sky.

As is so often the case, the Wikipedia article on the subject is excellent.

could you explain exactly HOW This info is “handy for people interested in viewing the night sky?”  I’m a bit curious how 12 hours of Earth time makes a practical difference in the precision of the pole, or the celestial coordinates of a star?



#8 Tony Flanders

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Posted 20 June 2020 - 01:36 PM

could you explain exactly HOW This info is “handy for people interested in viewing the night sky?”  I’m a bit curious how 12 hours of Earth time makes a practical difference in the precision of the pole, or the celestial coordinates of a star?

Sorry, my grammar may have been twisted.

 

What I meant is that whereas conventional days, which run from midnight to midnight, are handy for most purposes, astronomers prefer Julian days, which run from noon to noon. That way, an entire night's observing happens on a single date.



#9 csrlice12

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Posted 20 June 2020 - 02:39 PM

Sorry, my grammar may have been twisted.

 

What I meant is that whereas conventional days, which run from midnight to midnight, are handy for most purposes, astronomers prefer Julian days, which run from noon to noon. That way, an entire night's observing happens on a single date.

So that's why a work day lasts forever.....



#10 Andrekp

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Posted 20 June 2020 - 04:52 PM

Sorry, my grammar may have been twisted.

 

What I meant is that whereas conventional days, which run from midnight to midnight, are handy for most purposes, astronomers prefer Julian days, which run from noon to noon. That way, an entire night's observing happens on a single date.

ok, that makes more sense.  Thanks for clarifying.

 

For the record, there are actually three different uses of “Julian” when it comes to dates:

1.  The Astronomers who use noon to noon as you mentioned.

2.  The ordinal dates of the year, which are midnight to midnight.

3.  A standardly written date, noted as being Julian so as to specifically differentiate it from the same Gregorian date.
 

most people, if they use them at all, use the second type.  The military uses the second type.  Probably the third type is used by historians and academics.  I’m not sure anyone outside of Astronomy uses the first type (to be honest, I’d never ever heard of that prior to you bringing it up.  It seems pretty impractical, outside of the stated use-case.).
 

I would be curious, if you know, why using a single date is really that much simpler across the astronomical spectrum?  Seems to be that a lot of “observing” of various types also is going on during the day, not to mention the fact that the connected world has made dates and times far less local than they once were.  It seems like you would still run into cases of dates being straddled, despite noon being the switchover.  And I’d imagine that this is a universal time?  Meaning that you’d still have the Undiminished  problem of date straddling in other parts of the world.  Maybe there is another reason it is used thusly?


Edited by Andrekp, 20 June 2020 - 05:34 PM.


#11 kathyastro

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Posted 20 June 2020 - 05:27 PM

I’m not sure anyone outside of Astronomy uses the third type

Just for completeness, various Orthodox churches still use the Julian calendar.


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#12 Waddensky

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Posted 20 June 2020 - 06:25 PM

And I’d imagine that this is a universal time?  Meaning that you’d still have the Undiminished  problem of date straddling in other parts of the world.  Maybe there is another reason it is used thusly?

The Julian Day (JD) and Julian Day Number (JDN, also from noon to noon) are in UTC or TT. the Chronological Julian Day (CJD) and Chronological Day Number (CJDN) are in local time. It's just an easy way to calculate the elapsed days between two events without taking month length, calendar types and leap years into account.

 

The Julian Day (named after the father of Joseph Scaliger) has nothing to do with the Julian Calendar (named after Gaius Iulius Caesar).


Edited by Waddensky, 20 June 2020 - 06:27 PM.

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#13 Andrekp

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Posted 20 June 2020 - 07:08 PM

The Julian Day (JD) and Julian Day Number (JDN, also from noon to noon) are in UTC or TT. the Chronological Julian Day (CJD) and Chronological Day Number (CJDN) are in local time. It's just an easy way to calculate the elapsed days between two events without taking month length, calendar types and leap years into account.

 

The Julian Day (named after the father of Joseph Scaliger) has nothing to do with the Julian Calendar (named after Gaius Iulius Caesar).

Yes, but it being in universal time suggests against it being especially useful to astronomers Because of anything they do overnight, which was what Tony suggested, but rather is just a convention for timing purposes, so everyone is on the same page.




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