Converting RA/Dec into Alt/Az
McDowell / Darren John (COM)
Was wondering if there was a simple formula for converting RA and Dec
into Alt and Azimuth. This would be a great help at the 'scope during
observing since all star charts are in RA/Dec and I have a dobsonian
telescope.
Thanks in advance
Darren McDowell
t...@bga.com
Jim Robinson
>> "McDowell / Darren John (COM)" <u94...@student.canberra.edu.au> writes:
>> Hi all,
>>
>> Was wondering if there was a simple formula for converting RA and Dec
>> into Alt and Azimuth. This would be a great help at the 'scope during
>> observing since all star charts are in RA/Dec and I have a dobsonian
>> telescope.
>>
>> Thanks in advance
>>
>> Darren McDowell
>>
The formulas are fairly simple if you know the sidereal time. Here they
are straight from my programmable calculator
(1) compute the hour angle in degrees of the object you want to view:
H = ((LSH + LSM / 60) - (RAH + RAM / 60)) * 15
where LSH = the local sidereal hour
LSM = the sidereal minute
RAH = the right ascension hour
RAM = the right ascension minute
(2) compute the altitude of the object
ALT = invsin(sin (DEC) * sin(LAT) + cos(DEC) * cos(LAT) * cos(H))
where LAT is your latitude
(3) compute the azimuth of the object. The form given below is complicated
because my calculator does not allow logic branches.
AZ = 180 * (1 + sign(sin(H))) - sign(sin(H)) *
invcos((sin(DEC -
Jim Robinson
Jim Robinson
here's the rest.
The formulas are fairly simple if you know the sidereal time. Many
planetarium programs display the sidereal time. For a few hours
observing, it is adequate to set a regular clock to the local sidereal
time just before going out.
Here are the formulas straight from my programmable calculator:
(1) compute the hour angle in degrees of the object you want to view:
H = ((LSH + LSM / 60) - (RAH + RAM / 60)) * 15
where LSH = the local sidereal hour
LSM = the sidereal minute
RAH = the right ascension hour
RAM = the right ascension minute
(2) compute the altitude of the object
ALT = invsin(sin (DEC) * sin(LAT) + cos(DEC) * cos(LAT) * cos(H))
where DEC is the declination.
LAT is your latitude
(3) compute the azimuth of the object. The form given below looks needlessly
complicated because my calculator does not allow logic branches. I
offer it in this ugly form since me people have such calculators.
AZ = 180 * (1 + sign(sin(H))) - sign(sin(H)) *
invcos((sin(DEC) - sin(LAT) * sin(ALT)) /
cos(LAT) * cos(ALT))
where sign returns -1 if its argument is negative, 1 if positive. Basically,
the ugliness in the first line causes us to subtract the result of the
rest of the formula from 360 if sin(H) is positive.
I use these formulas with in combination with large az-alt setting circles to
get within a couple of degrees of the target object.
--
Jim Robinson
Robert Swarts
"McDowell / Darren John (COM)" <u94...@student.canberra.edu.au>
writes:
>
>Hi all,
>
>Was wondering if there was a simple formula for converting RA and Dec
>into Alt and Azimuth. This would be a great help at the 'scope during
>observing since all star charts are in RA/Dec and I have a dobsonian
>telescope.
>
Algorithms). There also are a number of one-shot conversion programs
available. See S&T web page software section, and the astro section at
any SIMTEL site.
Ramshorn also sells a continuously updating and speaking RA/Dec to
Alt/Az converter for $26.95 post paid. It includes a data base of the
Messier objects and several bright stars.
BS
Ramshorn at Yamhill -- home of the Azimouth and Julian Astro Utilities
Manny
If you send me a fax number, i'll send two programs for the conversion.
Manny
Bill Owen
> > Was wondering if there was a simple formula for converting RA and Dec
> > into Alt and Azimuth. This would be a great help at the 'scope during
> > observing since all star charts are in RA/Dec and I have a dobsonian
> > telescope.
The equations are straightforward. Let alpha be the R.A. and delta be the
declination of the object in question. Let phi be your (astronomical) latitude,
positive north, and let lambda be your longitude, positive EAST. Let theta be
the Greenwich sidereal time. Convert alpha and theta to degrees (i.e. multiply
decimal hours by 15.) Then:
Hour angle h = theta + lambda - alpha
Alt = arcsin (sin phi sin delta + cos phi cos delta cos h)
Az = atan2 (-cos delta sin h, cos phi sin delta - sin phi cos delta cos h)
where in the last equation atan2 is the C or Fortran four-quadrant inverse
tangent function, i.e. x = r cos theta and y = r sin theta implies that
theta = atan2 (y,x).
-- Bill Owen, w...@wansor.jpl.nasa.gov
P.S. I didn't mention above how to get the Greenwich sidereal time. First
get UT1 from WWV or such (the difference UT1-UTC is coded in the time ticks),
and find the number of seconds from 2000 Jan 01 12:00:00. This will be a
negative number for another four-plus years. Call this number S. The formula
for GST (measured in seconds) is
T = S / (36525*86400) (i.e. T is measured in centuries)
GST = 67310.54841 + 8640184.812866 T + 0.093104 T^2 - 0.0000062 T^3
Obviously you have to do this in double precision! There are ways of handling
that huge linear term and preserving precision; my favorite is to define
d = fractional part of (S/86400)
t = fractional part of T
and then replace the linear term by 86400 (d+t) + 184.812866 T.
Bill Owen
> (3) compute the azimuth of the object. The form given below is complicated
> because my calculator does not allow logic branches.
>
... the rest of the formula didn't make it. But the formula I posted earlier
is better for two reasons:
1. The inverse cosine function becomes difficult to handle computationally
when its argument approaches 1. A tiny change in the argument gives a huge
change in the result. Compare arccos(0.999990) with arccos(0.999991) and
see for yourself. The difference is 47 arc seconds.
2. There has to be extra logic involved in determining the quadrant of the
answer, since the arccos function is only defined over [0,180].
I show you a still more excellent way. And this works on your calculator just
as easily.
AZ = atan2 (-cos(DEC)*sin(H), cos(LAT)*sin(DEC)-sin(LAT)*cos(DEC)*cos(H)).
For a calculator such as my HP-15, put the expression before the comma into
the calculator's Y register, the expression after the comma into the X
register, and hit R->P to convert rectangular to polar. The "radius" on
output is the cosine of the altitude, and the polar angle is the desired
azimuth.
-- Bill Owen, w...@wansor.jpl.nasa.gov