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Converting RA/Dec to earth centered coordinates?



 
 
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  #11  
Old May 6th 11, 07:23 PM posted to sci.astro,sci.astro.amateur
Chris.B[_2_]
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Posts: 2,410
Default Converting RA/Dec to earth centered coordinates? (Correction)

How many times will this verbose boor receive a polite response to his
deliberate idiocy? 1000 times. 10,000 times? Do you value your time,
or ours, so cheaply?

Forget the telescope. The deliberate cretin does not believe in
telescopes. Magnification is the work of the devil. He made that up
too.

Tie the dunce firmly to a post with another tall, slender post set out
some distance before him. Place several accurate stop watches, with
seconds indication, just out of reach in front of him.

Point out a bright star accurately aligned with the distant post from
his fixed position. Now start the watches. Ensure he understands which
pattern the star belongs to so there is no confusion next time around,
and the next. Particularly when he is tired, bored, hungry and
thirsty. Martyrs are two a penny and no more valuable to anybody sane
on this earth.

Now leave him there for as long as it takes. He may be instantly
released, at any time, only on admission of his deliberate stupidity
and endless trolling. His freedom lies entirely in his own hands. No
further explanations are remotely necessary. Nor required. The stars
do not lie. They have not lied since classical times when the Earth's
rotation was judged to be 23:56:04 within the tolerance of their
instrumentation.

If you want to make it more entertaining you could offer food and
drink at 23:56:04 precisely. He has only 3 minutes and 50 seconds to
accept the meal but not a moment longer. Acceptance of the meal will
be an open admission of his own deliberate idiocy and his wasted years
of pathetic, infantile trolling. Refusal will mean another 23:56:04
tied to his dunce's post before the meal is offered again. He will
then have another 3 minutes and 50 seconds to accept or decline the
meal. Do the stars lie?

You can take a boor to knowledge but you cannot make him think.

http://en.wikipedia.org/wiki/Earth%27s_rotation
  #12  
Old May 6th 11, 07:55 PM posted to sci.astro,sci.astro.amateur
oriel36[_2_]
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Posts: 8,478
Default Converting RA/Dec to earth centered coordinates? (Correction)

On May 6, 5:59*am, oriel36 wrote:
On May 6, 1:34*am, Bill Owen wrote:


This absolutely blows me away that guys with NASA and Harvard monikers
attached to their posts are no better or worse than the other
nuisances ,it takes literally nothing to look at a telescope tracking
stellar circumpolar motion and conclude that the axis of the rotation
is not the telescope sweeping out the rotation of the Earth as it
turns around its mount but merely a homocentric convenience where the
number of returns of a star follow days and dates of the calendar
system.References for geocentricity don't even exist as the Earth's
dynamics are split into daily and annual cycles,normally referenced to
the central Sun as cause and effects,this business of dumping
everything into Right Ascension is generating horrifying results and
moreso in that your kind insist that it is the basis for both daily
and orbital dynamics.

Not enough intelligence to pull back,not enough courage to move in a
direction to set the matter straight,the result is ignominy for
everyone and make no mistake about it,despite the carefully
constructed language that operates on the basis of Ra/Dec reckoning,it
still amounts to a civilization ending conclusion that a 24 hour
rotation does not match a daylight/darkness cycle.You have no idea of
the implications attached to the attempt to create an imbalance
between primary cause and effect but I assure you that I do.



  #13  
Old May 6th 11, 10:06 PM posted to sci.astro
John Polasek
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Posts: 95
Default Converting RA/Dec to earth centered coordinates? (Correction)

On Mon, 02 May 2011 18:48:38 -0700, "W. eWatson"
wrote:

On 5/2/2011 2:28 PM, Steve Willner wrote:
In ,
"W. writes:
My xy plane is in the plane of the equator, and its projection into the
sky represents the celestial equator. Declination is measured +/- from
the celestial equator to each pole along great circle lines that pass
through each pole.

Somewhere on the celestial equator is point from where RA is measured.


To go from RA/Dec to an x,y,z unit vector is just simple trignometry:

x = cos(dec)*cos(RA)
y = cos(dec)*sin(RA)
z = sin (dec)

Make sure to convert RA/dec to degrees or radians (whatever units
your calculator or program uses). It's easy to forget to multiply
hours by 15 to get degrees.

The zero point of RA is the place where the ecliptic and celestial
equator intersect with the equinox heading north. Both this zero
point and the location of the celestial poles changes with respect to
the stars because of precession. If you want a _current_ x,y,z unit
vector, you need to start from current RA/Dec coordinates rather than
coordinates at a standard "ecliptic and equinox" (B1950 or J2000).
The Meeus book will tell you how to do that calculation.

Ah, the obliquity of the ecliptic (e) is what I need.


Don't see why you need that. Did you mean ecliptic coordinates
rather than celestial?

Whoops, I posed the question backwards. I have the x,y,z coordinates of
a vector and I want to convert them to ra/dec. In a unit sphere I think
of z as pointing through the north pole, x pointing south through 1,0,0,
and y pointing east through 0,1,0.

In my case, precession does not enter into matters. I'm constructing a
simulation that is mostly grounded in az/el and lat/lng. I wrote a
program that produces the path of a fake meteor moving in a straight
line. Time is not yet useful as a consideration yet.

The direction of the line points to the radiant point in the sky.
Meteors lie on a great circle, hence pass their plane passes through the
earth's center (spherical earth). My program has not yet needed ra/dec,
which is usually the measure of the radiant point in the sky and is
given in ra/dec. However,I have the data from a similar program, and
they provide the radiant as ra/dec. internally, my program seems sound
when I sort of run it backwards. I get agreeable results. I want to see
if the independent source and I agree.

You may never be happy with your equations as they stand because you
have chosen the X axis to point south and Y pointing east, so the
positive values of X will run backwards i.e. down.
You can still have a right-hand system with X pointing east and Y
pointing north.
Then envision a pitch over yaw gimbal set, like a theodolite, to do
your pointing and let Y be the "look vector". Now you can read the
declination D on the pitch axis and right ascension A on the yaw
axis, CCW positive. X and Y will then "lay out" normally.
This gimbal set P/Y is the only one that is congruent to the longitude
latitude geometry. The XYZ coordinates are the result of swinging the
"look" vector, y = (010), through D1 (declination on X axis) In
combination with swiveling R3 (right ascension on Z).
The validity of the physical result is ensured by the rigidity of the
gimbal set where the concept of temporal order is trivialized. It is
only in matrix multiplication that order is important. If we use row
vectors we get
U2D1A3 = V, U2 = (010)
or with column vectors
V = A3D1U2
The result I get is
X = - cosDsinA
Y = cosDcosA
Z = sinD
Thus you can invert for angles
A = atan(-x/y)
D = asin(z)
I am quite certain that you need to follow this scheme if you want to
stay out of trouble.
John Polasek

  #14  
Old May 7th 11, 02:29 AM posted to sci.astro
oriel36[_2_]
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Posts: 8,478
Default Converting RA/Dec to earth centered coordinates? (Correction)

On May 6, 11:06*pm, John Polasek wrote:

Then envision a pitch over yaw gimbal set, like a theodolite, to do
your pointing and let Y be the "look vector". Now you can read the
declination D *on the pitch axis and right ascension A *on the yaw
axis, CCW *positive. X and Y will then "lay out" normally.
This gimbal set P/Y is the only one that is congruent to the longitude
latitude geometry. The XYZ coordinates are the result of swinging the
"look" vector, *y = (010), through D1 (declination on X axis) In
combination with swiveling R3 (right ascension on Z).
The validity of the physical result is ensured by the rigidity of the
gimbal set where the concept of temporal order is trivialized.


How sweet,there is a good reason why I have yet to find individuals
capable of discussing the issue of planetary dynamics insofar as
empiricists insist on using the daily rotational characteristics to
define both daily and orbital characteristics off Right Ascension.When
I see the time lapse footage of Uranus where daily rotational
characteristics to the Sun run North/South and parallel with the
Equatorial rings,at least to the viewers perspective,while the planet
turns to the central Sun East/West on a separate travelling axis -

http://astro.berkeley.edu/~imke/Infr..._2001_2005.jpg

The necessity of two separate cycles,one where a planet turns to the
central Sun in terms of intrinsic rotation and variations in
latitudinal speeds and set off against the orbital turning to the
central Sun requires clear thinking.I know none of you can compete and
I suggest you do not bother,interpretative astronomy is at a
compositional level rather than a commentary level and however
colorful you write your descriptions to impress an audience,it is
charming to me for all the wrong reasons.


  #15  
Old May 7th 11, 10:14 PM posted to sci.astro,sci.astro.amateur
oriel36[_2_]
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Posts: 8,478
Default Converting RA/Dec to earth centered coordinates? (Correction)

Ultimately this is all forensics and no matter how interesting it may
be,sort of like stopping an intellectual runaway train,it doesn't
actually address the many positives that arise from disentangling the
Earth's daily and orbital dynamics from Right Ascension.If you all
would excuse me while I comment on some loose ends.

The arithmetical progression of 24 hour days forms the basis of the
calendar system hence the additional 24 hour rotational cycle of Feb
29th to square away the nearest proportion to a full 24 hour rotation
matching an orbital circuit calculated at 365 1/4 rotations per
circuit,in this case, 1461 rotations to 4 orbital circuits.This is
what makes Ra/Dec homocentric rather than geocentric as the equatorial
coordinate system using Ra/Dec maintains a steady progression of 24
hour days and rotations with the number of returns of a circumpolar
star match the number of rotations as days and dates within the
calendar system,people may fool themselves that there is a 3 minute 56
second divergence to 24 hours but this is mere trivia as all it takes
is the ability to count.

At one time I could say it amounted to people not doing their jobs but
presently I have taken a different view and really unfortunately it
would come down to criminal incompetence,it has to be considering the
braking mechanism for the number of rotations in an orbital circuit of
within context of the calendar system relies solely on cause and
effect,in this case ,24 hours of rotation for each corresponding
daylight/darkness cycle.

The bottom to all this was reached quite some time ago and
readers,indifferent or not,now wake up to astronomy as it should be
practiced and specifically in this thread where researchers are
really discussing homocentric ideologies rather than geocentric or to
put it in another way - only an astronomer would know the difference.

  #16  
Old May 8th 11, 04:14 AM posted to sci.astro
John Polasek
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Posts: 95
Default Converting RA/Dec to earth centered coordinates? (Correction)

On Mon, 02 May 2011 18:48:38 -0700, "W. eWatson"
wrote:

On 5/2/2011 2:28 PM, Steve Willner wrote:
In ,
"W. writes:
My xy plane is in the plane of the equator, and its projection into the
sky represents the celestial equator. Declination is measured +/- from
the celestial equator to each pole along great circle lines that pass
through each pole.

Somewhere on the celestial equator is point from where RA is measured.


To go from RA/Dec to an x,y,z unit vector is just simple trignometry:

x = cos(dec)*cos(RA)
y = cos(dec)*sin(RA)
z = sin (dec)

Make sure to convert RA/dec to degrees or radians (whatever units
your calculator or program uses). It's easy to forget to multiply
hours by 15 to get degrees.

The zero point of RA is the place where the ecliptic and celestial
equator intersect with the equinox heading north. Both this zero
point and the location of the celestial poles changes with respect to
the stars because of precession. If you want a _current_ x,y,z unit
vector, you need to start from current RA/Dec coordinates rather than
coordinates at a standard "ecliptic and equinox" (B1950 or J2000).
The Meeus book will tell you how to do that calculation.

Ah, the obliquity of the ecliptic (e) is what I need.


Don't see why you need that. Did you mean ecliptic coordinates
rather than celestial?

Whoops, I posed the question backwards. I have the x,y,z coordinates of
a vector and I want to convert them to ra/dec. In a unit sphere I think
of z as pointing through the north pole, x pointing south through 1,0,0,
and y pointing east through 0,1,0.

In my case, precession does not enter into matters. I'm constructing a
simulation that is mostly grounded in az/el and lat/lng. I wrote a
program that produces the path of a fake meteor moving in a straight
line. Time is not yet useful as a consideration yet.

The direction of the line points to the radiant point in the sky.
Meteors lie on a great circle, hence pass their plane passes through the
earth's center (spherical earth). My program has not yet needed ra/dec,
which is usually the measure of the radiant point in the sky and is
given in ra/dec. However,I have the data from a similar program, and
they provide the radiant as ra/dec. internally, my program seems sound
when I sort of run it backwards. I get agreeable results. I want to see
if the independent source and I agree.

See if this explanation regarding nomenclature will help clarify
things and perhaps bring an end to the random and inexplicable
outbursts of vituperation.
You have written a program that simply calculates the path of a meteor
in XYZ coordinates whose origin is in the equatorial plane. From the
XYZ coordinates you want to calculate the pitch and yaw or elevation
and azimuth as measured on an equatorial mount that would point at the
meteor.
You apparently raised the hackles of purists by using the terms right
ascension and declination which have specific astronomical meaning and
are measured from the point of Aries if I recall correctly, and
therefore has no place in your problem.
I think you will find that the equations I gave you will give you the
correct result.
John Polasek
 




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