| Re: circles on the sky [message #65871] |
Sun, 29 March 2009 16:18  |
wlandsman
Messages: 743
Registered: June 2000
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Senior Member |
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On Mar 27, 1:53 pm, Christopher Thom <ct...@oddjob.uchicago.edu>
wrote:
> Hi all,
>
> I'm drawing some points on a sky map. The map is <0.5deg across, so I
> thought a flat approximation would be ok...this may not be true. Around my
> central point, I draw a circle using a flat geometry relation [x =
> x0+r*cos(theta); y = y0 + r*sin(theta)], but i see a point outside the
> circle that I expect to be inside.
>
> I expect this point to be inside the circle, because the radius of the
> circle (in arcsec) is *greater* than the great-circle angular distance
> from the centre of the circle to the point.
>
> So...I'm thinking that my flat-geometry assumption is false. My question:
> can anyone point me towards forumlae/code that will calculate this circle
> on the sky (i.e. all points which have a fixed great-circle distance from
> the centre)? I'm using the astro library gcirc.pro to calculte my
> great-circle angular distances...I kind of want the "inverse" of that
> routine, I guess.
>
> Or...is there a better way to do it? [Or maybe my bug is elsewhere?]
>
> cheers
> chris
One thing that is unclear in this question is whether you are talking
about the surface of a sphere (in which case spherical trig formulae
like in gcirc.pro are appropriate) or a projection onto a flat map.
You begin by saying that you are drawing points on a map. In that
case you need to know what projection you are using to create the flat
map (e.g. gnomomic? Mercator?). Once you specify the projection
(e.g. with MAP_INIT) then you can use MAP_PROJ_FORWARD /
MAP_PROJ_INVERSE to convert between X,Y and spherical coordinates.
( In astronomy you would use the world coordinate system routines
wcssph2xy / wcsxy2sph ).
--Wayne
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| Re: circles on the sky [message #65874 is a reply to message #65871] |
Sat, 28 March 2009 08:54  |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Mar 27, 1:53 pm, Christopher Thom <ct...@oddjob.uchicago.edu>
wrote:
> Hi all,
>
> I'm drawing some points on a sky map. The map is <0.5deg across, so I
> thought a flat approximation would be ok...this may not be true. Around my
> central point, I draw a circle using a flat geometry relation [x =
> x0+r*cos(theta); y = y0 + r*sin(theta)], but i see a point outside the
> circle that I expect to be inside.
>
> I expect this point to be inside the circle, because the radius of the
> circle (in arcsec) is *greater* than the great-circle angular distance
> from the centre of the circle to the point.
>
> So...I'm thinking that my flat-geometry assumption is false. My question:
> can anyone point me towards forumlae/code that will calculate this circle
> on the sky (i.e. all points which have a fixed great-circle distance from
> the centre)? I'm using the astro library gcirc.pro to calculte my
> great-circle angular distances...I kind of want the "inverse" of that
> routine, I guess.
>
> Or...is there a better way to do it? [Or maybe my bug is elsewhere?]
>
> cheers
> chris
Would something like this work? (all quantities in radians)
phi = 2.*!pi*findgen(nphi)/(nphi-1)
x = radius*cos(phi)
y = radius*sin(phi)
dec = dec_center + y
RA = RA_center + asin(sin(radius)/cos(dec))
(totally untested, but that's what I get from spherical trig...)
-Jeremy.
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| Re: circles on the sky [message #65943 is a reply to message #65871] |
Tue, 31 March 2009 12:09 |
Christopher Thom
Messages: 66
Registered: October 2006
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Member |
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Quoth wlandsman:
> On Mar 27, 1:53 pm, Christopher Thom <ct...@oddjob.uchicago.edu>
> wrote:
>> Hi all,
>>
>> I'm drawing some points on a sky map. The map is <0.5deg across, so I
>> thought a flat approximation would be ok...this may not be true. Around my
>> central point, I draw a circle using a flat geometry relation [x =
>> x0+r*cos(theta); y = y0 + r*sin(theta)], but i see a point outside the
>> circle that I expect to be inside.
>>
>> I expect this point to be inside the circle, because the radius of the
>> circle (in arcsec) is *greater* than the great-circle angular distance
>> from the centre of the circle to the point.
>>
>> So...I'm thinking that my flat-geometry assumption is false. My question:
>> can anyone point me towards forumlae/code that will calculate this circle
>> on the sky (i.e. all points which have a fixed great-circle distance from
>> the centre)? I'm using the astro library gcirc.pro to calculte my
>> great-circle angular distances...I kind of want the "inverse" of that
>> routine, I guess.
>>
>> Or...is there a better way to do it? [Or maybe my bug is elsewhere?]
>>
>> cheers
>> chris
>
> One thing that is unclear in this question is whether you are talking
> about the surface of a sphere (in which case spherical trig formulae
> like in gcirc.pro are appropriate) or a projection onto a flat map.
> You begin by saying that you are drawing points on a map. In that
> case you need to know what projection you are using to create the flat
> map (e.g. gnomomic? Mercator?). Once you specify the projection
> (e.g. with MAP_INIT) then you can use MAP_PROJ_FORWARD /
> MAP_PROJ_INVERSE to convert between X,Y and spherical coordinates.
> ( In astronomy you would use the world coordinate system routines
> wcssph2xy / wcsxy2sph ).
I think i was pretty unclear about what i was trying to accomplish, and
included some confusing details. So, a condensed question, on what I'm
actually trying to accomplish:
Given a co-ordinate position (ra/dec or lat/long), a direction (e.g an
angle east of north, for instance), and a great circle angular distance,
how do I compute the coordinate of the final position?
cheers
chris
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