关于RADEC-ALTAZ的算法
本帖最后由 马铃鼠 于 2015-3-4 17:43 编辑关于赤道坐标系和地平坐标系的转换问题，查了很多资料，在《天文爱好者手册》和《天文学手册》中都只有类似的时角坐标和地平坐标转换的计算公式，在这个网站（http://www.stargazing.net/kepler/altaz.html）中有具体的算法，但是没怎么看懂。。。
偶然间发现了这里，貌似是NASA的很高大上的东西：The IDL Astronomy User's Library
http://idlastro.gsfc.nasa.gov/contents.html
http://rpackages.ianhowson.com/cran/astrolibR/man/altaz2hadec.html
这里的源码算法涉及了天文中很多方面的问题，其中有HADEC2ALTAZ Convert Hour Angle and Declination to Horizon (alt-az) coordinates（http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro）
虽然还不是我做需要的，但还毕竟是现成的算法，这里的代码看着像是VB，但好像又不是，貌似是什么IDL语言，总之是都没学过的。。。
在此特地复制粘贴过来，求大神指教。
[*]PRO hadec2altaz, ha, dec, lat, alt, az, WS=WS
[*]
[*];+
[*];NAME:
[*]; HADEC2ALTAZ
[*];PURPOSE:
[*]; Converts Hour Angle and Declination to Horizon (alt-az) coordinates.
[*];EXPLANATION:
[*]; Can deal with NCP/SCP singularity. Intended mainly to be used by
[*]; program EQ2HOR
[*];
[*]; CALLING SEQUENCE:
[*]; HADEC2ALTAZ, ha, dec, lat ,alt ,az [ /WS ]
[*];
[*]; INPUTS
[*]; ha -the local apparent hour angle, in DEGREES, scalar or vector
[*]; dec -the local apparent declination, in DEGREES, scalar or vector
[*]; lat -the local latitude, in DEGREES, scalar or vector
[*];
[*]; OUTPUTS
[*]; alt - the local apparent altitude, in DEGREES.
[*]; az- the local apparent azimuth, in DEGREES, all results in double
[*]; precision
[*]; OPTIONAL KEYWORD INPUT:
[*]; /WS - Set this keyword for the output azimuth to be measured West from
[*]; South. The default is to measure azimuth East from North.
[*];
[*]; EXAMPLE:
[*]; What were the apparent altitude and azimuth of the sun when it transited
[*]; the local meridian at Pine Bluff Observatory (Lat=+43.07833 degrees) on
[*]; April 21, 2002? An object transits the local meridian at 0 hour angle.
[*]; Assume this will happen at roughly 1 PM local time (18:00 UTC).
[*];
[*]; IDL> jdcnv, 2002, 4, 21, 18., jd; get rough Julian date to determine
[*]; ;Sun ra, dec.
[*]; IDL> sunpos, jd, ra, dec
[*]; IDL> hadec2altaz, 0., dec, 43.078333, alt, az
[*];
[*]; ===> Altitude alt = 58.90
[*]; Azimuthaz = 180.0
[*]
[*]; REVISION HISTORY:
[*]; WrittenChris O'Dell Univ. of Wisconsin-Madison May 2002
[*];-
[*]
[*]if N_params() LT 4 then begin
[*] print,'Syntax - HADEC2ALTAZ, ha, dec, lat ,alt ,az [ /WS ]'
[*] return
[*]endif
[*]
[*]d2r = !dpi/180.
[*]
[*]sh = sin(ha*d2r) & ch = cos(ha*d2r)
[*]sd = sin(dec*d2r) & cd = cos(dec*d2r)
[*]sl = sin(lat*d2r) & cl = cos(lat*d2r)
[*]
[*]x = - ch * cd * sl + sd * cl
[*]y = - sh * cd
[*]z = ch * cd * cl + sd * sl
[*]r = sqrt(x^2 + y^2)
[*]; now get Alt, Az
[*]
[*]az = atan(y,x) /d2r
[*]alt = atan(z,r) / d2r
[*]
[*]; correct for negative AZ
[*]w = where(az LT 0)
[*]if w ne -1 then az = az + 360.
[*]
[*]; convert AZ to West from South, if desired
[*]if keyword_set(WS) then az = (az + 180.) mod 360.
[*]
[*]
[*]END
额，貌似找到了。。。
来源也挺高大上。。。美国海军天文台什么的。。。
http://www.slac.stanford.edu/grp/eg/minos/dist/minossoft/releases/development/AstroUtil/
是verilog 硬件描述语言。在FPGA之类的运行的。看着挺简单的，就是几个数据的换算，应该都能看懂。具体变量代表的含义上面也都写着的啊 可以通过方位-仰角转换到时角-赤纬，然后再依据恒星时转换到赤经赤纬两步走的办法。算法可以参考天球坐标系统 http://zh.wikipedia.org/wiki/天球坐标系统 天球坐标转换
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