电子邮件:flashindream[AT]gmail[DOT]com
略懂Flash的天文爱好者
Calculating Stellar Positions
Post by indream, 2010-8-27, Views:Coordinate translation
Now you're ready to calculate, for example, what the night sky from Sirius or Epsilon Eridani looks like. To do this, you'll need to know two things about the stars you'll be considering:
- Their positions
- Their brightnesses
Let's tackle the positions first.
Positions
Step 1: Get Cartesian coordinates for the stars
First, I'll introduce a little terminology that should help keep some ideas straight:
- Reference star: The star that you'll be "traveling to" and calculating the positions of other stars from. Its Cartesian coordinates are xR, yR and zR.
- Target Star: Any star you're looking at from the reference star. Its Cartesian coordinates are xT, yT, and zT.
- Transformed Target Coordinates: The coordinates of the target star as seen from the reference rather than from Earth. They are x'T, y'T, and z'T.
Suppose you're interested in figuring out what the night sky looks like from Barnard's Star, the second closest star to the Sun. The reference coordinates are therefore those of Barnard's Star, which has
= 17.9636 hours,
= 4.668 degrees, and d = 1.82 pc:
- xR = -0.017 pc
- yR = -1.815 pc
- zR = +0.148 pc
Let's figure out where Proxima Centauri would be. Its coordinates (from the Calculating Stellar Positions page) are:
- xT = -0.472 pc
- yT = -0.361 pc
- zt = -1.151 pc
Step 2: Figure out the transformed target coordinates
This is easier than it might seem! To find out the transformed target coordinates, as seen from the reference star, just subtract the reference's coordinates from the target's coordinates.
For Proxima Centauri:
- x'T = -0.472 - (-0.017) = -0.455 pc
- y'T = -0.361 - (-1.815) = +1.454 pc
- z't = -1.151 - (0.148) = -1.299 pc
Step 3: If desired, convert the transformed Cartesian
coordinates to spherical coordinates
The transformed target coordinates may be all you need if you have a computer program that'll plot the stars. If you want to draw star charts by hand, especially if you want to use common map projections, it will be useful to convert the Cartesian coordinates to something else.
The most useful transformation is to something called general spherical coordinates, which are very similar to equatorial coordinates. In fact, they're essentially identical -- the main reason for using them instead of equatorial coordinates is to avoid confusion between Earth-based measurements (which use the equatorial system) and calculations like these which apply far from
Earth. The spherical coordinates are:
- Azimuth

- Altitude

- Distance r
is like right ascension: it measures the angle around the Cartesian coordinate system, starting at the x-axis, as seen from the reference star.
is like declination: it measures the angle "up" or "down". r is the target star's distance from the reference.
To transform Cartesian coordinates to spherical coordinates:
- Calculate two distances:
- r = sqrt(x'T2 + y'T2 + z'T2)
- rxy = sqrt ( x'T2 +
y'T2)
- Apply two formulas:
= tan-1 (z'T / rxy);
= tan-1 (y'T / x'T)
This gives two angles similar to right ascension and declination (or, for that matter, longitude and latitude), which is what many map projections will call for. If you're using a computer rather than a hand calculator, the inverse tangent function will probably return a value in radians rather than degrees -- you may want to convert. You will use the distance r to get the star's brightness later.
Make sure the angle
is in the correct quadrant. If you are using the standard ATAN function on most calculators and computers,
will be calculated correctly only if x'T is positive. If it's negative, you'll generally have to add 180 degrees to the result to get the correct angle. Also, if x'T is positive and y'T is negative, ATAN will calculate a negative value for the angle; in that case, add 360 to bring it into the range 0 - 360 degrees. If you are using a calculator or a programming language with an ATAN2 function, use it instead:
= atan2 (y'T, x'T)
ATAN2 automatically puts the angle into the correct quadrant, in the range 0 - 360 degrees.
As an example: let's find the general spherical coordinates for Proxima Centauri as seen from Barnard's Star. Recall its transformed target coordinates:
- x'T = -0.472 - (-0.017) = -0.455 pc
- y'T = -0.361 - (-1.815) = +1.454 pc
- z't = -1.151 - (0.148) = -1.299 pc
- r = sqrt( (-0.455)2 + (1.454)2 + (-1.299)2)
- r = 2.002 pc.
- rxy = sqrt( (-0.455)2 + (1.454)2)
- rxy = 1.524 pc.
= tan-1 (-1.299 pc / 1.524 pc);
= -40.44 degrees.
= tan-1 (1.454 pc / -0.455 pc);
= -72.62 degrees.
Ugh. x'T is negative and the angle is outside the
usual range 0 - 360. Add 180 to bring
to the correct quadrant:
= 107.37 degrees.
Brightnesses
You now have the position of a star, in two different types of coordinates. The next step is the brightness.
Step 1: Get the distance to the target star from the
reference
If you calculated spherical coordinates, you have the distance, r, already. If not, calculate it:
- r = sqrt(x'T2 + y'T2 + z'T2)
Step 2: Find the star's apparent magnitude
As seen from Earth, a target star has an apparent magnitude, V. From a different reference star, the target star will have a different apparent magnitude, V'. Once you know the star's distance from the reference, r, as well as its distance from the Sun, d, you can use either of two formulas, depending on whether you know the target's apparent magnitude V (as seen from Earth), or its absolute magnitude MV:
- Earth-based apparent magnitude: V' = V - 5 log10 (d/r)
- Absolute magnitude: V' = MV - 5 log10 (10/r)
As an example, what is Proxima Centauri's apparent magnitude from Barnard's Star? From Earth, V = +11.01 and d = 1.29 pc, and the transformed coordinates of Proxima Centauri are:
- x'T = -0.472 - (-0.017) = -0.455 pc
- y'T = -0.361 - (-1.815) = +1.454 pc
- z't = -1.151 - (0.148) = -1.299 pc
- r = sqrt( (-0.455)2 + (1.454)2 +
(-1.299)2) - r = 2.002 pc.
- V' = +11.01 - 5 log10 (1.29/2.002)
- V' = +11.96
Proxima Centauri is dimmer than it is from Earth.
天荒坪测试之旅
Post by indream, 2010-8-16, Views:在天荒坪的三个夜晚测试自动导星,天气很理想,测试结果不理想.拍出的照片很多都有拉线,暗场也懒得拍.3天测试发现很多问题,整个系统的改善现在也有了目标
火焰星云&马头星云
礁湖星云&三叶星云
Ω星云
哑铃星云
猎户座大星云
昴星团
双星团
结果不太理想
Post by indream, 2010-8-1, Views:图像插值走了不少弯路,虽然昨天就正确获得了插值后的图像(将8x8插值到44x44大小进行星点定位),导星的结果还是有较大拖线.今天反复调整控制命令,最后还是使用bang-bang控制获取到下图的结果.RA轴仍旧有拖线,实际是修正的来回摆动.
赤道仪:LXD75
控制器:Autostar
主镜:80ED/F7
导星镜:200mm/F3.3
曝光时间:600秒@ISO100(图像中部100%截取)

终于上ED了
Post by indream, 2010-6-30, Views:
收了一支二手英田80ED,终于配齐了一套全二手班.
Hot Articles
Friend Connect
New Articles
Archives
- 2010 August (5)
- 2010 June (7)
- 2010 May (10)
- 2010 April (2)
- 2010 March (4)
- 2010 February (3)
- 2010 January (3)
- 2009 December (2)
- 2009 November (7)
- 2009 October (8)
- 2009 September (6)
- 2009 August (11)
- 2009 July (9)
- 2009 June (2)
- 2009 May (9)
- 2009 April (8)
- 2009 March (6)
- 2009 January (2)
- 2008 December (6)
- 2008 November (4)
- 2008 October (1)
- 2008 August (3)
- 2008 July (2)
- 2008 June (9)
- 2008 May (8)
- 2008 April (20)
- 2008 March (4)
- 2008 January (1)
- 2007 December (3)
- 2007 November (7)
- 2007 October (6)
- 2007 September (1)
- 2007 August (10)
- 2007 July (11)
- 2007 June (5)
- 2007 April (4)
- 2007 March (10)
- 2007 February (4)
- 2007 January (10)
- 2006 December (17)
- 2006 November (11)
- 2006 October (6)
- 2006 September (19)
- 2006 August (22)
Favorite
Links
Statistics
- 文章总数:312
- 评论总数:194
- 引用总数:0
- 浏览总数:1747
- 留言总数:1
- 当前主题:sean
- 当前样式:Sean_Sim
