小明接到任务后就一直守在水缸旁边，时间长就觉得无聊，就跑到房里看小说了，每30分钟来检查一次水面高度。水漏得太快，每次小明来检查时，水都快漏完了，离要求的高度相差很远，小明改为每3分钟来检查一次，结果每次来水都没怎么漏，不需要加水，来得太频繁做的是无用功。几次试验后，确定每10分钟来检查一次。这个检查时间就称为采样周期

      开始小明用瓢加水，水龙头离水缸有十几米的距离，经常要跑好几趟才加够水，于是小明又改为用桶加，一加就是一桶，跑的次数少了，加水的速度也快了，但好几次将缸给加溢出了，不小心弄湿了几次鞋，小明又动脑筋，我不用瓢也不用桶，老子用盆，几次下来，发现刚刚好，不用跑太多次，也不会让水溢出。这个加水工具的大小就称为比例系数

      小明又发现水虽然不会加过量溢出了，有时会高过要求位置比较多，还是有打湿鞋的危险。他又想了个办法，在水缸上装一个漏斗，每次加水不直接倒进水缸，而是倒进漏斗让它慢慢加。这样溢出的问题解决了，但加水的速度又慢了，有时还赶不上漏水的速度。于是他试着变换不同大小口径的漏斗来控制加水的速度，最后终于找到了满意的漏斗。漏斗的时间就称为积分时间

      小明终于喘了一口，但任务的要求突然严了，水位控制的及时性要求大大提高，一旦水位过低，必须立即将水加到要求位置，而且不能高出太多，否则不给工钱。小明又为难了！于是他又开努脑筋，终于让它想到一个办法，常放一盆备用水在旁边，一发现水位低了，不经过漏斗就是一盆水下去，这样及时性是保证了，但水位有时会高多了。他又在要求水面位置上面一点将水凿一孔，再接一根管子到下面的备用桶里这样多出的水会从上面的孔里漏出来。这个水漏出的快慢就称为微分时间

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:

1. Their positions
2. 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:

1. Reference star: The star that you'll be "traveling to" and calculating the positions of other stars from. Its Cartesian coordinates are x_R, y_R and z_R.
2. Target Star: Any star you're looking at from the reference star. Its Cartesian coordinates are x_T, y_T, and z_T.
3. 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:

1. x_R = -0.017 pc
2. y_R = -1.815 pc
3. z_R = +0.148 pc

Let's figure out where Proxima Centauri would be. Its coordinates (from the Calculating Stellar Positions page) are:

1. x_T = -0.472 pc
2. y_T = -0.361 pc
3. z_t = -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:

1. x'_T = -0.472 - (-0.017) = -0.455 pc
2. y'_T = -0.361 - (-1.815) = +1.454 pc
3. 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:

1. Azimuth 
2. Altitude 
3. 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:

1. Calculate two distances:
   1. r = sqrt(x'_T² + y'_T² + z'_T²)
   2. r_xy = sqrt ( x'_T² +
      y'_T²)
2. Apply two formulas:
   1.  = tan^-1 (z'_T / r_xy);
   2.  = 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:

1. x'_T = -0.472 - (-0.017) = -0.455 pc
2. y'_T = -0.361 - (-1.815) = +1.454 pc
3. z'_t = -1.151 - (0.148) = -1.299 pc

• r = sqrt( (-0.455)² + (1.454)² + (-1.299)²)
• r = 2.002 pc.
• r_xy = sqrt( (-0.455)² + (1.454)²)
• r_xy = 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 M_V:

1. Earth-based apparent magnitude: V' = V - 5 log₁₀ (d/r)
2. Absolute magnitude: V' = M_V - 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:

1. x'_T = -0.472 - (-0.017) = -0.455 pc
2. y'_T = -0.361 - (-1.815) = +1.454 pc
3. 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.