Calculation of spherical distance between two points of spherical
In fact, this is a very simple question. Today, the reason why he raises him and makes a solution, because when the project is discussed yesterday, the project LEAD proposed that the spherical distance of two points on the earth spherical is difficult, actually It is a very simple three-dimensional geometry calculation. On the other hand, the right is to practice the C / C language and exercise your design ability.
1. submit questions
The two points of the spherical surface are known. For the convenience, the location of the unique identification point in latitude and longitude is required (related concepts, please refer to the relevant preparation knowledge), requiring their spherical distance.
2. Related preparatory knowledge
The preparation knowledge mentioned here is the related knowledge of the earth, such as shape and size, latitude, and longitude.
(1) Shape and size:
The global shape is an irregular sphere that is slightly flat. The Earth has an average radius of 6371 kilometers, 6378 kilometers of equatorial radiors, with a very radius of 6357 kilometers. The equator is about 40,000 meters long.
(2) Weft and latitude, warp and longitude
1 weft: weft is round, also known as latitude coils, lengths in length. The equator is the longest, gradually shortened by the equator to the two, and finally become a point. Weft latitude indicates something direction.
2 latitude: equator is zero latitude. The latitude of the Equator is called north latitude, using the "N" as a sample; the latitude of the equator is called South latitude, using the "S". North latitude, there are 90 ° in the south latitude.
3 Wide line: also called the meridian. The warp is a semicircle, all of the wires, etc. Line instructions in the North China direction.
4 longitude: Zero warping line is called this junior meridian. From the midnapthrics of the initiator, 180 degrees from the west, 180 ° in the east, with the "E" as a "E"; 180 ° in the west, with the "W".
Things 180 ° is connected to a warp.
Use 20 ° W and 160 ° e to divide the earth into east, west half a half.
3. solution
As shown in the figure above, first assume that the radius of the ball is R, the given 2 points are two points, first assume that A is in the northern hemisphere, b in the south hemisphere. This is only one of the cases, as in other cases, the same method can be calculated, only the same size is small. Of course, there is still a special case, you can't forget it.
It is assumed that the ball is point O, then the finally obtained ∠AOB radius is multiplied by the radius r of the ball.
The straight line of the pool and the pole of the Northern Pole is L, and the vertical lines of the L, the vertical points are d, c.
The parallel line of the Over Point C line BD, the B as the parallel line of the CD, these two parallel lines must intersect, the set point is E, which is easy to prove that BDCE is a rectangle.
Due to the latitude latitude latitude of the A and B, ∠OBD and ∠OAC are also known, is β, α, respectively, since radius r is known, so | bd | = r * COSβ, | AC | = r * COSα, | Od | = r * sinβ, | o | = r * sinα.
Since the length of points A, B is known, it is not difficult to find the value of the ∠ACE. Therefore, it is not difficult to use the cosine therapy in the triangular ACE to find the value of | AE |.
In the right triangular ABE, the value of AB is easily obtained. At this time, the triangular AOB is known, so the ∠aob can also be obtained with a cosine theorem, so that the spherical distance of AB is also solved.
Oh, the three-dimensional geometry of the semester, now it is still a little effort. Originally, I wanted to write the code today. Forget it, I'm so late, I will write it tomorrow.
Sleep, suddenly! (To be commitUed)
