以度为单位的Python haversine公式已经过时了

这是我所有的python代码，距离返回正确的距离还很远。我把哈弗森公式拆开，知道它在C的某个地方出错了。C是一个太大的数字，不允许D返回正确的距离

from math import sin, cos, atan2, sqrt, pi

First are my functions then my main part of the code
#-----FUNCTIONS------

#Header function
def Header():
    print("This program will calculate the distance between two geographic points!")

#get_location function
def Get_location():
    userLat = input("\n\n Please enter the latitude of your location in decimal degrees: ")
    userLon = input("Enter the longitude of the location in decimal degrees: ")
    return (userLat, userLon)


#Calculate distance function
#def Distance(lat1, lon1, lat2, lon2):
def Distance(location1, location2):
    radEarth = 6371 #km
    #location1 = Get_location()
    #location2 = Get_location()

    lat1 = location1[0]
    lon1 = location1[1]
    lat2 = location2[0]
    lon2 = location2[1]

    B = sin((lat1-lat2)/2)**2
    S = sin((lon1-lon2)/2)**2
    F = (cos(lat1))

    A = B + (F * (cos(lat2)) * S)


    C = 2 * (atan2(sqrt(A),sqrt(1-A)) * (180/pi))
    print(C)

    D = radEarth * C

    return D                                                       

This is the main part of my program

#-------MAIN---------

#Call header function
Header()

在用户继续时开始执行另一个循环：

doAnother = 'y'
while doAnother == 'y':

    #Collect location points from user
    location1 = Get_location()
    location2 = Get_location()
    print(location1)
    print(location2)
    #Calculate distance between locations
    distance = Distance(location1, location2)

    print('The distance between your two locations is: ' + str(distance))


    doAnother = raw_input('Do another (y/n)?'.lower())

#Display goodbye
print('Goodbye!')

---

看起来您正在实施所述的哈弗森公式。顺便说一句，我必须做同样的事情。你是对的，C中有一个问题

您的Python代码：

C = 2 * (atan2(sqrt(A),sqrt(1-A)) * (180/pi))

来自上面Javascript的URL的代码：

var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

问题是，你们用180/pi将C转换成度，但下一个计算D=radEarth*C只有在C是弧度时才有数学意义