python - 相对于行进的风向

Question

我正在尝试构建一个函数，以便在给定风向的情况下，我可以获得相对于行进方向的风向。我已经提供了一个长期这样做的例子。感觉臃肿和简陋。

我遇到了麻烦，因为风是基于矢量的，我需要找到一种方法来评估函数，同时考虑 360 度圆。

def relative_wind(wind_heading, rider_heading):

    if 320 < wind_heading < 60:
        wind_direction = 'north'
    elif 140 > wind_heading > 60:
        wind_direction = 'east'
    elif 230 > wind_heading > 140:
        wind_direction = 'south'
    else:
        wind_direction = 'west'

    if 320 < rider_heading < 60:
        rider_direction = 'north'
    elif 140 > rider_heading > 60:
        rider_direction = 'east'
    elif 230 > rider_heading > 140:
        rider_direction = 'south'
    else:
        rider_direction = 'west'

    if rider_direction == 'north':

        if wind_direction == 'north':
            relative_direction = 'tail'
        elif wind_direction == 'east':
            relative_direction = 'right'
        elif wind_direction == 'south':
            relative_direction = 'head'
        else:
            relative_direction = 'left'
        return relative_direction

    elif rider_direction == 'east':

        if wind_direction == 'north':
            relative_direction = 'left'
        elif wind_direction == 'east':
            relative_direction = 'tail'
        elif wind_direction == 'south':
            relative_direction = 'right'
        else:
            relative_direction = 'head'
        return relative_direction

    elif rider_direction == 'west':

        if wind_direction == 'north':
            relative_direction = 'right'
        elif wind_direction == 'east':
            relative_direction = 'head'
        elif wind_direction == 'south':
            relative_direction = 'left'
        else:
            relative_direction = 'tail'
        return relative_direction

    else:

        if wind_direction == 'north':
            relative_direction = 'head'
        elif wind_direction == 'east':
            relative_direction = 'left'
        elif wind_direction == 'south':
            relative_direction = 'tail'
        else:
            relative_direction = 'right'
        return relative_direction

print relative_wind(285, 285)

Answer 1

1. 你选择的角度有点奇怪:

   

   从勾勒出您的角度实际应该是什么样子开始可能是个好主意；以轴为中心的象限边界为 45、135、225 和 315 度。

2. 与其使用硬编码的航向，我建议您创建一个函数来获取您的航向和风向并返回相对风向:

   def relative_wind_direction(heading, wind_dir):
       """
       Return wind direction relative to plane heading, in [-180..180) degrees
       """
       return ((wind_dir - heading + 180) % 360) - 180
   

   正如@sjy 所建议的，这使用 % mod 运算符将结果固定在所需范围内。

3. 您现在可以编写如下函数

   def wind_aspect(heading, wind_dir):
       rel_dir = relative_wind_direction(heading, wind_dir)
   
       if rel_dir < -135:
           return "head"
       elif rel_dir < -45:
           return "right"
       elif rel_dir < 45:
           return "tail"
       elif rel_dir < 135:
           return "left"
       else:
           return "head"