比较两种罗盘方位

Question

我有两个指南针方位(0-360度)：

var routeDirection = 28
var windDirection = 289

我需要比较这些轴承，以确定骑手是否会得到一个

( a)横风

( b)尾风

( c)迎风

我试着把方位转换成指南针的方向。

  var compass = ['N', 'NNE', 'NE', 'ENE', 'E', 'ESE', 'SE', 'SSE', 'S', 'SSW', 'SW', 'WSW', 'W', 'WNW', 'NW', 'NNW', 'N'];
  var windDirection = compass[Math.round(bearing / 22.5)];

然后进行基本的字符串比较：

if (routeDirection=='N') && (windDirection=='S') {
 output = 'Headwind'
}

但很明显这既冗长又低效..。

Answer 1

我会直接比较路线方向和风向，并根据差异来确定风的类型：

if(routeDirection > windDirection)
    var difference = routeDirection - windDirection;
else
    var difference = windDirection - routeDirection;

// left/right not important, so only need parallel (0) or antiparallel (180)
difference = difference - 180;
//keep value positive for comparison check
if(difference < 0)
    difference = difference * -1;

if(difference <= 45) // wind going in roughly the same direction, up to you and your requirements
    output = "headwind";
elseif(difference <= 135) // cross wind
    output = "crosswind";
elseif (difference <= 180)
    output = "tailwind";
else
    output = "something has gone wrong with the calculation...";

上面的计算意味着你没有对每个指南针点进行比较，只对船的航向和风的相对差异进行比较，从而减少了细节。它还允许多角度比较，使用较小的度步骤和添加更多的其他条件。这也可以用switch()来完成，但是也会有类似的代码行计数。

Answer 2

如果你要直截了当的话，你就会有这个：

\ Head wind /
 \         /
  \       /
   \  |  /
    \ | /
Cross\|/Winds
     / \
    /   \
   /     \
  /       \
 /         \
/ Tail wind \

所以基本上..。首先，你旋转你的观点，这样你就可以在方位为0的情况下旅行：

var adjustedWindDirection = windDirection - routeDirection;

当然，轴承应该在0-360的范围内，所以再调整一次：

adjustedWindDirection = (adjustedWindDirection+360)%360;

现在我们需要找出这个方向的哪个象限：

var quadrant = Math.round((adjustedWindDirection-45)/90);

最后：

var winds = ["head","cross (left)","tail","cross (right)"];
var resultingWind = winds[quadrant];

完成了！