我有两个罗盘轴承(0-360度):
var routeDirection = 28
var windDirection = 289
我需要比较这些方位以确定骑手是否会得到
a)横风
b)尾风 c)头风我尝试将轴承转换为指南针方向,例如:
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'
}
但显然这是冗长而低效的......
答案 0 :(得分:1)
假设你正在直接上升,你就有了这个:
\ Head wind /
\ /
\ /
\ | /
\ | /
Cross\|/Winds
/ \
/ \
/ \
/ \
/ \
/ Tail wind \
所以基本上......首先你转动你的视角,这样你就可以在0号轴上旅行:
var adjustedWindDirection = windDirection - routeDirection;
当然,轴承应该在0-360范围内,所以再次调整:
adjustedWindDirection = (adjustedWindDirection+360)%360;
现在我们需要找出方向所在的象限:
var quadrant = Math.round((adjustedWindDirection-45)/90);
最后:
var winds = ["head","cross (left)","tail","cross (right)"];
var resultingWind = winds[quadrant];
完成!
答案 1 :(得分:1)
我会直接比较路线方向和风向,并根据差异确定风的类型:
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...";
上述计算意味着您没有对每个罗盘点进行比较,只对船头和风的相对差异进行比较,从而减少了冗长度。它还允许通过使用较小程度的步骤和添加更多的elseif来进行多角度比较。这也可以通过switch()完成,但会出现类似的代码行数。