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线上面的航点怎么计算呢,老大为什么不一步到位搞完呢 #9

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juny218 opened this issue May 30, 2024 · 4 comments

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@juny218
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juny218 commented May 30, 2024

如标题

@Char-Ten
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两点确定一条直线,你们的sdk还需要在线的中间打中继点吗?

@juny218
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juny218 commented May 31, 2024

是在,正射,或者倾斜射影需要密集拍摄照片

@juny218
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juny218 commented May 31, 2024

微信图片_20240531174909
要shi实现这种效果

@Char-Ten
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我们没有这种需求😂
我们就是洒农药,飞到哪洒到哪,不需要在中间打点。

不过对于你这种需求,我也不知道这些点之间的分布是否有要求。如果就按简单的按航线距离分布的话,你可以这样做:

  1. 通过cpRPA获取到航线的点集U
  2. U的起点开始,依次取两个点p1,p2,计算线段的角度deg
  3. 则第一个点NP1的坐标为:x = p1.x + d*cos(deg)y=p1.y+d*sin(deg),d是你的分布距离;
  4. 然后把p1换成NP1,重复上面的计算,可以得到第二个点NP2的坐标,依次类推,直到NPx超出这条线段

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