Geometry 垂直线（横向/纵向）

我试图实现所提出的关于在给定的垂直距离上找到一个点的方程。不幸的是，我收到的线不是直的。这是因为我将纵横坐标与常规x/y坐标混合，还是我做错了什么

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

现在已经更改了代码和变量名。仍在接收故障：-(

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

以下是变量值：

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

latDiff = -0.0046187639236450195
longDiff = -0.0058203935623168945

length = 0.0074303405980227239
uLat = -0.62160864131505777
uLong = -0.78332796263279647

newLat1 = 58.39273776863341
newLong1 = 15.558675147558933

newLat2 = 58.408404327886061
newLong2 = 15.546242974732632

---

更新：我得出的结论是，该故障是由于lat/long问题造成的。认为lat/long等同于正方形，而事实上并非如此，这似乎是合理的。尤其是在北欧工作时。

在您链接的公式中，最后4行有一个乘法序列作为：

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

[dy，dx，dy，dx]

=

[uLong，uLat，uLong，uLat]

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

无论您如何使用：

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

[dx，dy，dx，dy]

=

[uLat，uLong，uLat，uLong]

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

最后4行应该是：

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

double newLat1 = temp2.Latitude + (distPoint / 2) * uLong;   //dy appears first
double newLong1 = temp2.Longitude - (distPoint / 2) * uLat;  //then dx

double newLat2 = temp2.Latitude -  (distPoint / 2) * uLong;  
double newLong2 = temp2.Longitude + (distPoint / 2) * uLat;

---

在如此小的范围内，x/y和lat/long之间的差异是不相关的。您还做了其他错误；您应该做的是：

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + > longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLong;
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLat;

 double newLat2 = temp2.Latitude - (distPoint / 2) * uLong;
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLat;

---

也就是说，向量

（uLat，uLong）

是直线方向上的单位向量，因此垂直单位向量是

（uLong，-uLat）

-注意坐标交换的位置，除了一个被否定的位置。

这里的图像：较长的线应该是垂直的。对于我们这些视觉类型的人-谢谢你的图像！在这一点上，我认为图像可以通过注释哪些点（或它们的x-y值）变得更有用来自伪代码中的值，因为我不确定是否来自变量名。我认为您的问题不完整-或者结果未定义。您似乎在球面极坐标中指定了两个位置（但问题是2D，而不是3D？）。您似乎希望确定一个点的极坐标，该点与连接两个点的直线之间的距离为给定的距离。您没有指定球体的半径（这一点很重要，因为距离是给定的）。而且，通常有许多点（两条线的值）这是给定的直线距离。因此，您需要提供更多的解决方案条件。@svanerik：在您对Daniel Martin和LFSR Consulting的评论中，您提到他们的解决方案仍然不起作用。它怎么不起作用？您可以发布另一张图片吗？尽管CptSkippy的数字不正确，但我相信原则仍然有效相关：也就是说，我认为你必须考虑你的纬度（甚至比他的英国汉普郡的例子还要高）。请注意，此比率并非在所有位置都准确。事实上，纬度和经度之间的表面距离比率取决于您离赤道的距离，因此您的数字是错误的，除非询问者的地图位于同一纬度。这甚至在链接页面中有解释。我非常想因此而投反对票，但这是一个错误NSWER确实有一定的价值，因为它是ASK可能要考虑的东西。1度经度在赤道上大约69.13英里，在极点上有0。这不是一个简单的比率。如果你在汉普郡，在英国，或者在其他纬度相当的地方，这个等式是正确的。大多数人都不是。我的位置是在Sout。瑞典的h。嗨，这似乎是有道理的。我尝试过实施它，但没有效果。数字真的很小，这是原因吗？我在上面公布了数字。“最终结果”你的答案基本上等同于LFSR咨询公司的答案，但我更喜欢你的垂直矢量解释。这取决于你是想垂直于地球表面还是想垂直于坐标。记住，90度的纬度在赤道附近几千公里/英里，在两极附近几厘米/英寸。这使得直线的梯度更小如果你从局部墨卡托投影到1:1拉特长的投影。嗨，我试着实现了这个，但没有成功。我已经公布了问题中变量的值，这样你就可以看到它是否与此有关。谢谢！

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;