使用水平和垂直角度和斜距计算 3D 点坐标 [英] Calculate 3D point coordinates using horizontal and vertical angles and slope distance

问题描述

我正在尝试学习如何使用原点的 XYZ 坐标、水平和垂直角度以及 3d 距离来计算点的 XYZ 坐标.我可以简单地通过将点投影到 2D 平面上来进行计算，但在 3D 中是否有更直接的方法?

I am trying to learn how to calculate the XYZ coordinates of a point using the XYZ coordinates of an origin point, a horizontal and vertical angle, and 3d distance. I can make the calculations simply by projecting the points onto 2D planes, but is there a more straightforward way to do this in 3D?

我试图了解测量全站仪如何根据其测量位置、它测量到新点的 3d(斜率)距离以及测量到的新点的水平和垂直角度来计算新点位置地点.

I am trying to understand how a surveying total station calculates new point locations based on it's measured location, the 3d (slope) distance that it measures to a new point, and the measured horizontal and vertical angles that sight onto the new point location.

谢谢，

E

推荐答案

关于约定的说明:2D polar坐标经常使用(radius, theta)，其中theta是'水平'或'方位'角.它的范围从:theta=0，X 轴上的二维点(x,y) = (radius,0)，到:theta=2*PI 在 XY 平面上 - 随着 theta 的增加呈逆时针方向.现在混淆问题...

Just a note on conventions: 2D polar coordinates often use (radius, theta), where theta is the 'horizontal' or 'azimuth' angle. It ranges from: theta=0, the 2D point (x,y) = (radius,0) on the X-axis, to: theta=2*PI on the XY plane - an anti-clockwise direction as theta increases. Now to confuse matters...

3D 球面坐标(保持右手坐标系)经常使用坐标:(半径，θ，phi).在这种情况下，theta 用于垂直"或天顶"角度，范围从 theta=0(Z 轴)到 theta=PI(-Z 轴).phi 用于方位角.

3D spherical coordinates (maintaining a right-handed coordinate system) often use coordinates: (radius, theta, phi). In this case, theta is used for the 'vertical' or 'zenith' angle, ranging from theta=0 (the Z axis) to theta=PI (the -Z axis). phi is used for the azimuth angle.

其他文本将使用不同的约定 - 但这似乎受到物理学家和(某些)数学文本的青睐.重要的是您选择一个约定并始终如一地使用它.

Other texts will use different conventions - but this seems to be favoured by physicists and (some) mathematics texts. What matters is that you pick a convention and use it consistently.

按照这个:

radius:到点的距离.给定笛卡尔坐标中的一个点 (x,y,z)，我们有(勾股)半径:r = sqrt(x * x + y * y + z * z)，例如 0 <= radius <+无穷大

radius: distance to the point. given an point (x,y,z) in cartesian coordinates, we have the (pythagorean) radius: r = sqrt(x * x + y * y + z * z), e.g., 0 <= radius < +infinity

theta:天顶角，其中 theta=0 位于(+Z 轴)正上方，theta=PI 直接位于(-Z 轴)下方，theta=PI/2 是您认为 0 度的仰角"，例如，
0 <= theta <= PI

theta: the zenith angle, where theta=0 is directly above (the +Z axis), and theta=PI is directly below (the -Z axis), and theta=PI/2 is what you would consider an 'elevation' of 0 degrees, e.g.,
0 <= theta <= PI

phi:方位角，其中 phi=0 是右"(+X 轴)，当你'逆时针'，phi=PI/2(+Y轴)，phi=PI(-X轴)，phi=3*PI/2(-Y 轴)和 phi=2*PI - 相当于 phi=0(回到 +X 轴).例如，0 <= phi <2*PI

phi: the azimuth angle, where phi=0 is to the 'right' (the +X axis), and as you turn 'anticlockwise', phi=PI/2 (the +Y axis), phi=PI (the -X axis), phi=3*PI/2 (the -Y axis), and phi=2*PI - equivalent to the phi=0 (back to the +X axis). e.g., 0 <= phi < 2*PI

伪代码:(标准数学库三角函数)

Pseudo-code: (standard math library trigonometric functions)

从 (radius, theta, phi) 你可以找到点 (x,y,z) :

From (radius, theta, phi) you can find the point (x,y,z) :

x = 半径 * sin(theta) * cos(phi);
y = 半径 * sin(theta) * sin(phi);
z = 半径 * cos(theta);

相反，你可以找到一个 (radius, theta, phi) from (x,y,z) :

Conversely, you can find a (radius, theta, phi) from (x,y,z) :

radius = sqrt(x * x + y * y + z * z);
theta = acos(z/radius);
phi = atan2(y, x);

注意:在决赛中使用 atan2 很重要等式，不是 atan！

Note: it is important to use atan2 in the final equation, not atan!

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