基于角度和距离在空间中查找2D点 [英] Find a 2D point in space based on angle and distance

问题描述

好吧，所以我做了一个快速图表，解释我希望完成什么。可悲的是，数学不是我的forte，我希望你的一个向导可以给我正确的公式:)这是一个c ++程序，但真的我正在寻找公式，而不是c ++代码。

好吧，现在基本上，红色圆圈是我们的0,0点，我站在那里。蓝色圆圈是我们以上300单位，我认为是0度的角度。我想知道，我可以找到一个找到x，y的点在这个图表使用我选择的角度以及一定的距离，我选择。

我想知道如何找到x，y的绿色圆圈，这是说225度和500个单位。

所以我假设我有找出一个方法来转置一个圆圈的所有点，从0,0的500单位，而不是基于我想要的角度选择一个地方在那个圆圈上？

解决方案

飞机上的点可以用两个主要的数学表示来表示，即笛卡尔（因此 x，y ）和polar：使用离中心的距离和角度。通常 r 和希腊字母，但让我们使用 w 。

定义

在常见惯例下，r是从中心（0,0）到点的距离，
是逆时针注释

p>注意有关极坐标表示中的角度的一些事情：

• 角度也可以用弧度表示，其中π是相同的角度为180°，因此π/ 2 90°等。 π= 3.14（约）由2π=半径为1的圆的周长定义。




• 角度可以表示为整圆。整圆为2π或360°，因此+ 90°与-270°相同，+ 180°和-180°相同，以及3π/ 4和-5π/4,2π和0， 360°和0°等。你可以考虑[-π，π]（即[-180,180]）或[0,2π]（即[0,360]）之间的角度，或者根本不限制它们，




• 当你的点位于中心（r = 0）
  

• r根据定义总是为正。如果r是负数，您可以更改其符号并添加半圈（π或180°）以获取同一点的坐标。

您的图表上的点

• red：x = 0，y = 0或r = 0 w =任何值




• blue：x = 0，y = 300或r = 300 and w = 90°




• green：x = -400，y = -400或r = -565和w = 225°（近似值，我没有进行实际测量）

请注意，对于蓝点，您可以有w = -270°，绿色w = -135°等。

从一个表示转到另一个表示

最后，需要三角公式在表示之间来回转换。更容易的转换是从极地到笛卡尔：

  x = r * cos（w）
y = r * sin w）
  

由于cos²+sin²= 1，pythagoras等等，你可以看到x²+y²=r²cos²（w）+r²sin²（w）=r²，因此得到r，使用：

  r = sqrt（x²+y²）
  

角度，我们使用 cos / sin = tan 其中tan是另一个三角函数。从 y / x = r sin（w）/（r cos（w））= tan（w），你得到：

  w = arctan（y / x）[modπ] 
  

tan是以π为模的函数，而不是2π。 arctan简单地表示函数tan的倒数，有时写为tan ^ -1或atan。

通过反转切线，可以得到-π/ 2和π/ 2（或-90°和90°）：您需要最终将π添加到您的结果。这是对于[π/ 2，π]和[-π，π/ 2]（[90,180]和[-180，-90]）之间的角度。这些值由cos的符号表示：因为 x = r cos（w）你知道x在所有这些角度上都是负的。尝试看看这些角度在你的图上，这是真的很简单。因此：

  w = arctan（y / x）+（如果x <0，则为π）
  

最后，如果值为0，则不能除以x。在这种情况下，您有

• 如果y> 0， w =π/ 2




• 如果y < 0， w =-π/ 2

Ok.... so I made a quick diagram to sorta explain what I'm hoping to accomplish. Sadly math is not my forte and I'm hoping one of you wizards can give me the correct formulas :) This is for a c++ program, but really I'm looking for the formulas rather than c++ code.

Ok, now basically, the red circle is our 0,0 point, where I'm standing. The blue circle is 300 units above us and at what I would assume is a 0 degree's angle. I want to know, how I can find a find the x,y for a point in this chart using the angle of my choice as well as a certain distance of my choice.

I would want to know how to find the x,y of the green circle which is lets say 225 degrees and 500 units away.

So I assume I have to figure out a way to transpose a circle that is 500 units away from 0,0 at all points than pick a place on that circle based on the angle I want? But yeah no idea where to go from there.

解决方案

A point on a plane can be expressed in two main mathematical representations, cartesian (thus x,y) and polar : using a distance from the center and an angle. Typically r and a greek letter, but let's use w.

Definitions

Under common conventions, r is the distance from the center (0,0) to your point, and angles are measured going counterclockwise (for positive values, clockwise for negative), with the 0 being the horizontal on the right hand side.

Remarks

Note a few things about angles in polar representations :

• angles can be expressed with radians as well, with π being the same angle as 180°, thus π/2 90° and so on. π=3.14 (approx.) is defined by 2π=the perimeter of a circle of radius 1.

• angles can be represented modulo a full circle. A full circle is either 2π or 360°, thus +90° is the same as -270°, and +180° and -180° are the same, as well as 3π/4 and -5π/4, 2π and 0, 360° and 0°, etc. You can consider angles between [-π,π] (that is [-180,180]) or [0,2π] (i.e. [0,360]), or not restrain them at all, it doesn't matter.

• when your point is in the center (r=0) then the angle w is not really defined.

• r is by definition always positive. If r is negative, you can change its sign and add half a turn (π or 180°) to get coordinates for the same point.

Points on your graph

• red : x=0, y=0 or r=0 w= any value

• blue : x=0, y=300 or r=300 and w=90°

• green : x=-400, y=-400 or r=-565 and w=225° (approximate values, I didn't do the actual measurements)

Note that for the blue point you can have w=-270°, and for the green w=-135°, etc.

Going from one representation to the other

Finally, you need trigonometry formulas to go back and forth between representations. The easier transformation is from polar to cartesian :

x=r*cos(w)
y=r*sin(w)

Since cos²+sin²=1, pythagoras, and so on, you can see that x² + y² = r²cos²(w) + r²sin²(w) = r², thus to get r, use :

r=sqrt(x²+y²)

And finally to get the angle, we use cos/sin = tan where tan is another trigonometry function. From y/x = r sin(w) / (r cos(w)) = tan(w), you get :

w = arctan(y/x) [mod π]

tan is a function modulo π, instead of 2π. arctan simply means the inverse of the function tan, and is sometimes written tan^-1 or atan.

By inverting the tangent, you get a result betweeen -π/2 and π/2 (or -90° and 90°) : you need to eventually add π to your result. This is done for angles between [π/2,π] and [-π,π/2] ([90,180] and [-180,-90]). These values are caracterized by the sign of the cos : since x = r cos(w) you know x is negative on all these angles. Try looking where these angles are on your graph, it's really straightforward. Thus :

w = arctan(y/x) + (π if x < 0)

Finally, you can not divide by x if it is 0. In that corner case, you have

• if y > 0, w = π/2

• if y < 0, w = -π/2

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