绘图后得到距离 [英] Get distance mesure after drawing

问题描述

嗨我通过importin数据制作了一个2d自动绘图的应用程序。不是我的积分是绘制，我想计算两点之间的距离并自动显示它和屏幕。

我需要移动它，如果有必要绘制质量视图



我尝试过：

 double 
 
 dblDistX = Math.Abs​​（pt1.X  -  pt2.X）; 
 double 
 dblDistY = Math.Abs​​（pt1.Y  -  pt2.Y）; 

解决方案

看起来像一个好的开始;寻找X&的增量Y坐标。

  double  dblDistX = Math.Abs​​（pt1.X  -  pt2.X）; 
  double  dblDistY = Math.Abs​​（pt1.Y  -  pt2.Y）; 

现在我猜你想要的这两点之间的对角线距离（Z）。是时候打破HS几何书了，看看我的朋友毕达哥拉斯，看看他对定理有什么。

Pythagoreas写道：

在一个直角三角形：斜边的平方等于另外两边的平方和。

更新02/14/2019

这就是手写数学的方式：

z ² = x ² + y ²

z =√（x ² + y ² ）



我的原始答案包括此代码，该代码没有特定的语言。根据下面的评论，我被告知这适用于VB

 '  VB解决方案 
  double  Z =（dblDistX ^ 2 + dblDistY ^ 2）^。 5  

C＃是OP使用的，所以这里有2个版本; long long version和1 liner

  //   C＃version  
  //  长版 
  double  dblSqrdDistX = Math.Pow（dblDistX， 2 ）; 
  double  dblSqrdDistY = Math.Pow（dblDistY， 2 ）; 
  double  dblSqrdHypotenuseZ = dblSqrdDistX + dblSqrdDistY; 
  double  Z = Math.Sqrt（dblSqrdHypotenuseZ）; 
 
  //  简短版 
  double  Z = Math.Sqrt（Math.Pow（dblDistX， 2 ）+ Math.Pow（dblDistY， 2 ））; 

Hi iam make an app for 2d automatic drawing by importin data. Not my points are draw, i want to calculate distance between two points and show it automaticaly and the screen.
I would need to move it if necessary for drawing quality view

What I have tried:

double

 dblDistX = Math.Abs(pt1.X - pt2.X);
 double
 dblDistY = Math.Abs(pt1.Y - pt2.Y);

解决方案

Looks like a good start; finding the deltas for the X & Y coordinates.

double dblDistX = Math.Abs(pt1.X - pt2.X);
double dblDistY = Math.Abs(pt1.Y - pt2.Y);

Now I'm guessing that you want the diagonal distance (Z) between these 2 points. Time to break out the HS Geometry book and look up my buddy Pythagoreas and see what he has for theorems.

Pythagoreas wrote:

In a right angled triangle: the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Update 02/14/2019
This is how it would be in handwritten math:
z² = x² + y²
z = √(x² + y²)

My original answer included this code which was in no specific language. From the comments below I was informed that this would work for VB

' VB Solution
double Z = (dblDistX^2 + dblDistY^2)^.5

C# is what the OP is using so here are 2 versions of it; the long long version and the 1 liner

// C# version
// long version
double dblSqrdDistX = Math.Pow(dblDistX,2);
double dblSqrdDistY = Math.Pow(dblDistY,2);
double dblSqrdHypotenuseZ = dblSqrdDistX + dblSqrdDistY;
double Z = Math.Sqrt(dblSqrdHypotenuseZ);

// short version
double Z = Math.Sqrt(Math.Pow(dblDistX,2) + Math.Pow(dblDistY,2));

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