在来自A的给定的距离计算沿A-B线的一个点 [英] Calculate a point along the line A-B at a given distance from A
问题描述
我要疯了相当特林计算点沿给定线A-B，在从A一定距离，这样我可以在两个特定点之间画就行了。这听起来在一开始很简单，但我似乎无法得到它的权利。更糟糕的是，我不明白的地方我已经错了。几何（和一般的数学）不是我的套件。
I'm going quite mad tring to calculate the point along the given line A-B, at a given distance from A, so that I can "draw" the line between two given points. It sounded simple enough at the outset, but I can't seem to get it right. Worse still, I don't understand where I've gone wrong. Geometry (and math in general) is NOT my strong suite.
我看过类似的问题，并因此有回答了。其实我抬起当前实现CalculatePoint功能的直接的的Mads Elvheim 的的回答：<一href=\"http://stackoverflow.com/questions/1800138/given-a-start-and-end-point-and-a-distance-calculate-a-point-along-a-line\">Given起点和终点，和距离，计算沿线点（加上后来评论的修正 - 如果我没有理解错了他），因为我上独立试图解决的问题是让我无处，除第一类ex preSS票frusterpationland。
I have read similar questions and there answers on SO. In fact I lifted my current implementation of CalculatePoint function directly from Mads Elvheim's answer to: Given a start and end point, and a distance, calculate a point along a line (plus a correction in a later comment - if I understand him correctly) because my indepedent attempts to solve the problem were getting me nowhere, except a first class express ticket frusterpationland.
下面是我的更新时间： code（请参阅编辑笔记后的底部）：
Here's my UPDATED code (please see the EDIT notes a bottom of post):
using System;
using System.Drawing;
using System.Windows.Forms;
namespace DrawLines
{
public class MainForm : Form
{
// =====================================================================
// Here's the part I'm having trouble with. I don't really understand
// how this is suposed to work, so I can't seem to get it right!
// ---------------------------------------------------------------------
// A "local indirector" - Just so I don't have go down and edit the
// actual call everytime this bluddy thing changes names.
private Point CalculatePoint(Point a, Point b, int distance) {
return CalculatePoint_ByAgentFire(a, b, distance);
}
#region CalculatePoint_ByAgentFire
//AgentFire: Better approach (you can rename the struct if you need):
struct Vector2
{
public readonly double X;
public readonly double Y;
public Vector2(double x, double y) {
this.X = x;
this.Y = y;
}
public static Vector2 operator -(Vector2 a, Vector2 b) {
return new Vector2(b.X - a.X, b.Y - a.Y);
}
public static Vector2 operator *(Vector2 a, double d) {
return new Vector2(a.X * d, a.Y * d);
}
public override string ToString() {
return string.Format("[{0}, {1}]", X, Y);
}
}
// For getting the midpoint you just need to do the (a - b) * d action:
//static void Main(string[] args)
//{
// Vector2 a = new Vector2(1, 1);
// Vector2 b = new Vector2(3, 1);
// float distance = 0.5f; // From 0.0 to 1.0.
// Vector2 c = (a - b) * distance;
// Console.WriteLine(c);
//}
private Point CalculatePoint_ByAgentFire(Point a, Point b, int distance) {
var vA = new Vector2(a.X, a.Y);
var vB = new Vector2(b.X, b.Y);
double lengthOfHypotenuse = LengthOfHypotenuseAsDouble(a,b);
double portionOfDistanceFromAtoB = distance / lengthOfHypotenuse;
var vC = (vA - vB) * portionOfDistanceFromAtoB;
Console.WriteLine("vC="+vC);
return new Point((int)(vC.X+0.5), (int)(vC.Y+0.5));
}
// Returns the length of the hypotenuse rounded to an integer, using
// Pythagoras' Theorem for right angle triangles: The length of the
// hypotenuse equals the sum of the square of the other two sides.
// Ergo: h = Sqrt(a*a + b*b)
private double LengthOfHypotenuseAsDouble(Point a, Point b) {
double aSq = Math.Pow(Math.Abs(a.X - b.X), 2); // horizontal length squared
double bSq = Math.Pow(Math.Abs(b.Y - b.Y), 2); // vertical length squared
return Math.Sqrt(aSq + bSq); // length of the hypotenuse
}
#endregion
//dbaseman: I thought something looked strange about the formula ... the question
//you linked was how to get the point at a distance after B, whereas you want the
//distance after A. This should give you the right answer, the start point plus
//distance in the vector direction.
//
// Didn't work as per: http://s1264.photobucket.com/albums/jj496/corlettk/?action=view¤t=DrawLinesAB-broken_zps069161e9.jpg
//
private Point CalculatePoint_ByDbaseman(Point a, Point b, int distance) {
// a. calculate the vector from a to b:
double vectorX = b.X - a.X;
double vectorY = b.Y - a.Y;
// b. calculate the length:
double magnitude = Math.Sqrt(vectorX * vectorX + vectorY * vectorY);
// c. normalize the vector to unit length:
vectorX /= magnitude;
vectorY /= magnitude;
// d. calculate and Draw the new vector, which is x1y1 + vxvy * (mag + distance).
return new Point(
(int)((double)a.X + vectorX * distance) // x = col
, (int)((double)a.Y + vectorY * distance) // y = row
);
}
// MBo: Try to remove 'magnitude' term in the parentheses both for X and for Y expressions.
//
// Didn't work as per: http://s1264.photobucket.com/albums/jj496/corlettk/?action=view¤t=DrawLinesAB-broken_zps069161e9.jpg
//
//private Point CalculatePoint_ByMBo(Point a, Point b, int distance) {
// // a. calculate the vector from a to b:
// double vectorX = b.X - a.X;
// double vectorY = b.Y - a.Y;
// // b. calculate the length:
// double magnitude = Math.Sqrt(vectorX * vectorX + vectorY * vectorY);
// // c. normalize the vector to unit length:
// vectorX /= magnitude;
// vectorY /= magnitude;
// // d. calculate and Draw the new vector, which is x1y1 + vxvy * (mag + distance).
// return new Point(
// (int)( ((double)a.X + vectorX * distance) + 0.5 )
// , (int)( ((double)a.X + vectorX * distance) + 0.5 )
// );
//}
// Didn't work
//private Point CalculatePoint_ByUser1556110(Point a, Point b, int distance) {
// Double magnitude = Math.Sqrt(Math.Pow(b.Y - a.Y, 2) + Math.Pow(b.X - a.X, 2));
// return new Point(
// (int)(a.X + distance * (b.X - a.X) / magnitude + 0.5)
// , (int)(a.Y + distance * (b.Y - a.Y) / magnitude + 0.5)
// );
//}
// didn't work
//private static Point CalculatePoint_ByCadairIdris(Point a, Point b, int distance) {
// // a. calculate the vector from a to b:
// double vectorX = b.X - a.X;
// double vectorY = b.Y - a.Y;
// // b. calculate the proportion of hypotenuse
// double factor = distance / Math.Sqrt(vectorX*vectorX + vectorY*vectorY);
// // c. factor the lengths
// vectorX *= factor;
// vectorY *= factor;
// // d. calculate and Draw the new vector,
// return new Point((int)(a.X + vectorX), (int)(a.Y + vectorY));
//}
// Returns a point along the line A-B at the given distance from A
// based on Mads Elvheim's answer to:
// http://stackoverflow.com/questions/1800138/given-a-start-and-end-point-and-a-distance-calculate-a-point-along-a-line
private Point MyCalculatePoint(Point a, Point b, int distance) {
// a. calculate the vector from o to g:
double vectorX = b.X - a.X;
double vectorY = b.Y - a.Y;
// b. calculate the length:
double magnitude = Math.Sqrt(vectorX * vectorX + vectorY * vectorY);
// c. normalize the vector to unit length:
vectorX /= magnitude;
vectorY /= magnitude;
// d. calculate and Draw the new vector, which is x1y1 + vxvy * (mag + distance).
return new Point(
(int)(((double)a.X + vectorX * (magnitude + distance)) + 0.5) // x = col
, (int)(((double)a.Y + vectorY * (magnitude + distance)) + 0.5) // y = row
);
}
// =====================================================================
private const int CELL_SIZE = 4; // width and height of each "cell" in the bitmap.
private readonly Bitmap _bitmap; // to draw on (displayed in picBox1).
private readonly Graphics _graphics; // to draw with.
// actual points on _theLineString are painted red.
private static readonly SolidBrush _thePointBrush = new SolidBrush(Color.Red);
// ... and are labeled in Red, Courier New, 12 point, Bold
private static readonly SolidBrush _theLabelBrush = new SolidBrush(Color.Red);
private static readonly Font _theLabelFont = new Font("Courier New", 12, FontStyle.Bold);
// the interveening calculated cells on the lines between actaul points are painted Black.
private static readonly SolidBrush _theLineBrush = new SolidBrush(Color.Black);
// the points in my line-string.
private static readonly Point[] _theLineString = new Point[] {
// x, y
new Point(170, 85), // A
new Point( 85, 70), // B
//new Point(209, 66), // C
//new Point( 98, 120), // D
//new Point(158, 19), // E
//new Point( 2, 61), // F
//new Point( 42, 177), // G
//new Point(191, 146), // H
//new Point( 25, 128), // I
//new Point( 95, 24) // J
};
public MainForm() {
InitializeComponent();
// initialise "the graphics system".
_bitmap = new Bitmap(picBox1.Width, picBox1.Height);
_graphics = Graphics.FromImage(_bitmap);
picBox1.Image = _bitmap;
}
#region actual drawing on the Grpahics
private void DrawCell(int x, int y, Brush brush) {
_graphics.FillRectangle(
brush
, x * CELL_SIZE, y * CELL_SIZE // x, y
, CELL_SIZE, CELL_SIZE // width, heigth
);
}
private void DrawLabel(int x, int y, char c) {
string s = c.ToString();
_graphics.DrawString(
s, _theLabelFont, _theLabelBrush
, x * CELL_SIZE + 5 // x
, y * CELL_SIZE - 8 // y
);
}
// ... there should be no mention of _graphics or CELL_SIZE below here ...
#endregion
#region draw points on form load
private void MainForm_Load(object sender, EventArgs e) {
DrawPoints();
}
// draws and labels each point in _theLineString
private void DrawPoints() {
char c = 'A'; // label text, as a char so we can increment it for each point.
foreach ( Point p in _theLineString ) {
DrawCell(p.X, p.Y, _thePointBrush);
DrawLabel(p.X, p.Y, c++);
}
}
#endregion
#region DrawLines on button click
private void btnDrawLines_Click(object sender, EventArgs e) {
DrawLinesBetweenPointsInTheString();
}
// Draws "the lines" between the points in _theLineString.
private void DrawLinesBetweenPointsInTheString() {
int n = _theLineString.Length - 1; // one less line-segment than points
for ( int i = 0; i < n; ++i )
Draw(_theLineString[i], _theLineString[i + 1]);
picBox1.Invalidate(); // tell the graphics system that the picture box needs to be repainted.
}
// Draws all the cells along the line from Point "a" to Point "b".
private void Draw(Point a, Point b) {
int maxDistance = LengthOfHypotenuse(a, b);
for ( int distance = 1; distance < maxDistance; ++distance ) {
var point = CalculatePoint(a, b, distance);
DrawCell(point.X, point.X, _theLineBrush);
}
}
// Returns the length of the hypotenuse rounded to an integer, using
// Pythagoras' Theorem for right angle triangles: The length of the
// hypotenuse equals the sum of the square of the other two sides.
// Ergo: h = Sqrt(a*a + b*b)
private int LengthOfHypotenuse(Point a, Point b) {
double aSq = Math.Pow(Math.Abs(a.X - b.X), 2); // horizontal length squared
double bSq = Math.Pow(Math.Abs(b.Y - b.Y), 2); // vertical length squared
return (int)(Math.Sqrt(aSq + bSq) + 0.5); // length of the hypotenuse
}
#endregion
#region Windows Form Designer generated code
/// <summary>
/// Required method for Designer support - do not modify
/// the contents of this method with the code editor.
/// </summary>
private void InitializeComponent() {
this.picBox1 = new System.Windows.Forms.PictureBox();
this.btnDrawLines = new System.Windows.Forms.Button();
((System.ComponentModel.ISupportInitialize)(this.picBox1)).BeginInit();
this.SuspendLayout();
//
// picBox1
//
this.picBox1.Dock = System.Windows.Forms.DockStyle.Fill;
this.picBox1.Location = new System.Drawing.Point(0, 0);
this.picBox1.Name = "picBox1";
this.picBox1.Size = new System.Drawing.Size(1000, 719);
this.picBox1.TabIndex = 0;
this.picBox1.TabStop = false;
//
// btnDrawLines
//
this.btnDrawLines.Location = new System.Drawing.Point(23, 24);
this.btnDrawLines.Name = "btnDrawLines";
this.btnDrawLines.Size = new System.Drawing.Size(77, 23);
this.btnDrawLines.TabIndex = 1;
this.btnDrawLines.Text = "Draw Lines";
this.btnDrawLines.UseVisualStyleBackColor = true;
this.btnDrawLines.Click += new System.EventHandler(this.btnDrawLines_Click);
//
// MainForm
//
this.AutoScaleDimensions = new System.Drawing.SizeF(6F, 13F);
this.AutoScaleMode = System.Windows.Forms.AutoScaleMode.Font;
this.ClientSize = new System.Drawing.Size(1000, 719);
this.Controls.Add(this.btnDrawLines);
this.Controls.Add(this.picBox1);
this.Location = new System.Drawing.Point(10, 10);
this.MinimumSize = new System.Drawing.Size(1016, 755);
this.Name = "MainForm";
this.SizeGripStyle = System.Windows.Forms.SizeGripStyle.Hide;
this.StartPosition = System.Windows.Forms.FormStartPosition.Manual;
this.Text = "Draw Lines on a Matrix.";
this.Load += new System.EventHandler(this.MainForm_Load);
((System.ComponentModel.ISupportInitialize)(this.picBox1)).EndInit();
this.ResumeLayout(false);
}
private System.Windows.Forms.PictureBox picBox1;
private System.Windows.Forms.Button btnDrawLines;
#endregion
}
}
如果它是一个有点长，但这是一个
Sorry if it's a bit long, but this an is SSCCE exhumed from my real project, which is an implementation of the A* shortest route algorithm to run the MazeOfBolton... i.e. a maze runner.
我真正想要做的是pre-计算篱笆（即一个缓冲 MBR ）围绕在迷宫两个特定点（原点和目标）（基质），以使得围栏之内的所有点是从两点之间的直线，一个给定的距离之内，以快速消除上百个的，成千上万的可能的路径这是从球门标题了。
What I actually want to do is pre-calculate a "fence" (i.e. a buffered MBR) around two given points (origin and goal) in the maze (a matrix), such that all points within the "fence" are within a given distance from "the straight line between the two points", in order to quickly eliminate the hundreds-of-thousands of possible paths which are heading away from the goal.
请注意，这个编程挑战年前关闭，所以有一个与竞争性plagerism在这里没有问题。不，这不是功课，其实我是一个专业的程序员......我只是WAAAAY在这里我的安乐窝，即使是相对简单的几何形状。叹了口气。
Note that this programming challenge closed years ago, so there's no issue with "competitive plagerism" here. No this is not homework, in fact I'm a professional programmer... I'm just WAAAAY out of my comfort zone here, even with relatively simple geometry. Sigh.
所以......请任何人都可以给我任何指针帮我把CalculatePoint功能的正确：在从A给定距离计算沿线AB点
So... Please can anyone give me any pointers to help me get the CalculatePoint function to correctly: Calculate a point along the line A-B at the given distance from A?
在此先感谢您的慷慨...即使在遥远阅读本。
Thanks in advance for your generosity... even in reading this far.
干杯。基思。
编辑：我刚刚更新了发布源$ C $ C监守：
I just updated the posted source code becuase:
（1）我才意识到这不是自给。我忘了单独的MainForm 设计师 cs文件，我已经追加到贴code的底部。
(1) I just realised that it wasn't self contained. I forgot about the seperate MainForm.Designer.cs file, which I've appended to the bottom of the posted code.
（2）的最新版本包含了我到目前为止已经试过，用的photobucket链接什么的每一次的失败看起来像一个画面......他们都是一样的。伊？ WTF？
(2) The latest version includes what I've tried so far, with a photobucket link to a picture of what each failure looks like... and they're all the same. Huy? WTF?
我想我的问题可能是在其他地方一样，那是$ P $由其他人pviously错过了，因为我忘了张贴设计器生成的code一些时髦的Windows窗体设置......除了everythingelse（在我的实际项目）描绘正是我期望它，所以为什么要计算点有什么不同。我不知道！？！？！？我是pretty沮丧，我越来越胡思乱想，所以我想我会离开这个另一天; - ）
I suppose my problem may be elsewhere, like some funky windows form setting that was previously missed by everyone else because I forgot to post the designer-generated code... Except everythingelse (in my actual project) paints exactly where I expect it to, so why should a calculated point be any different. I don't know!?!?!? I'm pretty frustrated and I'm getting cranky, so I think I'll leave this for another day ;-)
正好说明我们有多么经常低估要花多少力气才能使一台计算机做什么......哪怕只是画一个简单的线......它甚至不是一条曲线，更何况是一个大圆或横向墨卡托或任何幻想......只是一个简单的bluddy线！？！？！？ ; - ）
Goes to show how much we routinely underestimate how much effort it'll take to make a computer do ANYthing... even just draw a simple line... it's not even a curve, let alone a great circle or a transverse mercator or anything fancy... just a simple bluddy line!?!?!? ;-)
再次干杯。基思。
推荐答案
计算矢量AB
首先定义从A点矢量（1，-1），从B.点B（2,4）减去一个
矢量是VAB（1,5）。
First define the vector from point A(1,-1) to point B(2,4) substracting A from B. The vector would be Vab(1,5).
计算AB的长度
使用勾股定理计算矢量AB的长度。
Use Pythagorean theorem to calculate the length of vector AB.
|Vab| = SQRT(1²+5²)
总长（四舍五入）5.1
The Length is (rounded) 5.1
计算单位矢量
通过其长度除以向量获得的单位矢量（长度为1的向量）。
Divide the vector by its length to get the unit vector (the vector with length 1).
V1(1/5.1,5/5.1) = V1(0.2, 0.98)
长度计算矢量4
现在V1乘以你想要的长度，例如4，拿到佛蒙特州
Now multiply V1 with the length you want, for example 4, to get Vt.
Vt(0.2*4,0.98*4) = Vt(0.8,3.92)
计算目标点
添加矢量Vt的A点得到点T（目标）。
Add the vector Vt to point A to get point T (target).
T = A + Vt = T(1.8,2.92)
编辑：回答你的修改
该方法LengthOfHypotenuse应该看起来像
The method LengthOfHypotenuse should look like that
- 在计算BSQ固定错误
- 和去除多余Math.Abs打电话，因为2战俘总是正
- 删除加入0.5，不知道你为什么会需要一个
-
你至少应该用一个浮点数作为返回值（双精度或十进制也将工作）
- fixed an error on calculating bSq
- and removed redundant Math.Abs call, because a pow of 2 is always positive
- removed the addition of 0.5, don't know why you would need that
you should at least use a float as return value (double or decimal would work also)
//You should work with Vector2 class instead of Point and use their Length property
private double LengthOfHypotenuse(Point a, Point b) {
double aSq = Math.Pow(a.X - b.X, 2); // horizontal length squared
double bSq = Math.Pow(a.Y - b.Y, 2); // vertical length squared
return Math.Sqrt(aSq + bSq); // length of the hypotenuse
}
该方法绘制（A点，B点）应该看起来像：
The method Draw(Point a, Point b) should look like that:
-
修正DrawCell（）调用
Corrected DrawCell() call
private void Draw(Point a, Point b) {
double maxDistance = LengthOfHypotenuse(a, b);
for (int distance = 0; distance < maxDistance; ++distance) {
var point = CalculatePoint(new Vector2(a), new Vector2(b), distance);
DrawCell(point.X, point.Y, _theLineBrush);
}
}
您CalculatePoint（A点，B点，INT距离）方法：
Your CalculatePoint(Point a, Point b, int distance) method:
-
动了一些计算到Vector2类
Moved some calculations into Vector2 class
private Point CalculatePoint(Vector2 a, Vector2 b, int distance) {
Vector2 vectorAB = a - b;
return a + vectorAB.UnitVector * distance;
}
我已经扩展了Vector类为您丢失的运营商（学分AgentFire）添加
I have extended the Vector class for you to add the missing operators (credits to AgentFire)
//AgentFire: Better approach (you can rename the struct if you need):
struct Vector2 {
public readonly double X;
public readonly double Y;
public Vector2(Point p) : this(p.X,p.Y) {
}
public Vector2(double x, double y) {
this.X = x;
this.Y = y;
}
public static Vector2 operator -(Vector2 a, Vector2 b) {
return new Vector2(b.X - a.X, b.Y - a.Y);
}
public static Vector2 operator +(Vector2 a, Vector2 b) {
return new Vector2(b.X + a.X, b.Y + a.Y);
}
public static Vector2 operator *(Vector2 a, double d) {
return new Vector2(a.X * d, a.Y * d);
}
public static Vector2 operator /(Vector2 a, double d) {
return new Vector2(a.X / d, a.Y / d);
}
public static implicit operator Point(Vector2 a) {
return new Point((int)a.X, (int)a.Y);
}
public Vector2 UnitVector {
get { return this / Length; }
}
public double Length {
get {
double aSq = Math.Pow(X, 2);
double bSq = Math.Pow(Y, 2);
return Math.Sqrt(aSq + bSq);
}
}
public override string ToString() {
return string.Format("[{0}, {1}]", X, Y);
}
}
这篇关于在来自A的给定的距离计算沿A-B线的一个点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
