绘制以直线结尾的弧的公式，Y是X的函数，起始斜率，结束斜率，起点和弧半径？ [英] Formula to draw arcs ending in straight lines, Y as a function of X, starting slope, ending slope, starting point and arc radius?

问题描述

我正在寻找一个数学公式，在一个图形上绘制Y作为X的函数，在指定的起点（X的值，甚至更好的X和Y坐标）之前会有一定的斜率，那么之后它将绘制一个指定半径的圆弧，该圆弧在到达第二个指定的斜率时结束，从该点开始将是该第二个斜坡的另一条直线。

<我知道，因为Y是X的函数，所以斜率参数需要大于-90，小于90。我并不担心在（或超出）这些极端情况下的任何不当行为。其实，我会更加高兴一个公式，它需要一个起点和终点（2D坐标），以及起止斜率;并且在它们之间会有两个弧线（需要时它们之间有一条直线），将两条直线无缝地连接起来（显然，终点的X必须大于起点的X;我不在乎什么当情况并非如此时）。但我想象这样一个公式可能比我先问的难得多。

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ps：by弧我的意思是一个圆的部分;如图所示，如果图的两个轴具有相同的比例，则圆弧对于相同半径的圆将具有正确的高宽比。

解决方案我看到它是这样的：

1. compute P0

   
   
   

   作为行的交点 A + t * dA 和 B - t * dB




2. compute P1
   $ b

   它是翻译行 A-> P0 和的交集。 B-> P0 通过半径 r 垂直。有两种可能性，所以选择正确的（这导致圆形部分的角度较小）。

   
   
   




3. compute P2，P3

   
   
   

   只是行 A-P0 和 B- P0 和垂直线从 P1 加入它




4. 曲线
   
   

     //一些常量首先
    da = P2-A; 
    db = B-P3; 
    a2 = atan2（P2.x-P1.x，P2.y-P1.y）; 
    a3 = atan2（P3.x-P1.x，P3.y-P1.y）; 
    if（a2> a3）a3- = M_PI * 2.0; 
    dang = a3-a2; 
    
    // now（x，y）= curve（t）...其中t =< 0,3> 
    if（t <= 1.0）
    {
    x = A.x + t * da.x; 
    y = A.y + t * da.y; 
   } 
    else if（t <= 2.0）
    {
    t = a2 +（（t-1.0）* dang）; 
    x = P1.x + r * cos（t）; 
    y = P1.y + r * sin（t）; 
   } 
    else 
    {
    t = t-2.0; 
    x = P3.x + t * db.x; 
    y = P3.y + t * db.y; 
   } 
     

I'm looking for a math formula that on a graph plotting Y as a function of X, before a specified starting point (a value of X, or even better, X and Y coordinates) will have a certain slope, then after that it will draw an arc of a specified radius that will end when it reaches a second specified slope, and from the point on will be another straight line of that second slope.

I'm am aware that because it's Y as a function of X, the slope parameters would need to be bigger than exactly -90 and smaller than exactly 90 degrees; i'm not worried about any misbehavior at (or beyond) those extremes.

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Actually, i would be even happier with a formula that takes a starting and ending points (2d coordinates), and starting and ending slopes; and will have two arcs in between (with a straight line between them when needed), connecting the two straight lines seamlessly (obviously the X of the ending point needs to be bigger than the X for the starting point; i don't care what happens when that isn't the case). But i imagine such a formula might be much harder to come up with than what i asked first.

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ps: by "arc" i mean segment of a circle; as in, if both axes of the graph have the same scale, the arcs will have the correct aspect ratio for a circle of the same radius.

解决方案

Well I see it like this:

1. compute P0

   as intersection of lines A + t*dA and B - t*dB

2. compute P1 (center of circle)

   it is intersection of translated lines A->P0 and B->P0 perpendicular by radius r. There are 2 possibilities so choose the right one (which leads to less angle of circular part).

3. compute P2,P3

   just an intersection between lines A-P0 and B-P0 and perpendicular line from P1 to it

4. the curve

   // some constants first
   da=P2-A;
   db=B-P3;
   a2=atan2(P2.x-P1.x,P2.y-P1.y);
   a3=atan2(P3.x-P1.x,P3.y-P1.y);
   if (a2>a3) a3-=M_PI*2.0;
   dang=a3-a2;
   
   // now (x,y)=curve(t) ... where t = <0,3>
   if (t<=1.0)
    {
    x=A.x+t*da.x;
    y=A.y+t*da.y;
    }
   else if (t<=2.0)
    {
    t=a2+((t-1.0)*dang);
    x=P1.x+r*cos(t);
    y=P1.y+r*sin(t);
    }
   else
    {
    t=t-2.0;
    x=P3.x+t*db.x;
    y=P3.y+t*db.y;
    }
   

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