Javascript 以角度将线绘制为y值

Javascript 以角度将线绘制为y值,javascript,math,Javascript,Math,我计算了一个方向(如果你愿意,你可以称它为向量,但它实际上是一个斜率…),我想从给定y值的函数中得到一个x值 基本上,我试图画一条从x,y值到x,y值的线,我有方向 假设坡度/方向为1/4(上升超过行程),起点为200,终点y值为250,如何找到x 我知道这确实是高中的基础代数,但由于某些原因,我无法将其概念化…使用 这里有(包括)使用说明 这里是(包括)如果端点是A(x1,y1)和B(x2,y2),则定义为: m = ( y2 - y1 ) / ( x2 - x1 ) 因为你有坡度,你需要至

我计算了一个方向(如果你愿意,你可以称它为向量,但它实际上是一个斜率…),我想从给定y值的函数中得到一个x值

基本上,我试图画一条从x,y值到x,y值的线,我有方向

假设坡度/方向为1/4(上升超过行程),起点为200,终点y值为250,如何找到x

我知道这确实是高中的基础代数,但由于某些原因,我无法将其概念化…

使用

这里有(包括)

使用说明

这里是(包括)

如果端点是A(x1,y1)和B(x2,y2),则定义为:

m = ( y2 - y1 ) / ( x2 - x1 )
因为你有坡度,你需要至少有三个坐标才能计算出剩下的一个

根据你的问题,我假设你想计算y2。因此,需要有x1、y1和x2

例如:

m = 1/4
A(1,1)
B(9,y2)
---
y2 = ?

m = ( y2 - y1 ) / ( x2 - x1 ) 
y2 - y1 = m * ( x2 - x1 )
y2 = m * ( x2 - x1 ) + y1
y2 = 1/4 * ( 9 - 1 ) + 1
y2 = 3
如果端点为A(x1,y1)和B(x2,y2),则定义为:

m = ( y2 - y1 ) / ( x2 - x1 )
因为你有坡度,你需要至少有三个坐标才能计算出剩下的一个

根据你的问题,我假设你想计算y2。因此,需要有x1、y1和x2

例如:

m = 1/4
A(1,1)
B(9,y2)
---
y2 = ?

m = ( y2 - y1 ) / ( x2 - x1 ) 
y2 - y1 = m * ( x2 - x1 )
y2 = m * ( x2 - x1 ) + y1
y2 = 1/4 * ( 9 - 1 ) + 1
y2 = 3

Bresenham直线算法:

void DrawLineLCD(int x1,int y1,int x2,int y2,int nState)
{
    unsigned int nTmp;
    unsigned int nAlt=0;
    int x,y;        // where is the current pixel.
    int dx;     // dx is the delta for x
    int dy;     // dy is the delta for y
    int StepVal=0;  // variable for figuring out when to increment the other 
axis.
    if (x1>x2 && y1>y2)
    {
        nTmp=x2;
        x2=x1;
        x1=nTmp;

        nTmp=y2;
        y2=y1;
        y1=nTmp;

        dx=x2-x1;   // dx is the delta for x
        dy=y2-y1;   // dy is the delta for y
    }else
    {
        dx=x2-x1;   // dx is the delta for x
        dy=y2-y1;   // dy is the delta for y

        if (dy<0)
        {   
            dy=-dy;

            nTmp=y2;
            y2=y1;
            y1=nTmp;

            nAlt=1;
        }else
            if (dx<0)
            {   
                dx=-dx;

                nTmp=x2;
                x2=x1;
                x1=nTmp;

                nAlt=1;
            }
    }

    if (nAlt)
    {

        if(dx>=dy)       // The slope is less than 45 degres
        {
            y=y2;
            for(x=x1; x<=x2; x++)
            {
                // Call your function to draw a pixel here.
                SetPixelLCD(x,y,nState);
                StepVal+=dy;
                if(StepVal>=dx) // Increment y if enough x steps
                                // have been taken.
                {
                    y--;
                    StepVal-=dx;    // Reset StepVal, but  
                    // not to 0.  This gives even slopes.
                }
            }
        }
        else    // The slope is greater than 45 degrees, just like 
                // above, but with y instead of x.
        {
            x=x2;
            for(y=y1; y<=y2; y++)
            {
                // Call your function to draw a pixel here.
                SetPixelLCD(x,y,nState);
                StepVal+=dx;
                if(StepVal>=dy)
                {
                    x--;
                    StepVal-=dy;
                }
            }
        }
        return;
    }

    if(dx>=dy)       // The slope is less than 45 degres
    {
        y=y1;
        for(x=x1; x<=x2; x++)
        {
            // Call your function to draw a pixel here.
            SetPixelLCD(x,y,nState); 
            StepVal+=dy;
            if(StepVal>=dx) // Increment y if enough x steps
                            // have been taken.
            {
                y++;
                StepVal-=dx;    // Reset StepVal, but  
                // not to 0.  This gives even slopes.
            }
        }
    }
    else    // The slope is greater than 45 degrees, just like 
            // above, but with y instead of x.
    {
        x=x1;
        for(y=y1; y<=y2; y++)
        {
            // Call your function to draw a pixel here.
            SetPixelLCD(x,y,nState);
            StepVal+=dx;
            if(StepVal>=dy)
            {
                x++;
                StepVal-=dy;
            }
        }
    }
    return;
}
void DrawLineLCD(intx1、inty1、intx2、inty2、intnstate)
{
无符号整数nTmp;
无符号整数nAlt=0;
int x,y;//其中是当前像素。
int dx;//dx是x的增量
int dy;//dy是y的增量
int-StepVal=0;//用于计算何时递增另一个变量的变量
轴
如果(x1>x2&&y1>y2)
{
nTmp=x2;
x2=x1;
x1=nTmp;
nTmp=y2;
y2=y1;
y1=nTmp;
dx=x2-x1;//dx是x的增量
dy=y2-y1;//dy是y的增量
}否则
{
dx=x2-x1;//dx是x的增量
dy=y2-y1;//dy是y的增量
如果(dy=dy)//斜率小于45度
{
y=y1;
对于(x=x1;x=dx)//如果有足够的x步,则增加y
//已经被带走了。
{
y++;
StepVal-=dx;//重置StepVal,但是
//不为0。这将提供均匀的坡度。
}
}
}
否则//坡度大于45度,就像
//上面,但用y代替x。
{
x=x1;
对于(y=y1;y=dy)
{
x++;
StepVal-=dy;
}
}
}
返回;
}

Bresenham直线算法:

void DrawLineLCD(int x1,int y1,int x2,int y2,int nState)
{
    unsigned int nTmp;
    unsigned int nAlt=0;
    int x,y;        // where is the current pixel.
    int dx;     // dx is the delta for x
    int dy;     // dy is the delta for y
    int StepVal=0;  // variable for figuring out when to increment the other 
axis.
    if (x1>x2 && y1>y2)
    {
        nTmp=x2;
        x2=x1;
        x1=nTmp;

        nTmp=y2;
        y2=y1;
        y1=nTmp;

        dx=x2-x1;   // dx is the delta for x
        dy=y2-y1;   // dy is the delta for y
    }else
    {
        dx=x2-x1;   // dx is the delta for x
        dy=y2-y1;   // dy is the delta for y

        if (dy<0)
        {   
            dy=-dy;

            nTmp=y2;
            y2=y1;
            y1=nTmp;

            nAlt=1;
        }else
            if (dx<0)
            {   
                dx=-dx;

                nTmp=x2;
                x2=x1;
                x1=nTmp;

                nAlt=1;
            }
    }

    if (nAlt)
    {

        if(dx>=dy)       // The slope is less than 45 degres
        {
            y=y2;
            for(x=x1; x<=x2; x++)
            {
                // Call your function to draw a pixel here.
                SetPixelLCD(x,y,nState);
                StepVal+=dy;
                if(StepVal>=dx) // Increment y if enough x steps
                                // have been taken.
                {
                    y--;
                    StepVal-=dx;    // Reset StepVal, but  
                    // not to 0.  This gives even slopes.
                }
            }
        }
        else    // The slope is greater than 45 degrees, just like 
                // above, but with y instead of x.
        {
            x=x2;
            for(y=y1; y<=y2; y++)
            {
                // Call your function to draw a pixel here.
                SetPixelLCD(x,y,nState);
                StepVal+=dx;
                if(StepVal>=dy)
                {
                    x--;
                    StepVal-=dy;
                }
            }
        }
        return;
    }

    if(dx>=dy)       // The slope is less than 45 degres
    {
        y=y1;
        for(x=x1; x<=x2; x++)
        {
            // Call your function to draw a pixel here.
            SetPixelLCD(x,y,nState); 
            StepVal+=dy;
            if(StepVal>=dx) // Increment y if enough x steps
                            // have been taken.
            {
                y++;
                StepVal-=dx;    // Reset StepVal, but  
                // not to 0.  This gives even slopes.
            }
        }
    }
    else    // The slope is greater than 45 degrees, just like 
            // above, but with y instead of x.
    {
        x=x1;
        for(y=y1; y<=y2; y++)
        {
            // Call your function to draw a pixel here.
            SetPixelLCD(x,y,nState);
            StepVal+=dx;
            if(StepVal>=dy)
            {
                x++;
                StepVal-=dy;
            }
        }
    }
    return;
}
void DrawLineLCD(intx1、inty1、intx2、inty2、intnstate)
{
无符号整数nTmp;
无符号整数nAlt=0;
int x,y;//其中是当前像素。
int dx;//dx是x的增量
int dy;//dy是y的增量
int-StepVal=0;//用于计算何时递增另一个变量的变量
轴
如果(x1>x2&&y1>y2)
{
nTmp=x2;
x2=x1;
x1=nTmp;
nTmp=y2;
y2=y1;
y1=nTmp;
dx=x2-x1;//dx是x的增量
dy=y2-y1;//dy是y的增量
}否则
{
dx=x2-x1;//dx是x的增量
dy=y2-y1;//dy是y的增量
如果(dy=dy)//斜率小于45度
{
y=y1;
对于(x=x1;x=dx)//如果有足够的x步,则增加y
//已经被带走了。
{
y++;
StepVal-=dx;//重置StepVal,但是
//不为0。这将提供均匀的坡度。
}
}
}
否则//坡度大于45度,就像
//上面,但用y代替x。
{
x=x1;
对于(y=y1;y=dy)
{
x++;
StepVal-=dy;
}
}
}
返回;
}
给定一个点
(x1,y1)
和一个斜率
m
,则直线上的任何其他点由

y = y1 + m*(x-x1)  // point is (x,y)
给定一个点
(x1,y1)
和一个斜率
m
,则直线上的任何其他点由

y = y1 + m*(x-x1)  // point is (x,y)

200的起点,即x或y?200的起点,即x或y?