距离一条线 - Shadertoy

问题描述 投票:0回答:1

所以我是一个很新的人,而且我一直在玩一些东西,但有一点我无法弄清楚如何计算一个点到一条线的距离。我可以用铅笔和纸自己轻松地做到这一点,但是当我真正尝试将它应用于shadertoy时,不知何故我一直搞砸了。这就是我所拥有的:

void mainImage( out vec4 fragColor, in vec2 fragCoord )
{
    vec2 uv = fragCoord/iResolution.xy;

    uv.x -= 0.5; //Puts the origin at the center of the screen

    float r; //Red value
    float g; //Green value
    float b; //Blue value

    float lm = 1.1; //slope
    float lb = 0.5; //intercept

    //Slope/intercept declarations manipulate line
    //second line
    float lmp = 0.0-(1.0/lm); //calculates the slope of the perpendicular line
    float lbp = lb + uv.y + lmp*(uv.x); //and the intercept

    //Intersection
    float ix = (lbp-lb)/(lm-lmp); //Intersection Y
    float iy = lm*(ix)+lb; //Intersection X based off of the first equation
    //distance
    float dist = sqrt(pow((uv.x - ix),2.0)+pow((uv.y - iy),2.0));

    if (dist < 0.05){

        r = 1.0;
        g = 1.0;
        b = 1.0;
    }

    fragColor = vec4(r,g,b,1.0); //supposed to draw a line
}

现在来自Flyguy的“Neontoy”项目不仅有效,而且比我的短得多。

float dfLine(vec2 start, vec2 end, vec2 uv)
{
    start *= scale;
    end *= scale;

    vec2 line = end - start;
    float frac = dot(uv - start,line) / dot(line,line);
    return distance(start + line * clamp(frac, 0.0, 1.0), uv);
}
glsl shader
1个回答
0
投票

如果你有一条线,由一个点(O)和一个方向(D)给出,那么线上最近的点到点p可以按如下方式计算

X = O + D * dot(P-O, D);

2个矢量的点积等于2个矢量之间的角度的余弦乘以两个矢量的幅度(长度)。

enter image description here

dot( A, B ) == | A | * | B | * cos( alpha ) 

VD的点积等于线(OD)和矢量V = P - O之间的角度的余弦,乘以V的量(长度),因为Dunit vectorD的长度是1.0) ,

在你的情况下,该行由Linear equation以下列形式给出:

f(x) = lb + lmp * x;

线上的点是(0.0,lb),方向是(1.0,lmp)。

将其应用于您的代码会产生以下片段着色器:

float dfLine(vec2 O, vec2 dir, vec2 P)
{
    vec2 D = normalize(dir);
    vec2 X = O + D * dot(P-O, D);

    return distance(P, X);
}

void mainImage( out vec4 fragColor, in vec2 fragCoord )
{
    vec2 uv = fragCoord/iResolution.xy;
    uv.x -= 0.5; //Puts the origin at the center of the screen

    float lm = 1.1; //slope
    float lb = 0.5; //intercept

    float dist = dfLine(vec2(0.0, lb), vec2(1.0, lmp));

    float onLine = step(dist, 0.05);   // 1.0 if on line, else 0.0
    vec3  color  = onLine * vec3(1.0); 

    fragColor = vec4(color, 1.0);
}
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