directx-11 - 在 directx 11 中使用实例渲染来渲染粗线

Question

我想在 directx11 中实现粗线。 我想，我可以使用实例渲染技术为每条线渲染“高质量”几何图形，如下图所示:

P1 和 P2 表示不等距线顶点，以“D3D11_PRIMITIVE_TOPOLOGY_LINELIST”形式给出。线条粗细被存储到常量缓冲区中。每条线都有相同的粗细。 实例几何体还有一个索引缓冲区，其中包含有关如何将顶点连接到三角形的信息(图中顶点为 I0 - I11)。

如果我将 P1 和 P2 位置以及 SV_VertexID 放入一个顶点着色器线程中，我就可以计算 I0 - I11 顶点的每个位置。所以这不是问题。

问题是:是否可以使用实例渲染技术来实现这一点？

如果是这样:这样使用它是个好主意吗？或者是否有更多的性能方法来实现粗圆线？例如使用几何着色器或仅使用该几何体进行 1000 次绘制调用...

Answer 1

我尝试了很多使用实例化渲染的想法，但现在我将想法改为geometryshader，并且它非常容易实现。 作为输入，它变成一条线(2 个顶点)，输出是 30 个三角形。

这里是像素着色器输入结构

struct PS_INPUT
{
    float4 Pos : SV_POSITION;
};

这里是几何着色器:

#include <psiPosition.hlsli>

//#pragma warning (disable:3206)
//#pragma warning (disable:3554)

static const float PI = 3.1415926f;
static const float fRatio = 2.0f;
static float fThickness = 0.01f;

void addHalfCircle(inout TriangleStream<PS_INPUT> triangleStream, int nCountTriangles, float4 linePointToConnect, float fPointWComponent, float fAngle)
{
    PS_INPUT output = (PS_INPUT)0;
    for (int nI = 0; nI < nCountTriangles; ++nI)
    {
        output.Pos.x = cos(fAngle + (PI / nCountTriangles * nI)) * fThickness / fRatio;
        output.Pos.y = sin(fAngle + (PI / nCountTriangles * nI)) * fThickness;
        output.Pos.z = 0.0f;
        output.Pos.w = 0.0f;
        output.Pos += linePointToConnect;
        output.Pos *= fPointWComponent;
        triangleStream.Append(output);

        output.Pos = linePointToConnect * fPointWComponent;
        triangleStream.Append(output);

        output.Pos.x = cos(fAngle + (PI / nCountTriangles * (nI + 1))) * fThickness / fRatio;
        output.Pos.y = sin(fAngle + (PI / nCountTriangles * (nI + 1))) * fThickness;
        output.Pos.z = 0.0f;
        output.Pos.w = 0.0f;
        output.Pos += linePointToConnect;
        output.Pos *= fPointWComponent;
        triangleStream.Append(output);

        triangleStream.RestartStrip();
    }
}

[maxvertexcount(42)]
void main(line PS_INPUT input[2], inout TriangleStream<PS_INPUT> triangleStream)
{
    PS_INPUT output= (PS_INPUT)0;

    int nCountTriangles =6;

    float4 positionPoint0Transformed = input[0].Pos;
    float4 positionPoint1Transformed = input[1].Pos;

    float fPoint0w = positionPoint0Transformed.w;
    float fPoint1w = positionPoint1Transformed.w;

    //calculate out the W parameter, because of usage of perspective rendering
    positionPoint0Transformed.xyz = positionPoint0Transformed.xyz / positionPoint0Transformed.w;
    positionPoint0Transformed.w = 1.0f;
    positionPoint1Transformed.xyz = positionPoint1Transformed.xyz / positionPoint1Transformed.w;
    positionPoint1Transformed.w = 1.0f;

    //calculate the angle between the 2 points on the screen
    float3 positionDifference = positionPoint0Transformed.xyz - positionPoint1Transformed.xyz;
    float3 coordinateSysten = float3(1.0f, 0.0f, 0.0f);

    positionDifference.z = 0.0f;
    coordinateSysten.z = 0.0f;

    float fAngle = acos(dot(positionDifference.xy, coordinateSysten.xy) / (length(positionDifference.xy) * length(coordinateSysten.xy)));

    if (cross(positionDifference, coordinateSysten).z < 0.0f)
    {
        fAngle = 2.0f * PI - fAngle;
    }

    fAngle *= -1.0f;
    fAngle -= PI *0.5f;

    //first half circle of the line
    addHalfCircle(triangleStream, nCountTriangles, positionPoint0Transformed, fPoint0w, fAngle);
    addHalfCircle(triangleStream, nCountTriangles, positionPoint1Transformed, fPoint1w, fAngle + PI);

    //connection between the two circles
    //triangle1
    output.Pos.x = cos(fAngle) * fThickness / fRatio;
    output.Pos.y = sin(fAngle) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint0Transformed;
    output.Pos *= fPoint0w; //undo calculate out the W parameter, because of usage of perspective rendering
    triangleStream.Append(output);

    output.Pos.x = cos(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness / fRatio;
    output.Pos.y = sin(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint0Transformed;
    output.Pos *= fPoint0w;
    triangleStream.Append(output);

    output.Pos.x = cos(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness / fRatio;
    output.Pos.y = sin(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint1Transformed;
    output.Pos *= fPoint1w;
    triangleStream.Append(output);

    //triangle2
    output.Pos.x = cos(fAngle) * fThickness / fRatio;
    output.Pos.y = sin(fAngle) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint0Transformed;
    output.Pos *= fPoint0w;
    triangleStream.Append(output);

    output.Pos.x = cos(fAngle) * fThickness / fRatio;
    output.Pos.y = sin(fAngle) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint1Transformed;
    output.Pos *= fPoint1w;
    triangleStream.Append(output);

    output.Pos.x = cos(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness / fRatio;
    output.Pos.y = sin(fAngle + (PI / nCountTriangles * (nCountTriangles))) * fThickness;
    output.Pos.z = 0.0f;
    output.Pos.w = 0.0f;
    output.Pos += positionPoint1Transformed;
    output.Pos *= fPoint1w;
    triangleStream.Append(output);
}

我知道它是极其硬编码的，但至少它可以工作;) 现在是一个具有较粗线条的立方体图片(第一张是透视投影，第二张是正交投影)

我希望我能帮助别人解决这个问题。如果有人对实现粗线有更好的想法，请发表评论。