OpenGL Grass rendering

Question

我一直在研究渲染草地的不同技术。我决定使用几何着色器生成的草，所以当我渲染它们为GL_POINTS时，我可以在飞行中生成三角形风扇，但我没有看到我想看到的性能。在10万个草叶的情况下，我的帧率可能是20-50，而且我有一个不错的GPU。我想知道如果我的方法是错误的，或者如果我达到了我的GPU的限制，或者如果我做的事情不正确，或者如果他们是一个更快的方式(我的目标是单独的刀片，我可以理想地操纵顶点)。纹理我使用256x256

我的渲染步骤是:

创建VAO和VBO，存储位置和绑定一次:

   float[] GrassLocations= new float[100000];
   int vaoID = createVAO();
 . //bind VBO to VAO
   storeDataInAttributeList(0, 3, GrassLocations,0,0);

我然后渲染:

    GL30.glBindVertexArray(VAO);
    GL20.glEnableVertexAttribArray(0);
    GL13.glActiveTexture(GL13.GL_TEXTURE0);
    GL11.glBindTexture(GL11.GL_TEXTURE_2D, texture);
    GL11.glDrawArrays(GL11.GL_POINTS, 0, 100000);
    GL20.glDisableVertexAttribArray(0);
    GL30.glBindVertexArray(0);

然后My Vertex Shader:

#version 400

layout (location = 0) in vec3 VertexLocation;

uniform float time;
out vec3 offsets;
out vec3 Position;
out vec3 Normal;
out vec2 TexCoord;
out float visibility;
uniform mat4 transformationMatrix;
uniform mat4 viewMatrix;
uniform mat4 MVPmatrix;
uniform mat4 modelViewMatrix;
const float density = .007;
const float gradient = 1.5;
out float Time;
void main()
{
Time = time;
vec4 worldPosition = transformationMatrix * vec4(VertexLocation,1.0);
vec4 positionRelativeToCam = modelViewMatrix* vec4(VertexLocation,1.0); 
Normal = vec3(0,1,0);
Position = vec3(  worldPosition );
gl_Position = MVPmatrix* vec4(VertexLocation,1.0);
float distance = length(positionRelativeToCam.xyz);
visibility = exp(-pow((distance * density), gradient));
visibility = clamp(visibility,0.0,1.0);
offsets = offset;

}

我去掉了顶点着色器，只留下GL_POSITION，仍然不是问题。My Geometry Shader:

#version 400
layout( points ) in;
layout( triangle_strip, max_vertices = 10 ) out;
float Size2=1;   // Half the width of the quad

in vec3 Position[];
in vec3 Normal[];
in vec3 offsets[];
out vec3 position;
out vec3 normal;
in float Time[];
out vec2 TexCoord;
out vec3 color;
const  float width = 5;
void main()
{
position = Position[0];
normal = Normal[0];
color = offsets[0];
gl_Position =  (vec4(-Size2*width,-Size2,0.0,0.0) + gl_in[0].gl_Position);
TexCoord = vec2(0.0,0.0);
EmitVertex();
gl_Position =  (vec4(Size2*width,-Size2,0.0,0.0) + gl_in[0].gl_Position);
TexCoord = vec2(1.0,0.0);
EmitVertex();
gl_Position = (vec4(-Size2*width+(Time[0].x),10,0.0,0.0) + 
gl_in[0].gl_Position);
TexCoord = vec2(0.0,.25);
EmitVertex();
gl_Position = (vec4(Size2*width+(Time[0].x),10,0.0,0.0) + 
gl_in[0].gl_Position);
TexCoord = vec2(1.0,.25);
EmitVertex();
///////////////////////////////////////////////////
gl_Position =  (vec4(-Size2*width+(Time[0].x)*2,15,0.0,0.0) +   
gl_in[0].gl_Position);
TexCoord = vec2(0.0,.50);
EmitVertex();
gl_Position =  (vec4(Size2*width+(Time[0].x)*2,15,0.0,0.0) +   
gl_in[0].gl_Position);
TexCoord = vec2(1.0,.50);
EmitVertex();   
///////////////////////////////////////////////////
gl_Position =  (vec4(-Size2*width+(Time[0].x)*3,25,0.0,0.0) + 
gl_in[0].gl_Position);
TexCoord = vec2(0.0,.75);
EmitVertex();
 gl_Position =  (vec4(Size2*width+(Time[0].x)*3,25,0.0,0.0) + 
gl_in[0].gl_Position);
TexCoord = vec2(1.0,.75);
EmitVertex();  
 ///////////////////////////////////////////////////
gl_Position =  (vec4(-Size2*width,Size2*7,Time[0].x,0.0) +   
gl_in[0].gl_Position);
TexCoord = vec2(0.0,1.0);
EmitVertex();
gl_Position = (vec4(Size2*width,Size2*7,Time[0].x,0.0) +     
gl_in[0].gl_Position);
TexCoord = vec2(1.0,1.0);
EmitVertex();    
}

和我的片段着色器:(这是在一个延迟的引擎，我也尝试过向前渲染，我不认为性能打击是在这里)

#version 400
in vec2 TexCoord;
layout (binding=0) uniform sampler2D SpriteTex;
in vec3 color;
in vec3 normal;
in vec3 position;
layout( location = 0 ) out vec4 FragColor;
void main() {
vec4 texColor =  texture(SpriteTex,TexCoord);
vec4 posColor = vec4(position.xyz,0);

gl_FragData[1] =  posColor;
gl_FragData[2] =  vec4(normal,1);

if(texColor.a<.5){
discard;
}

gl_FragData[0] = texColor;
}

Answer 1

您需要的是一种称为实例化的技术。我所链接的教程对于弄清楚如何实例化是非常棒的。

我可能会建议你避免使用几何着色器(因为几何着色器在其目的是扩展顶点数量时通常不能很好地缩放)，而只是定义一个包含绘制单个叶片(或补丁)草所需的所有顶点的缓冲区，然后使用实例来重新绘制该对象数千次