CODEWITHSUNDEEP
c++opengl
nullpotent
nullpotent

Reputation: 9260

Dynamic 2d shadows - Blending issue

Good day my dear community.

I'm working on dynamic shadows for a game I shall work on, but as it usually happens I bring you a problem, in hope (I'm certain actually) that someone will help.

This is where I am right now:enter image description here

Notice the red square, I want it to gradually fade away as the light source moves out of the sight. I do check if a point of a polygon is inside circle's radius, but that of course doesn't solve it; as I said I want it to fade gradually until it completely blacks out If the light is too far away.

There's one idea on my mind but I hope for a better one. I will not talk about it since it's really the last option and I find it to be a 'brute force' technique.

This is how I render my light:

    glBegin(GL_TRIANGLE_FAN);
    {
        Graphics::Instance()->SetColor(r_,g_,b_,intensity_);
        glVertex2f(posX_,posY_);

        glColor4f(0.f, 0.f, 0.f, 0.0f);

        for (angle_=0.0; angle_<=3.14159265*2; angle_+=((3.14159265*2)/64.0f) )
        {
            glVertex2f(range_*(float)cos(angle_) + posX_,
                       range_*(float)sin(angle_) + posY_);
        }

        glVertex2f(posX_+range_, posY_);
    }

And this is how I blend it: 

glBlendFunc(GL_SRC_ALPHA, GL_ONE);
l0->Render();

glBlendFunc(GL_SRC_ALPHA,GL_ONE_MINUS_SRC_ALPHA);
l0->ProjectShadow(*mmm);
l0->ProjectShadow(*bb);

That is all. If I didn't made myself clear or If I missed to post relevant code, please do say so and don't downvote.

Upvotes: 2

Views: 727

Answers (2)

Jon Cage
Jon Cage

Reputation: 37490

How about calculating the range to the center of your red square from the light sources center? Normalise that value to a suitable range and adjust the transparency or colour of the red square? Something like this:

double Range(double x1, double y1, double x2, double y2)
{
    double xDist = x1-x2;
    double yDist = y1-y2;
    return math::sqrt(xDist*xDist+yDist*yDist);
}

double CalcIntensity(double lightX, double lightY, double lightRadius, double objectX, double objectY)
{
    double range = Range(lightX, lightY, objectX, objectY);
    double intensity;
    if( range > lightRadius )
    {
        intensity = 0.0;
    }
    else
    {
        intensity = range/lightRadius;
    }
    return intensity;
}

Then just call CalcIntensity and feed in the reletive positions of the light and the square and the radius of the light.

[Edit] ...or this would be a slightly more optomised version if you're not pre-checking it's within the lights radius:

double CalcIntensity(double lightX, double lightY, double lightRadius, double objectX, double objectY)
{
    double intensity = 0.0;
    double xDist = lightX-objectX;
    if( xDist < lightRadius )
    {
        yDist = lightY-objectY;
        if( yDist < lightRadius )
        {
            double range = math::sqrt(xDist*xDist+yDist*yDist);
            intensity = range/lightRadius;
        }
    }

    return intensity;
}

Upvotes: 3

Jonas B&#246;tel
Jonas B&#246;tel

Reputation: 4482

Well the light is at full brightness at posX,posY and fully "depleted" (i.e. black) at range. The values between are interpolated linearly. Therefore a point at any position with distance d from the light source is lit by rgb * (d/range).

If you now calculate the distances d_i of each vertex v_i of your red square, you can apply shadowing to each vertex color c_i by multiplying (d_i/range) to it. If you want the whole square to appear in the same color regardless of further-away vertices just use the distance of its center for every vertex, that is set the color only once before you draw the square.

Upvotes: 1

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