着色器太阳位置 - glsl
我正在尝试制作一个着色器来模拟太阳位置及其在物体中反射的光。
为了模拟太阳轨迹，我有一个计时器，并且灯光位置定义为：
fvLightPosition.x=-cos(Time)*speed;
fvLightPosition.y=sin(Time)*speed;
fvLightPosition.z=100.0;
计时器是一个变量 float Time0_X
我几乎得到了正确的轨迹，除了它是颠倒的，并且不知何故看起来有点诡异的。谁能告诉我如何以正确的方式模拟太阳轨迹？
我正在使用 RenderMonkey 来制作着色器。
I am trying to make a shader that simulates the sun position and the light that it reflects in a object.
To simulate the sun trajectory I have a timer, and the light position is defined by:
fvLightPosition.x=-cos(Time)*speed;
fvLightPosition.y=sin(Time)*speed;
fvLightPosition.z=100.0;
The timer is a variable float Time0_X
I almost get the right trajectory, except that is upside down, and somehow look a bit weird. Can anyone give me a tip how to simulate the sun trajectory in the right way?
I am using RenderMonkey to make the shader.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
确保缩放
Time值，使其位于0和2*PI弧度（分别对应于 0 度和 180 度）之间。如果您的原始Time值范围从0到某个数字MAX，您可以按如下方式进行缩放：(Time / MAX) * 2PI。我不确定速度指的是什么，但您很可能不应该乘以它。
cos和sin值代表太阳与原点之间向量的 X 和 Y 分量，因此您应该乘以太阳距地球中心的距离。场景，通常是恒定的。Make sure that you scale your
Timevalue so that it lies between0and2*PIradians (which corresponds to 0 and 180 degrees respectively). If your originalTimevalue ranges from0to some numberMAX, you can do the scaling like this:(Time / MAX) * 2PI.I'm not sure what
speedrefers to, but you should most likely not be multiplying by that. Thecosandsinvalues represent the X and Y components of the vector between the sun and the origin, so you should multiply by the distance of the sun from the centre of your scene, which is usually constant.