我正在创建一个程序生成平面和紫外线的游戏,它的工作大多很好,但有一个地方,紫外线似乎扭曲在一个非常混乱的方式。
简介: uv被映射到“全局”的uv坐标中,而不是局部的,因为纹理必须与相邻的平面对齐。每个纹理都有一个2*2的平铺,我引入了一个2的uv标度,这意味着每个平面"v“坐标将垂直地映射在纹理的0,0.5范围内,以确保水平"u”坐标不会超出0,1的范围。(水平噪声在某种程度上是不可预测的)。代码:
Vector2[] CalculateU(float[] leftXMap, float[] rightXMap) {
int length = leftXMap.Length;
float totalHeight = (length - 1) * 1; // distance between two vertices vertically is 1
float uvheight = totalHeight * uvScale; // uvScale is 2, purpose is to map a plane's v(vertically), into a range of [0, 0.5]..
Vector2[] UMap = new Vector2[length * 2];
for(int i = 0; i < length; i++) {
float left = leftXMap[i];
float right = rightXMap[i];
// make left and right positive.
while (left < 0) {
left += uvheight;
}
while(right < 0) {
right += uvheight;
}
float leftu = (left % uvheight) / uvheight;
float leftv = (i / (float)(length-1)) / uvScale;
float rightu = (right % uvheight) / uvheight;
float rightv = leftv; //(i / (float)length) / 2f;
UMap[i * 2] = new Vector2(leftu, leftv);
UMap[i * 2 + 1] = new Vector2(rightu, rightv);
}
}解释:函数的参数,是生成平面的噪声图。噪音是x坐标。而y坐标是0,1,2,.的简单值。(长度-1)
扭曲的纹理:

放大:

发布于 2018-04-04 20:19:47
我设法解决了这个问题,当左噪声映射值为负值,右噪声映射为正值时,UV地图就会变得混乱。
因为我总是尝试将所有的负值映射到正值,所以左噪声映射,也就是我的左顶点x坐标,将被映射到正值。但是,它将大于正确的噪声映射(右顶点x坐标)。然后它搞砸了紫外线。
我解决了它完全没有映射,并突然意识到,紫外线的地图值可能是负的,我根本不需要地图!
https://stackoverflow.com/questions/49619992
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