编辑:最后基于附加问题信息的示例
对我来说听起来像是线性插值 - 如果你从二维的角度来看它,你有两点:
(x1, y1) = point one on heightmap
(x2, y2) = point two on heightmap
(x1,x2)
和一个未知高度之间的某个点:
pu = (xu, yu)
LERP 的通用公式是:
pu = p0 + (p1 - p0) * u
在哪里:
p0
= 第一个值
p1
= 第二个值
u
= % 你的未知点介于两者之间(p0,p1)
在这里,我们会说p0
==y2
和p1
== y1
。x1
现在我们需要确定未知点之间的“多远” x2
- 如果您知道两个高度图点的角度,这很容易:
u = ang(xu) - ang(x1) / (ang(x2) - ang(x1))
或者,您可以将角度投影到以Max(y1,y2)
这种方式获得“未知 x 位置”,然后计算上述值。
所以,让我们尝试一个人为的例子:
p1 = point one in map = (1,2) therefore ang(p1) ~ 57 degrees
p2 = point two in map = (2,4) therefore ang(p2) ~ 114 degrees
请注意,这里的“x 轴”是沿着球体表面的,“y 轴”是距球心的距离。
pu = object location = py @angle 100 degrees ~ 1.74 radians
px = (1.74 rad - 1 rad ) / (2 rad - 1 rad) = 0.74 / 1.0 = 0.74 => 74%
py = y0 + (y1 - y0) * u
= 2 + (4 - 2) * 0.74
= 2.96
希望我没有在某处丢掉或放错标志...... :)
好的,您的示例代码 - 我已经对其进行了一些调整,这就是我想出的:
首先,让我们定义一些我自己的助手:
public static class MathHelper
{
public const double TwoPi = Math.PI * 2.0;
public static double DegToRad(double deg)
{
return (TwoPi / 360.0) * deg;
}
public static double RadToDeg(double rad)
{
return (360.0 / TwoPi) * rad;
}
// given an upper/lower bounds, "clamp" the value into that
// range, wrapping over to lower if higher than upper, and
// vice versa
public static int WrapClamp(int value, int lower, int upper)
{
return value > upper ? value - upper - 1
: value < lower ? upper - value - 1
: value;
}
}
我们的测试设置:
void Main()
{
var random = new Random();
// "sea level"
var baseDiameter = 10;
// very chaotic heightmap
heightmap = Enumerable
.Range(0, 360)
.Select(_ => random.NextDouble() * baseDiameter)
.ToArray();
// let's walk by half degrees, since that's roughly how many points we have
for(double i=0;i<360;i+=0.5)
{
var angleInDegrees = i;
var angleInRads = MathHelper.DegToRad(i);
Console.WriteLine("Height at angle {0}°({1} rad):{2} (using getheight:{3})",
angleInDegrees,
angleInRads,
heightmap[(int)angleInDegrees],
getheight(angleInRads));
}
}
double[] heightmap;
还有我们的“getheight”方法:
// assume: input angle is in radians
public double getheight(double angle)
{
//find out angle between 2 heightmap point
double dTheta = MathHelper.TwoPi / (heightmap.Length);
// our "offset" will be how many dThetas we are
double offset = angle / dTheta;
// Figure out two reference points in heightmap
// THESE MAY BE THE SAME POINT, if angle ends up
// landing on a heightmap index!
int lowerAngle = (int)offset;
int upperAngle = (int)Math.Round(
offset,
0,
MidpointRounding.AwayFromZero);
// find closest heightmap points to angle, wrapping
// around if we go under 0 or over max
int closestPointIndex = MathHelper.WrapClamp(
lowerAngle,
0,
heightmap.Length-1);
int nextPointIndex = MathHelper.WrapClamp(
upperAngle,
0,
heightmap.Length-1);
//find heights
double height1 = heightmap[closestPointIndex];
double height2 = heightmap[nextPointIndex];
// percent is (distance from angle to closest angle) / (angle "step" per heightmap point)
double percent = (angle - (closestPointIndex * dTheta)) / dTheta;
// find lerp height = firstvalue + (diff between values) * percent
double lerp = Math.Abs(height1 + (height2 - height1) * percent);
// Show what we're doing
Console.WriteLine("Delta ang:{0:f3}, Offset={1:f3} => compare indices:[{2}, {3}]",
dTheta,
offset,
closestPointIndex,
nextPointIndex);
Console.WriteLine("Lerping {0:p} between heights {1:f4} and {2:f4} - lerped height:{3:f4}",
percent,
height1,
height2,
lerp);
return lerp;
}