使用波形顶点着色器计算球体上的法线

问题描述

我一直在努力为一个球体获得正确的法线，我正在使用一个顶点着色器。该算法可简化为

vert.xyz + = max（0，sin（时间+ 0.004 * vert.x））* 10 * normal.xyz

这会导致波形在球体上滚动。
为了使我的法线正确，我需要改变它们。我可以在给定的x，y，z处取切向量，得到一个垂直向量（0，-vert.z，vert.y），然后用perp向量交叉切线。

虽然我一直在讨论数学问题，但在这一点上它已成为一种个人的仇恨。我已经为衍生工具解决了数百次，但我一直不明确。我怎样才能得到切线？
分解上面的线，我可以做出一个数学函数
$ b $ f（x，y，z）= max（0，sin（时间+ 0.004 *其中Norm（..）是Normalize（（x，y，z）+（x，y，z））

，z） - CenterOfSphere）

解决方案

如果你的曲面点是非线性分布的，或者存在一些数学奇点，或者如果你犯了一个数学错误（在99.99％的情况下），那么通过函数派生得到切线/法线有时会失败。无论如何，您始终可以使用几何方法：

1。你可以很容易地得到切线

• U（x，y，z）= f （x，y，z）-f（x，y，z）;


• = f（x，y + d，z）-f（x，y，z）;


• 其中 d 是足够小的一步


• 和 f（x，y，z）是当前曲面点计算
li>

• 不确定为什么使用3个输入变量我只用2


• 但是因此如果移位点与unshifted


• 使用它而不是 = f（x，y，z + d）-f（x，y，z）;


• 最后不要忘记将 U，V 大小标准化为单位矢量

2。下一步

• 那么您可以简单地（x，y，z）= f（x + d，y，z）-f（x，y）解决U，V代数


• ，通过将 f（x，y，z）替换为完整的等式


• 表面点方程


• 并简化
  

[笔记]

• 有时可以很好地选择 d 可以简化标准化以乘以常量


• 你应该像这样添加法线可视化：


• 实际查看实际发生的情况（针对调试目的）

I've been trying to get the correct normals for a sphere I'm messing with using a vertex shader. The algorithm can be boiled down simply to

vert.xyz += max(0, sin(time + 0.004*vert.x))*10*normal.xyz

This causes a wave to roll across the sphere.In order to make my normals correct, I need to transform them as well. I can take the tangent vector at a given x,y,z, get a perpendicular vector (0, -vert.z, vert.y), and then cross the tangent with the perp vector.

I've been having some issue with the math though, and it's become a personal vendetta at this point. I've solved for the derivative hundreds of times but I keep getting it incorrect. How can I get the tangent?Breaking down the above line, I can make a math function

f(x,y,z) = max(0, sin(time + 0.004*x))*10*Norm(x,y,z) + (x,y,z)

where Norm(..) is Normalize((x,y,z) - CenterOfSphere)

After applying f(x,y,z), unchanged normals

What is the correct f '(x,y,z)?

I've accounted for the weirdness caused by the max in f(...), so that's not the issue.

Edit: The most successful algorithm I have right now is as follows:

Tangent vector.x = 0.004*10*cos(0.004*vert.x + time)*norm.x + 10*sin(0.004*vert.x + time) + 1

Tangent vector.y = 10*sin(0.004*vert.x + time) + 1

Tangent vector.z = 10*sin(0.004*vert.x + time) + 1

2nd Tangent vector.x = 0

2nd Tangent vector.y = -norm.z

2nd Tangent vector.z = norm.y

Normalize both, and perform Cross(Tangent2, Tangent1). Normalize again, and done (it should be Cross(Tangent1, Tangent2), but this seems to have better results... more hints of an issue in my math!).

This yields this

解决方案

Get tangent/normal by derivate of function can sometimes fail if your surface points are nonlinearly distributed and or some math singularity is present or if you make a math mistake (which is the case in 99.99%). Anyway you can always use the geometric approach:

1. you can get the tangents easy by

• U(x,y,z)=f(x+d,y,z)-f(x,y,z);
• V(x,y,z)=f(x,y+d,z)-f(x,y,z);
• where d is some small enough step
• and f(x,y,z) is you current surface point computation
• not sure why you use 3 input variables I would use just 2
• but therefore if the shifted point is the same as unshifted
• use this instead =f(x,y,z+d)-f(x,y,z);
• at the end do not forget to normalize U,V size to unit vector

2. next step

• if bullet 1 leads to correct normals
• then you can simply solve the U,V algebraically
• so rewrite U(x,y,z)=f(x+d,y,z)-f(x,y,z); to full equation
• by substituting f(x,y,z) with the surface point equation
• and simplify

[notes]

• sometimes well selected d can simplify normalization to multipliyng by a constant
• you should add normals visualization for example like this:
• to actually see what is really happening (for debug purposses)

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