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# Normal of sine

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### #1alkisbkn  Members   -  Reputation: 411

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Posted 09 August 2012 - 08:15 PM

Hello,

I have been developing a simple ocean wave "heightfield" composed out of 6 sine waves. Now, I need to find the normal of each vertex, and I am stuck basically. How I calculate my waves:
[source lang="cpp"]float2 dir = float2(1.0f, 0.0f);float dir_vertex = dot(dir, wave_vertex.xz); float wave1 = 25.0f * pow(sin( (2 * PI / 800) * dir_vertex - time * 2 * PI / 45.0f - 105.0f),2);// .... and so on for the others ...vertex.y = wave1 + wave2 + wave3 + wave4 + wave5 + wave6;[/source]
(I know looks ugly, but it works)

Is there a computationally cheap way to find the normal (this is for an iOS game).

Thanks a bunch!

### #2jefferytitan  Crossbones+   -  Reputation: 2219

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Posted 09 August 2012 - 08:32 PM

If my maths does not fail me, the slope of the normal = the derivative, and the derivative of sin(x) is cos(x).

### #3Álvaro  Crossbones+   -  Reputation: 13637

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Posted 09 August 2012 - 08:43 PM

Since you seem to have your surface expressed as y = F(x,z), you can compute two tangent vectors as (1,Fx(x,z),0) and (0,Fz(x,z),1), where Fx and Fz are the derivatives of F with respect to x and z respectively. You can then compute the cross product of these two vectors to get a normal vector. And then you probably need to normalize the resulting vector.

Post again if you try to do that for a while and don't succeed.

### #4alkisbkn  Members   -  Reputation: 411

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Posted 10 August 2012 - 05:29 AM

Hi and thanks!

It does make sense now. However, I don't think I am getting the correct results. How I calculate my wave, tangent, bitangent and normal:
[source lang="c++"]float2 dir = float2(1.0f, 0.0f);float dir_vertex = dot(dir, wave_vertex.xz); float wave1 = 25.0f * pow(sin( (2 * PI / 800) * dir_vertex - time * 2 * PI / 45.0f - 105.0f),2);float3 tan1 = float3(1.0, 2 * cos( (2 * PI / 800) * dir_vertex - time * 2 * PI / 45.0f - 105.0f), 0.0);float3 bitan1 = float3(0.0, 2 * cos( (2 * PI / 800) * dir_vertex - time * 2 * PI / 45.0f - 105.0f), 1.0);float3 normal1 = normalize(cross(bitan1, tan1));[/source]
Maybe my derivation is wrong. A power exponent goes in front of the base number when we derive, correct? Also the constant = 1, thus the amplitude is removed from the tan1 and bitan1 equations. I have attached a screenshot of the current output (for 2 perpendicular waves). To my understanding, when the wave is at full amplitude, the normal should be (0,1,0) (green colour), however it doesn't seem so in the screenshot.

Moreover, to get the final vertex normal, I do:
normal = normalize(normal1 + normal2);

### #5haegarr  Crossbones+   -  Reputation: 4420

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Posted 10 August 2012 - 07:44 AM

The way you compute wave1(x,z) uses dir_vertex, which itself is a function of (x,z). The way you compute tan1 and bitan1 neglect this fact! Notice that alvaro hinted at computing the derivates dF(x,z)/dx and dF(x,z)/dz.

In fact, the formula you use is something like
wave1(x,z) := ( sin( fs * ( dirx * x + dirz * z ) + ft * t + p ) )2
where fs denoting a spatial frequency, ft denoting a temporal frequency, and p denoting a phase.

For the purpose of spatial derivative the term ft * t + p =: ct is a constant, and fx := fs * dirx, fz := fs * dirz can be used for simplicity. Then IIRC the derivatives are
tan1 := d wave(x,z) / d x = ( sin( fx * x + fz * z + ct ) )2 / d x = 2 sin( fx * x + fz * z + ct ) * cos( fx * x + fz * z + ct ) * fx
and
bitan1 := d wave(x,z) / d z = ( sin( fx * x + fz * z + ct ) )2 / d z = 2 sin( fx * x + fz * z + ct ) * cos( fx * x + fz * z + ct ) * fz
due to the chain rule. Please check this twice.

Edited by haegarr, 10 August 2012 - 07:59 AM.

### #6alkisbkn  Members   -  Reputation: 411

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Posted 10 August 2012 - 09:45 AM

Hi haegarr,

I am now testing using 1 wave only and by implementing the equation you mentioned before. The problem is that the normal seems to always be (0,1,0) using this implementation:
[source lang="cpp"]float2 dir = float2(1.0, 0.0);float dir_vertex = dot(dir, wave_vertex.xz); float fs = 2 * PI / 800.0f;float t = time * 2 * PI / 45.0f; float wave1 = 30.0f * pow(sin(fs * dir_vertex + t), 2); float fx = dir.x * fs;float fz = dir.y * fs; float3 tan1 = float3(1.0, 2 * sin(fx * wave_vertex.x + fz * wave_vertex.z + t) * cos (fx * wave_vertex.x + fz * wave_vertex.z + t) * fx, 0.0); float3 bitan1 = float3(0.0, 2 * sin(fx * wave_vertex.x + fz * wave_vertex.z + t) * cos (fx * wave_vertex.x + fz * wave_vertex.z + t) * fz, 1.0); float3 normal1 = normalize(cross(bitan1, tan1));[/source]
resulting in a green fragment color.

Edited by alkisbkn, 10 August 2012 - 09:46 AM.

### #7haegarr  Crossbones+   -  Reputation: 4420

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Posted 10 August 2012 - 12:51 PM

Well, using the correspondence
2 * sin(k) * cos(k) = sin(2k)
simplifies the implementation. In your given case there is fz = 0.0, giving
tan1 = [ 1, sin( 2 * fx * wave_vertex.x + 2 * t ), 0 ]
bitan1 = [ 0, 0, 1 ]

Here tan1 shouldn't be constant over x, should it? Hmmm. Perhaps using a graph plotter with that sine function gives an answer...

### #8alkisbkn  Members   -  Reputation: 411

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Posted 10 August 2012 - 01:22 PM

This appears to be tricky afterall. Shouldn't the vertex and direction z be affecting the tan1 as well?

Edit:
I changed the wave function to something more simple for now, I basically removed the power to 2.
So, the derivative of:

float wave1 = 30.0f * sin(fs * dir_vertex + t);


should now be:
float tan1y = cos(fs * dir.x * wave_vertex.x)
float bitan1y = cos(fs * dir.y * wave_vertex.z);

(is t in the derivative equation? I think not, as it's constant)

However this too doesn't give me a correct result.

Edited by alkisbkn, 10 August 2012 - 01:45 PM.

### #9alkisbkn  Members   -  Reputation: 411

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Posted 10 August 2012 - 06:23 PM

Okay, it seems that the correct derivative of "wave1" equation is the following:
float3 tan1 = float3(1.0, 30.0 * fs * cos(fs * waveVertex.x + t), 0.0);
float3 bitan1 = float3(0.0, 30.0 * fs * cos(fs * waveVertex.x + t), 1.0);

for tan and bitan. Then I do
normal = normalize(cross(bitan1, tan1));

For 1 wave it seems to be correct, however how should I go about adding many wave's normals together? Just summing them and normalizing the result?

### #10Álvaro  Crossbones+   -  Reputation: 13637

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Posted 10 August 2012 - 09:16 PM

Normals don't add up, but derivatives do. So compute the derivatives, add them together and then use the cross product to compute the normal.

### #11alkisbkn  Members   -  Reputation: 411

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Posted 11 August 2012 - 06:16 AM

Thanks everyone, I think it is working correctly now. Below I am showing how I am doing it:
float height = 50.0 * sin(fs * waveVertex.x + t * _Time.y) +
125.0 * sin(fs * waveVertex.z + t * _Time.y + PI/4.0) +
135.0 * sin(-fs * waveVertex.x/4.0 - t * _Time.y * 0.75 + PI/2.0) +
165.0 * sin(-fs * waveVertex.z/8.0 - t * _Time.y * 0.5 + PI);
height *= 0.25;

float derivative = 150.0 * fs * cos(fs * waveVertex.x + t * _Time.y) +
125.0 * fs * cos(fs * waveVertex.z + t * _Time.y + PI/4.0) +
135.0 * -fs * cos(-fs * waveVertex.x/4.0 + t * _Time.y * 0.75 + PI/2.0) +
165.0 * -fs * cos(-fs * waveVertex.z/8.0 + t * _Time.y * 0.5 + PI);

float3 tan1 = float3(1.0, derivative, 0.0);
float3 bitan1 = float3(0.0, derivative, 1.0);

normal = normalize(cross(bitan1, tan1));

And this gives me the result as you see in the screenshot.

#### Attached Thumbnails

Edited by alkisbkn, 11 August 2012 - 06:39 AM.

### #12alkisbkn  Members   -  Reputation: 411

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Posted 11 August 2012 - 03:44 PM

Which still doesn't look correct :S

### #13haegarr  Crossbones+   -  Reputation: 4420

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Posted 12 August 2012 - 03:50 AM

If you have a height field y(x,z) and compute the derivatives
t(x,z) := d y(x,z) / d x
b(x,z) := d y(x,z) / d z
then t denotes the rate of change of y when stepping in direction of x, and analogously b denotes the rate of change of y when stepping in direction of z. Written as vectors, you'll choose the unit length 1 for the step in direction of x or z, resp., and t or b for the corresponding change of y, i.e.
T(x,z) := [ 1; t(x,z); 0 ]
B(x,z) := [ 0; b(x,z); 1 ]

The cross product of these 2 vectors gives you the normal. However, you have 2 possibilities T x B and B x T, but only one is valid for your purpose (depending on your choice of co-ordinate system). Above you've chosen
N(x,z) := B(x,z) x T(x,z) = [ -t(x,y); 1; -b(x,z) ]
It has to be normalized, of course.

So what does this mean w.r.t. some picked slopes?
a) For a flat height field you'll get
y(x,z) = c ==> t(x,z) = 0; b(x,z) = 0 ==> N(x,z) = [ 0; 1; 0 ]
b) For a wave in x direction only you'll get at a position with slope of 45° (i.e. increasing)
t(x,z) = 1; b(x,z) = 0 ==> N(x,z) = [ -1; 1; 0 ]
c) For a wave in z direction only you'll get at a position with slope of -45° (i.e. decreasing)
t(x,z) = 0; b(x,z) = -1 ==> N(x,z) = [ 0; 1; 1 ]

This seems me okay.

Now, there are some mistakes in your pre-previous post:

1. Notice that normals are not invariant to non-uniform scaling.

E.g. scaling the height by a factor of 0.25 is non-uniform, because it has an effect on y only but not on x and z. If you scale y(x,z) by s and compute N as above, you'll see that
N'(x,z) = [ -t(x,y) * s; 1; -b(x,z) * s ] != N(x,z) * s

2. The 1st sin() is scaled by 50, while the 1st cos() is scaled by 150.

3. The argument in the 3rd sin() has a frequency of -fs/4, but the factor of the 3rd cos() shows -fs only.

4. The argument in the 4th sin() has a frequency of -fs/8, but the factor of the 4th cos() shows -fs only.

5. The wave function depends on both x and z position, but you compute only one derivative and use it for both directions.

Edited by haegarr, 12 August 2012 - 03:52 AM.

### #14alkisbkn  Members   -  Reputation: 411

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Posted 12 August 2012 - 06:32 AM

haegarr,

thank you for your help! I had 2 mistakes in my current code (the points from 2 - 5, I had corrected them before):
I was calculating the normal as:
normal = normalize(float3(derivativeX, 1, derivativeZ));
where it should have been with minus sign, as you said in your post.

The second mistake, was the scaling of the final accumulated height. I didn't expect it to have such a big impact!

So, thank you for your help, here is a screenshot of how the water looks now

#### Attached Thumbnails

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