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I understand the idea, you must place an hemisphere over the normal, then you need to sum every incoming radiance. Then write the radiance back at the corresponding pixel in the cube map. However I didn't understand how author iterated over the hemisphere and the direct translation from hlsl code to glsl didn't worked as I expected. It smooth the original cube map a lot. As if I dropped resolution from 1024px to 32px.

I changed the hemisphere iteration as below:

    normal = normalize(normal);
vec3 up = vec3(0.0, 1.0, 0.0);
vec3 right = normalize(cross(normal, up));

int index = 0;
vec3 irradiance = vec3(0.0 ,0.0 ,0.0);

for (float longi = 0.0; longi <= 90.0; longi += 3.0)
{
for (float azi = 0.0; azi <= 360.0; azi += 3.0)
{
vec3 sampleVec = (tra * trl * vec4(normal, 1.0)).xyz;
irradiance += texture(iChannel0, sampleVec).rgb * dot(sampleVec, normal);
index++;
}
}

fragColor = vec4((PI * irradiance / float(index)), 1.0);

Generated irradiance map seems to me too bright. Also I don't understand why we are averaging the summed radiance by dividing it to "index" ? Aren't we after the total incoming radiance to a point ?

Edited by afraidofdark

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6 hours ago, afraidofdark said:

I didn't understand how author iterated over the hemisphere

Now I understand how the author traversed over the hemisphere. However there is still a problem. In monte carlo integration, you consider the integration out put as the average of random samples. This is why we divide the sum by "index". However, this part isn't clear to me yet. If I want to approximate x^2 integral over 0 to 5 period, does this method suggest me to take for example 100 random number in that interval and average them, and expect this outcome as my integration result ?

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37 minutes ago, afraidofdark said:

If I want to approximate x^2 integral over 0 to 5 period, does this method suggest me to take for example 100 random number in that interval and average them

I have lived enough enlightenment today thanks to scratchapixel ! I've seen the light ! (Literally) Since I used a cube map to look up incoming radiance, I tie my hands to use an approximation of the actual integral. Because for incoming radiance, I don't have a function which I can integrate over spherical coordinates !

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So the full monte carlo formula is essentially (1 / NumSamples) * Sum(f(x) / p(x), NumSamples), where f(x) is the function you're integrating, and p(x) is probability distribution function (PDF) evaluated for x. The PDF is basically the likelihood of a particular sample being chosen. For "uniform" sampling schemes where samples are distributed evenly across the whole domain (for instance, the surface of a sphere) the PDF is the same for all samples, and so you can pull it out of the sum.

Now for integrating irradiance for a point with normal N, you'll need integrate over the surface of the hemisphere that surrounds N. Uniformly sampling a hemisphere oriented around Z = 1 (the canonical "upper" hemisphere) is pretty simple:

// Returns a direction on the hemisphere around z = 1
Float3 SampleDirectionHemisphere(float u1, float u2)
{
float z = u1;
float r = std::sqrt(std::max(0.0f, 1.0f - z * z));
float phi = 2 * Pi * u2;
float x = r * std::cos(phi);
float y = r * std::sin(phi);

return Float3(x, y, z);
}

As input you need u1 and u2, which are two random variables in the range [0, 1]. These can be generated from a random number generator like rand(), or can be generated by a sequence that produces more a optimal sample distribution that avoids clustering (this is known as Quasi-Monte Carlo sampling).

To make this work for our case, we need to sample the hemisphere surrounding the surface normal. This means we need a transformation that can go from tangent space -> world space. If you're doing normal mapping then you probably already have such a matrix, and if not you can build one by starting with the normal as your Z basis and then generating a perpendicular vector to use as the X basis (after that you can use a cross product to generate the Y basis). Then you just transform the hemisphere sample direction by this matrix, and you now have a world space direction.Now if you recall from earlier, we need to divide samples by the PDF of each sample. For uniform hemisphere sampling the PDF is constant for all samples, and it ends up being one over the surface area of a unit hemisphere (1 / (2 * Pi)). So our algorithm ends up looking like this:

float3 irradiance = 0.0f;
for(uint i = 0; i < NumSamples; ++i)
{
float u1 = RandomFloat();
float u2 = RandomFloat();
float3 sampleDirInTangentSpace = SampleDirectionHemisphere(u1, u2);
float3 sampleDirInWorldSpace = mul(sampleDirInTangentSpace, tangentToWorld);
}

float hemispherePDF = 1.0f / (2 * Pi);
irradiance /= (NumSamples * hemispherePDF);

Just be aware that if you use this result to light a surface using a Lambertian BRDF, make sure that you include the 1 / Pi term that's part of that BRDF. A lot of people like to bake the 1/ Pi into their light sources which is fine, you just have to be careful to make sure that you're doing it somewhere otherwise your diffuse lighting can be too bright.

Once you have this working, you can improve the quality by being smarter about choosing your sample directions. The first way to do that is to use QMC techniques like I mentioned earlier, for instance using Halton sequence or using stratified sampling. You can also just pick values that are evenly spaced out over the [0, 1] domain, which is basically stratified sampling without any jitter (this is essentially what you're doing in the code you posted above). The other way to is to importance sample your function by choosing samples that match the shape of the function being sampled. For irradiance, the common way to do this is to choose sample directions that are proportional to the cosine of the angle between that direction and the surface normal.

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• Here is the original blog post.
Edit: Sorry, I can't get embedded LaTeX to display properly.
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Source files are on GitHub.
Shortcut to sterp implementation.
Shortcut to code used to generate animations in this post.
An Alternative to Slerp
Slerp, spherical linear interpolation, is an operation that interpolates from one orientation to another, using a rotational axis paired with the smallest angle possible.
Quick note: Jonathan Blow explains here how you should avoid using slerp, if normalized quaternion linear interpolation (nlerp) suffices. Long store short, nlerp is faster but does not maintain constant angular velocity, while slerp is slower but maintains constant angular velocity; use nlerp if you’re interpolating across small angles or you don’t care about constant angular velocity; use slerp if you’re interpolating across large angles and you care about constant angular velocity. But for the sake of using a more commonly known and used building block, the remaining post will only mention slerp. Replacing all following occurrences of slerp with nlerp would not change the validity of this post.
In general, slerp is considered superior over interpolating individual components of Euler angles, as the latter method usually yields orientational sways.
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Mathematically, slerp takes the “shortest rotational path”. The quaternion representing the rod’s orientation travels along the shortest arc on a 4D hyper sphere. But, given the rod’s elongated appearance, the rod’s moving end seems to be deviating from the shortest arc on a 3D sphere.
My intended effect here is for the rod’s moving end to travel along the shortest arc in 3D, like this:

The difference is more obvious if we compare them side-by-side:

This is where swing-twist decomposition comes in.

Swing-Twist Decomposition
Swing-Twist decomposition is an operation that splits a rotation into two concatenated rotations, swing and twist. Given a twist axis, we would like to separate out the portion of a rotation that contributes to the twist around this axis, and what’s left behind is the remaining swing portion.
There are multiple ways to derive the formulas, but this particular one by Michaele Norel seems to be the most elegant and efficient, and it’s the only one I’ve come across that does not involve any use of trigonometry functions. I will first show the formulas now and then paraphrase his proof later:
Given a rotation represented by a quaternion R = [W_R, vec{V_R}] and a twist axis vec{V_T}, combine the scalar part from R the projection of vec{V_R} onto vec{V_T} to form a new quaternion: T = [W_R, proj_{vec{V_T}}(vec{V_R})]. We want to decompose R into a swing component and a twist component. Let the S denote the swing component, so we can write R = ST. The swing component is then calculated by multiplying R with the inverse (conjugate) of T: S= R T^{-1} Beware that S and T are not yet normalized at this point. It's a good idea to normalize them before use, as unit quaternions are just cuter. Below is my code implementation of swing-twist decomposition. Note that it also takes care of the singularity that occurs when the rotation to be decomposed represents a 180-degree rotation. public static void DecomposeSwingTwist ( Quaternion q, Vector3 twistAxis, out Quaternion swing, out Quaternion twist ) { Vector3 r = new Vector3(q.x, q.y, q.z); // singularity: rotation by 180 degree if (r.sqrMagnitude < MathUtil.Epsilon) { Vector3 rotatedTwistAxis = q * twistAxis; Vector3 swingAxis = Vector3.Cross(twistAxis, rotatedTwistAxis); if (swingAxis.sqrMagnitude > MathUtil.Epsilon) { float swingAngle = Vector3.Angle(twistAxis, rotatedTwistAxis); swing = Quaternion.AngleAxis(swingAngle, swingAxis); } else { // more singularity: // rotation axis parallel to twist axis swing = Quaternion.identity; // no swing } // always twist 180 degree on singularity twist = Quaternion.AngleAxis(180.0f, twistAxis); return; } // meat of swing-twist decomposition Vector3 p = Vector3.Project(r, twistAxis); twist = new Quaternion(p.x, p.y, p.z, q.w); twist = Normalize(twist); swing = q * Quaternion.Inverse(twist); } Now that we have the means to decompose a rotation into swing and twist components, we need a way to use them to interpolate the rod’s orientation, replacing slerp.
Swing-Twist Interpolation
Replacing slerp with the swing and twist components is actually pretty straightforward. Let the Q_0 and Q_1 denote the quaternions representing the rod's two orientations we are interpolating between. Given the interpolation parameter t, we use it to find "fractions" of swing and twist components and combine them together. Such fractiona can be obtained by performing slerp from the identity quaternion, Q_I, to the individual components. So we replace: Slerp(Q_0, Q_1, t) with: Slerp(Q_I, S, t) Slerp(Q_I, T, t) From the rod example, we choose the twist axis to align with the rod's longest side. Let's look at the effect of the individual components Slerp(Q_I, S, t) and Slerp(Q_I, T, t) as t varies over time below, swing on left and twist on right:
And as we concatenate these two components together, we get a swing-twist interpolation that rotates the rod such that its moving end travels in the shortest arc in 3D. Again, here is a side-by-side comparison of slerp (left) and swing-twist interpolation (right):

I decided to name my swing-twist interpolation function sterp. I think it’s cool because it sounds like it belongs to the function family of lerp and slerp. Here’s to hoping that this name catches on.
And here’s my code implementation:
public static Quaternion Sterp ( Quaternion a, Quaternion b, Vector3 twistAxis, float t ) { Quaternion deltaRotation = b * Quaternion.Inverse(a); Quaternion swingFull; Quaternion twistFull; QuaternionUtil.DecomposeSwingTwist ( deltaRotation, twistAxis, out swingFull, out twistFull ); Quaternion swing = Quaternion.Slerp(Quaternion.identity, swingFull, t); Quaternion twist = Quaternion.Slerp(Quaternion.identity, twistFull, t); return twist * swing; } Proof
Lastly, let’s look at the proof for the swing-twist decomposition formulas. All that needs to be proven is that the swing component S does not contribute to any rotation around the twist axis, i.e. the rotational axis of S is orthogonal to the twist axis. Let vec{V_{R_para}} denote the parallel component of vec{V_R} to vec{V_T}, which can be obtained by projecting vec{V_R} onto vec{V_T}: vec{V_{R_para}} = proj_{vec{V_T}}(vec{V_R}) Let vec{V_{R_perp}} denote the orthogonal component of vec{V_R} to vec{V_T}: vec{V_{R_perp}} = vec{V_R} - vec{V_{R_para}} So the scalar-vector form of T becomes: T = [W_R, proj_{vec{V_T}}(vec{V_R})] = [W_R, vec{V_{R_para}}] Using the quaternion multiplication formula, here is the scalar-vector form of the swing quaternion: S = R T^{-1} = [W_R, vec{V_R}] [W_R, -vec{V_{R_para}}] = [W_R^2 - vec{V_R} ‧ (-vec{V_{R_para}}), vec{V_R} X (-vec{V_{R_para}}) + W_R vec{V_R} + W_R (-vec{V_{R_para}})] = [W_R^2 - vec{V_R} ‧ (-vec{V_{R_para}}), vec{V_R} X (-vec{V_{R_para}}) + W_R (vec{V_R} -vec{V_{R_para}})] = [W_R^2 - vec{V_R} ‧ (-vec{V_{R_para}}), vec{V_R} X (-vec{V_{R_para}}) + W_R vec{V_{R_perp}}] Take notice of the vector part of the result: vec{V_R} X (-vec{V_{R_para}}) + W_R vec{V_{R_perp}} This is a vector parallel to the rotational axis of S. Both vec{V_R} X(-vec{V_{R_para}}) and vec{V_{R_perp}} are orthogonal to the twist axis vec{V_T}, so we have shown that the rotational axis of S is orthogonal to the twist axis. Hence, we have proven that the formulas for S and T are valid for swing-twist decomposition. Conclusion
That’s all.
Given a twist axis, I have shown how to decompose a rotation into a swing component and a twist component.
Such decomposition can be used for swing-twist interpolation, an alternative to slerp that interpolates between two orientations, which can be useful if you’d like some point on a rotating object to travel along the shortest arc.
I like to call such interpolation sterp.
Sterp is merely an alternative to slerp, not a replacement. Also, slerp is definitely more efficient than sterp. Most of the time slerp should work just fine, but if you find unwanted orientational sway on an object’s moving end, you might want to give sterp a try.

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PROBLEM - To do the map explained in the introduction, I looked for some tools but I did not find any good, so I would ask to be told APIs to do this, if someone has already played table RPG knows more or less how I want to do, I will leave an image of the roll 20 that is more or less similar, and some more images of how I want it to stay on the cell phone for those who never played understand , I just need to know a good tool to do this.
I want the master himself to put the image that will be used when starting the map (photo or maybe tiled map).
It can be solutions for web to and call the website into application, literally anything that does what I want will help.
If you can help me, I'll be grateful, sorry for the awful English.

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