Compute sphere tangent for normal mapping

Question

I'm implementing normal mapping on my toy pathtracer. I need to compute the tangent (and bitangent) of any sphere point in order to create the matrix that will transform tangent space to world space.

More specifically, I want the tangent that runs along the u texture coordinate, and the bitangent that runs along the v coordinate.

Here's how I implemented it:

• At the intersection point P, compute the spherical coordinates (theta, phi) of P.
• Add a tiny delta to the angle that relates to the u texture coordinate (phi), finding another point Q on the sphere.
• Compute Q - P and normalize it to have the tangent.
• The bitangent is then computed by doing the cross product of the tangent and the normal at the point P.

Relevant code on the sphere collision function:

    hitRecord.point = ray.pointAtParameter(t);

    Vec3 localCoord = hitRecord.point - center;
    float theta = asin(localCoord.y / radius); //[-pi/2,pi/2]
    float phi = atan2(localCoord.z, localCoord.x); //[-pi,+pi]

    float deltaPhi = phi + FLOAT_BIAS;
    Vec3 deltaPoint(radius * cos(theta) * cos(deltaPhi),
        radius * sin(theta),
        radius * cos(theta) * sin(deltaPhi));

    //Assume local space to be aligned with world space
    //So this is also the tangent in world space
    hitRecord.tangent = deltaPoint - localCoord;
    hitRecord.tangent.normalize();

    hitRecord.normal = localCoord / radius;
    hitRecord.bitangent = hitRecord.tangent.cross(hitRecord.normal);
    hitRecord.bitangent.normalize();

This seems to work, but all this trigonometry is not light and makes me wonder if there is a better way of doing it. Does anyone has any ideias/pointers?

Answer 1

You can do this a little bit more efficiently and accurately by calculating derivatives of $P$ with respect to $u, v$—or the derivatives with respect to $\phi, \theta$, equivalently (up to a scalar multiple).

Start by expressing $P$ in terms of the spherical coordinates: $$\begin{aligned} P_x &= r \cos \theta \cos \phi \\ P_y &= r \sin \theta \\ P_z &= r \cos \theta \sin \phi \end{aligned}$$

Now you can get the tangent as $T = \partial P / \partial \phi$, which works out as: $$\begin{aligned} T_x &= -r \cos \theta \sin \phi \\ T_y &= 0 \\ T_z &= r \cos \theta \cos \phi \end{aligned}$$

Once you normalize it, the $r \cos \theta$ factors drop out and you're left with just $T = (-\sin \phi, 0, \cos \phi)$. Pretty simple!

You can similarly get the bitangent as $B = \partial P / \partial \theta$, or you can just calculate it as the cross product $N \times T$, which is probably faster. (Note it should be $N \times T$, not $T \times N$, assuming you want $T, B, N$ to form a right-handed basis in that order.)

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The above approach will work for pretty much any kind of surface that can be parameterized, not just spheres. However, in the case of spheres with the usual $\phi, \theta$ coordinates, there's an even quicker shortcut that you can do with just vector math, no trig at all. Let $C$ be the center of the sphere, and $A$ be a vector representing the axis of the spherical coordinates—i.e. pointing from the center toward the north pole. Then you can just calculate $T = \text{normalize}\bigl(A \times (P - C) \bigr)$, then $B = N \times T$ as before.