# sphere tangent in a specific point

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I sucefully implemented normal mapping for triangles in my raytracer, and was thinking to add it to spheres as well, but I don't know how to compute the tangent of the intersection point on the sphere. I was not able to find useful resources, except a post here on gamedev. Since the approach suggested was not trivial and I don't even know if it is suited for texture mapping, and since I'm mainly targeting polygons anyway, I would like to ask for a (relatively) easy way to accomplish this. Otherwise I will skip tangents for sphere (wich are used only for test scenes untill a proper geometry loader is ready). Thank you.

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Can't you just take the partial derivatives to derive the tangent vectors?

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You can compute a tangent space given any normal using the following procedure

given that normal is not lying on the y axis. project onto xz plane, rotate 90 degrees around y axis and normalise, then take cross of this with the normal to give the basis.

i.e. along the lines of

normal = y given that y.x!=0&&y.z!=0
double iy = 1/sqrt(y.x*y.x+y.z*y.z);
vec x (-y.z*iy,0,y.x*iy);
vec z = x cross y;

then you can build the tangent from the 3 vectors

may need to switch signs depending on your system. This is what i do in my ray tracer for whatever type of object it is if it cannot be precomputed like with a triangle or other.

i find this is best done in object space so that the same object will appear the same when normal mapped independant of its rotation, remember theres an infinite number of tangent spaces given only a normal.

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Here's a formal way that works in all cases (from the journal of graphics tools):

given u = (x,y,z)

v' = ( 0,-z, y ) if x is the smallest
( -z, 0, x ) if y is the smallest
( -y, x, 0 ) if z is the smallest
v = normalize(v')
w = u X v

Note that v,w won't be continuous on on a sphere for example as that just isn't possible.

Cheers!

Eric Penner

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Quote:
 Original post by luca-deltodescoYou can compute a tangent space given any normal using the following proceduregiven that normal is not lying on the y axis. project onto xz plane, rotate 90 degrees around y axis and normalise, then take cross of this with the normal to give the basis.i.e. along the lines ofnormal = y given that y.x!=0&&y.z!=0double iy = 1/sqrt(y.x*y.x+y.z*y.z);vec x (-y.z*iy,0,y.x*iy);vec z = x cross y;then you can build the tangent from the 3 vectorsmay need to switch signs depending on your system. This is what i do in my ray tracer for whatever type of object it is if it cannot be precomputed like with a triangle or other.i find this is best done in object space so that the same object will appear the same when normal mapped independant of its rotation, remember theres an infinite number of tangent spaces given only a normal.

Thank you for answering. Just a question: what is the vector y? Is this method suited for texturing? I ask because I don't see references to texturing axis...

Quote:
 Here's a formal way that works in all cases (from the journal of graphics tools):given u = (x,y,z)v' = ( 0,-z, y ) if x is the smallest( -z, 0, x ) if y is the smallest( -y, x, 0 ) if z is the smallestv = normalize(v')w = u X v

I suppose that u is the intersection point. But since it is a point (in world space) this means that I have to use (point - sphereposition) instead?

I thought a bit about that, and wonder if the following could work:
Let north be the vector that points to the north pole (set in the constructor and used for texturing).
- vector tonorth = point - sphereposition + north * radius)
- vector bitangent = normal.Cross(north);
- vector tangent = normal.Cross(bitangent);

It seems to easy, but if I don't miss something it could work (at least with a couple of minor changes)... what do you think about this?

[Edited by - cignox1 on January 10, 2008 6:25:43 AM]

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Apparently the method I tried works. Thank you all, and if you think that my method is for some reason wrong, please tell me!

Thank you again!

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