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28 May 2015

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I think that the answer to exercise 3 a) in the old computer graphics exam as seen here: http://imgur.com/gjNRKBF is wrong. At 1) it says: 'The diffuse term is ld * Kd * cos( dot(nl, n) )' and at 2) it says 'Specular term is ks * ls * cos^s( dot(r,nv) )'. But if the angle between two normalised vectors u and v is a then cos(a) = dot(u,v). Given the info in the slides this means that it should be: 'The diffuse term is ld * Kd * dot(nl, n)' and the 'Specular term is ks * ls * dot(r,nv)^s'.

Secondly the specular term is defined as: S(@) = I * K * cos ( @ ) ^n. where @ is the angle between the reflection R and View point vector V as described in this picture: http://www.viz.tamu.edu/…/parke/ends489f00/notes/JOB0325.GIF . r=d−2(d⋅n)n is how you calculate the reflection r of vector d for a surface with normal n. However the answer as seen here: http://imgur.com/gjNRKBF uses the view vector v and the normal to calculate the reflection which makes no sense. ´r is the normalized reflected light vector refl, computed by refl = v – 2* dot(n,v) * n´ is therefore wrong.

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