Forward Scattering - The Weblog of Nicholas Chapman

The weblog of Nicholas Chapman

Using a a single linearly interpolated sample to evaluate a weighted sum of two texels<br>Posted 13 Aug 2026

f(floor(x))<br>|-- f(x) f(floor(x)+1)<br>| \--*---|<br>| |<br>floor(x) ^ floor(x) + 1

Bilinear sampling at point x in gives

f_bilinear(x) = f(floor(x))(1 - alpha) + f(floor(x)+1) alpha

where alpha = x - floor(x)<br>e.g.<br>f_bilinear(x) = f(floor(x))(1 - (x - floor(x))) + f(floor(x) + 1) (x - floor(x))<br>= f(floor(x))(1 - x + floor(x)) + f(floor(x) + 1) (x - floor(x))

Suppose we want to evaluate<br>v = a f(x_0) + b f(x_0+1)

for some x_0, a, b.

Consider<br>C f_bilinear(x)<br>= C f(floor(x))(1 - x + floor(x)) + f(floor(x) + 1) (x - floor(x))

= f(floor(x))(C - C.x + C.floor(x)) + f(floor(x) + 1) (C.x - C.floor(x))

defining x_0 = floor(x):

= f(x_0)(C - C.x + C.x_0) + f(x_0 + 1) (C.x - C.x_0)

= f(x_0)(C - C.x + C.x_0) + f(x_0 + 1) (C.x - C.x_0)

= (C - C.x + C.x_0) f(x_0) + (C.x - C.x_0) f(x_0 + 1)

So for this to equal v = a f(x_0) + b f(x_0+1), we want

a = C - C x + C x_0<br>and<br>b = C x - C x_0

but<br>a + b<br>= C - C x + C x_0 + C x - C x_0<br>= C

consider

b = C x - C x_0<br>C x = b + C x_0<br>x = (b + C x_0)/C

x = (b + (a + b)x_0)/(a + b)<br>x = b/(a + b) + (a + b)x_0/(a + b)<br>x = b/(a + b) + x_0

alternatively:<br>x = b/(a + b) + (a + b)x_0/(a + b)<br>= (b + (a + b)x_0) / (a + b)

So we have

a f(x_0) + b f(x_0+1) = (a + b) f_bilinear(b/(a + b) + x_0)

Checking:

(a + b) f_bilinear(b/(a + b) + x_0) =

[if a >= 0 and b >= 0, then b/(a + b) 0 or b > 0, then (a + b) > 0, so b/(a + b) is defined and b/(a + b) > 0.]<br>[also since it is

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