Alpha vs Transparency¶

OpenEXR images use alpha\(\alpha\) instead of transparency.\(T=1-\alpha\): if\(\alpha=1\) (so\(T=0\)) then all lightis absorbed; if\(\alpha=0\) (so\(T=1\)), then all light istransmitted and the material is transparent. Substituting\(T\) into(1) and(2) combined alpha of two objects is given by thescreen compositing operation:

where\(\alpha_c\) is the combined alpha value, and\(\alpha_1\) and\(\alpha_2\) are the alpha values of the two samples.

Throughout this document,\(T\) is preferred to\(\alpha\) whereit gives simpler equations.

Computing\(t\)¶

OpenEXR images samples store the\(\alpha\) of the entire sample.This gives us the total transmission of the sample, not thetransmission per unit length\(t\). The total transmission\(T\)and the sample length\(n\) can be used to compute the transmission per unit length:

Sample Properties¶

The following are assumptions made by the formulas derived here:

• Samples haveconstant optical density and colour: if a sample haslength\(n\) and a subsample of length\(n'\) is extractedfrom it, the RGBA colour of the subsample will be the same regardlessfrom where in the sample it is extracted. In particular, if a sampleis split into\(k\) subsamples of equal length, each subsamplewill have the same RGBA colour.

• Sample attenuation isnon-scattering andpixel independent:light travelling through the sample is either absorbed ortransmitted; it is not reflected back down the sample or scatteredinto neighbouring samples. Scattering causes point lights to appearblurred when passing through fog, and also tends to make fog lookmore optically dense than it really is, since detail is lost veryquickly, even though light is being transmitted. This effect is notmodelled with OpenEXR volumetric samples, and must be approximated byincluding the scattered light within the volume or applied as apost-process. Since the light attenuation profile throughout a pixelwill not follow the Beer-Lambert equation(3), extra samplesmust be used to model the pixel.

• Sample behaviour isunit andscale independent: if a sampleis divided into\(k\) subsamples of equal length, then scalingthe depth channels\(z_\textsf{front}\) and\(z_\textsf{back}\) of the deep image will not change thecolour of the samples. When merging two deep images, prescaling thedepth of each image by the same amount then merging the images isidentical to merging the original images then scaling the depth ofthe result. This scaling property allows any unit to be used to storedepth, and the unit to be changed by scaling the depth channelswithout modification of the RGBA channels.

• Sample behaviour isposition independent: moving a sample indepth will not change the RGBA values obtained by subdividing it. Ifa sample is\(10\) units long and a subsample of\(2\) unitsextracted from it, the subsample will have the same RGBA valuesregardless of the position in depth of the sample. Shifting twoimages by adding a constant\(c\) to the depth channels of eachimage then merging them is identical to merging the original imagesthen shifting the result.

• Sample splitting isnon-destructive: using the equations derivedhere, the samples of a pixel can be arbitrarily subdivided.Recombining the subsamples with theover compositing operation willyield the RGBA colour of the original sample (assuming sufficientnumerical precision).

Alpha of a Subsample¶

When a subsample of length\(n'\) is extracted from an originalsample of length\(n\), the alpha value\(\alpha'\) forsubsample must be computed.

The total transmission\(T'=t^{n'}\). Substituting for\(t\)computed for the entire sample length using(5) gives:

Here,\(n\) refers to thelength of the original sample, butsamples are specified in OpenEXR with a front and back depth. Thus, if asample\(T\) with front and back depths\(z_\textsf{front}\)and\(z_\textsf{back}\) is split at point\(z\) in space,the transmission\(T'\) of the front subsample is given by

Since\(({z-z_\textrm{front}})/({z_\textsf{back} - z_\textsf{front}})\)is theratio of the original sample length to the extracted samplelength, we can substitute this ratio for\(r\) for the remainder.This is equivalent to considering subdivision of a sample of unitlength.

Colour of a Subsample¶

A sample with RGB information is modelled as a cylinder. The alphachannel(s) of the sample are represented by some non-scattering,absorbent material within it, which attenuates the light passing throughthe sample. The colour channel(s) of the sample is represented byinfinitely many light sources, which emit coloured light (but don’tabsorb anything), with identical intensities\(\mathbf{c}\), evenlyspaced through the sample:

A light source at distance\(x\) into the sample will be attenuatedby the amount of absorber between it and the front of the sample.According to Beer-Lambert, the total transmission of that part of thesample\(t\) is given by

The light which reaches the front of the sample from the individuallight source at distance\(x\) is\(\mathbf{c} T^{x}\). If therewere a finite number\(N\) sources, the total light reaching thefront of the sample would be:

where\(k/N\) is the position of light\(k\), equal to\(x\). As\(N\) tends to infinity, this becomes

The value\(\mathbf{C}\) is the sample’s RGB value, as stored in theOpenEXR image. Rearranging gives us the colour of eachlight (or perhaps the ‘instantaneous colour’ of the sample)

If the entire sample is reduced to a subsample length\(r\), only asubsection of lights are included, and we must compute the new RGBcolour\(\mathbf{C'}\) for the subsample. Hence:

Substituting for\(\mathbf{c}\) from(7) gives

Noting that\(T-1=-\alpha\) and\(T^{r}-1=T'-1=-\alpha'\) andmultiplying top and bottom by\(-1\) gives:

This is exactly equivalent tounpremultiplying the colour by theoriginal alpha value, computing the new alpha value, andpremultiplying by the new value.

Splitting Transparent Samples¶

When\(\alpha=0\) (or alternatively,\(T=1\)), unpremultiplying in(8) would require a division by zero.A special case is derived for\(\alpha=0\). The sample modelof the colour provided by many light sources still applies, but thereis now no attenuation. Therefore, each light contributes equally to theobserved colour\(\mathbf{C}\). The ‘instantaneous’ colour equalsthe final colour, (assuming sample length is normalised to 1)

A subsection\(\mathbf{C'}\) of length\(r\) is given by

That is, if\(\alpha=0\), then scaling a sample’s length by\(r\) scales the colour by the same amount.

Merging when One Sample is Transparent¶

Now suppose one of the samples (say,\(b\)) has no alpha, so\(\alpha_b=0\) and\(T_b=1\). From(9)\(\mathbf{c_b}=\mathbf{C_b}\). The combined colour\(\mathbf{C_c}\) is attenuated according to\(T_a\), thetransmission of\(a\) alone, and the combined colour is given by:

and substituting into(6) gives:

Merging Two Transparent Samples¶

Where both samples are transparent, the colours simply add together:

Merging Solid Samples¶

For numerical stability, we must give a sensible value for\(\mathbf{C_c}\) when\(\alpha_b=1\). Equation(11)is undefined, since\(\log(1-\alpha_b)=\log(0)=-\infty\). We followthe initial model of light sources in an absorbing material, but now theabsorbing material absorbs all light. Thus, we must be observing onlyits closest light source (that at\(x=0\)) in the solid sample: thesample will absorb light from its other sources. Thus, we can treat\(b\) as an infinitely thin ‘discrete’ sample at the front of thesample, which will block all light behind it; we can simply composite\(b\) over\(a\). Therefore, if\(\alpha_b=1\) and\(\alpha_a<1\), the total observed colour will be\(\mathbf{C_b}\). Transposing\(a\) and\(b\) gives\(\mathbf{C_c}=\mathbf{C_a}\) if\(\alpha_a=1\)

Following the definitions above, when\(\alpha_a=1\) and\(\alpha_b=1\), then\(\mathbf{C_c}=\mathbf{C_a}+\mathbf{C_b}\).However, for stability, it makes sense to define

This approach gives more stable results when there is random samplingerror in the depth channel. Assume that the front depth of sample\(A\) differs by some small amount from that of\(B\), so that\(z_a=\delta+z_b\). Only the front sample will be visible: if\(\delta<0\), then\(\mathbf{C_c}=\mathbf{C_a}\); if\(\delta>0\), then\(\mathbf{C_c}=\mathbf{C_b}\). Setting\(\mathbf{C_c}=\mathbf{C_a}+\mathbf{C_b}\) when\(\delta=0\)gives an image which is (potentially) twice as bright. If\(\delta\)is random noise due to sampling, we will get a random pattern of brightpixels whenever\(\delta=0\). Using the mean colour seems a saferoption. However, this isnot associative: when combining three solidsamples\(d\),\(e\) and\(f\) does not give the sameresult as combining first\(d\) and\(e\), then combining theresult with\(f\).