Area weighting of radiant exposure ($\overline{H}={\sum }_j{A}_j{H}_j/{\sum }_j{A}_j$)

Area weighting of radiant exposure combines patch-average exposures in proportion to the receiving areas they represent. For disjoint patches with areas $A_j$ and mean radiant exposures $H_j$, the coefficients ${\alpha }_j=A_j/{\sum }_k{A}_k$ satisfy ${\sum }_j{\alpha }_j=1$, and the combined spatially averaged radiant exposure is ${\sum }_j{\alpha }_j{H}_j$. This follows by adding the received energies $A_j{H}_j$, rather than averaging a list of values without its geometry.^[ref]

An original coordinate example explains a calibration issue in image-based measurements. Let a rectangular receiving strip have length $\ell$ and constant width. Stipulate $H(x)=H_{0}x/\ell$ for $0\le x\le \ell$. Suppose a calibrated image coordinate $u\in [0,1]$ maps to physical position $x=\ell u^2$. Uniform sampling in $u$ is therefore nonuniform in physical area.

The physical area average isAn unweighted average over increasingly dense, uniformly spaced $u$ samples would instead converge to ${\int }_0^1{H}_0{u}^{2}du=H_0/3$. The factor $2u$ is the change of variables formula's area correction for the chosen mapping; the constant strip width and $\ell$ cancel in the normalized average.

Thus a Jacobian area correction for radiant-exposure maps can matter even if every local exposure value is correct. An optical camera measurement needs its spatial calibration and the receiving-surface map, not just pixel values. Overlapping patches must not be counted as disjoint areas, and local point samples need an explicit quadrature model before they can stand for patch averages.

A Jacobian area correction for radiant-exposure maps converts a coordinate-space integral into one over the physical receiving surface. For a one-to-one planar calibration map $\mathbf{x}=f(u,v)$,The Jacobian determinant supplies physical area per coordinate-area increment. This change of variables formula preserves the radiant exposure definition as incident energy per receiving area.^[ref]

For an original two-dimensional example, let $0\le u,v\le 1$ and map $x=au$, $y=bv(1+u)$, with $a=2\mathrm{mm}$ and $b=3\mathrm{mm}$. Then $|\det Df|=ab(1+u)$. Stipulate $H(f(u,v))=H_0(1+u)$ with $H_0=2\mathrm{J}{\mathrm{m}}^{-2}$. The physical receiving area is $3ab/2=9\mathrm{m}{\mathrm{m}}^2$, andIts spatially averaged radiant exposure is $28/9\approx 3.111\mathrm{J}{\mathrm{m}}^{-2}$. Uniformly averaging the exposure values over $u,v$ without the determinant gives only $3\mathrm{J}{\mathrm{m}}^{-2}$.

The correction applies to exposure densities represented in distorted coordinates. If each detector sample already reports energy integrated over its physical patch, summing those calibrated energies requires no second multiplication by patch area. Otherwise the area is counted twice. An optical camera reduction must specify whether its values represent energy, irradiance, exposure or a further processed signal.