The basic concept of the energy fraction is to express the noise exposure E produced at the observer position from a flight path segment P₁P₂ (with a start-point P₁ and an end-point P₂) by multiplying the exposure E_∞ from the whole infinite path flyby by a simple factor — the energy fraction factor F:

Since the exposure can be expressed in terms of the time-integral of the mean-square (weighted) sound pressure level, i.e.

$E=\mathrm{const}\times \int p^2\left(\tau \right)\mathrm{d\tau }$

(E-2)

to calculate E, the mean-square pressure has to be expressed as a function of the known geometric and operational parameters. For a 90° dipole source,

$p^2=p_2^p\times \frac{d_2^p}{d^2}\times {\sin }^2\psi =p_2^p\times \frac{d_4^p}{d^4}$

(E-3)

where p² and p_p² are the observed mean-square sound pressures produced by the aircraft as it passes points P and P_p.

This relatively simple relationship has been found to provide a good simulation of jet aircraft noise, even though the real mechanisms involved are extremely complex. The term d_p²/d² in equation E-3 describes just the mechanism of spherical spreading appropriate to a point source, an infinite sound speed and a uniform, non-dissipative atmosphere. All other physical effects — source directivity, finite sound speed, atmospheric absorption, Doppler-shift etc. — are implicitly covered by the sin²ψ term. This factor causes the mean square pressure to decrease inversely as d⁴; whence the expression ‘fourth power’ source.

Introducing the substitutions

$d^2=d_2^p+q^2=d_2^p+{\left(V\times \tau \right)}^2$ and ${\left(\frac{d}{d_p}\right)}^2=1+{\left(\frac{V\times \tau }{d_p}\right)}^2$

the mean-square pressure can be expressed as a function of time (again disregarding sound propagation time):

$p^2=p_2^p\times {\left(1+{\left(\frac{V\times \tau }{d_p}\right)}^2\right)}^{-2}$

(E-4)

Putting this into equation (E-2) and performing the substitution

$\alpha =\frac{V\times \tau }{d_p}$

(E-5)

the sound exposure at the observer from the flypast between the time interval [τ₁,τ₂] can be expressed as

$E=\mathrm{const}\times p_2^p\times \frac{d_p}{V}\times \underset{{\alpha }_1}{\overset{{\alpha }_2}{\int }}\frac{1}{{\left(1+{\alpha }^2\right)}^2}\mathrm{d\alpha }$

(E-6)

The solution of this integral is:

$E=\mathrm{const}\times p_2^p\times \frac{d_p}{V}\times \frac{1}{2}\left(\frac{{\alpha }_2}{1+{\alpha }_2^2}+\arctan {\alpha }_2-\frac{{\alpha }_1}{1+{\alpha }_2^1}-\arctan {\alpha }_1\right)$

(E-7)

Integration over the interval [–∞,+∞] (i.e. over the whole infinite flight path) yields the following expression for the total exposure E_∞:

$E_{\infty }=\mathrm{const}\times \frac{\pi }{2}\times p_2^p\times \frac{d_p}{V}$

(E-8)

and hence the energy fraction according to equation E-1 is

$F=\frac{1}{\pi }\left(\frac{{\alpha }_2}{1+{\alpha }_2^2}+\arctan {\alpha }_2-\frac{{\alpha }_1}{1+{\alpha }_2^1}-\arctan {\alpha }_1\right)$

(E-9)