Atmospheric CO2 – Geometrical Construction

While many approaches to collecting the desired information of interest are under consideration, the structure of the measurement environment is identical for each. The basic approach is that a small diameter laser beam of the desired wavelength(s) is transmitted normal to the reflecting surface (ground). The beam is reflected back and is gathered by a telescope and directed to a photodetector. It is assumed the transmitter (TX) and receiver (RX) are on the same plane.

Optical Path Volume

What is actually measured is the current from a photodetector which is – to 1st-order – linearly proportional to the number of photons of a particular range of wavelengths striking the photodetector. One of the scaling factors necessary to extract the information desired is the optical path volume in both travel directions: i.e., density is number per volume.

The optical path volume is derived from the product of cross-sectional area of the beam and its length. Within this volume are a number of attenuation elements, each with an effective absorption or scattering volume.

The project optical depth has been formulated through the simplified expression $\vartheta\;=\;n_a\,\sigma_a\,Z$ where both $n$ _a and $\sigma$ _a have been assumed constant and independent of total path length Z.

This assumption is not particularly valid as $n$ _a is dependent on altitude (and so is $\sigma$ to a lesser extent). However, using the average value within the total path length allows its use herein.

$\begin{displaymath}\vartheta\;=\;\int_0^Z\,K\, d z\;\;=\;\;\eta_{CO2}\,\sigma_{CO2}\,\int_0^Z\,d z\;\;\Rightarrow\;\;\eta_{CO2}\,\sigma_{CO2}\,Z\end{displaymath}$

The expression inherently suggests a cylindrical-shaped optical path by implying a constant cross-sectional area. However, the laser has small but significant divergence. The radius of the beam (assumed normal to the reflecting surface and symmetrical about the axis) is dependent on altitude H such that the cross-sectional area of the transmitted optical path is:

$A[z]\;=\;\pi\,r^2\;=\;\pi\,\left(\,H\,\mathit{tan}\theta\,\right)^2\;=\;\pi\,\left(\,H\,\theta\,\right)^2$

where it is assumed that the beam divergence angle $\theta$ is small enough that $\mathit{tan}\theta\,\approx\,\theta$ (divergence is on order of 0.15 mrad)

The reflected/return optical path is not subject to beam divergence.

This example has a path length very much greater than the transmit diameter which results in a significant field of view (FOV) at the reflecting surface. A cone-shaped transmit path is the result … which has 1/3 the volume of a cylinder of equal length and base cross-section area. Furthermore, the receive path starts from the FOV and is received by a finite diameter telescope, resulting in a truncated cone-shaped return volume – a different volume than the transmit path.

The physical situation is similar to:

Cartoon of Idealized Optical Path Geometry

One might quickly assume that for h = 10 km (and Z = 2 h), the terminating radii would be insignificant … I originally thought so.

Assuming the telescope diameter is less than the FOV (r₂ $\le$ r₁), the volume limits fit between the cone with zero terminating diameter and a cylinder of uniform diameter – such that:

$\begin{displaymath}\left(\,V_{cone}\;=\;\frac{\,\pi\,}{3}\,h\,r^2\,\right)\;\le\;V_\vartheta\,\le\,\left(\,V_{cylinder}\;=\;\frac{\,\pi\,}{3}\,3\,h\,r^2\;=\;\pi\,h\,r^2\,\right)\end{displaymath}$

The volume of a truncated cylinder is calculated as

$\begin{displaymath}\begin{align}V_{cone,t}\;=\;&\frac{\,\pi\,}{3}\,h\,\left(\,r_1^2\,+\,r_1\,r_2\,+\,r_2^2\,\right)\\\\\Rightarrow\;\;&\frac{\,\pi\,}{3}\,h\,\left(\,(h\,\theta)^2\,+\,(h\,\theta)\,r_2\,+\,r_2^2\,\right)\end{align}\end{displaymath}$

with r₁ the beam radius (h $\theta$ ) at the reflecting surface and r₂ being the fixed receiving telescope lens radius.

The transmit volume V_T is clearly not identical to the receive volume V_R

But is the difference significant? … keeping in mind the desired measurement accuracy is $\pm$ 0.25%.

A pictorial representation of the volumes of interest is shown:

The radial scale in this illustration is multiplied by 1000 related to the length; the beam radius at the reflecting surface is h $\theta$ where $\theta$ is the beam divergence angle: 0.2E-3 rad for this illustration. The lens diameter is 0.4 m (~ 16”) .

Since h tan θ ~ h θ, radius at surface = 10e3 x 0.2e-3 = 2m … beam diameter at surface = 4 m

The thin BLK line is the ray path representation.

The cylindrical area is based on a uniform cross-section defined by the beam area at the ideal reflecting surface. The GRN cone represents the beam when transmitted by a laser with opening less than 1 cm over a distance of 10,000 m with beam divergence $\theta$ . The RED truncated cone represents a telescope with a field-of-view matching the beam area at the reflecting surface and gathered by a lens of 0.4 m diameter.

Much of the previous analyses have been based on equal transmit and receive volumes. This implies

$V_R\;=\;V_T\;=\;V_{cyl}\;=\;\pi\,h\,r_1^2$ so that $V_{tot}\;=\;\dfrac{\,V_T\,+\,V_R\,}{2}\;=\;\dfrac{\,2\,V\,}{2}\;=\;\pi\,h\,r^2$

The more accurate representation would be:

$\begin{displaymath}\begin{align}V_{tot}\;=\;V_T\,+\,V_R\;\;=\;\;&\frac{\,\pi\,}{3}\,h\,r^2\,\right)\,+\,\frac{\,\pi\,}{3}\,h\,\left(\,r_1^2\,+\,r_1\,r_2\,+\,r_2^2\,\right)\\\\=\;\;&\frac{\,\pi\,}{3}\,h\,\left(\,2\,r_1^2\,+\,r_1\,r_2\,+\,r_2^2\,\right)\;\;\neq\;\;\pi\,h\,r_1^2\\\end{align}\end{displaymath}$

The absorption coefficient $\alpha$ is derived from the product of the number of absorbers and the effective absorber cross-sectional area. By assuming the species absorption cross-section is constant, the optical depth is more properly expressed as

$\vartheta\;=\;\alpha\,\dfrac{\,\pi\,H\,}{6}\,\left[\,2\,\left(\,\theta\,H\,\right)^2\,+\,\theta\,H\,r_2\,+\,r_2^2\,\right]$

where r1 = $\theta\,H$ – representing the beam cross-sectional area as a function of distance z over height H. In this calculation, parameter r₂ represents the fixed radius of the receiving telescope lens.

Assuming a uniform planar distribution, the absolute quantity of absorbing gas is necessarily different in the transmit and receive optical depth paths due to differing volumes

The return path volume is greater than the transmit volume as a function of receiving lens diameter

Comparison of TX and RX Volumes vs Receive Lens Radius
RED: q = 0.17 mrad ; BLU: q = 0.20 mrad

Consider a cylindrical volume formed of a cross-section area equal to the beam diameter at the reflecting surface. The optical path volume for $\theta$ = 0.2 mrad would be $V_{cyl}\;=\;\pi\,z\,r^2\;=\;125,664\,m^3$ where r is a fixed quantity of some dimension: $r\;= \;(\,\theta\,H\,) \;\rightarrow\;2 \,m$ in this example

Beam divergence angle will have a significant influence on the measurement.

Using the expanded representation of the optical volume as a cone during transmission (TX) and a truncated cone on the receive path (RX) gives results of

$\begin{displaymath}\begin{align}V_{TX}\;&=\;\frac{\,\pi\,}{3}\,H\,(\,\theta\,H\,)^2\;=\;41,888\,m^3\\\\V_{RX}\;&=\;\frac{\,\pi\,}{3}\,H\,\left[\,(\,\theta\,H\,)^2\,+\,(\,\theta\,H\,)\,r_2\,+\,r_2^2\,\right]\;=\;47,778\,m^3\end{align}\end{displaymath}$

where a telescope lens diameter of 0.5 m has been assumed.

The cylinder average volume is twice the volume divided by 2: the single direction optical path. The volumes of the two differing paths are significantly different:

The calculation of an absorbing gas is highly dependent on volume.

The desired quantity measured is number of attenuation elements. The differential method of measure compares a measure of all attenuation elements within an optical volume to a measure of all attenuation elements except gas absorption in order to derive the gas absorption value.

Not only are two accurate measures of the number of attenuation elements necessary, but a measure of two path volumes to better accuracy is necessary.

The difference between the TX and RX path volumes is more significant than expected. The sensitivity of the measurement to the absolute volume of the optical path makes geometrical knowledge paramount.

Even in this idealized “best-case” illustration, the sensitivity to the volume of the optical path is high.

Consider a system with divergence $\theta$ ~ 0.17E-3 rad and lens radius r2 ~ 0.1 m (8” lens). For a fixed distance h = 10 km, the volume sensitivity with respect to beam divergence angle and lens radius is:

$\begin{displaymath}\begin{align}\mathbb{S}_V^\theta\;&=\;\frac{\,\theta\,}{\,V\,}\,\frac{\,\partial V\,}{\,\partial \theta\,}\bigg|_{z=10km}\;\;\Rightarrow\;\;1.94\big|_{r2\,=\,5\,cm}\\\\\mathbb{S}_V^{r2}\;&=\;\frac{\,r2\,}{\,V\,}\,\frac{\,\partial V\,}{\,\partial r2\,}\bigg|_{z=10km}\;\;\Rightarrow\;\;0.062\big|_{\theta\,=\,0.17\,mrad}\end{align}\end{displaymath}$

Sensitivity of volume to beam divergence for various lens radii r₂

1^st-Order Volume Sensitivity to Lens Radius and Divergence Angle

Sensitivity Plane: BLU is sensitivity: PNK is S = 1

2D Detail

Only a small range of parameter variation (PNK region) is allowable for maintaining the volume sensitivity to less than 1 (where a change of 1 in volume causes change of 1 in measurement).

Based on these parameters, the sensitivity of volume with respect to divergence angle is 1.94 for this example system and only 0.062 for lens radius.

That’s all for now.

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