Background information

I carried out an experiment where a pyranometer (on a flat surface) was used to measure the irradiance at the ground due to solar radiation over a range of solar zenith angles (SZA).

[To be clear, the definition of irradiance is: "the radiant flux (power) received by a surface per unit area. The SI unit of irradiance is the watt per square metre (W/m^2)" (from Wikipedia)]

The lab script I was given used the formula $$\tau = \frac{I_\lambda}{I_{\lambda 0}} = e^{-u \sec(Z)}$$ where

  • $\tau$ is the transmissivity of the atmosphere
  • $I_\lambda$ is the "intensity" at the surface
  • $I_{\lambda 0}$ is the "intensity" at the top of atmosphere (TOA)
  • $u$ is the optical depth of the vertical column (see diagram below)
  • $Z$ is the SZA
  • thus $u \sec(Z)$ is the optical depth experienced by a ray of sunlight propagating at an angle $Z$ to the vertical

This is a form of the Beer-Lambert-Bouguer law, then transmissivity is inversely proportional to the exponential of the optical depth, and also equal to the ratio of the irradiance at the start of the path to the irradiance at the end of the path, and I am satisfied this is correct. The diagram below (that I have made) illustrates this law (ignore the lowercase $z$).

enter image description here

The lab script (perhaps rather sloppily) uses the word "intensity" instead of irradiance, but the demonstrators did agree that we're measuring in watts per square metre.

The problem (TL;DR version)

In my analysis, I have used $I_{\lambda 0} = I_{SC} \cos(Z)$, where $I_{SC}$ is the solar constant (~ 1360 W/m^2), as the TOA irradiance.

This has been marked as wrong; apparently I should've used $I_{\lambda 0} = I_{SC}$, but this seems obviously wrong to me.

Who is correct?

The problem (long version)

Explanation 1

As $I_{\lambda 0}$ is the irradiance at the TOA, and irradiance is power per unit area, it changes over the course of a day as the SZA varies, following a cosine law with respect to SZA, but the solar constant is a constant (1360 W/m^2), therefore $I_{\lambda 0} \neq I_{SC}$, instead $I_{\lambda 0} = I_{SC} \cos(Z)$

Response 1 (from marker A)

enter image description here

But the problem I see with this is that for $I_2$, it seems to assume the pyranometer is facing directly towards the sun, which is incorrect, as the pyranometer was facing vertically upwards (normal to the surface of the planet) in the experiment.

This means $I_2 = \frac{I_\lambda}{\cos(Z)}$, leading to my initial conclusion.

Also the diagram is way out of proportion.

Explanation 2

In the case of a very optically thin atmosphere, $u \approx 0$, so our Beer-Lambert law gives $$\frac{I_\lambda}{I_{\lambda 0}} = 1 = const$$

If $I_{\lambda 0} = I_{SC} = const$, as the marker claims, then the expression above would be false, because $I_\lambda$ (the irradiance measured at the ground by the pyranometer) varies with the cosine of the SZA.

Therefore, we need to divide the left hand side by $\cos(Z)$ for that expression to be correct/constant.

Response 2 (from marker B)

The pyranometer has a $2\pi$ field of view due to its hemispherical glass dome.

Consider a parallel beam incident on a normal surface, with intensity $F_0$. When the beam is not normal to the surface, the intensity perpendicular to the surface is now reduced and is $F_0 \cos(Z)$. The cosine corrected response of the pyranometer means that the output of the pyranometer is $F_0$ for the case of normal incidence and for the case where the angle is $Z$.

As the pyranometer manual states, this correction is valid up until $Z$ = 80 deg. You have misinterpreted what the manual refers to as a ‘cosine corrected response’.

This manual states, "Ideally a pyranometer has a directional response which is exactly the same as the cosine-law," which seems to be in direct contradiction with what he is saying.

Also, he seems to suggest that the output voltage will be identical for any SZA(less than 80 deg) assuming no atmosphere. But the output voltage is proportional to the irradiance in W/m^2 (17.01e-6 V/W/m^2 is printed on the side of the pyranometer). Surely this means the output voltage is proportional to $F_0 \cos(Z)$ if the irradiance is proportional to $F_0 \cos(Z)$?

As for the dome, the explanation of its purpose given in the manual is, “The rise of temperature in the thermopile is easily affected by wind, rain and thermal radiation losses to the environment (for example, a 'cold' sky) and the delicate black coating must be protected. Therefore the detector is shielded by a dome. This dome allows equal transmittance of the direct solar radiation component for every position of the sun in the hemisphere above the detector.”

To me, this indicates that the hemispherical shape is such that the direct beam from the sun is always normal to the dome surface, so no refraction occurs and the transmitted beam carries on straight to the thermopile. But the angle the beam makes with the thermopile still determines the irradiance.

This is the pyranometer used, Kipp & Zonen CMP3

enter image description here

Unfortunately, the markers are either telling me to take it up with each other or no longer responding to my latest emails, hence I have taken it here.


2 Answers 2


A few things seem incorrect about this setup:

  1. A notation detail. The $λ$ subscript in $I_λ$ and $I_{λ0}$ suggests spectral irradiance, i.e. irradiance at certain wavelengths $λ$. For the described setup, in which the whole spectrum is considered, $I$ and $I_0$ are more appropriate instead.

  2. As the OP explains correctly, a comparison of $I_0$ and $I$ measured at identically inclined surfaces is needed (see for example this, where the term "clearness index" is used for transmissivity). And because the height of the atmosphere is much smaller than the Earth's radius, it can be assumed that the angle of incidence of the solar beam is the same at both the top of the atmosphere (TOA) and the Earth's surface (see plane-parallel atmosphere). Instead, with the described setup, the solar constant is the irradiance from radiation incident perpendicularly upon a surface at the TOA, while $I$ is measured by the pyranometer at a horizontal surface at the ground.

  3. It is definitely incorrect that the pyranometer measures anything other than the Global Irradiance, i.e. the irradiance resulting from solar radiation incident on a horizontal surface from all directions. The pyranometer's "directional response" describes how large the error is when the direction of incoming radiation departs from being normal to the sensor's surface.


$I_{\lambda_0}$ is the intensity of light at the top of the atmosphere - this is the solar constant.

It is then modified by the absorption in the atmosphere to give you $I_{\lambda}$ (which ignores scattering).

The irradiance that you measure is then $I_{\lambda} \cos Z$ (assuming that you mean the irradiance on a horizontal flat surface).

  • $\begingroup$ In your notation scheme (which is perhaps what the lab script was using), would you agree that the my pyranometer irradiance measurements would have to be divided by $\cos(Z)$ in order to be used in the Beer-Lambert law given? Also, I've updated the question now. If you have the time, could you give your views on the responses my markers gave? I understand that it's a very long question, so no worries if not! $\endgroup$
    – binaryfunt
    Aug 21, 2016 at 17:48
  • $\begingroup$ @BinartFunt In your first sentence, you say the aim is to measure the irradiance - that is $I_{\lambda}\cos Z = I_{\lambda_0} \exp(-u \sec Z) \cos Z$. As far as I know, that is what your device measures. If your device somehow knows what angle the light is coming from, then perhaps it corrects for the $ \cos Z$; that will clearly be in the instrument manual. $\endgroup$
    – ProfRob
    Aug 21, 2016 at 18:53
  • $\begingroup$ As far as I can tell, the pyranometer measures $I_\lambda \cos(Z)$ in the way you've put it above. Marker B seems to think it measures $I_\lambda$ then, but he doesn't say that the pyranometer detects the angle of the light, he says the curved dome somehow means identical beams of radiation from any angle give the same irradiance measurements. Perhaps this point should be answered in another question $\endgroup$
    – binaryfunt
    Aug 22, 2016 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.