<p>According to <a href="https://books.google.co.uk/books?hl=en&amp;lr=&amp;id=tF1-AV_bqDkC&amp;oi=fnd&amp;pg=PA8&amp;ots=JCSNCY8WqR&amp;sig=ZrryQEeWhymiOSashcjF2tCmoGs&amp;redir_esc=y#v=onepage&amp;q&amp;f=false" rel="nofollow noreferrer">Ultraviolet Radiation, Human Health, and the Urban Forest</a> it would seem at least the general shape you obtained is correct:
<a href="https://i.sstatic.net/XpoxH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XpoxH.png" alt="UVB irradiance on south-facing surfaces with clear skies at solar noon on the 21st of each month for the five cities listed in figure 5" /></a></p>
<p>Here they tracked the irradiance at noon over different months of the year not hours of the day, but the basic change is the same: high to low to high Solar Zenith Angle when irradiating a vertical surface leads to an irradiance graph with 2 &quot;shoulders&quot; on both sides of the maximum solar elevation point. The irradiance is lower at the central point because the direct rays are parallel to the target surface and the only contribution are the scattered and reflected rays from the environment and the rest of the sky, and then as the SZA progresses more and more toward 90 degrees (or elevation toward 0 degrees) the increasing atmospheric absorbance comes into play and reduces the irradiation long before the rays become exactly perpendicular to the target surface.</p>
<p>But if the goal of the UV Index was to provide safety data for the general public, I would say it should've been measured assuming the absolute worst-case scenario of some unprotected skin being exposed to exactly perpendicular solar rays at <em>any</em> hour of the day (this is possible because our bodies have complex surfaces and it's always possible to have at least a little skin that's perpendicular to almost any ray direction in all of 3D space). So they should've used a detector that is rotated to face the sun at all hours of the day, and the resulting graph I suspect would not look like the official daily UV Index graph but also not like the &quot;2 shoulders&quot; one we have here. It would probably look more like the &quot;IND&quot;, &quot;SYR&quot; and &quot;INL&quot; variants in the image above: a sort of &quot;Gaussian with a wider top&quot;.</p>
<p>LE: Here we go, it should look like the <strong>D</strong>irect <strong>N</strong>ormal <strong>I</strong>rradiance curve, with no cosine correction for horizontal receiving surfaces: <a href="https://www.researchgate.net/figure/Direct-normal-irradiance-vs-solar-time-at-Izana-Tenerife-Kaashidhoo-Maldives-and_fig1_236314638" rel="nofollow noreferrer">https://www.researchgate.net/figure/Direct-normal-irradiance-vs-solar-time-at-Izana-Tenerife-Kaashidhoo-Maldives-and_fig1_236314638</a></p>

CC BY-SA 4.0
2021-05-18T21:35:33.207Z
2021-05-26T12:40:07.020Z
2021-05-26T12:40:07.020Z
Don Joe
<p>To answer my own question: the principle is sound however the calculation does not work out, because it assumes only direct UV radiation, and not the real-world scenario of scattered UV radiation.</p>

CC BY-SA 4.0
2018-07-05T21:22:50.513Z
2018-07-05T21:22:50.513Z
rogerwilco
<p>UVvertical Index = UVdirect Index + UVdiffuse</p>

<p>UVdirect Index (vertical) = cos(altitude angle)*(UVhorizontal Index)/(sin(altitude angle)+coefReflection)</p>

<p>UVdiffuse (vertical) = (UVhorizontal Index)/(sin(altitude angle)+coefReflection)*coefReflection*0,55+0,437*cos(altitude angle)<em>cos(45)+0,313</em>(cos(altitude angle)*cos(45))^2</p>

<p>Reflection Coefficient varies during the year, its average value is 0.1.</p>

CC BY-SA 4.0
2020-05-09T18:26:19.253Z
2020-05-09T18:26:19.253Z
lkn
<p>Due to extensive damage to my face because of an accident, I am currently undergoing skin therapy (laser/IPL/medication), that makes my skin extremely sensitive to UV radiation. So I need to stay out of the sun, at the very least between 11:00 and 16:00, need to reapply sun screen every two hours, and generally be careful with exposure. Since that makes me hyper aware to the sun, perceived warmth, UV index and all that, it made me reasearch the subject.</p>

<p>I live in The Netherlands and the meteorology institute published real time UV measurements (as UV index) at <a href="https://www.rivm.nl/Onderwerpen/Z/Zonkracht" rel="nofollow noreferrer">https://www.rivm.nl/Onderwerpen/Z/Zonkracht</a>
However the UV index is defined as the amount energy that is radiated on the flat surface of the earth. As we humans generally have more vertical surfaces, especially sensitive surfaces like the face, would it not make more sense to not just measure in the horizontal plane? Especially during the evening and the morning the sun is at a low angle, so the (relative) energy on a vertical plane would be a lot higher than the above graph suggests.</p>

<p>I made a small program that parses this image, and recalculates the UV index for the vertical plain (divide the UV index by the tangent of (90 degrees minus current zenith angle). This would be the output (blue line is vertical UV index):
<a href="https://i.sstatic.net/SDpj1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/SDpj1.png" alt="UV index in vertical plane"></a></p>

CC BY-SA 4.0
2018-07-05T15:25:11.333Z
One thing you have not considered is that at low angles (sunrise & sunset) light from the sun travels through a greater thickness of atmosphere than it does at noon. Because of this the higher frequencies of light, blue, violet & UV tend to be reflected more by the apparent thicker atmosphere. It's why sunrises & sunsets tend to be red, orange & yellow, which have the lower frequencies of light. Because of this less UV will strike any surface during low sun angles.
Fred
CC BY-SA 4.0
2018-07-05T15:56:48.040Z
Hello Fred, I did consider this, but this is already part of the/measurement of the horizontal plane (the base graph). I do not need to compensate for that again, when switching to the vertical plane (the blue line).
rogerwilco
CC BY-SA 4.0
2020-03-21T20:18:58.273Z
Do you think that your calculation overestimated or underestimates UV when the sun is at the angles lower than 45 degrees? I found this calculation easy and pretty good, it really does make sense. Overestimation isn't problem for me, sun safety should be everyone's priority. I made an excel calculation for personal use, with the outputs being the optimal exposure time for vitamin D production and maximal exposure time for skin damage. I can translate it and share it with you, I think you could find it useful.
lkn
<sun>
calculate UV index in vertical plane
rogerwilco
# calculate UV index in vertical plane

(Asked by: username_0 on 2018-07-05)

Due to extensive damage to my face because of an accident, I am currently undergoing skin therapy (laser/IPL/medication), that makes my skin extremely sensitive to UV radiation. So I need to stay out of the sun, at the very least between 11:00 and 16:00, need to reapply sun screen every two hours, and generally be careful with exposure. Since that makes me hyper aware to the sun, perceived warmth, UV index and all that, it made me reasearch the subject.

I live in The Netherlands and the meteorology institute published real time UV measurements (as UV index) at <https://www.rivm.nl/Onderwerpen/Z/Zonkracht>
However the UV index is defined as the amount energy that is radiated on the flat surface of the earth. As we humans generally have more vertical surfaces, especially sensitive surfaces like the face, would it not make more sense to not just measure in the horizontal plane? Especially during the evening and the morning the sun is at a low angle, so the (relative) energy on a vertical plane would be a lot higher than the above graph suggests.

I made a small program that parses this image, and recalculates the UV index for the vertical plain (divide the UV index by the tangent of (90 degrees minus current zenith angle). This would be the output (blue line is vertical UV index):
[![UV index in vertical plane](https://i.sstatic.net/SDpj1.png)](https://i.sstatic.net/SDpj1.png)

**Tags:** sun

> **Comments:**
> **username_1**: One thing you have not considered is that at low angles (sunrise & sunset) light from the sun travels through a greater thickness of atmosphere than it does at noon. Because of this the higher frequencies of light, blue, violet & UV tend to be reflected more by the apparent thicker atmosphere. It's why sunrises & sunsets tend to be red, orange & yellow, which have the lower frequencies of light. Because of this less UV will strike any surface during low sun angles.  
> **username_0**: Hello username_1, I did consider this, but this is already part of the/measurement of the horizontal plane (the base graph). I do not need to compensate for that again, when switching to the vertical plane (the blue line).  
> **username_2**: Do you think that your calculation overestimated or underestimates UV when the sun is at the angles lower than 45 degrees? I found this calculation easy and pretty good, it really does make sense. Overestimation isn't problem for me, sun safety should be everyone's priority. I made an excel calculation for personal use, with the outputs being the optimal exposure time for vitamin D production and maximal exposure time for skin damage. I can translate it and share it with you, I think you could find it useful.  

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## 3 answers

### Answer by username_3 on 2021-05-18. Score: 5

According to [Ultraviolet Radiation, Human Health, and the Urban Forest](https://books.google.co.uk/books?hl=en&lr=&id=tF1-AV_bqDkC&oi=fnd&pg=PA8&ots=JCSNCY8WqR&sig=ZrryQEeWhymiOSashcjF2tCmoGs&redir_esc=y#v=onepage&q&f=false) it would seem at least the general shape you obtained is correct:
[![UVB irradiance on south-facing surfaces with clear skies at solar noon on the 21st of each month for the five cities listed in figure 5](https://i.sstatic.net/XpoxH.png)](https://i.sstatic.net/XpoxH.png)

Here they tracked the irradiance at noon over different months of the year not hours of the day, but the basic change is the same: high to low to high Solar Zenith Angle when irradiating a vertical surface leads to an irradiance graph with 2 "shoulders" on both sides of the maximum solar elevation point. The irradiance is lower at the central point because the direct rays are parallel to the target surface and the only contribution are the scattered and reflected rays from the environment and the rest of the sky, and then as the SZA progresses more and more toward 90 degrees (or elevation toward 0 degrees) the increasing atmospheric absorbance comes into play and reduces the irradiation long before the rays become exactly perpendicular to the target surface.

But if the goal of the UV Index was to provide safety data for the general public, I would say it should've been measured assuming the absolute worst-case scenario of some unprotected skin being exposed to exactly perpendicular solar rays at *any* hour of the day (this is possible because our bodies have complex surfaces and it's always possible to have at least a little skin that's perpendicular to almost any ray direction in all of 3D space). So they should've used a detector that is rotated to face the sun at all hours of the day, and the resulting graph I suspect would not look like the official daily UV Index graph but also not like the "2 shoulders" one we have here. It would probably look more like the "IND", "SYR" and "INL" variants in the image above: a sort of "Gaussian with a wider top".

LE: Here we go, it should look like the **D**irect **N**ormal **I**rradiance curve, with no cosine correction for horizontal receiving surfaces: <https://www.researchgate.net/figure/Direct-normal-irradiance-vs-solar-time-at-Izana-Tenerife-Kaashidhoo-Maldives-and_fig1_236314638>


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### Answer by username_0 on 2018-07-05. Score: 1

To answer my own question: the principle is sound however the calculation does not work out, because it assumes only direct UV radiation, and not the real-world scenario of scattered UV radiation.


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### Answer by username_2 on 2020-05-09. Score: -1

UVvertical Index = UVdirect Index + UVdiffuse

UVdirect Index (vertical) = cos(altitude angle)\*(UVhorizontal Index)/(sin(altitude angle)+coefReflection)

UVdiffuse (vertical) = (UVhorizontal Index)/(sin(altitude angle)+coefReflection)\*coefReflection\*0,55+0,437\*cos(altitude angle)*cos(45)+0,313*(cos(altitude angle)\*cos(45))^2

Reflection Coefficient varies during the year, its average value is 0.1.


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