# Debunked: Dane Wigington's Claims That UV is "Off The Charts"

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#### Jay Reynolds

##### Senior Member.
Dane, will he be your downfall?

#### David Fraser

##### Senior Member.
Have we identified which piece if kit he used? I know he said"Omega Instruments" but I can't seem to find a meter that gives UV A/B ( at least not a manual)

#### Mick West

Staff member
His column shows UVA, UV A/B, and UVC. They say they had two meters, so how are they are getting three numbers? Presumably something must measure UV C as well as something else.

It is possible that instead of the A/B v.s A+B thing they just have a couple of meters that are just stunningly wrong. There are meters that are listed as UV A/B that measure UV A+B. THe numbers make a lot more sense with the divide error though.

It's also possible that they have different units on different meters, and just got the conversion wrong.

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#### Jay Reynolds

##### Senior Member.
Have we identified which piece if kit he used? I know he said"Omega Instruments" but I can't seem to find a meter that gives UV A/B ( at least not a manual)
Check this out, Dave. Dane's last show he said they had to use a UV A/B meter and a UVA meter. This company sells a total meter and a UVB meter which can help determine the percentage UVB:
http://www.solarmeter.com/faqs.html

Q.How can the Model 6.0 UVB meter help?

A.Two ways: Determine % UVB and check acrylic transmission. You will need both Model 5.0 (total UV) and Model 6.0 (UVB) to calculate % UVB. Eg: Model 5.0 reads 17.3 - Model 6.0 reads 1.30. Percent UVB = 1.30 / 17.3 = 0.075 = 7.5%. To check acrylic transmission, take UVB reading w/out acrylic and with acrylic. Most UVB should pass thru acrylic, or it needs replacement. As acrylic ages, it blocks/absorbs UVB quickly vs. UVA.

Turns out amateur and professional hereptologists are acutely aware of UVB to keep their cherished and expensive pets in good condition. Because Vitamin D is a major requirement for reptile health, and is not normally available indoors supplemental lighting needs to be supplied. But how much is enough and how much do various reptiles require without being fired or turned rickety?
http://www.uvguide.co.uk/uvinnature.htm

http://www.uvguide.co.uk/usinguvmeter.htm
http://www.uvguide.co.uk/whatreptilesneed.htm

It seems they have compiled a database of worldwide UVB readings which I am trying to get into. This would be an amateur project not government connected, much as the chemtrails folks could have begun. Looks interesting:
http://groups.yahoo.com/group/UVB_Meter_Owners

it requires a relatively simple account and membership to view, but my slow connection is timing out at this time and I haven't gotten in to see it yet. Anybody else?

If the database is what they say, it should be an independent way to see amateur UVB readings which could debunk directly using data what Dane is claiming with data.

#### Mick West

Staff member

There's 103 UVB readings, taken between 2004 and 2005, ranging from 12 to 502, with the "clear and about noon" figures ranging from about 200 to 502 uW/CM2.

Group description says:
The 6.2 is:
http://solarmeter.com/model62.html
280-320 nm.

#### Attachments

• UVB-Yahoo.xlsx
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#### Mick West

Staff member

This is what he needs to post:

Anacortes, Washington, USA
March Point, Approx 2pm

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#### Mick West

Staff member
Their highest reading: 545 uW/cm2, posted May 2005, from Haleakala on Maui.

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#### Mick West

Staff member
Some more historical UVB readings from Oregon, Paulina Peak, about 180 miles North of Dane and Francis in Shasta:
http://www.oregonphotos.com/UVB-Oregon.html

Note the description of 400+ readings being "Dangerously high", and yet Dane is claiming readings of 10,000+

#### Leifer

##### Senior Member.
UV meters can get expensive.....but modern units are out there, with NIST Cert.
His old Omega unit was likely +10% or more off from NIST standards when brand new.....no telling how off it may be now.

UV meters, Ozonometers....and Pyranometers...
http://solarlight.com/product_application/atmospheric-instrumentation/

Work being done using these type instruments. Are scientists with better equipt. getting frightening readings ?
http://www.pmodwrc.ch/annual_report/annualreport2011.pdf

(from here)
http://www.pmodwrc.ch/

#### Cairenn

##### Senior Member.
If the readings were 10,000 +, folks would be getting sunburns walking to their cars, animals would have burned noses and there would be a huge increase in cataracts and melanomas. Sometimes one has to accept your work is flawed. In physics, my lab mate and I got a different rate for gravity. I believe our lab report said "it seem that a black hole for a different universe intersected with the location of our experiment, while we were conducting it"

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#### Mick West

Staff member
More issues, here's his readings again.

He's saying this shows a an average of around 16 mw/cm2 for UV A+B, and about 7 for UVA (so around 9 for UVB)

Now, in space, sunlight is a total 1366 w/m2, which is 136.6 mW/cm2. That's for ALL the radiation, visible, infrared, and ultraviolet.

UV makes up around 10% of that. about 14 mw/cm2
http://en.wikipedia.org/wiki/Ultraviolet#Sources
Hence Dane's figure for the total UV at sea level is greater than the level in space, before the atmosphere blocks any.

Hence it's wrong. It's literally impossible. The only way you could get readings that high would be if A) there was no atmosphere, and B) You were near the orbit of Venus.

Actual total UV that hits the ground is normally around 3.2 mW/cm2. Most of this is UVA. Hence his UVA readings also seem to be entirely wrong.

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#### Mick West

Staff member
And yet more perspective on the matter, because I find it interesting.

The above graph shows the difference in the solar radiation in space (yellow), and the solar radiation that reaches sea level (red). The various dips show absorption bands where something in the atmosphere absorbs a particular wavelength. These are mostly from H20 (water) but also O2 (regular oxygen), and O3 (ozone) which in particular absorbs the shorter UV wavelength.

Now this graph can be a little difficult to understand in the context of the mW/cm2 readings discussed above. The horizontal scale here is nm (nanometers) and the vertical is W/m2/nm which is watts per square meter per nanometer. And 1 W/m2 = 0.1 mw/cm2

But to calculate the actual level for a particular type of radiation, you need to "integrate", meaning you have to find the area under the graph over the range of wavelengths that type of radiation covers.

Now we are just looking at UV here, and the upper range of UV (the dashed line separating it from Visible Light) is at 400 nm, so we can see that the amount of UV light at the top of the atmosphere (the yellow) could roughly be approximated by a triangle with the base at 250nm to 400 nm, and a height of 1.4. Giving us an area of 150*1.4/2 = 105 W/m2, or 10.5 mW/cm2. Something of an underestimate from the 14 mW/cm2 in wikipedia, but in the same general ballpark.

But what if we were to have a whopping 9 mW/cm2 of UVB, as Dane suggests is happening? Well, that's 90 W/m2, and UVB is actually quite a narrow band, which goes from 280 to 320, so just 40 nm base (compared to the 200 nm spread for all UV), 90/40 = 2.25 W/m2/nm

So in order to get the readings Dane is claiming, all that UVB needs to be crammed into a region just 40 nm wide, which needs an area on the graph which would look like:

Which yet again is obviously totally impossible. The fact that his total UV was higher than UV in space should be enough to prove it wrong, but if you look at just the UVB, its four times the amount of UVB in space. Physically impossible.

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Debunked.

#### Steve Funk

##### Senior Member.
Now, in space, sunlight is a total 1366 w/m2, which is 136.6 mW/cm2. That's for ALL the radiation, visible, infrared, and ultraviolet.

I think it would actually be 13,660 mW/cm2 There are 10000 cm2 in one meter squared (100x100). Then divide by 1000 to convert to milliwatts.

#### Mick West

Staff member
Debunked.

Indeed. But the question remains how to communicate this debunking. I would not be surprised if Dane actually persists with it indefinitely. He's still claiming that the sludge test in his pond was valid.

It's yet another example of "bunk by complexity". I need to come up with some infographic that neatly explains it.

#### Mick West

Staff member
I think it would actually be 13,660 mW/cm2 There are 10000 cm2 in one meter squared (100x100). Then divide by 1000 to convert to milliwatts.

Upside down, watts are on the top, so you multiply by 1000 to convert to mw, and divide by 10,000 to convert to cm2^-1

1 W/m2 = 1,000mw/10,000cm2
1 W/m2 = 0.1 mw/cm2
1,366 W/m2 = 136.6 mw/cm2

http://www.wolframalpha.com/input/?i=1366 watts/m2 in mw/cm2

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#### TomC

##### Member
Indeed. But the question remains how to communicate this debunking. I would not be surprised if Dane actually persists with it indefinitely. He's still claiming that the sludge test in his pond was valid.

It's yet another example of "bunk by complexity". I need to come up with some infographic that neatly explains it.

Something like....
"Are uv levels high, or did someone make a mistake?
Chemtrailers claim uv levels are ....
But the maximum total uv emitted by the sun is ....
Link to calculation of max possible uv, with sources]"

The tricky part is making the calculations simple to understand.

#### Mick West

Staff member
Something like:

The problem here is that some people are going to go "OMG, chemtrails are magnifying the UVB to four times what astronauts get!"

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#### blargo

##### Member
Something like:

The problem here is that some people are going to go "OMG, chemtrails are magnifying the UVB to four times what astronauts get!"

That is the problem Mick, everything is Chemtrails. Today I learned that the rain at Woodstock was from Chemtrails, I kid you not. I would go with the less personal one.

#### Mick West

Staff member
That is the problem Mick, everything is Chemtrails. Today I learned that the rain at Woodstock was from Chemtrails, I kid you not. I would go with the less personal one.

Yes, I've removed the other one.

#### Steve Funk

##### Senior Member.
Upside down, watts are on the top, so you multiply by 1000 to convert to mw, and divide by 10,000 to convert to cm2^-1

Sorry, Brainfart.

#### Mick West

Staff member
Sorry, Brainfart.

Hey, I forgot to divide by two when calculating the area of a triangle in the first draft.

#### Jay Reynolds

##### Senior Member.
I believe that the solar irradiance above the atmosphere is called the "Solar Constant".

In this paper from Israel, they developed a solar UVB constant of about 19 watts/sq. meter, so Dane's figure of 90 watts/sq meter is definitely off.

#### Mick West

Staff member
I believe that the solar irradiance above the atmosphere is called the "Solar Constant".

In this paper from Israel, they developed a solar UVB constant of about 19 watts/sq. meter, so Dane's figure of 90 watts/sq meter is definitely off.

My 2.5 (25) figure was just a conservative estimate from the Solar Radiation Spectrum chart, above. 1.9 means he was even more wrong. I've also see figures of 21.51 given. I think 19.02 might be for a lower altitude.

 19.02 is for a narrower range of UVB, 290-320nm, 21.51 is for the full range 280-320nm.

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#### Steve Funk

##### Senior Member.
Mick, do you have a link for the graph in #55?

#### Steve Funk

##### Senior Member.
Seems like there was a separate thread somewhere for Dane;s claim that atmospheric oxygen levels are dropping at a dangerous rate, but I can't find it.

#### Steve Funk

##### Senior Member.
Got it, thanks. I'm guessing Dane might be a speaker at Saturday's event.

My chemistry is very rusty, but is there any lab or theoretical evidence that aluminum oxide, a stable molecule except in an acid environment, would react with ozone in the same way that chlorine and fluorine ions do?

#### Mick West

Staff member
Got it, thanks. I'm guessing Dane might be a speaker at Saturday's event.

My chemistry is very rusty, but is there any lab or theoretical evidence that aluminum oxide, a stable molecule except in an acid environment, would react with ozone in the same way that chlorine and fluorine ions do?

Seems like there's a couple of things going on, looking specifically at mineral dust there is some direct reaction, but also the surface area allows for other reactions. See attached.

#### Attachments

• acp-10-3999-2010.pdf
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• acp-4-1301-2004.pdf
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#### Steve Funk

##### Senior Member.
From Sullivan, "Ozone Decomposition Kinetics on Alumina . . .:
Given that alumina is both a principal mineral in dust
aerosol and one of the more reactive components for ozone
destruction, we chose to begin our mineral dust studies with
alumina before moving on to authentic dust samples.

So it appears that aluminum oxide, in an optimum environment, could contribute to ozone destruction, if they were spraying it.

#### Mick West

Staff member
Yes, although water seems to be an important factor. I've not read them in depth.

#### Attachments

• NU80r.zip
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#### David Fraser

##### Senior Member.

His UVA meter reads up to 390nm
http://www.omega.com/manuals/manualpdf/M5083.pdf

Yet his UV A/B reads to 400nm
http://www.generaltools.com/assets/images/manuals/UV513AB_Manual.pdf

As UVA seems to be 315-400nm 10nm is nearly an 11.8% difference in readable range. That is extremely significant.

Dane this is why a scientific paper requires that you specify what instruments and makes are used. It allows for repeatability as well as verifying data. If it was a paper it would be dismissed entirely. Try it and see.

#### Mick West

Staff member
Sticking to his guns I see:

Which are wrong. It does not matter how good you think the meters are if they show more radiation than is in space. And my initial estimate of 14 mW/cm2 turns out to be wrong, it's more like 10 or 11. See below.

"doesn't look like much" is hardly a strong argument here. Let's find some references:

American Conference of Government Hygienists Guidelines for Exposure to UV light Related to Typical conditions of exposure:
http://www.dymax.com/pdf/literature/lit023_guidelines_for_uv_exposure.pdf
Pacific University Oregon:
8% of 1367 is 109, or 10.9 mW/cm2. So you can't be getting 18.

Back to Dane:
His linked text says the solar constant has been updated from 1,353 to 1,367 W/m2, an increase of less than 1%. This makes zero practical difference to the figures.

The official figures from which these numbers are derived are the ASTM Reference Air Mass 1.5:
http://rredc.nrel.gov/solar/spectra/am1.5/

• Extraterrestrial: 10.33 mW/cm2 (i.e., the value in space, just above the atmosphere)
• Global Tilt: 4.64 (standard total average for hemisphere facing the sun)
• Direct+Circumsolar: 3.07 (at an angle selected to give an average in the US)
The last two figures are for the sunlight reaching the ground. But the bottom line here is the value for UV in space, just over 10. Dane is claiming 18 on the ground.

If you break down the extraterrestrial UV in UVA and UVB, you get:
• UVB in Space = 2.07 mW/cm2
• UVA in Space = 8.18 mW/cm2
So Dane is actually claiming a UVB on earth of 4.5x that of the UV in space.

I have attached a spreadsheet that shows the working for these figures, with a detailed graph of the UV portion of the spectrum.

So it's irrefutable that the figures Dane is quoting are wrong.

The readings are wrong. The question is why they are wrong.

Now based on what's been said, I think that my earlier suggestion for why they were wrong (A/B vs. A+B) is itself wrong. I'd still like to see a video of the actual reading though. But I'll accept that these are the numbers you are getting out of the meters.

So why is it wrong? I think either the meter is not doing what you think it is doing, or there's some other user error.

Have a look at the spec sheet for the \$800 HHUV254SD UVA/UVB meter.

and also:
First a minor quibble, 4% of full scale reading means 4% of 20 mW/cm2, so it has a claimed accuracy of +/- 0.8.

But the interesting thing here is the measurement range of 240 to 390 nm. Now we have:
• UVC - 200-280
• UVB - 280-320
• UVA - 320-400
So I suspect that what is going on here is that the meters simply measure one very narrow band of radiation, and then extrapolate this out to the full range. Giving vast room for errors. The A/B meter is far cheaper, and probably even more prone to error. But again, I'd like to see it in use.

But bottom line: clearly the figures are wrong. They are greater than the figures for extraterrestrial radiation.

#### Attachments

• ASTMG173-UV-Graph.xlsx
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#### Mick West

Staff member
I've left a note for Dane with the more accurate numbers:

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