Friday, August 28, 2015

Dear Bishop Hill: read your links. Also, take a look at this graph.

Bishop Hill thinks they've caught Nicholas Stern in a contradiction, saying one thing in 2009 and another in 2015. So let's take a look, using BH's own links.

Stern 2009:

Lord Stern said that although robust expansion could be achieved until 2030 while avoiding dangerous levels of greenhouse gas emissions, rich nations may then have to consider reining in growth...."At some point we would have to think about whether we want future growth. We don't have to do that now."

(Emphasis added.) That would be the second sentence of the article BH linked to.

And Stern 2015:
...Professor Stern, the chair of the Grantham research institute on climate change and the environment, said that it was a false dichotomy to posit growth against climate action. “To portray them as in conflict is to misunderstand economic development and the opportunities that we now have to move to the low-carbon economy,” he said. “To pretend otherwise is diversionary and indeed creates an ‘artificial horse race’ which can cause real damage to the prospects for agreement.” Green parties in Europe have often argued that decarbonisation requires an end to the model of economic growth “at all costs”. But Stern said that there was now “much greater understanding of how economic growth and climate responsibility can come together and, indeed, how their complementarity can help drive both forward”.

(Emphasis added.) In both cases Stern appears to be focusing on the short to medium term, and in both cases saying there's not a conflict between economic growth and addressing climate change.

In BH's telling, Stern said in 2009 they had to stop growing (BH gave no time frame so one would assume it was immediate) but that Stern in 2015 is saying grow away. Alarmist hypocrisy!!!

As for whether there's a difference over what to do in 2030, who knows - Stern wasn't being asked recently about policies 15 years from now, but I don't see a necessary difference in his statements. Even if there was a difference, BH somehow finds it unforgivable that someone could change their mind on a peripheral issue (what policies should be in place in 2030, as opposed to policies today).

Finally, BH might want to take a look at a graph at renewable power prices. Any graph really, but here's one:



This is new information available to Stern in 2015 and not in 2009, and I could see it having an effect on someone thinking about long-term compatibility of growth with limiting carbon emissions. I remember the debate 5 years ago over whether the long-term decline in solar costs would continue. Now we have the result.

Inability of denialists to adjust opinions to new facts is matched only by their inability to accept long-established ones.

760 comments:

«Oldest   ‹Older   601 – 760 of 760
Everett F Sargent said...

BBD,

So, a line drawn between the Earth and the Sun, or better yet, a sphere with a radius of one AU (Note to self: What is an AU?), intersected by a plane at an angle of, oh, say 50 degrees, that incident TSI, in space, at that angle, is defined as ...

~1400 W/m^2 * cos(theta)

Boy, that's some really tough maths you found on the internetwebs, I could never have figured that one out for myself. And you say that this there really tough maths is in something called "books" and "lecture" notes and peer (Note to self: Is peer the same thing as potty?) reviewed "papers" and that these so called "books" and "lecture" notes and "papers" are widely available on the internetwebs. I've actually learned something on this blog for the 1st time in my life, and I didn't even have to use those so called "books" and "lecture" notes and "papers" and something you call "knowledge" by looking for those things on the internetwebs using something called a "search" engine (Note to self: I only have a web browser, so how do I search for a "search" engine?), well WOW!

But, you also say that there is this something called "calculus", what is this "calculus" thing that you speak of, never heard of that word before, as I only gradumitated, err grad-ge-ated, the 6th grade as a double naught spy.

So the Earth has an "atmosphere" you say, didn't know that either, boy, do I really fell stoopit now. Clouds you say, yes I've seen them there clouds in the sky all the time, they look like aminables even. Optical depth you say, why that sounds so complicated, I don't think my fragile little mind can handle all this complex stuff that you speak of, with links even.

But you say there is this thing called measurements, instrumentation, observations and empirical "data" what is this thing called "data" that you speak of? So that if I were to continuously make these so called "measurements" on the Earth's surface and collect the "data" that you speak of, say at a totally random site, like say the UK, and I took these so called "measurements" and "data" 24/7 over a minimum of one seasonal cycle, which you call a "year" that I would get something called a "sum" that I could take the "average" of, oh boy that sounds so complicated, in fact so complicated that it really does hurt my very fragile little mind.

So, you say that other people, who call themselves "scientists" and "engineers" and "technicians" have done these so called "experiments" on Earth's surface countless times, they even have something called "theories" and that these so called "theories" are something you call "testable" whatever that means.

So this thing that you call the "bottom line" is that the "average" per annum (Note to self: What is an annum?) of this there place you call the "UK" is only about 100 W/M^2. So this thing that you did, I call that ciphering, I done lurned that in the 1st grade at Oxford (I'm an eatin' man), with an areal density of say 25% (as an example, as I really don't have a clue without some actual boots-on-the-ground "data") and an efficiency of say 20% (again, don't have a clue as I would need some boots-on-the ground "data") that would look something like this ...

25% times 20% times 100 W/m^2 is equal to ... 50000%%-W/m^2

WOW! That is a very large number. I'm such a double naught ciphering spy.

What did you say? I have TWO extra double naughts and that the actual cipher is ...

0.25 times 0.20 times 100W/m^2 is equal to ... 5.0 W/m^2

WOW! That is still a large number. I'm still a double naught ciphering spy from Oxford.

Thanks BBD, I done lurned me some of that there them real double naught ciphering spy ciphering.

Blogger profile said...

"But you say there is this thing called measurements, instrumentation, observations and empirical "data" what is this thing called "data" that you speak of?"

EXACTLY! And href="http://lasp.colorado.edu/data/sorce/total_solar_irradiance_plots/images/tim_level3_tsi_24hour_3month_640x480.png">here</a< is the data that proves that 1400 is correct and ACTUALLY MEASURED!

I *keep* telling buddy dumdum that I've not made the data up, but he appears to think that saying that the data from that measurement is not valid for the purpose I wish to put it to is "wrong" or something.

Sheesh!

But I guess he just can't help but imagine a fictional world where such things are valid arguments against a statement of what power production you can get from Solar PV.

But no, he really doesn't get it. Indeed when BPL tried to use maths and equations and values for fluxes and so on to show that 100W/m^2 was wrong, Buddy dumdum just held his fingers in his ears and thought it specious!

So I have ACTUAL DATA that proves my answer is correct and gives a different answer to Mackay, and BPL has ACTUAL MATHS WORKED OUT HIMSELF that proves Mackay incorrect as well.

There really isn't any other argument to make than that. If data and equations both prove Mackay was incorrect, then what is there other than denial of reality?

Blogger profile said...

Oh, "Serge", a tip for you. Proving that something is 100W by saying it's 100W and then proving that 10% of that is 10W, doesn't prove it's 100W. You've missed a quite important step.

BBD said...

Everett

BP doesn't understand sarcasm either ;-)

I think we may have got off on the wrong foot a while back. Even so, I am filled with a warm, nostalgic glow when I recall our chats about Hansen and SLR.

Those were the days, eh?

Blogger profile said...

Hey, Serge, Your maths, such as it is is

TSI at UK latitude = 1400*cos(theta)

You don't actually calculate what that is.

Average annual at the UK = 100

There's a heck of a lot of missing maths there.

Because, as said before, for the UK, theta is on average 50.

Therefore you have

1400*cos(50)=900

And then leap straight to

900 * some factor I can't work out = 100

what is the "some factor"?

I mean, you didn't even go as far as I did.

I ACTUALLY WORKED OUT MORE THAN YOU DID.

Ergo, as far as this "maths" and "calculus" you talk of with disgusted surprise, I beat you by, oh, about 1000%.

Blogger profile said...

"BP doesn't understand sarcasm either ;-)"

Oh, I do, Buddy dumdum. It means "I was sarcastic, so do not draw any conclusion from what I said".

I *also* understand maths. I did some.

So far neither of you have managed a single equation.

As for your nuttery about TSI, well WTF can I say except that you prove my point. You don't recognise physics even when it is biting you in the arse.

Wrong, insane and tedious.

BBD said...

I *also* understand maths. I did some.

So far neither of you have managed a single equation.


Merely a description of the correct approach and of your fundamental error.

Since you are something of a mathematician, you will be keen to proceed with the full calculation.

Note that in the correct formalism, S is held to be approximately 1000 W/m^2 not 1400 W/m^2.

I've explained why.

Have fun.

Blogger profile said...

More unreferenced loudmouthing from Buddy Dumdum.

Show your working.

Blogger profile said...

But lets take your 1000 and see where it goes.

1000*cos(50)=650W/m^2

6.5 times what you claim to be correct.

MASSIVE fail, even if we used the UNSUPPORTED 1000, rather than the MEASURED record of solar output.

Remember: your asinine claim is that it's supposed to be 100. You're wrong even by your "proof" I'm wrong, and a damn sight more wrong to boot!

BBD said...

More unreferenced loudmouthing from Buddy Dumdum.

It's just standard educational material from NASA, BP.

The problem is of course that you are in error using the 1400 W/m^2 solar constant value as that is TOA only.

Show your working.

The point of this exercise is that you show yours.

Correct formalism and value for S supplied free of charge as a courtesy.

BBD said...

1000*cos(50)=650W/m^2

6.5 times what you claim to be correct.

MASSIVE fail, even if we used the UNSUPPORTED 1000, rather than the MEASURED record of solar output.


The next baby step in your understanding of this topic is to distinguish between the idealised noon summer maximum value (~643 W/m^2) and the annual average.

Blogger profile said...

"It's just standard educational material from NASA, BP. "

HA, yeah, right. 'cept it's not EVIDENCED DATA that you're relying on, it contains NO referenced and measured data like this.

And it uses the EXACT SAME calculation as I've used ALL ALONG!

1400*cos(50).

You know that calculation you claimed at one time was an error!

This sort of serial dishonesty is unacceptable, particularly as you never acknowledge that you have repeatedly been shown to be bullshitting.

Blogger profile said...

"The next baby step in your understanding of this topic is to distinguish between the idealised noon summer maximum value (~643 W/m^2) and the annual average."

Which you've never supplied a factor for.

EVER.

WRT calculations, I've already linked to scores of times now, but I appreciate that your ranting rageposting makes the thread extremely difficult to follow.

Blogger profile said...

"The point of this exercise is that you show yours. "

I have, MULTIPLE TIMES.

You? NEVER.

All I've done is quantify which is why I am so often met with such hysterical denialism by ideologically-motivated but fact-free ranters.

You need to RTFL matey, instead of spewing out so much bullshit. Mind you, this couldn't get any worse, so I don't suppose anything you do really matters anymore

BBD said...

LYou know that calculation you claimed at one time was an error!

No, what I said was that your usage of solar constant was in error. You need to use the correct value for terrestrial S, which is ~1000 W/m^2.

It's all in the links. You just need to stop shouting and read.

Blogger profile said...

"No, what I said was that your usage of solar constant was in error."

BZZZT! WRONG!

Blogger BBD said...

"You have not found an error in my calculation."

Yes I have, you idiot.

9/9/15 2:54 PM


1400*cos(50)=900

IS ABSOLUTELY CORRECT.

You are as dishonest as the worst of the other lot.

BBD said...

*Your* calculation is incorrect because you use the solar constant instead of the terrestrial value for S.

As I have pointed out repeatedly, you are working with Plane B when you should be using Plane A.

See fig. 2.3.

Text:

- plane A, a horizontal plane at the point P on the Earth’s surface

- plane B, a surface parallel to plane A but on the edge of the Earth’s atmosphere, often referred to as the horizontal plane

- plane C, a surface perpendicular to the Sun’s rays, often referred to as the normal plane

Blogger profile said...

"*Your* calculation is incorrect because you use the solar constant instead of the terrestrial value for S."

No, MY calculation is CORRECT.

Now, I really don't care whether you have an emotional response to the actual data or not - they are what they are. However, I do have a finite and now exhausted amount of patience for ludicrous nonsense like this.

I have provided data and links supporting what I wrote. Nobody else has, so as far as I am concerned, this discussion is at an end.

Blogger profile said...

I quote your source:
"ISC = the solar constant (1367 W/m2)"

Correcting for 50N, that's just under 900W/m^2!

Even if you use your unsourced claim of 1000, IT IS STILL NOWHERE NEAR 100!

Blogger profile said...

As I said, nothing is relevant here except real world data. I don't care where you went wrong and it doesn't matter. It's a rabbit hole. We are discussing actual power per unit area detected directly from the sun. Nothing else.

BBD said...

Quoting me back at myself only serves to underline how correct I was in the first place.

All the real world power per unit area data from large SPV arrays proves that the 10 W/m^2 annual average figure is about right.

After all, if it were 900 W/m^2, the arrays are really, seriously fucked if the best they can manage is 10 W/m^2.

So it's up to you really. Either SPV technology is utterly useless at converting solar energy into electricity, or MacKay is correct.

BBD said...

Should be >>>> All the real world power per unit area data from large SPV arrays proves that the 100 W/m^2 annual average figure is about right.

BBD said...
This comment has been removed by the author.
Blogger profile said...

" Quoting me back at myself only serves to underline how correct I was in the first place."

So I'm right and you're an idiotic fact-resistant denier.

Fair enough!

Blogger profile said...

"All the real world power per unit area data from large SPV arrays proves that the 10 W/m^2 annual average figure is about right."

Unfortunately, Mackay is making a mistake - a methodological error.

You have to calculate what is possible, not measure what has happened. Otherwise there would be no possible progress, since once you measure what is possible and claim it the limit, you cannot break that limit.


I have provided data - empirical evidence in support of what I have said.

Blogger profile said...

"All the real world power per unit area data from large SPV arrays proves that the 100 W/m^2"

Yet he calculates from the 100W/m^2 that you claim is power per unit area from data found in real actual large SPV arrays from what you continually claim is a FAR LARGER figure than what the real actual large SPV arrays actually produce. 10W.

So your claim is:
SPV actual records show they get 100W/m^2 ground radiance.
SPV actual records show that they get around or less than 5W/m^2, therefore his 10W figure is generous and "proof" he's trying to put the best light on SPV.
And calculating 10% efficiency, accorded to be a low figure that we'd have to put up with when spamming SPV over the planet, from the 100W that the SPV data shows as ground intensity gives 10W, a figure greater than the SPV data shows as their SPV power production.

SPV ground x a lowball 10% = 10W/m^2 = SPV actual figures from SPV = 4-5W/m^2

And you think your maths should be accepted???
TSI is shown here to be 1400W/m^2
Any further argument on this point needs to be backed up by actual data. Otherwise, the matter is closed. I've only got so much patience.

Blogger profile said...

Interested readers are invited to compare the bilge above from Budy Dumdum with an actual reference: Real Scientific Data

Blogger profile said...

"if the best they can manage"

Plsplspls show your evidence that 10W/m^2 is the best they can manage.

Blogger profile said...

Here's proof 900 is correct:

1/7 occupancy of SPV (Mackay's figure) -> 128
8% efficiency (many of those places are old, SPV has moved on a lot) -> 10.29

BOOM.

Please read the links provided. I doubt that The Laboratory for Atmospheric and Space Physics publishes 'bogus' data on their home site run by them. You are behaving like our resident clown and it is extremely disappointing.

BBD said...

Rant, rant, rant.

Still can't tell the difference between TOA and surface.

Still denying empirical evidence about large scale SPV power per unit area.

Still incapable of reading a link.

Still thread bombing like a chimp flinging shit.

Still barking mad.

BBD said...

I knew you'd flip into a frenzy when confronted with this, so I'm going to repeat it:

All the real world power per unit area data from large SPV arrays proves that the 100 W/m^2 annual average figure is about right.

After all, if it were 900 W/m^2, the arrays are really, seriously fucked if the best they can manage is 10 W/m^2.

So it's up to you really. Either SPV technology is utterly useless at converting solar energy into electricity, or MacKay is correct.

Cue frantic thread bombing to bury the inconvenient facts in 3, 2, 1...

Anonymous said...

"So it's up to you really. Either SPV technology is utterly useless at converting solar energy into electricity, or MacKay is correct."

We've got both kinds, country AND western!

Jeez, you really are a boob.

Blogger profile said...

It's all complete and utter bollocks spouted by someone who has very clearly never even looked at the link.

That plays badly here, IIRC.


(cue Buddy Dumdum pretending that he's being hard done by multiple times)

Blogger profile said...

Hey, Buddy dumdum, why do you pretend that a measure of what someone built is the limit of what can be built?

And do you not understand how "I don't have to prove I'm right, you have to prove you're wrong" DOES NOT WORK?

Blogger profile said...

"We've got both kinds, country AND western!

Jeez, you really are a boob."

Indeed

Buddy Dumdum is VERY reminiscent of Shrub or any Creationist.

Blogger profile said...

"Still denying empirical evidence about large scale SPV power per unit area."

Who is denying the recorded output of large scale SPV, or their county record of land purchase?

Mackay is making a methodological error.

YOU are incompetent.

BBD said...

So if we have 20% efficient panels and solar irradiance is 900 W/m^2, I should be looking at 180 W/m^2 power per unit area.

:-)

You are scuppered on that, BP.

Mind you, facts don't really stand in your way, do they?

Shit flinging in 3, 2, 1...

Blogger profile said...

" So if we have 20% efficient panels and solar irradiance is 900 W/m^2, I should be looking at 180 W/m^2 power per unit area."

Yup.

Maths. Did you get someone to work that out for you? You've never actually DONE a sum before.

"You are scuppered on that, BP. "

What? Do you think that if it's 100W/m^2 that I should be looking at 10W/m^2 scuppers you?

If it's 320, that we would be looking at 32, it scuppers both of us?

HOW does it scupper?

BBD said...

How does it scupper? Are you kidding me?

Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2.

Then post the link.

If you should happen to fail, then just post the highest PPUA for an SPV farm you can find.

Then we'll do some more sums ;-)

This is just funny.

Blogger profile said...

Where Buddy dumdum went wrong was in not understanding the way large scale SPV farms work. They don't use 100% occupancy of solar panels because they are too expensive. There are spacing constraints because of the need to expand in the future after profits generate enough cashflow to loan more.

So you get between 5 - 10W/m2 power per unit area BUT CANNOT APPLY IT AS A LIMIT.

That's really all there is to it.

Goodness only knows why you are fighting an endless, hopeless battle against well established matters of fact. #

Go to the pub FFS.

Blogger profile said...

"Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2."

Go find me a large scale SPV installation anywhere in the world that uses 20% efficient panels at full occupancy. TRY to remember that if it casts a shadow, then it will block the light from the solar panel in the shade, BUT ONLY BECAUSE IT IS INTERCEPTING IT ITSELF. So, really, the shadow effect only puts a lowered per-panel-area efficiency, but retains the GROUND AREA efficiency of full occupation.

BBD said...

Stop being evasive.

Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2.

Then post the link.

If you should happen to fail, then just post the highest PPUA for an SPV farm you can find.

Blogger profile said...

"Then we'll do some more sums ;-)"

Do one. Just one to start with. What you REALLY need to do is more than one sum.

1400*cos(50)=900. <- WE ARE HERE.
900*A=B
...
Y*Z=100 <- You claim THIS

Provide your workings out to get from 900 to 100. Explain their proofs, otherwise any other figure could be used just as validly and the result 100 is unspported assertion.

Blogger profile said...

" Stop being evasive."

I asked for hard data from a large scale SPV installation

That is not what I asked for.

You sound stressed Buddy DumDum.

A predicate solution is not disproven in the absence of the predicate.

Ask an adult what that means.

BBD said...

I'm not interested in you wittering.

Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2.

Then post the link.

If you should happen to fail, then just post the highest PPUA for an SPV farm you can find.

Come on.

Blogger profile said...

" I'm not interested in you wittering. "

It doesn't matter what you claim I'm doing, it's irrelevant. As most of your posts have been.

Do you know of ANY large-scale SPV that uses 20% efficient panels and 100% occupancy? Otherwise all you have is "argument from personal incredulity", a fallacious argument.

You have absolutely no regard for the truth at all.

Either that or you don't know what 'empirical evidence' actually means.

Blogger profile said...

You stated that these data were horse shit and that my 900W/m2 estimate was wrong.

Now you need to show why the data are horse shit, or admit that *you* were wrong.

Since the data show very clearly that you *were* wrong, you are in the horse shit up to your neck. The more you wriggle, the deeper you will sink and the fouler the smell will get.

You know me. You know this is just going to get worse. So admit your error.

Blogger profile said...

1/ where is the demonstration that the data linked above are horseshit?

- Either provide it or admit error.
2/ Where is your acknowledgement that you don't know the difference between proving and proclaiming?

-- Admit your error
We are going nowhere until we resolve this.

I want some answers now.

Blogger profile said...

Come on.

With your new found love of mathematics, why not try proving your claims with it?


1400*cos(50)=900. <- WE ARE HERE.
900*A=B
...
Y*Z=100 <- You claim THIS

Get from B to 100.

Everett F Sargent said...

Y*Z=100 <- You claim THIS

Yes!

That is based on a per annum basis, that means the AVERAGE taken over 24 hours per day (which includes those times when the Sun is below the horizon, most people refer to that as "nighttime") 365 days per year.

Your number is NOT that number.

Your number is the MAXIMUM possible instantaneously derived value under PERFECT clear sky conditions with the SUN at it's MAXIMUN two angled position (azimuth/zenith).

Solar zenith angle
https://en.wikipedia.org/wiki/Solar_zenith_angle


"The solar zenith angle is the angle measured from directly overhead to the geometric centre of the sun's disc, as described using a horizontal coordinate system. The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc. If we write θs for the solar zenith angle, then the solar elevation angle αs = 90° – θs."

Solar azimuth angle
https://en.wikipedia.org/wiki/Solar_azimuth_angle

"The solar azimuth angle is the azimuth angle of the sun. It defines in which direction the sun is, whereas the solar zenith angle or its complementary angle solar elevation defines how high the sun is. ...

However, despite tradition, the most commonly accepted convention for analyzing solar irradiation, e.g. for solar energy applications, is clockwise from due north, so east is 90°, south is 180° and west is 270°.

This is the definition used by NREL in their solar position calculators and is also the convention used in the formulas presented here. ... "

That's why, upthread, Jethro Bodine mentioned "calculus" as in an "integration" is required to calculate something called the "sum" (not to be confused with the Sun), this then gets divided by the length of time used in the limits of the "integral" to derive the "mean" or "average" over that integration "period" most people use a time "period" something called a "year" or so I've been told.

So to repeat, you use a single MAXIMUM possible instantaneous clear sky value, the Sun at it's highest clear sky position.

We use the annual "average" (BEFORE weighting it by the instillation total footprint and BEFORE weighting it by the solar PV efficiency (which itself (efficiency) is a function of the Sun's position).

Most people "should" understand this key difference, I mean seriously, even Jethro Bodine understands this key difference.

Those that don't understand this key difference are called DENIERS.

In fact, an AGW DENIER would "claim" that solar PV collects ZERO W/m^2 by "claiming" well lookie here, at midnight the solar PV collects ZERO W/m^2, proof that solar PV is totally worthless. See how that works? You are, in fact, doing the exact opposite, but it is still DENIAL of the worst kind.

Finally.

Why do you think we all want to use the annual "average" as we do have a reason for doing so. It might have something to do with something called the "baseline" conditions, others might refer to this as the "baseload" conditions.

Blogger profile said...

"Y*Z=100 <- You claim THIS

Yes!"

Yes. Now how do you GET it, "Serge"?

Or do you not know?

You have ZERO proof that it's 100. NONE of what you claim is supported by ACTUAL calculation. You propose that it's multiplied by some factor, then some other factor, then another. And maybe a few more. However, WHAT ARE THOSE FACTORS?

900*.45*0.7*0.86=272.

That's not 100.

Blogger profile said...

"Your number is the MAXIMUM possible instantaneously derived value under PERFECT clear sky conditions with the SUN at it's MAXIMUN two angled position (azimuth/zenith)."

So my figure is right.

Prove it must be reduced and by what (else you can't and haven't proven squat).

Everett F Sargent said...

We have proved it. Solar PV "scientists" and "engineers" and "technicians" have "proved" it.

Those people have the wareZ to "prove" it.

You do not have the ability to "prove" that your number is a per annum number.

Until YOU can PROVE that your number is a per annum "average" ...

Or are you "claiming" that your number is, in fact, a per annum number?

As in, the Sun, as seen from the Earth's SURFACE, is always at it's MAXIMUM zenith and MAXIMUM azimuth angles 24 HOURS per DAY and 265 DAYS per YEAR.

That would "appear" to be your "claim" and if so we have already "proved" that that is DEFINITELY and OBJECTIVELY NOT the case.

Thus, we HAVE PROVED you are WRONG.

Consider this a "formal" proof of your abject WRONGNESS!

Blogger profile said...

" We have proved it. Solar PV "scientists" and "engineers" and "technicians" have "proved" it."

No you haven't. Solar PV engineers haven't even TRIED to prove it, it was NOT what they were doing. They installed solar PV, NOT proved that 100W is correct.

Stop trying to gallop away from your problem. From now on, no matter how far or fast you go, it will remain in front of you. You know me.
1/ where is the demonstration that the data linked above are horseshit?

- You made the claim. You will either substantiate it or withdraw it.

2/ Where is your acknowledgement that you don't know the difference between proving and proclaiming?

- Admit your error

The more you wriggle, the worse it looks.

We are going nowhere until we resolve this.

Admitting your errors will be much less damaging than continuing to wriggle. You know how this works. You know me.

Blogger profile said...

"Consider this a "formal" proof of your abject WRONGNESS!"

Fucking priceless!

You hven't a clue how to get from 900 to 100.

And you thing "Well, you do some maths and get 100. PROVED!"

Everett F Sargent said...
This comment has been removed by the author.
BBD said...

Ten comments full of rubbish!

I sense desperation. You *really* are trying to bury this one!

Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2.

Then post the link.

If you should happen to fail, then just post the highest PPUA for an SPV farm you can find.

Why the problem?

If you were correct you should be able to find the evidence in far less time that it took you to write all the bullshit you've stuffed into the ten comments preceding this one.

Everett F Sargent said...

But, using something called the "scientific method" and "testability" thereof we have PROVED your WRONGNESS!

Test: Is the Sun, as seen from the Earth's surface, at it's MAXIMUM zenith angle and MAXIMUM azimuth angle 24 hours per day and 365 days per year? Yes or No.

Simplest "I'm a nincompoop" Test: Is the Sun a FIXED object in the sky as seen from the Earth's surface? Yes or No.

This is a test that even a moran or most other species on this planet could pass!

BBD said...

EFS

I mean seriously, even Jethro Bodine understands this key difference.

That old style charm must be a killer with the laydeez.

Blogger profile said...

"Go and find me an example of a large-scale SPV installation anywhere in the world with a power per unit area of 180 W/m^2."

Keep trying that one, retarded buddy dumdum.

If you can't find a place that shows it is not possible, then you have failed.

But you think that if you make a claim, I have to prove it wrong, if I make a claim, I have to prove me right.

Just like creatards do.

And it doesn't work AT ALL.

Blogger profile said...

"But, using something called the "scientific method" and "testability" thereof we have PROVED your WRONGNESS!"

HAHAHA!

No you haven't. That is just absurd. You have absolutely no regard for the truth at all.

Either that or you don't know what 'empirical evidence' actually means. Or proof. Or "scientific method" or "testability".

The scientific method isn't going "900 is wrong, what you need to do is do some deductions that someone else did, and the end result is 100!".

That's called TALKING SHITE.

Blogger profile said...

900W.

You get 100 HOW?

BBD said...

Here's you (unwisely) agreeing with the implications of your incorrect calculation upthread:

[BBD:] "So if we have 20% efficient panels and solar irradiance is 900 W/m^2, I should be looking at 180 W/m^2 power per unit area."

[You:] Yup.


But when asked to provide some actual data supporting this claim... nothing except wittering.

Asked again... more wittering.

And again... still more wittering.

But no data.

Let's have a look at some solar irradiance data by location / latitude*.

Well, well, well.

Everything below 280 W/m^2 and averaging between London and Edinburgh = 100 W/m^2 for the UK.

Looks like 900 W/m^2 might be the result of a botched calculation, but I believe I have mentioned this already once or twice.

From now on, I'm going to quote a few real world figures from time to time.

First installment = << 100W/m^2 annual average:

Anchorage

Edinburgh

Oslo

Dublin



*******************************************

*Caption and sources:

Variation of average sunshine with latitude and with time of year. (a) Average power of sunshine falling on a horizontal surface in selected locations in Europe, North America and Africa. These averages are whole-year averages over day and night. (b) Average solar intensity in London and Edinburgh as a function of time of year. (Average powers per unit area are sometimes measured in other units, for example kWh per year per square metre; for the reader who prefers those units, the following equivalence may be useful: 1 W=8.766 kWh per year.) Sources: NASA's Surface meteorology and Solar Energy (eosweb.larc.nasa.gov; www.africanenergy.com/files/File/Tools/AfricaInsolationTable.pdf; www.solarpanelsplus.com/; solar-insolation-levels/lightbucket.wordpress.com/2008/02/24/insolation-and-a-solar-panels-true-power-output/.)

Blogger profile said...

"[BBD:] "So if we have 20% efficient panels and solar irradiance is 900 W/m^2, I should be looking at 180 W/m^2 power per unit area."

[You:] Yup.

But when asked to provide some actual data supporting this claim... nothing except wittering."

What?

TSI=1400W/m^2. Cos(50)=0.64. Multiply by 0.2.

WHAT THE HELL NEEDS PROVING????

As I said above, I did not fabricate the empirical data I simply presented it. AND I provided sources, which I have already linked to twice but will do so again..

Or are you accusing that Uiversity of fabricating the data? Because if not, you are making a specious argument. The data are what they are, as I keep on telling you. Your denial of matters of fact is irritating.

I didn't fabricate the data and the sources are there for you to check. What's more, I just linked to them for you.

You have lost this argument.

BBD said...

Look at the irradiance data and stop wittering

*You* have lost the argument.

And for the absolute last fucking time, 1400 W/m^2 is not TSI.

It's the solar constant. Sort your muddled mess out FFS.

************************************

Annual average surface irradiance between 100 - 120 W/m^2

Hamburg

London

Brussels

BBD said...

TSI=1400W/m^2. Cos(50)=0.64. Multiply by 0.2.

1400 cos(50) = 899.9

899.9 * 0.2 = 179.98

Locations with average annual surface irradiance below 100W/m^2

Anchorage

Edinburgh

Oslo

Dublin


I'm not surprised you are having trouble finding an SPV farm with PPUD of 180 W/m^2.

;-)

Blogger profile said...

So far, all you have is:
1) Mackay says it's 100.
2) 1400*cos(50) DOES equal 900
3) But that's not the right number
4) Therefore it must be 100, like Mackay says.

You two retards aren't even trying, are you.

Buddy dumdum never was serious, they just have insanity running the show.

"Serge" just thinks this is a way to get back at me for. Well, who knows. Mother issues? Wasn't picked for the football team at school? He's not happy, that's all we have now.

But you're pretending that "the scientific method" involves:

1) We don't do a calculation
2) Someone took some measurements
3) The answer is therefore 100

You took readings from solar panels. Whoopdedoo.

You think that it's 100 because "that's what the SPV places say!" Except they don't. You say "it's 100 because that's what is measured!" Except it isn't.

You go:

Oh, you have to reduce your number. And get 100. Then stop.

If it's 100, and they use 10% efficient solar PV as per Mackay, why is it none of the ones you linked to got 10W?

Blogger profile said...

"Locations with average annual surface irradiance below 100W/m^2 "

Locations with average annual surface irradiance above 200W/m^2:

Edinburgh
Glasgow.
Stornoway
Leeds
Leicester
Manchester
Rhyll
Cardiff
Bristol
London
Exeter
Dover.

Blogger profile said...

"TSI=1400W/m^2. Cos(50)=0.64. Multiply by 0.2.

1400 cos(50) = 899.9

899.9 * 0.2 = 179.98"

Yes. Your problem?

BBD said...

This is me:

Empirical evidence (post after post)

This is you:

NA-NA-NA-NA!

It's pathetic.

Locations with annual average surface irradiance between 120 W/m^2 and 140 W/m^2:

Munich

Paris

Bern

Blogger profile said...

" This is me:

Empirical evidence (post after post)"

Eeeeh. Not so much:

Blogger BBD said...

Rant, rant, rant.

Still can't tell the difference between TOA and surface.

Still denying empirical evidence about large scale SPV power per unit area.

Still incapable of reading a link.

Still thread bombing like a chimp flinging shit.

Still barking mad.

11/9/15 7:17 AM

BBD said...

And you are just making shit up wholesale, which is also pathetic.

Locations with average annual surface irradiance above 200W/m^2:

Edinburgh
Glasgow.
Stornoway
Leeds
Leicester
Manchester
Rhyll
Cardiff
Bristol
London
Exeter
Dover.


Lies, all of it. Just complete bullshit.

You. Are. Insane.

Blogger profile said...

You want empirical evidence?

See Fig 2.4

BBD said...

More empirical evidence for you to blank out with shut-eyed denial:

Locations with annual average surface irradiance between 140 W/m^2 and 160 W/m^2:

Toulouse

Montpelier, VT

Pittsburgh, PA

Seattle, WA

Chicago, IL

Fargo, ND

Portland, OR



Nutter.

Blogger profile said...

Lies, all of it. Just complete bullshit.

You. Are. Insane.

11/9/15 11:45 AM


Is this ore of your "Empirical evidence (post after post)"?

And since we haven't GOT a future here yet, how the hell can you have empirical evidence of the future???

Blogger profile said...

Nutter.

11/9/15 11:46 AM


Yet more "Empirical evidence", right?

And you are *still* raving counterfactually that the 900W/m2 figure for NH mid-latitude TSI is wrong despite being shown over and over again actual figures from actual satellites proving that it is a reasonable estimate.

Now that's denial on stilts, that is.


And next, show me some hard data from large-scale installations to back up your claim that large scale SPV cannot have a power per unit area of more than 10W/m2.

Show me the data. I want to see hard data.

Now.

Defend your claims.

I can't believe you are this stupid, BBD.

You should have stopped a long time ago.

Thanks for trashing your reputation here some more.

Now, go to the pub.

Blogger profile said...

Go and google this topic. Then post a link to a reputable source that contradicts the following two statements:

1/ At TOA, the solar constant is approximately 1366W/m2.
2/ The UK is not at 50N

Everett F Sargent said...

"TSI=1400W/m^2. Cos(50)=0.64. Multiply by 0.2."

That is NOT a per annum number, that IS an instantaneous number (in outer space at one AU no less).

The simplest nincompoop question remains ...

Is the Sun a FIXED object in the sky as seen from the Earth's surface? Yes or No.

But you seem determined, so just go out and buy some solar PV and find out for yourself, then in 30 years you can report back the single second of the single minute of the single hour of the single day of the single year when you had the MAXIMUM.

However, you will only have PROVED your point with respect to the MAXIMUM.

Per annum != MAXIMUM


BBD said...

Ha ha ha!

You *still* don't get it.

H0 = Plane C (in space)

H1 = Plane B (TOA)

But we are talking about Plane A at the Earth's surface.

Clownshoe. At least try.

Blogger profile said...

" "TSI=1400W/m^2. Cos(50)=0.64. Multiply by 0.2."

That is NOT a per annum number, that IS an instantaneous number (in outer space at one AU no less)."

If that's true, what is the correction for getting to the ground?

BBD said...

Whoops, clipped the top of the last comment.

You want empirical evidence?

See Fig 2.4


Ha ha ha!

You *still* don't get it.

H0 = Plane C (in space)

H1 = Plane B (TOA)

But we are talking about Plane A at the Earth's surface.

Clownshoe. At least try.

BBD said...

If that's true, what is the correction for getting to the ground?

We've been through this.

If you read the thread properly instead of burying it in bullshit and you might learn something.

Blogger profile said...

"You *still* don't get it.

H0 = Plane C (in space)"

But atmospheric absorption is about 22%, and 7% is really scattered back anyway. So we now have

1400*cos(50)=900
900*(1-.22+.07)=785

We're still one hell of a long way away from 100...

BBD said...

More wittering from BP.

You use a value of 1000 W/m^2 for S, idiot, *not* the solar constant. You've been told often enough.

*********************

Nearly there! The entry after this will show locations at which SPV (according to you) would not need to create energy in order to have a power per unit area of 180 W/m^2!!!

Locations with annual average surface irradiance between 160 W/m^2 and 180 W/m^2:

Madrid

Rome

Charlotte, NC

Kansas City, KS

Nashville, TN

Omaha, NE

Billings, MT

Louisville, KY

Blogger profile said...

" More wittering from BP."

By which alchemy you think that the problem doesn't exist?


1400*cos(50)=900
900*(1-.22+.07)=785

We're still one hell of a long way away from 100...

Proclaiming it is "Wittering" doesn't prove that 785 == 100.

BBD said...

Let's just review the empirical evidence that you have blanked so far:

Locations with annual average surface irradiance below 100 W/m^2:

Anchorage

*Edinburgh*

Oslo

Dublin



Locations with annual average surface irradiance between 100 W/m^2 and 120 W/m^2:

Hamburg

*London*

Brussels



Locations with annual average surface irradiance between 120 W/m^2 and 140 W/m^2:

Munich

Paris

Bern



Locations with annual average surface irradiance between 140 W/m^2 and 160 W/m^2:

Toulouse

Montpelier, VT

Pittsburgh, PA

Seattle, WA

Chicago, IL

Fargo, ND

Portland, OR



Nearly there! The entry after this will show locations at which SPV (according to you) would not need to create energy in order to have a power per unit area of 180 W/m^2!!!

Locations with annual average surface irradiance between 160 W/m^2 and 180 W/m^2:

Madrid

Rome

Charlotte, NC

Kansas City, KS

Nashville, TN

Omaha, NE

Billings, MT

Louisville, KY

Blogger profile said...

" Let's just review the empirical evidence that you have blanked so far:"
...since when was New York in the UK??

Facts you've blanked for what seems like eighteen months:

1400*cos(50)=900
900*(1-.22+.07)=785

We're still one hell of a long way away from 100...

Blogger profile said...

"Locations with annual average surface irradiance below 100 W/m^2:

Anchorage

*Edinburgh*"

WRONG!

Tip the surface up.

Blogger profile said...

Not to mention one of them isn't in the UK. A 50% miss rate of an ENTIRE COUNTRY and retard extraordinare Buddy DumDum thinks he's smart.

S
M
R
T

One suspects.

But here we are, still at 785W/m^2. And that's NOWHERE NEAR 100.

BBD said...

By which alchemy you think that the problem doesn't exist?


1400*cos(50)=900
900*(1-.22+.07)=785

We're still one hell of a long way away from 100...

Proclaiming it is "Wittering" doesn't prove that 785 == 100.


Well, first we have to get the calculation right, so instead of using the solar constant (in space), we use the correct value for an idealised clear sky noon summer surface irradiance of 1000 W/m^2,

S = 1000 W/m^2

1000 cos(50) = 643

Now that gets you the maximum possible instantaneous and transient value for surface irradiance under idealised conditions.

Then you return to the overlooked issue of getting an annual average including night, cloud and the months DJF MAM and SON.

When the averaging is done, you find that the maximum annual average surface irradiance for the hottest places on Earth is below 280 W/m^2.

The UK, (50N - 59N latitude), with it's maritime climate, averages 100 W/m^2.

No alchemy required.

Blogger profile said...

"S = 1000 W/m^2

1000 cos(50) = 643"

So? 643 is still not 100.

All you've proven is that a different number that you want to use STILL proves Mackay wrong.

Blogger profile said...

And please source your proof for it to be 1000.

Is it a global average? If so, you don't know it applies to the UK.

Also, remember that 100 you want to prove is right?

Locations with annual average surface irradiance between 120 W/m^2 and 140 W/m^2:

Munich 48.1N

Paris 48.9N

Bern 46.9N



Locations with annual average surface irradiance between 100 W/m^2 and 120 W/m^2:

Hamburg 53.6N

*London* 51.3N

Brussels 50.85N

So your list, which you apparently accept refute Mackay's 100.

Blogger profile said...

"The UK, (50N - 59N latitude)"

Why use 59N?

We have more land for it at 50N than 59N.

Blogger profile said...

So lets take 120W/m^2 for the UK.

Area: 2.43 x10^11m
5% of that: ~5x10^10m.
120W/m^2: 6x10^12W.
16% SPV panel: 10^12W.

UK power requirements: 205GW. Left: 0.795x10^12W.

You know, I think it can be done...

BBD said...

All you've proven is that a different number that you want to use STILL proves Mackay wrong.

If you stopped blanking almost everything I write, you wouldn't keep going round in circles.

Repeat:

S = 1000 W/m^2

1000 cos(50) = 643

Now that gets you the maximum possible instantaneous and transient value for surface irradiance under idealised conditions.

Then you return to the overlooked issue of getting an annual average including night, cloud and the months DJF MAM and SON.

When the averaging is done, you find that the maximum annual average surface irradiance for the hottest places on Earth is below 280 W/m^2.

The UK, (50N - 59N latitude), with it's maritime climate, averages 100 W/m^2.

* * *

Why use 59N?

We have more land for it at 50N than 59N.


I'm *not* using 59N. It's a range. Risible, from someone who has used *only* 50N latitude for the UK down the entire thread.

It's why Edinburgh and London have slightly different annual average surface irradiance. Look up. I even asterisked them in the location list.

Blogger profile said...

"Then you return to the overlooked issue of getting an annual average including night, cloud and the months DJF MAM and SON.

When the averaging is done,"

Hang on. Do the averaging. NOT claim the result. DON'T CHEAT.

Also, isn't cloud also part of that "reflected" and "absorbed". Clouds are part of the earth's albedo, remember.

You're not double-dipping, are you?

NAUGHTY!

So, come on, do the maths.

Don't just go "therefore it can be calculated...". You get no marks unless you show all your working!

Blogger profile said...

"I'm *not* using 59N. It's a range."

Then don't include it in the calculations of the UK average.

Duh.

Blogger profile said...

Hottest place on earth.

peak 1000W/m^2

Pretty much cloudless. 1000*1.0=1000

88% direct sunlight and scatter from the sky.1000*.88=880

Equal day and night, negligible season variation. 880*1/2=440

This isn't 280.

THIS is why we don't accept "It can therefore be proved that... ANSWER!".

Blogger profile said...

FUCK!

The little weasel bastard has ME at it now!

Hottest place on earth.

peak 1400W/m^2 TSI=1400 Latitude=0. Cos(0)=1.

Pretty much cloudless. 1400*1.0=1400

88% direct sunlight and scatter from the sky.1400*.88=1232

Equal day and night, negligible season variation. 1232*1/2=616

This isn't 280.

THIS is why we don't accept "It can therefore be proved that... ANSWER!".

BBD said...

Then don't include it in the calculations of the UK average.

Duh.


Ha ha ha!

The annual average irradiance for the UK isn't based on 50N or on 59N it's, well, averaged across the range.

At least try, BP.

Blogger profile said...

"The annual average irradiance for the UK isn't based on 50N or on 59N it's, well, averaged across the range."

Yeeees.

But if you're including 59N, then you're using it. You just claimed that you weren't using it.

Here is an example.

Series of numbers. 4, 6, 2, 19, 3, 4, 2, 1, 4, 4, 2.

One of the numbers doesn't belong. 19. So you throw it away as anomalous.

You don't then calculate the average as 51/11. You calculate it as 32/10.

Likewise if you're not using 59N, it shouldn't be included in the average.

More maths failure, buddy dumdum? Hardly a surprise any more, really.

Blogger profile said...

So many maths fails.
Pitiful.

BBD said...

You don't understand what 'average' means, do you?

"Then you return to the overlooked issue of getting an annual average including night, cloud and the months DJF MAM and SON.

When the averaging is done,"

Hang on. Do the averaging. NOT claim the result. DON'T CHEAT.


You are the one trying to use the idealised transient maximum value as the annual average, not me.

The annual average *does* include JJA. It looks like this:

DJF MAM JJA SON

I pointed out that *you* excluded DJF MAM and SON and clouds and the night.

Which brings us back to the empirical data that you have been blanking, including the UK annual average surface irradiance of 100 W/m^2.

At least try, BP.

BBD said...

But if you're including 59N, then you're using it. You just claimed that you weren't using it.

That is a feeble nit-pick BP.

I wrote:

I'm *not* using 59N. It's a range. Risible, from someone who has used *only* 50N latitude for the UK down the entire thread.

Which I will happily clarify:

I'm *not* only using 59N. It's a range. Risible, from someone who has used *only* 50N latitude for the UK down the entire thread.

* * *

Annual average surface irradiance for the UK is still 100 W/m^2



Blogger profile said...

Still galloping off on Mr Gish because you can't face the problem of mathematics.

1400*cos(50)=900
900*(1-.22+.07)=785

We're still one hell of a long way away from 100...

Blogger profile said...

" But if you're including 59N, then you're using it. You just claimed that you weren't using it.

That is a feeble nit-pick BP."

Nope. It's finding you out in incompetence at mathematics, like your claim "the hottest places max out at 280" when, for example at the galapagos, 0.7S, it would be an annualized average of 616W/m^2.

THAT is why we need to continue with the mathematics WITHOUT SKIPPING TO THE END.


1400*cos(50)=900
900*(1-.22+.07)=785


So we have the effective ground irradiance at 50N being 785W/m^2 and nowhere near 100W/m^2.

Everett F Sargent said...

BBD,

"Equal day and night ...

Well at least BP admits to ... wait for it ... The Dark Side Of The Earth!

So BP now has a square wave, I think he should publish his work in "The Urinal of Professional Nincompoops" or some such.

700+ comments just to get to "It gets dark at night."

BBD said...

My God you've cracked it, Everett!

We stop farting around with evidence and quote some Floyd lyrics:

So you run and you run to catch up with the sun but it's sinking
Racing around to come up behind you again.
The sun is the same in a relative way but you're older,
Shorter of breath and one day closer to death.


Or is it seven? I've lost count.

Blogger profile said...

"700+ comments just to get to "It gets dark at night.""

You mean if you neglect, for example:

Blogger Blogger profile said...

A 1kW system will take up 6-6.7 sq m and in the UK in Devon/Cornwall produce 1.05kW peak, averaged over the year at that latitude. Shetland about 750W peak.

Tracking means you get about 10 hours sunlight (reduction from 12 being due to extinction at low altitude angles) a day, so tracking you get about 1050*10/24 average watts from the Devon site per 6-6.7sq m panel.

65-73W per square meter. Call it 70W.

Not tracking but tilted: 70% of that, 49W.
Not tracking, flat on the ground, 70% of THAT. 35W

3/9/15 10:25 AM


You know, like 650 posts ago...

Which also happened to be the first time I did the calculation.

BPL did one much earlier. But you ignored that and rubbished it because it wasn't 100W/m^2 but 180+.

Blogger profile said...

Now, please stop wriggling, it;s taken over 200 posts to get from the first line to the second one. We're still nearly 8 times too high.

1400*cos(50)=900
900*(1-.22+.07)=785


So we have the effective ground irradiance at 50N being 785W/m^2 and nowhere near 100W/m^2.

BBD said...

This is you.

I know that The Management would prefer a scalpel to a bludgeon, but sometimes we have to work with what's on the table.

Barton Paul Levenson said...

I'm going to do a web page on this, then post the URL here. Please stand by.

Blogger profile said...

So, two more lame attempts to keep away from the next step from:


1400*cos(50)=900
900*(1-.22+.07)=785

to the ground truth.

This is you >.<

Barton Paul Levenson said...

Aw, hell, I'll just do it here.

Let's find the average insolation (all through the day, all through the year--the annual average) at the ground, for the whole planet. Then let's find it for Britain (latitude 50 North).

The Solar constant is 1360 W/m^2 at top of atmosphere.
The albedo is 30% (unfortunately, that's as precise as we can get. Modern estimates range from 28% to 33%).
That means 952 W/m^2 is absorbed.

Because Earth is a sphere, half faces away, and the facing half is tilted away from normal, so we need to reduce this by a factor of 1/4. 238 W/m^2 is absorbed by the climate system, on average, through the whole year, over the whole globe.

Visual optical thickness averages 0.24, so transmissivity to the ground is exp(-0.24) or 0.7866... Reducing the mean illumination at the ground to 187 Watts per square meter. Over the whole globe.

Now. We can't just take a mean latitude to see what the figure at the equator is, because each latitude band is smaller as you move toward the poles, by a 1 - sin(theta) law on a perfect sphere. To simplify, we'll deal with just one hemisphere, broken up into ten latitude zones (each 9 degrees "high").

Low High Mean Area Fraction
81 90 85.5 0.012311659
72 81 76.5 0.036631824
63 72 67.5 0.060049992
54 63 58.5 0.081989530
45 54 49.5 0.101910213
36 45 40.5 0.119321529
27 36 31.5 0.133794753
18 27 22.5 0.144973505
9 18 13.5 0.152582529
0 9 4.5 0.156434465

To add up to an average of 187 over the whole globe, we have to multiply the cosine of the mean of each band by a constant. The constant turns out to be the top-of-atmosphere absorbed fraction, 238. Is this a coincidence? I'm not sure.

But if you multiply 238 by cosine (50 degrees) you get a mean ground illumination of 153 Watts per square meter for Britain, not 100. Perhaps Mckay was depending on cloudiness being much higher over Britain than the global average? Although it would be have to be something like, well, 53% higher throughout the year. Maybe that's plausible.

Barton Paul Levenson said...

Sorry, I forgot you had to multiply the constant times the cosine of the mean latitude _times the area fraction._ But I put down the right answer anyway. Sorry to mislead.

Barton Paul Levenson said...

In short, I contend that mean annual ground insolation at the equator is 238 W/m^2, which translates to 153 W/m^2 at 50 degrees North (or South). At 20% efficiency, this means a flat field of panels in Britain should be able to generate 30.6 W/m^2 on average (varying by time, season, and weather, of course).

Barton Paul Levenson said...

With 14% filling factor, that would be 4.3 W/m^2. But with 47% filling factor, it would be 14.4 W/m^2.

Blogger profile said...

BPL, there are at least two corrections necessary there.

1) INdirect light. That reduction from passing through the atmosphere also illuminates the sky. About 7% of the sunlight quotient is scattered that way, and being isotropic, doesn't drop the same way at high off-axis angles. Its total effect may be to increase the value by 10% or so. Clouds do the same thing, with more absorption than scattering.

2) Albedo is based off the ground cover. However, if you cover the ground with solar panels, it no longer has the albedo of the average earth ground. 90% absorption is approximately correct for dark solar panels.

However, as has been shown many many times, even a lowball estimate means we're using 1% of the land to manage 100% of ALL power needs. Even without efficiencies, reductions and changing heating requirements.

We even have "zero power" homes that produce as much power as they get from the solar panels on the roof built in.

Blogger profile said...

" With 14% filling factor, that would be 4.3 W/m^2. But with 47% filling factor, it would be 14.4 W/m^2."

There's no design need for 47%, though. Especially since 1% of the land is all we need.

BBD said...

BPL

But if you multiply 238 by cosine (50 degrees) you get a mean ground illumination of 153 Watts per square meter for Britain, not 100.

The very southernmost tip of Cornall (Lizard Point) is just below 50N latitude but let's take 50N as the bottom of the UK.

Going north, John O'Groats is ~58N latitude.

We need to be average, here.

BBD said...

The spread between London and Edinborough is clear in MacKay (2013O) fig. 1.

Blogger profile said...

Nuh uh uuh, Buddy Dum Dum.

You haven't managed a SINGLE EQUATION yet!

You've only ever leapt to the conclusion and gone "You get there with the corrections!".

YOU Are still at "785 watts per square meter". 115 gone, 685 to go...

BBD said...

BP

Now that Barton PL has posted an estimate of 153 W/m^2 for 50N latitude but *not* the UK average, are you going to rip into him too?

After all, 153 W/m^2 even just for 50N is a very long way from 900 W/m^2.

You can see how an annual average for 50N - 58N for the UK as a whole might be, say, 100 W/m^2.

Take it up with BPL if you don't trust me.

Blogger profile said...

"Take it up with BPL if you don't trust me. "

Take what up with him? The claim YOU made on it? No you're responsible for those. Not BPL.

"After all, 153 W/m^2 even just for 50N is a very long way from 900 W/m^2. "

Well, that's 100% your and "Serge"'s fault.

All we've manged so far is to get from

1400*cos(50)=900
900*(1-.22+.07)=785

You've still not managed to get anywhere from there.

That we've not finished is not MY fault, but your the result of your rabid avoidance of any actual effort proving your position.

Like creotards and deniers, it's all about everyone else doing the work to save you face.

Barton Paul Levenson said...

Putting it into equations:

Ig = 0.25 S (1 - A) exp(-Ï„VIS)

N

Ig = k Σ (cos(θMEAN) fA)

i=1

And regional Ig = k cos(θMEAN)

It may be way too late to make any peace in this free-for-all, but I think there's been some misunderstandings on both sides. I don't think BP ever really confused solar constant S with ground illumination Ig; but he was trying to get BBD to provide the intermediate terms. BBD was trying to point out that TSI and Ig are different terms, and you can't go directly from TSI to ground illumination in Britain. Neither was listening to the other. It's hard to listen carefully to someone you find antipathetic, and no one was taking pains not to offend.

This _is_ a war, but it's war between those denying the science for ideological and financial reasons, and those who at least accepts the basic facts that the world is warming, we're doing it, and it's extremely dangerous even in the short term. Can we all start over, or am I being my usual naive self again?

Blogger profile said...

"and you can't go directly from TSI to ground illumination in Britain."

Uh, you can.

You did, even.

Deduct the amount reflected or absorbed in the atmosphere. It all came from the top of the atmosphere.

"Neither was listening to the other. "

WRONG.

I was listening. It was, however, irrelevant to do dumdum's work for it. Especially since its incapacity (the reason why he could not work out if Mackay were actually right) was demonstrated in its extreme measures to avoid doing anything that was explicitly validated and instead leapt right to the conclusion after waffling about "you deduct for some things, and then this gets the answer I said!!!".

I was listening.

BBD said...

BPL

Neither was listening to the other.

Come on.

And it all ends up with an UK annual average irradiance of ~100 W/m^2.

Blogger profile said...

"Can we all start over, or am I being my usual naive self again?"

No, you're being orthogonal.

"Hey, who cares if there are lies being said! We're all fighting for the right cause!".

The gestation of every ill thought and disasterous movement starts exactly that way.

Blogger profile said...

See Buddy DumDum's response.

Despite your calculation that it's 153, dumdum knows that you haven't finished deducting from the value until you get 100.

Because.

Um.

Er.

It isn't listening to you.

Everett F Sargent said...

BPL,

Yes, it is a war, for the reasons that you state. I too would like to do a reboot.

As to MacKay's Figure 1 (BBD has linked to, caption may already have been mentioned verbatim upthread, but tl;dr, (meaning I no longer remember)) ...

"Variation of average sunshine with latitude and with time of year. (a) Average power of sunshine falling on a horizontal surface in selected locations in Europe, North America and Africa. These averages are whole-year averages over day and night. (b) Average solar intensity in London and Edinburgh as a function of time of year. (Average powers per unit area are sometimes measured in other units, for example kWh per year per square metre; for the reader who prefers those units, the following equivalence may be useful: 1 W=8.766 kWh per year.) Sources: NASA's Surface meteorology and Solar Energy (eosweb.larc.nasa.gov; www.africanenergy.com/files/File/Tools/AfricaInsolationTable.pdf; www.solarpanelsplus.com/; solar-insolation-levels/lightbucket.wordpress.com/2008/02/24/insolation-and-a-solar-panels-true-power-output/.) (Online version in colour.)"

I'm currently trying to reconstruct that figure from this site (NASA link) ... or some facsimile thereof (London looks close while Anchorage appears to be low).

https://eosweb.larc.nasa.gov/sse/
https://eosweb.larc.nasa.gov/cgi-bin/sse/retscreen.cgi?email=skip@larc.nasa.gov
https://eosweb.larc.nasa.gov/cgi-bin/sse/grid.cgi

We can, of course, question NASA's numbers or any numbers from MacKay's references/links too. However, MacKay's numbers do appear to be in the ballpark, so to speak. Note: MacKay's numbers may date to the time of his book publication (Version 3.5.2. November 3, 2008 according to the online PDF so the current version of the NASA datasets may now differ somewhat). YMMV

Kevin O'Neill said...

Averaged over the year, the UK receives less than 90W/m^2. Peak insolation is in June when it receives 162 w/m^2. This is in contrast to December (20) and January (25).

Everett F Sargent said...

The https://eosweb.larc.nasa.gov/* website has data circa 1983-2005 (which implies that MacKay's data, if taken from this website, has not changed at all).

A more recent NASA dataset can be found here (circa 1983-2015-near real time or ~32 years of full years of data) ...

http://power.larc.nasa.gov/ ... specifically ...
http://power.larc.nasa.gov/cgi-bin/cgiwrap/solar/building.cgi?na ... or ...
http://power.larc.nasa.gov/cgi-bin/cgiwrap/solar/agro.cgi?email=agroclim@larc.nasa.gov

For Anchorage, AK I have 110 W/m^2 (1983-2005) and 109 W/m^2 (1983-2015).

BBD said...
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BBD said...

So, in summary:

MacKay = Correct

BBD = Correct

Thank you for your patience.

Blogger profile said...

" So, in summary:

MacKay = Correct

BBD = Correct"

So in summary, 153==100???

10W/m^2 == 30W/m^2 (or 10W even an OVERESTIMATE????"

Yeah, right.

Blogger profile said...

"However, MacKay's numbers do appear to be in the ballpark, so to speak"

His proof you provide are method errors. They are used to "prove" we can't do better than 10W/m^2 when they provide no such information.

As for a reboot, we haven't even finished POST.

1400*cos(50)=900
900*(1-.22+.07)=785

So we're currently at UK ground level peak.

What's the next step?

Blogger profile said...

" Averaged over the year, the UK receives less than 90W/m^2."

You are wrong. Look at the above data.

BBD said...

|You are wrong. Look at the above data.

Oh shut up, you nutter.

The data above comprehensively demonstrate that you have your head up your backside.

Everybody's had enough.

Give it a rest.

Blogger profile said...

Yeah, just the exact same of empty bs you always manage, you retard.

Blogger profile said...

There is NO REASON to not use 100% renewables.

It can manage it with less than 2% of the land being used, and would generate a generous excess to boot.

It would be, even then, half the price of UNCOSTED BACKUP nuclear which wouldn't produce an excess overall, but be unable to follow the load variation, so would still require other methods of generation or dumping excess that has to be purchased to produce.

The UK uses 71% of its land for farming.

under 2% for generation is peanuts.

And, unlike nuclear, we can (and are) starting now.

Kevin O'Neill said...

Blogger Profile - I took the numbers from one of the many solar calculator sites on the web, though I forget which - I can dredge it up out of browser history if it's important.

I used London as the basis for the insolation - though it's a little south of the average latitude for the entire United Kingdom, so it's biased a little high. But there really isn't much difference between north and south in the case of the UK.

If you'd like to check yourself try this one:
http://www.efficientenergysaving.co.uk/solar-irradiance-calculator.html

BBD said...

Yeah, just the exact same of empty bs you always manage, you retard.

I filled this thread with references and empirical data.

You denied the lot and blanked it out wholesale. But it was, and still is, all there.

You were wrong. Fact. End of story. Now give it a rest. Everybody's had enough of your madness.

Blogger profile said...

It would cost less than £30Bn to do. It's £30Bn at 2014 prices. 100% of demand by Wind, 100% of demand by Solar.

The price of Hinkley Point for its 3200MW (2240MW actual), costing £24Bn. And doesn't manage to replace our power requirements.

Blogger profile said...

"Blogger Profile - I took the numbers from one of the many solar calculator sites on the web"

Yeah, so how do you know they are right? Like I said above, look at the data on this page.

Where is BPL's problem?

Blogger profile said...

"I filled this thread with references and empirical data. "

So did I.

If the claim of empircal data and references were sufficient, it's sufficient for me. If it's not for me, then it isn't for you.


You denied the lot and blanked it out wholesale. But it was, and still is, all there.

And you're doing the same thing here. BPL calculates 158W/m^2 and you claim it's 100 WHEN CLAIMING YOU BOTH AGREE.

Blogger profile said...

Read the wrong number for Hinkley.

Getting it from Nuclear would cost 182/67 for the merely nameplate capacity equal. Given 70% of nameplate, that makes it 116Bn for nuclear that is generated by two renewables for £30Bn.

BBD said...

And you're doing the same thing here. BPL calculates 158W/m^2 and you claim it's 100 WHEN CLAIMING YOU BOTH AGREE.

What is it with you?

Why will you not simply read the bloody thread?

BPL made a mistake. He gave the value for 50N only, which is why it is too high for a true UK average. As I pointed out at the time:

[BPL:] But if you multiply 238 by cosine (50 degrees) you get a mean ground illumination of 153 Watts per square meter for Britain, not 100.

[Me:] The very southernmost tip of Cornall (Lizard Point) is just below 50N latitude but let's take 50N as the bottom of the UK.

Going north, John O'Groats is ~58N latitude.

We need to be average, here.

* * *

Now that Barton PL has posted an estimate of 153 W/m^2 for 50N latitude but *not* the UK average, are you going to rip into him too?

After all, 153 W/m^2 even just for 50N is a very long way from 900 W/m^2.

You can see how an annual average for 50N - 58N for the UK as a whole might be, say, 100 W/m^2.

This is all simple stuff BP, but I am certain at this point that it goes right over your head.

Blogger profile said...

"What is it with you?

Why will you not simply read the bloody thread?

BPL made a mistake. He gave the value for 50N only"

Calculate it for 60N, dipshit.

Everett F Sargent said...

What's the next step?

Go to preschool where the tetrapod blocks are numbered 1 thru 4.
Go to kindergarten where the cubes are numbered 1 thru 6.
Go to elementary school where the blocks have symbols like + and - and * and / and =.
Go to junior high school where you had sex with Mr. Garrison when he was a man.
Go to high school where you had sex with Mrs. Garrison when she was a woman.
Go to college where you had sex with Mr. Garrison when he was a man (once again).
While you are in college, take courses in things that contains the words "calculus" and "statistics" and "applied mathematics" and "tensors" and ... and then you would never be as smart as Richard Tol as he is Earth's gift to the Universe (but we do think that Ralph Kramden could get him to the Moon).

Oh, wait, you didn't do any of those things, but other people apparently have (sans the Mr./Mrs./Mr. Garrison parts), and they say "Wait a minute there Bucko, not so fast." and "We do have theories that kind of get complicated and appear to give us reasonable answers." and "We need data on the ground to better inform our theories." and so they ponder and calculate and program and create online tools (one online tool that they did not create was you, but yes, you are an online tool).

Other people, those who could also create those online tools, and others who could not create those online tools, say "Other people have been there done that and still others have also been there and done that, as a check, which makes still others go, thanks, no need to reinvent the wheel, but we could download the public domain source codes to sort of see how difficult it does get ... "

Long story short?

You ask a simpleton question and demand a simpleton answer. Earth science appears to defy simpleton questions and simpleton answers, other than, humanity is totally hosing Earth on a global scale.

Long story even shorter?

Try to learn something from sources other than social media (e. g. this blog). There is no simpleton answer to your simpleton question.

But in the spirit of this "never ending thread" could you repeat ...

constant*constant=constant

... an infinite number of times?

Blogger profile said...

So I'm going to have to wait until you've done all this remedial adult education and THEN you think you might be able to manage the next line?

You're not going to bother with that, though, are you.

Because you really don't give a rats ass about facts, truth or rationality.

Why?

Because you're just an ignorant blowhard on the internet.

Blogger profile said...

"Simpleton"????

So simple it's for a simpleton, but not simple enough for you.

Right.

You're too much of a simpleton to work out your problem with this stance.

Blogger profile said...

Obviously Dumdum is STILL to pig ignorantly stupid to work out what it would be, but the result of multiplying 238 by Cosine of 60 is 119.

THIS IS BIGGER THAN 100!

Retard really doesn't want to admit this because they're a shithead. A complete and UTTER shithead.

The average of a number that varies between 150 and 120 is 100...

Frigging retard.

Barton Paul Levenson said...

BBD: And it all ends up with an UK annual average irradiance of ~100 W/m^2

BPL: I don't understand how. If the southernmost part is at 50 N, and the northernmost part is at 58 N, then the average is 54 N (ignoring what fraction is north and south of the middle line), and 238 cos 54 is 140 W/m^2. There is still a term missing:

M = f L

where M (Mckay) = 100, L (Levenson) = 140, and f = about 0.714. I would like to know what accounts for the value of f. Is it, in fact, above-average cloudiness? Or pollution? Or something else? Is it more than one factor? K is saying 90 is the average, which would give f = 0.643. I'm not saying either Mckay or NASA (or wherever K got his figures) is necessarily wrong, but I'd like to know the source of the discrepancy. Did I make a mistake in my calculations? If so, where? The only other thing I can think of is the departure of the Earth from perfect sphericity, but that can't give a very big factor.

BBD said...

BPL

I don't know either. I suspect the difference between your calculation and the measured surface irradiance might be cloud. The UK is after all subject to a maritime climate and this would be the simplest and arguably most likely explanation for lower than expected actual surface irradiance averaged over the year.

BBD said...
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Blogger profile said...
This comment has been removed by a blog administrator.
BBD said...
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Blogger profile said...
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BBD said...
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EliRabett said...

Take a timeout gentlebunnies. It's Saturday night

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