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The free to ALL "shape" issue pdf. - Derek.
#1
Hi All,
Please feel free to download and distribute as you see fit, for free, the below pdf and excel workbook.
I will try to answer any arising questions you may have on THIS THREAD

The piece in the pdf is titled,

The importance, relevance, and revolution in our studies of a planetary climate system
via the physical, and mathematical differences between a disc, a hemisphere and a sphere.


Later addition - The pdf I have put here is best read AFTER reading the articles I refer the reader to at the start of the piece. Particularly the Alan Siddons articles.
However for clarity it may be worth me briefly stating some of the relevant points here
just in case the reader does not get time, or have immediate access to the referred to articles.

The formula that the pdf is based upon AGW's, K&T type plots, and climate modeling's use and interpretation of,
is the below formula.

(P / 5.6704)^0.25 * 100 = K.

In this thread at Jeff Id's the Air Vent blog where the pdf and excel sheet is kindly being "blog reviewed", I have posted (slightly edited and corrected here) the below.

I will happily admit that I do not like equations, yes, they are “short hand”, but they can also be very, very misleading.
My usual “defense” is that I try to “walk” myself verbally through an equation.
So, in this case what would my verbal walk through the equation be?

The equation.

(P divided by 5.6704) raised to the power 0.25, and then multiplied by 100 = K.

Verbal “Walk through”
P is the power of a beam in Watts per meter squared received at a black body’s surface,
this is divided by 5.6704, and raised to the power 0.25, the resulting figure is then multiplied by 100.
The end figure is the temperature expected at the black body’s surface for the power of beam it is receiving in degrees Kelvin.

OK, so what does it mean? Well firstly the equation does not infer any shape whatsoever.
P is merely a rate for an area equivalent to a square meter, not a shape as such.
Yes, the answer applies to all the surface area receiving the same rate, but at the end of the day,
the equation is (merely) a “point” calculation.
This in itself I suggest gives credence to the points raised in my piece,
which is about the effect of incorrectly applying “shape” using the equation.
There is a mathematical “quirk” between the “shapes” as incorrectly applied, and
this mathematical “quirk” does produce a “33 degrees effect” that
does not necessarily exist in reality upon planet earth.

What does dividing P actually do.?
In the case of the equations use when comparing the equations results to earth, or a black body receiving energy from a single source,
dividing P moves the object away from the source.
This must be the case as that is the only way the W/m2 received by the black body could be reduced, with a single and constant source of P.

If we agree that earth is nominally 93.5 million miles from the sun, then dividing P by two doubles the distance between the sun and earth, and
if P is divided by 4, then the distance between the earth and the sun is quadrupled (187 million miles, and 374 million miles respectively).
So, I agree (at least partially) with Jeff Id, when he posted in comment 51 that my use of 682 is incorrect, but that also means that the 341 figure is also incorrect, and incorrectly used.

I do not see the point in considering a disc 187 million miles from the sun, or a disc 374 million miles from the sun, when
earth is only 93.5 million miles from the sun, I suggest we are all in error, and it is not a small error.

The other point I would like to raise is that the often quoted “disc” is not a disc at all really, and neither is it flat.
The equation is for an area receiving the same rate of input, so
from a point source of P the shape would have to be a saucer to maintain a constant R and therefore input.
Also, the sun is not a point, nor is it’s effective surface of emission either flat or constant.
One look at any image of the sun and it’s solar flares confirms this.
So, the often quoted flat disc is
a) more of a saucer shape “mirroring” the sun’s surface, and
b) the surface of the saucer is constantly in a flux moving to and fro to maintain a constant W/mm2 input from the source.
In short, when the equation is used to compare to the earth, then
the resulting “shape” is more like a very wobbly saucer shape that is most commonly [b]374 million miles from the sun…


I have recently posted this elsewhere which maybe of help.
In short, with regard to the equation,

(P/5.6704)0.25*100=K

1) The shape of the source (or combined sources of) P determines the shape of the area for which K is calculated by the equation.

2) Dividing P merely increases the distance between the source of P and the black body surface area temperature calculated.

3) Misinterpreting dividing P to assigning to the surface area that the temperature is being calculated for to another "shape"
is the error by which an imaginary "33 degrees greenhouse effect" has been "created".


I hope the above helps people understand the pdf better.

For those who are not sure of who I am,

I am an English factory floor worker,
who has never been paid a single penny by anyone,
nor received a single penny from any other source,
in regards of my AGW skepticism or various related writings.



Attached Files
.pdf   Derek - Free to all pdf to end AGW scam Saturday 18_12 version..pdf (Size: 1.14 MB / Downloads: 506)
.xlsx   Derek - Excel sheet calculator, disc, sphere, hemisphere temps 18-12.xlsx (Size: 2.04 MB / Downloads: 342)
.xls   Dividing P = Distance from P.xls (Size: 66.5 KB / Downloads: 374)
#2
Accompanying excel sheet in Microsoft Office 2003.

When you convert 2007 to 2003 the file size increases quite a bit, it almost doubles, so apologies to those on 2003,
the sheet has been split into two.
That said, I would of thought it will not be too difficult to download both and copy the sheets into one workbook.


Attached Files
.xls   Derek - Excel sheet calculator, disc, sphere, hemisphere temps 18-12 P1.xls (Size: 1.83 MB / Downloads: 432)
.xls   Derek - Excel sheet calculator, disc, sphere, hemisphere temps 18-12 P2.xls (Size: 1.72 MB / Downloads: 420)
#3
As a direct result of the "blog review" at Jeff I'd the air vent blog mentioned in the first post,
(Copy of tAV thread in word document form attached to this post, "blog review" up to Jan 1st 2011..)
I have further posted there at comment 60 the below (again edited and corrected here).

In short, when the equation is used to compare to the earth, then the resulting “shape” is more like a very wobbly saucer shape that is most commonly 374 million miles from the sun…

I think it is safe to say “we” need a better “model”….


My tentative, first suggestion.
Imagine a meter square at eye level about ten feet in front of you. The meter square is vertical, or perpendicular to your view of it.
If the square is tilted away from you (vertically) does it reduce in apparent height to you, yes.
If the square is then also tilted away from you (horizontally), it’s visible area to you is further reduced.
Your “view” would be of the effective area exposed to P.
Given a rate and time (W/m2 for example) for P, and an effective area exposed (Latitude and longitude “corrected”), then a volume (Joules) per square would be quite easy to calculate.

To apply this to a lit hemisphere of known size (earth), a known number of miles from a source of P (the sun),
would only require each “square” to be corrected for actual area on a hemisphere’s surface.
So, this would be a grid of degree squares (180 x 180) corrected for latitude and longitude inclination (from source of P),
(0 for latitude and longitude is only a line - with possibly a square overlapping a quarter of the four "0" squares to produce the correct "maxium")
and then corrected for actual area on a hemisphere’s surface.
Given the rate, time and area are thus calculable, then a volume (Joules) answer is relatively easily obtainable.
This would be a “model” for a result for a hemisphere “shape” that was it’s actual distance from the source of P.
A vast improvement over the present “disc” “calculations” that are apparently (when P is divided by 4, 374 million miles from the sun….).

This “hemisphere model” approach would be quite versatile, and it would seem to me to be capable of far more than merely a “snap shot” view,
it could easily be doubled into a “sphere”, and / or “rotated” for example, and
incorporate many things not presently included in modeling, such as
surface heating, retention and varying later release….Oh, and night of course…..


Thread closed, but considered comments are invited for discussion in Dereks Forum.
Please raise any questions or queries regarding what I have not explained clearly enough in the piece or accompanying excel workbook there.


Attached Files
.docx   FREE to all pdf Air vent blog review..Dec 2010 to Jan 1st 2011.docx (Size: 93.2 KB / Downloads: 580)


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