the Air Vent

Because the world needs another opinion

Concerned Scientist Kenji knows — Unvetted

Posted by Jeff Id on January 30, 2017

So, to the couple hundred readers who still stop by, Anthony Watts has a fun activity going on.  Kenji, the full fledged yet unvetted member of the union of concerned scientists, is going to Washington.  Kenji has been “concerned” for a half a decade at this point — 35ish dog years depending on which linear or nonlinear dog to human conversion you prefer. Heck I don’t know Kenji, but judging by the muzzle, he’s experienced nearly half a tenth of a degree of average global temperature rise, and in California — that’s food-dish spilling “serious crap”.

But let’s not talk methane….  This is about saving the planet’s fuzzy green ass and that is a serious matter.

Help send Kenji to the “Scientists March on Washington” event!

Look at the high forehead and messy hair.  A clear sign of inspired genius.  Long hours of studying nature with little room for grooming.  With his overwhelmingly efficient thermal coating, Kenji is a natural representative for those who recognize CO2 traps heat, yet somehow he can still find his food bowl!   Adaptation folks —it’s the name of the game!

Anyway I chucked in, you should too.  The fuzzy kid might need a kevlar coat to walk today’s Washington DC crowds.  He does have that cute but angry look of a snowflake though so hopefully the rest won’t bother to read Anthony’s sign.

 


7 Responses to “Concerned Scientist Kenji knows — Unvetted”

  1. If he makes it, he better be careful as he may find himself the next leader of the group with that forehead. Clearly a visionary thinker, and head and shoulders above the rest of the cast of characters.

  2. Well he will need to be vetted before he goes “…he’ll need to get a health certificate from the vet to fly…” 😉

  3. Ghowe said

    He has very small hands.

  4. page488 said

    Heard somewhere that the March for Science will be on Earth Day. I thought, originally, that it was to be a march OF scientists, but, now it seems to have morphed into a march “in support of open research and funding,” so I guess anybody can go. Ahhhhhhhhhhhhhhhhhh. Also, it’s about money. Who knew? Who could have guessed? Anyone who has done even a little research into the funding supplied to some of the big names (and even some of the smaller ones) knows that the “climate people” have been raking it in over the years. Curious to see who attends. We know Kenji will be there; keep him close, especially if the protesters come fro Berkeley; they’ve been known to toss a bottle or two!

    Great post – I needed a laugh this morning!

  5. hunter said

    My guess is that Kenji has been to the vet more than once and has been fully “vetted”. ;^)

  6. Leitwolf said

    Hi everyone! I know, this may not specifically belong here, but yet I do search for help. I am trying to determine earths surface temperature based on the properties of water, i.e. ocean surface with specific attention to hemispheric absorptivity and emissivity. My problem is that these attempts notoriously give a temperature of 288°K.

    I would love to give you my calculations, but I do not know how to upload pics or files (an excel sheet) here, so I would need advise with that too. Let me just describe how I get to these results.

    I started with that information on the emissivity of water presented here:
    https://scienceofdoom.com/2010/12/27/emissivity-of-the-ocean/

    Then I wanted to have a function which would be in line with the measured data quoted there, so that i would not have guess the specific data points from the chart and insert those values in my excel sheet for every single degree from 0 to 90. The trial & error approach gave me those parameters: starting with 1.3105 and incrementing it at the power of 1.032 yields 99.94 at 90°. 100 – these results almost perfectly draws the curve I searched for.

    This graph is actually very similar to the albedo or reflectivity of water, only inverted of course. And as I need absorptivity, I actually have to invert it.

    As I was mainly interested in the geometric aspect, I ignored these minor differences and just weighted this function
    a) according to the distribution of solar radiation, which apparently is one directional, only a small fraction of it will hit at the very low angles. For example the 80 to 90° range only accounts for 1 – sin(80)^2 = 3% of total light received. Thus low absorptivity at these angles play a minor role, and total hemispheric absorptivity would then be 93.95%.
    b) according to the surface of a hemisphere, thereby representing all the possible directions radiation may go from the surface. The formula hereto would be 1 – cos(x) with x being the specific angle. In this case the 80 to 90° range would account for 17.3% of the total surface of hemisphere, and low emissivity at there angles would have a much higher weight. Total hemispheric emissivity would then be 84.39%.

    If we assume that a perfect black body would have 280°K at earths position, then water should be (93,95 / 84,39) ^0.25 * 280 = 287.61°K

    This figure is almost exactly the temperature earth has, and that would mean the oceans were causing the “greenhouse effect”, which then of course would not even exist. I must be wrong! What is my mistake?

  7. russellseitz said

    As a well -muzzled illiterate, Kenji enjoys a freedom from censorship at WUWT that many bona fide climate scientists envy.

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