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OK.
First what I hope is an uncontrovesial graph
Temperature 0K to 10K – Quanta emitted LINEAR (because it fits the plot better but 10^4 works too) with respect to the temperature
The core A is actively stabilized from 10K to 0K and emitts 100 to 0 quanta
The shell B is actively stabilised from 0K to 10K and emits 0 to 100 quanta
Apart from the 10^4 quanta vs temperature hopefully there is nothing “wrong” with this

http://1.bp.blogspot.com/-73PK0_vgbFg/UWTfiCNpjCI/AAAAAAAABEU/kT5cBjLFOMo/s1600/quanta+vs+temp.jpg

Now the other plots. This assumes the slayer premise that quanta are not utilised when the reach a colder body than what they were emitted from. i.e. when A temperature is below B temperature A quanta have no effect on B (graph shows this as null)

There is also the plot showing the warmist view that all quanta reach and add to the energy in the target
You will note that when looking at net quanta from A to B (slayer mode) at the point A temperature = B Temperature the net quanta in B goes from 0 to -50 so it will begin cooling rapidly.
The warmist mode shows no discontinuity!

Perhaps you could show some plots which better represent your thoughts?

http://4.bp.blogspot.com/-7guLeJTj4U0/UWVPQPzXIkI/AAAAAAAABEs/LNqxJKnDIC0/s640/slay+flow.jpg

http://4.bp.blogspot.com/-yG2VdvcxuDE/UWVPSP4_apI/AAAAAAAABE0/A-9nt-8A330/s1600/warm+flow.jpg

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One Comment

  1. Posted 2013/06/07 at 11:01 | Permalink | Reply

    thefordprefect says:
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    2013/06/07 at 4:36 AM
    Greg House says:
    So, my first guess would be: you made it up. As a second guess, though, maybe this: you made it up but do not know it (any more). I do not know what is worse.
    ——-
    Why not replicate my experiment – you will get the same result. Your position seems to be GHG theory is wrong therefore you can never prove it to be correct. This is not scientific – where is the scepticism?

    thefordprefect says:
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    2013/06/07 at 4:59 AM
    Joseph E Postma says:
    2013/06/06 at 1:41 PM
    You have to obey the laws of thermodynamics. You can’t create more heat than what you put in for work in the first place; it is possible to convert the input into heat more efficiently, but unlike the GHE, you can not actually create more power than the input. The GHE doubles the input.

    All the energy going into the resistores is turning into heat there is no motion created, there is no light created. 100% of the volts time current is emerging as heat. Much is lost through conduction to the atmosphere through 20cm of insulation some is emerging as IR through the window. BUT IR+Conduction= total input.

    There is no way you can convert power to energy more eficiently in a PASSIVE system.

    Fotget that you believe that the 2nd law is violated and that this cannot be happening. It is happening and must therefore be explained.

    The real GHG operation on a steel green house does not change the generated power. When you first put on your steel greenhose at absolute zero escaping radiation is zero. as the steel increases in temperature the emiytted radiation increases. It increases outwards and inwards. This backradiation allows the same power to create a greater temperature on the core.
    The power to heat the core has come from the power that was stoped escaping to space when the steel was first erected. – No extra power is needed to warm the core !

    The power stored in the increase in temperature of the core will be emitted again when the steel is removed.
    With the stell in place there is a power debt to space.

    wWhen the steel is removed this debt is repayted in full and the total power *time with the shell placed then removed will be the same as the total power * time without the shell

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