New end Improved version:

I admit NOTHING but gratitude, for your input to my quest to understand human behavior. :-)

I have some numbers for the ‘nuclear heating’ problem, thanks to gbaikie’s post, but first let me explain about the IMPORTANT concepts that I gleaned from our information exchange.

My philosophy for understanding the roots of scientific progress

Imagination gives you a goal to reach for.
Questions and Mathematics gives you the ‘road map’ so you can figure out how to reach the goal.
Effort is needed in order to create progress.
So in other words: p=Ig/(Q*M)*E (needs refinement :-)

Is my formula correct?
I can test it by taking your assertions that:

“The radioactive materials released have had essentially ZERO effect on the atmospheric temperature.” And that:“The effect of billions of lightbulbs burning across america is FAR greater, and is still negligeable!”
Or my assertion that; the extra radioactivity that has been released into the atmosphere during our ‘nuclear age’, may have increased the kinetic energy of the atmospheric molecules.

And plug them into the formula.

These ‘assertions’ constitute the ‘Ig’ part of the formula.

The Q part of the formula is the process of obtaining information needed to do the proper Mathematics

Progress in this case, is for both of us to admit that; at least, some parts of the other persons ‘assertions’ are correct.

For this ‘nuclear heating’ situation, I obtained most of the Q part from here.

To find the needed information and do the required calculations takes some Effort and time.

I did not find any reference to global atmospheric radioactivity in the above link, but I did find out that Canada will be ‘producing’ (I assume this to mean mining and refining) about 18500 tons of Uranium by the year 2006 and this will represent about half of the world production. Therefore the world production will be twice that or about 37000 tons. When it comes to mathematics, I can only work from the information that I have. Since this information is a projection to the year 2006, the calculation will also be a projection to that year. As I understand it, the fission from -U235 atom releases about 1 w/s of energy.
From an Elementary Physics book that was printed in 1964 I obtained the information that: It has been shown that radium in equilibrium with its products emits heat at the rate of about 135 calories per hour per gram. About 4.5 per cent of this heat is due to < i > b < /i > rays, 5.5per cent. to < i > y < /i> rays, and the reminder to a rays.
Since there is 1000 grams in a kilogram and there is 1000 kilograms in a ton. Each ton emits 135*1000=135000*1000=135000000 calories of heat.
Multiplying that figure by 37000 gives us 4995000000000 calories per hour. So in a 24 hour day, that radiation gives out 11988000000000 calories or about 501757740000000 joules of heat energy and x number of a,b, & y emissions that need some place to go and something to do. (I wonder where they go and what they do. Neither mass nor energy can be destroyed, therefore they must ‘go’ somewhere and since they have a ‘charge’, they must ‘do’ something.)

Now how many “ billions of lightbulbs burning across america” are there and how many calories of heat do they emit in one hour or one day?

Hmmm……Lack of information STILL prevents progress.

Some information for you.
SODIUM: Atomic number = 11, density= 0.971 g/cc, melting point= 97.8 degrees C, boiling point= 883 deg. C.
NITROGEN gas: Atomic number=7, density=1.2506 g/cc, melting point is -209.886 deg. C, boiling point= -195.79 degrees C.

Anything else Master*****r? ;-)

Cheers
Johannes