I recall there was a documented case in 1962 of some proto-space explorers who were bombarded by cosmic rays. They affected them each in entirely different ways: it caused one of them to spontaneously combust, another’s flesh became extremely calloused and discolored, another’s entire body became pliable, and a fourth just disappeared.
Quite a fantastic account! 
Link to column: Cosmic Rays – CKDH
Rather interesting bit of stuff regarding the fellow Ed that Cecil was attempting to EDucated. What I want to know is how “conservation” that Ed touted would produce energy? Certainly it will reduce consumption, but that isn’t production. It would seem that it was Ed that “missed the boat” versus Cecil.
Hormesis.
See http://www.sciam.com/article.cfm?colID=5&articleID=00053833-173D-1F30-9AD380A84189F2D7
And it will eventually lead to their Doom!
V. Von D.
Why would lead shielding significantly reduce a plane’s carrying capacity? It’s not like they have to install thick metal plating, for crissakes. I am inclined to think that lead foil would provide a pretty good protective benefit at low cost and would hardly affect the plane’s overall weight at all.
CDorian is on the money. There was an article in Physics Today a while back (sorry, no time to find the cite), suggesting that there IS in fact a threshold below which radiation exposure is harmless.
Lead shielding would probably only make the problem worse. Cosmic rays would collide with lead nuclei producing a large number of low-energy secondary particles–the whole zoo. For charged particles, the lower the energy, the greater the damage. What you want is for the high-energy particles to pass through you without stopping.
I’m a practicing cosmic-ray astrophysicist, and I actually used the cosmic-ray scene from the Fantastic Four origin story as the opening slide in my qualifying exam talk.
Actually it is exactly like they have to install think metal plating. A thin foil will not block very much radiation. You basically need to get a lot of material between you and the radiation source so you ensure that most of the rays hit a big heavy atom nucleolus.
Cecil mentioned power plant workers in his piece, and I recalled that even coal fired power plants present some radioactivity to workers. Now that many coal fired plants have converted to natural gas, do they exude radioactivity like the coal did?
I know some of those guys in formerly-coal plants.
Why don’t you ask Anthracite? She’s hip to all that coal stuff.
“Many” coal power plants have converted to gas? What country are you writing from, BTW, before I comment specifically? Because it’s certainly not the US. Is it Italy?
Estimates on radiation exposure from coal power plants vary wildly, mainly because (surprise!) the composition and makeup of coal varies wildly from basin to basin, seam to seam, and even within the same mine, and is impacted by cleaning, blending, etc. The work I’m doing with mercury tracking and removal all has a direct impact on being able to predict and measure the level of radioactive materials in coal.
Natural gas plants generally do not emit measureable amounts of radiation. If we really want to get into the subject I can dig up some ranges of levels of radionucleides in coals.
The column: Is exposure to cosmic rays during cross-country flight dangerous?
Regarding discussion of radiation shielding, I would suggest looking over this material. It is actually discussing radiation exposure and hazards for astronauts regarding the Apollo missions, but it is informative.
http://www.clavius.org/envrad.html
You might scroll down to the statement about six feet of lead. You might also read the linked radiation primer page.
Essentially, if you’re going to use lead, you need a whole lot of it, because of the Bremsstrahlung (secondary effect radiation caused by high energy particles striking atom nuclei).
In Cecil’s column, he writes:
I beg to differ! If Uranium-235 didn’t exist, they could still synthesize Plutonium-239 and Plutonium-240 in breeder reactors using good old-fashioned Uranium-238.
The basic equation for shielding is X=Y/10^n where
X=exposure after shielding
Y=exposure before shielding
n=the number of tenth thicknesses of the shield
The tenth thickness of lead is 2 inches.
Now for some examples…
Assume the initial exposure level is 10 mrem/hour
2inches lead
10mrem/hr divided by 10^1=1mrem/hr
1inch lead=3.16 mrem/hr
1/2 inch lead=5.62 mrem/hr
1/4 inch lead=7.5 mrem/hr
Lead is so passe when it comes to modern radiological shielding. The current standard is Demron.