Rumor Mills Reading Room

Posted By: ipia <Send E-Mail>
Date: Thursday, 15 July 2004, 4:07 p.m.

TruthPassion has requested a debate on depleted uranium munitions. Lets have one! I love a debate.

If you are a soldier in Iraq, soon to be one (that would be everyone under the age of 20 when the draft reappears), or know someone who is a soldier in Iraq I suggest you may want to follow the debate. As the headline suggests, many of you who have been exposed to the uranium dust in Iraq will not be among the living in ten years. My generation was exposed to many agents in Viet Nam, including agent orange. A dear friend (Marine) of mine has just come down with a rare form of cancer. It is 28 years since his exposure. The doctors describe it as Agent Orange cancer, caused by his exposure to agent orange in Nam. It has an induction time of up to 28 years. Agent Orange was known to be carcinogenic during the war yet soldiers were routinely drenched in the stuff. During the time of the Viet Nam war many disinformation scientists were employed to claim this stuff is safe. To allow otherwise would have cut into the profits of corporations and raise serious questions about chemical agriculture. Unfortunately the danger from Agent Orange is trivial compared to that of depleted uranium.

In 2004 we again have legions of disinformation scientists proclaiming that depleted uranium is safe to eat, drink, and inhale. Besides all of the usual reasons of corporate greed there are very deep, even hidden (occult) reasons that American soldiers are being dusted with uranium dust. For instance: the leaders of the NWO want our soldiers to be disabled in the next few years. There are statistics that 10 years after the 1991 Gulf War well over one quarter of our soldiers are disabled and unable to defend the US. With the widespread use of depleted uranium it is likely that in ten years that less than half of our fighting men will actually be able to offer resistance to a "real" enemy.

Well, doesn't depleted uranium save lives because it is a superior penetration round?

Let me address that by talking about something that has been going on for a long time. Corporate profits are have always been promoted over the safety of the American soldier. During WWII the Germans called our best tanks "zippos". Why is that? It was a play on the advertising for the American cigarette lighter: they light up every time. Every time a German shell hit one of our tanks they exploded. The Germans couldn't understand why we would field such junk in a time of war. We won tank battles with air superiority and mass attacks (human wave) not with superior tanks. The American tanks had inferior armor, inferior (caliber) guns, and gasoline engines. The shells from American tanks bounced off the German armor while the American tanks blew up with the slighest stress. The German 88's sliced through our armor as if it wasn't there. We could have easily changed our designs to put diesel engines in our tanks and increase the armor and install large enough guns to challenge. However that would have hurt profits. Every time a tank blew up the corporations could make another junk tank just like it.

When Viet Nam began our soldiers went to war with bolt action rifles initially while the communist world had switched to AK47's. Eventually the M-16 was designed to provide automatic weapon fire to every infantryman. However it too was junk, the weapons manufactured in the first 5 or 6 years jammed everytime a drop of water hit them. Many soldiers threw them away and adopted shotguns or AK47's that they had captured. Many soldiers never came back from Nam because their M-16 had failed them. But the M-16 made a lot of profit for the corporations.

When the M-1 tank first rolled off the assembly line it averaged 25 miles before it broke down. Congress cheered when it managed 50 miles. However there was a problem, no American manufacurer could be found to make the first big guns. They had to buy cannons from Germany to mount on the first M-1 tanks.

So all of this talk about German tanks; I wonder what the big German tanks use for armor rounds these days, is it uranium? No it is tungsten. Again the German designers are concerned about their personnel and not about profits. For all I know tungsten is superior to uranium, or probably can be made superior. Molten tungsten is denser than uranium. So you see, there is no logical need for an uranium round unless you want to destroy the soldiers using the rounds. American depleted uranium shells do not save lives in combat situations. You have to find other reasons for its use.

Now consider some "information" from a previous post:

"The second reason most have a problem is because they do not understand the nature of radioactive isotopes. For instance, did you know that although uranium 235 has a half-life of 4.5 billion years, since most of it was created at the same time, when the earth was formed about 4 billion and some change years ago, "

Actually this statement is confused. U238 has a half-life of 4.5 billion years. U235 has a half-life of 7.13e8 years. Most theories of cosmology have the heavies forming in novas and supernovas which precede the formation of the earth by billions of years; uranium was created prior to the formation of the earth. For U238, the half-life of 4.5 billion years means that half of the amount present then is still with us. This meaning of half-life holds for all first order decay processes. Every scientist recognizes the U238 half-life of 4.5 billion years because that isotope is used in radioactive dating, not the U235 isotope. In other words, confusing these half-lives is not a small error, and it was made by a person without a technical background.

"then most uranium 235 is at the tail end of its 4.5 billion year radioactive half-life, which means that it will become less radioactive rather than more so..."

No. Half-live means that for either U235 or U238 originally present in the earth essentially half is still with us. The less radioactive statement is misleading because the daughter nuclei of U238 are pretty hot as a rule and radioactivity can actually increase as this stuff decays. However for chemically separated uranium, relatively pure uranium, depleted or not, the clock is set to zero because you now have 100 percent (radioactive) uranium. That is, it is irrelevant how much uranium has decayed in nature if you have depleted uranium. Its radioactivity is in fact determined by the radioactive properties of (pure)U238. More on that further on.

"It is possible that these aerosols may be toxic, but their toxicity is not related to their radioactivity, it is related to their nature as heavy metals.."

No. The problem is the alpha particle that U238 emits, which is very carcinogenic when inhaled.

"Now, understand this, because it is important. Three facts;

1) "naturally occuring uranium contains 235 U at 0.71%".

2)Naturally occuring uranium is more common than once thought as oxides and is more common than mercury, silver, molybdenum, arsenic and cadmium.

3) Depleted Uranium contains 235 U at the rate of 0.2%, or less than 1/3 as naturally occuring Uranium. So, at the very least, unless it is contained in..."

No. Hg, Ag, Mo, As, and Cd do not emit alpha particles. Also, the concentration of 235 is irrelevant to the radioactivity of either naturally occuring uranium or depleted uranium. Here's why:

Lets suppose we have one gram of naturally occuring uranium metal, with its 0.7% U235. Then essentially 99.35% is U238.

The half-life of 4.5 billion years means the rate constant is 0.693/4.5e9 = 1.54e-10 1/y
converting to seconds

Rate constant = 1.54e-10/3.16e7 = 4.88e-18 1/s


1 gram U x 6.02 atoms/238 gram U = 2.53e21 atoms U (per gram U)

2.53e21 atoms U x 4.88e-18 1/s = 12,347 atomic decays/s

So we have just calculated the radioactivity of the U238 in one gram of natural uranium.

For the U235 in this one gram sample we have

In one gram, .7% is the fraction 0.007 so there is 0.007 gram of U235 in one gram of natural uranium metal.

Then rate constant = 0.693/ (7.13e8 y x 3.16e7 s/y) =

3.07e-17 1/s

0.007 g U235 x 6.02e23 atoms U235/g U235 x 3.07e-17 1/s =

550 decays/s

This means that the U238 contributes 12,000 decays per second and the U235 contributes 550 decays per second. If all of the U235 is removed then you still have 11,500 decays per second. We can conclude that over 95% of the radioactivity comes from U238 and if you remove all of the U235 in uranium you haven't decreased the radioactivity significantly. Therefore, the quantitiy of U235 present in depleted uranium is irrelevant.

Well how much of this depleted uranium is dangerous to inhale? Consider inhaled plutonium, known to cause lung cancer due to alpha emission. In fact, one millionth of a gram is considered to be the amount that induces cancer. This is why plutonium is called the most poisonous substance known to man. Inhaling 10e-6 gram will kill you. How does this compare to U238?

Well the alpha emission is in inverse proportion to the half-life, as we have calculated previously.

Then 1/2 life U238 divided by 1/2 life Pu239 =

4.5e9 / 2.44e4 = 184,426 ; this means we need 184,426 as much U238 to cause the same alpha damage.


1e-6 g x 184,426 = .184 gram U238 inhaled is as destructive as one millionth gram of plutonium.

How much is that? With a density of 17 g/ml, one gram of U238 is about the size of a small drop of water (from an eyedropper). 0.18 gram is then less than one fifth as big as a drop of water. You inhale this much, you are gone. A typical tank shell weight 4,000 grams. This is a conservative estimate. You can get lung cancer with less.

So here is my opening round in the debate. Other significant points deserve separate posts. I would be happy to explain the calculations if they are unclear, what is needed is that everyone educate themselves to be able to do this on their own without having to depend on experts.

Please pardon the typos and the grammar, this is a hurried post.