WTC Dust Study Feb 29, 2012 by Dr. James Millette

ergo said:
This is incorrect. You don't appear to have even a wikipedia level of knowledge on the subject.
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[snort][snicker][lol]

This from someone who doesn't understand center of mass, a debris field the size of the moon, the difference between in and on, what simple a simple term like exponential means, and the list goes on and on...

really? You are calling someone else ignorant? W/out a'wikipedia level of knowledge on the subject.' Really?

I go away for about 8 months and I come back and we still have the typical ole ergo....
 
Redwood said:
Which part is incorrect? That it reacts faster? Or that it initiates at a lower temperature? Or that its actual energy yield is lower than standard thermite?
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I think you know which one. But the discussion is probably getting off topic here. You can defend your claim in this thread.
 
ergo said:
I think you know which one. But the discussion is probably getting off topic here. You can defend your claim in {"Thermite Goes Bang and Fizzle"}.
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That thread has gone dormant, and I see no need to revive it. I assume you dispute my claim that nano-thermite produces less energy than mundane thermite. Wrong! It's easily understood why: Aluminum is very reactive in air; a fresh scrubbed surface of aluminum will produce a surface layer of aluminum oxide, about 4 nm (nano!) thick, in less than 1 ns (nano!) of time.

In common use, this loss is inconsequential; a 40 micron (40,000nm) particle of aluminum, as might be found in common thermite, or aluminum paint, has lost only about 0.01% of its mass to oxidation. But for a 40 nm sphere (the most volumetrically efficient shape), the loss is about 50%! For microwelding (the most promising application for nanothermite, IMHO), this is immaterial; you're interested in producing iron, not a maximum energy yield. For producing energy, the loss is serious.

You can compensate by using a non-stoichiometric formula, doubling the amount of nano-aluminum. This would mean even more aluminum oxide product than is produced by common thermite, but Harrit, Jones, & Co. have failed to show any aluminum oxide product. A nanothermite formula optimized for energy production would produce about six times as much alumina, by volume, as iron! See the current discussion in the "Chris Mohr Rebuttal" thread.
 
In reviewing my "back of the envelope" calculations, there may be a subtle error, and the loss of aluminum to oxidation of a 40nm sphere may be "only" 25% or so. I'll leave it as a classroom exercise to detect the possible error. But the fact remains, that as one goes more and more "nano", oxidation losses become more severe.
 
Redwood said:
In reviewing my "back of the envelope" calculations, there may be a subtle error, and the loss of aluminum to oxidation of a 40nm sphere may be "only" 25% or so. I'll leave it as a classroom exercise to detect the possible error. But the fact remains, that as one goes more and more "nano", oxidation losses become more severe.
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Let's see.

The volume of a sphere is 4/3 Pi r3 = 4/3 Pi d3/8 = 1/6 Pi d3
With r, the radius, being half the diameter, d.

A sphere of diameter 40 nm thus has a volume of 33,493.33 nm3
If the top 5nm are oxidized, the radius of the inner, elemental sphere is less by 5 nm, and its diameter by 10 nm, so the diameter is 30 nm, and the volume 14,130 nm3, that's 42.2% of the 40 nm sphere. The remainder, 57.8% (by volume*), is then inert oxide.

For a 40,000 nm sphere with a 5 nm oxide layer, the elemental proportion is 39,9903/40,0003 = 99,925 %; conversely, oxide is only 0,075 of such a micro-sphere.



*) ETA: aluminium has a density of 2.70 g/cm3, alumina approx. 4.0 g/cm3. Let 1 cm3 worth of 40-nm-sized Al-spheres consist of Al and Al-oxide in the volume proportions given above, then this sphere has a mass of 2.7g*42.2% + 4.0g*57.8% = 3,4514 g: 1,1394 g Al (33%), 2,312 g Al2O3 (67%). Such Al nanospheres of 40 nm diameter have only one-third actual aluminum b weight!
Al2O3 is 53% by weight Al and 47% by weight O, so elemental composition of the spheres by weight is 31.5% O, 68.5% Al, and slightly less than half (48%) of the Al is elemental.
 
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Oystein said:
Let's see.

The volume of a sphere is 4/3 Pi r3 = 4/3 Pi d3/8 = 1/6 Pi d3
With r, the radius, being half the diameter, d.

A sphere of diameter 40 nm thus has a volume of 33,493.33 nm3
If the top 5nm are oxidized, the radius of the inner, elemental sphere is less by 5 nm, and its diameter by 10 nm, so the diameter is 30 nm, and the volume 14,130 nm3, that's 42.2% of the 40 nm sphere. The remainder, 57.8% (by volume*), is then inert oxide.

For a 40,000 nm sphere with a 5 nm oxide layer, the elemental proportion is 39,9903/40,0003 = 99,925 %; conversely, oxide is only 0,075 of such a micro-sphere.



*) ETA: aluminium has a density of 2.70 g/cm3, alumina approx. 4.0 g/cm3. Let 1 cm3 worth of 40-nm-sized Al-spheres consist of Al and Al-oxide in the volume proportions given above, then this sphere has a mass of 2.7g*42.2% + 4.0g*57.8% = 3,4514 g: 1,1394 g Al (33%), 2,312 g Al2O3 (67%). Such Al nanospheres of 40 nm diameter have only one-third actual aluminum b weight!
Al2O3 is 53% by weight Al and 47% by weight O, so elemental composition of the spheres by weight is 31.5% O, 68.5% Al, and slightly less than half (48%) of the Al is elemental.
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At nanometer scale, small differences in the size of aluminum spheres makes a big difference in the percentage lost to oxidation, and hence the actual energy produced in a reaction. Process control at the NWO's secret nanothermite production facility must be really tight! Imagine having to nanosieve tons and tons of the stuff.
 
Sunstealer said:
You may want to have a look at this paper. http://www.wydawnictwa.ipo.waw.pl/cejem/2-2010/full/klapotke.pdf

Page 6 (120) Table 4 and Page 10 (124) Table 6.
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Yup. You surely do lose a lot to surface oxidation as you nano-size aluminum. The 70 nm aluminum gel combustion velocity of 895 m/sec isn't bad, about the velocity of a .30-06 bullet, but what is the velocity of the emitted vaporized gel?

Nanoscale aluminum might be useful in a substitute formula for lead ammunition primers, but IMHO not in a thermite formula, rather mixed with a good oxidizing agent. Problem is in finding an oxidizer that doesn't have the same problem of producing corrosive products, as the old chlorate primers did. They worked just fine and they were lead free, but immediate cleaning of the firearm was mandatory after use.

There's also the problem of any new formula having the right sensitivity.
 
chrismohr said:
Tri,

Click the link, then hit the magnifying glass to see a pretty mind-boggling post from a 9/11 activist! I think it was a screen save thing.
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I did. Then I scrolled down and saw an article that caught my eye, and clicked on it, like it would do something.
 
If Tri still can't see this, can someone with better computer skills pitch in here? Again, I just clicked the link, hit the magnifying glass button, and read the article.
 
triforcharity said:
I did. Then I scrolled down and saw an article that caught my eye, and clicked on it, like it would do something.
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chrismohr said:
If Tri still can't see this, can someone with better computer skills pitch in here? Again, I just clicked the link, hit the magnifying glass button, and read the article.
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There seems to be some miscommunication here. If tri is going to the .png URL, then it is no wonder that clicking on a link in the .png wouldn't do anything, since it's just a graphic; we're trying to get him to read the text in the .png. If he isn't going to the .png URL, then it is no wonder that he can't see the .png. I think that one of those conditions must apply, but I really can't tell which one.
 
triforcharity said:
I did. Then I scrolled down and saw an article that caught my eye, and clicked on it, like it would do something.
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No, the link is only to an image.

As I think you saw, because it's a very lengthy image (the entire article) it may end up scrunched over on the left hand side (depending on the browser you're using), but if you hover your mouse cursor over it you'll see it turn into a magnifying glass icon. Left-click, the image should expand and you can read the article.

What you can't do, though, is click any of the links, because what you're looking at is still just an image, not the original web page. So if you click on something it'll revert to its original scrunched-down size over on the left (just click it again to bring it back, though).
 
I think tri was just narrating how he got fooled by a link in the image, trying to click on it to follow the link, which wasn't a link because it was just the picture of a link. I giggled :D
 
Oystein said:
Let's see.

The volume of a sphere is 4/3 Pi r3 = 4/3 Pi d3/8 = 1/6 Pi d3
With r, the radius, being half the diameter, d.

A sphere of diameter 40 nm thus has a volume of 33,493.33 nm3
If the top 5nm are oxidized, the radius of the inner, elemental sphere is less by 5 nm, and its diameter by 10 nm, so the diameter is 30 nm, and the volume 14,130 nm3, that's 42.2% of the 40 nm sphere. The remainder, 57.8% (by volume*), is then inert oxide.

For a 40,000 nm sphere with a 5 nm oxide layer, the elemental proportion is 39,9903/40,0003 = 99,925 %; conversely, oxide is only 0,075 of such a micro-sphere.



*) ETA: aluminium has a density of 2.70 g/cm3, alumina approx. 4.0 g/cm3. Let 1 cm3 worth of 40-nm-sized Al-spheres consist of Al and Al-oxide in the volume proportions given above, then this sphere has a mass of 2.7g*42.2% + 4.0g*57.8% = 3,4514 g: 1,1394 g Al (33%), 2,312 g Al2O3 (67%). Such Al nanospheres of 40 nm diameter have only one-third actual aluminum b weight!
Al2O3 is 53% by weight Al and 47% by weight O, so elemental composition of the spheres by weight is 31.5% O, 68.5% Al, and slightly less than half (48%) of the Al is elemental.
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Show off! :D
 
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