Bob B.
Bob the Excel Guru?
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Post by Bob B. on Apr 6, 2011 8:47:29 GMT -4
What if we just detonated 19 kg of TNT on the lunar surface? How big a crater would that make? I am sure there must be detailed tables relating crater size to impactor size and velocity and hence impact energy, but I don't know where they are offhand. I played around with this several years ago and what I was able to find out is that the crater diameter is proportional to the cube of the kinetic energy. According to my old notes, I had the empirical formula Diameter = 0.0062E 1/3but I don't know where I got the 0.0062 factor. I may have found it somewhere on the web, or I may have calculated it from a table of data. And, of course, this only gives diameter and not the crater depth. If we apply this formula my example, we get Diameter = 0.0062 x 89676000 1/3 = 2.775 m Interestingly, this is almost exactly the diameter that I used in my calculation for the area over which I distributed the energy.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Apr 6, 2011 9:06:24 GMT -4
I’ve thought of something else that should be taken into consideration. After the exhaust gas leaves the nozzle, it expands and cools. This means that the internal energy (enthalpy) of the gas decreases. I presume this energy is converted to kinetic energy (as happens in the divergent section of the engine nozzle). Therefore, the kinetic energy of the gas impacting the lunar surface is probably higher than I used in my calculation.
I think my analysis sufficiently debunks the notion that there should be a large blast crater, nonetheless, I think further study is necessary to produce something more definitive. I’ll probably develop this into a more in-depth analysis that I’ll eventually add to my web page.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Apr 6, 2011 10:09:45 GMT -4
According to my old notes, I had the empirical formula Diameter = 0.0062E 1/3but I don't know where I got the 0.0062 factor. I may have found it somewhere on the web, or I may have calculated it from a table of data. And, of course, this only gives diameter and not the crater depth. OK, I think I see where the 0.0062 number comes from. If the energy of the blast is 1 3 megatons, the diameter of the crater works out to be 1000 m. A 2 3 MT blast yields a 2000 m crater, a 3 3 MT blast a 3000 m crater, a 4 3 MT blast a 4000 m crater, and so on. I believe this relationship was worked out empirically from experience with nuclear bomb detonations. (FYI, a megaton is 4.2x10 15 joules.)
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Post by chew on Apr 6, 2011 11:18:13 GMT -4
Does everybody have me on ignore? I didn't know this forum had an ignore feature. Chapter six, section 6.09 of the epic "The Effects of Nuclear Weapons" discusses nuclear detonation crater formation. www.fourmilab.ch/etexts/www/effects/Crater dimensions scale to the 0.3 of the energy of the explosion. A 1 kiloton surface explosion will form a crater 60 feet in radius and 30 feet deep. So 19 kg of TNT will form a crater .72 m in radius and .36 m deep. But this is a concentrated source of the energy so it won't translate well to a dispersed released.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Apr 6, 2011 11:53:26 GMT -4
Does everybody have me on ignore? It wasn't a matter of ignoring, it was a matter of not wanting to study a 644 page document that, as you indicate, is not really applicable to the particular problem. In the Apollo case the surface is being eroded away by a fluid stream, it is not being blasted out as if in an explosion.
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Post by chew on Apr 6, 2011 11:58:44 GMT -4
Does everybody have me on ignore? It wasn't a matter of ignoring, it was a matter of not wanting to study a 644 page document Fair enough. Although I did mention which chapter cratering was in.
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Post by slang on Apr 6, 2011 18:50:12 GMT -4
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Post by chew on Apr 6, 2011 20:59:07 GMT -4
Heh. Second link says a crater 2.5 feet deep should have been made. Good thing the Moon isn't a giant dust ball.
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Post by ka9q on Apr 7, 2011 8:31:39 GMT -4
(FYI, a megaton is 4.2x10 15 joules I think the formal definition for nuclear arms control treaties is 1 "ton TNT" = 1 million kilocalories (The kilocalorie is the "food" calorie). 1 kcal is 4,186.8 joules. That's where your 4.2 comes from. The actual energy content of a ton of real TNT is somewhat more than "1 ton TNT". Again, it's just a convention.
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Post by ka9q on Apr 7, 2011 8:36:15 GMT -4
In the Apollo case the surface is being eroded away by a fluid stream, it is not being blasted out as if in an explosion. This is entirely true. I had been hoping that the crater from the 19 kg TNT explosion would be fairly small. Then one could make a somewhat intuitive argument that it's unreasonable to expect the LM to make a considerably larger crater with its DPS engine by releasing an equivalent amount of energy much more slowly. But it looks as though you can't really avoid the more detailed analysis.
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Post by Apollo Gnomon on Apr 7, 2011 11:00:04 GMT -4
I don't think you can directly compare an explosion with rocket thrust. The kinetic energy equation is m*v 2/2 en.wikipedia.org/wiki/Kinetic_energy-- velocity is much greater in an explosion than in slow burning, and the expanding gasses have more kinetic energy to transfer to the surroundings. Thought experiment: Imagine 1 kg of natural gas simmering a pot of soup, then imagine that 1 kg of natural gas burning all at once due to gas leak. To do a fair comparison of soil displacement capacity you'd have to calculate the gross kinetic energy of the exhaust gasses and back-calculate an appropriate amount of explosive giving the same total kinetic energy. The figures bandied about here have suggested a crater depth ranging from about a quarter inch to just under one inch. That's consistent with the pictures showing the area under the various landers. In one case one can even see a bit of "donut" shape to the crater, consistent with the centerpoint of the exhaust thrust pressing the dust down rather than accelerating it away. Sorry I don't remember which pictures show this. I've sorta had any interest in this subject ground out of me by the Delusional Idiots.
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Post by ka9q on Apr 7, 2011 14:04:38 GMT -4
Yeah, you're right. I know there's a big difference between a bomb and what's basically a hot air generator, but I was just hoping that we might be able to make some sort of simple, intuitive argument just from the known and limited rocket energy to do cratering. But there's too much energy for that simplistic analysis.
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Post by Glom on Apr 8, 2011 18:51:40 GMT -4
Also, there was some discolouration of the soil directly under the engine bell so some of the energy went was absorbed chemically rather than kinetically.
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Post by jrkeller on May 1, 2011 18:42:09 GMT -4
jrkeller is still active on BAUT. Fred Barely. I don't see him around that often. I'm still around. By the timeI notice an HB, they usually get themselves banned and their threads are locked.
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Post by jrkeller on May 1, 2011 18:48:38 GMT -4
I did some calculations years ago, but approached it from a heat transfer method and I used a lot of conservative assumptions* Basically, I used a method that is called the semi-infinite method and assumed that once the temperature of the rock reached it's melting point, the exhaust gasses would carry it away. I remember that it was less than an inch.
* For the HB crowd. Conservative assumptions are worst than reality. For example, Bob B. assumes that all the kinetic energy goes into the soil, which we know is not true, but that what one knows what the worst cold ever be.
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