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Review of Gravy's film at ABOVE TOP SECRET

realcddeal:

Aside from the fact that you are trying to use the units of pounds per square inch and pounds per inch as if they were interchangeable, a linear shaped charge expends its energy in all directions when fired, it is also not efficient. Unless you are accounting for this, your numbers aren't worth much more than a guess.

You can't be serious.

But just in case you are: what I am saying is there is 309,000 plus pounds of force for every square inch to be sheared in the situation of a one pound charge of C-4 and the 56 sq. inch cross section.

Shaped charges don't expend their energy in all directions.

Why don't you tell us what the efficiency is then since you are trying to pour water on the argument? Do you know? I think you are the one guessing.
 
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Shaped charges don't expend their energy in all directions.

Yes they do, it's just that the cavity focuses more of the energy on it's side... If shaped charges didn't expend all of it's energy in all directions, then you would be able to stand right next to a shaped-charge... If you want to try that, by all means, because I wouldn't...
 
Yes they do, it's just that the cavity focuses more of the energy on it's side... If shaped charges didn't expend all of it's energy in all directions, then you would be able to stand right next to a shaped-charge... If you want to try that, by all means, because I wouldn't...

Technically you are right, but most of it goes in the direction intended. Even if it were only 50% efficient, and this is very unlikely, a one pound charge still has over seven times the amount of energy needed for the cross section you discussed, a two pound charge would have over fourteen times the energy necessary, even at 50% efficiency, for that cross section.

The general point here is that one to two pound charges were more than sufficient to take out the core columns of the Twin Towers.
 
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I did not run away from a televised debate with you. I never accepted it. I simply don't like the way Ron Wieck runs his show and do think you two would come up with some way of having an advantage. That isn't running away. Most people consider it smart not to go into a situation where their adversary has an upper hand. I debate with you here all the time.

I'll have to get back to you tomorrow on the other things as I actually have to go to work tomorrow. Do you? I really am wondering when you sleep. Are you on the night shift for tour guides this weekend?


Someone on the LC forum brought you to my attention by claiming that there was a fantasist Mark was afraid to debate. I replied that there were no doubt fantasists who Mark regarded as too ridiculous or loathsome to debate, but as the entire evil movement is built on bogus science, distorted quotes, and outright falsehoods, there couldn't possibly be anyone whose forensic prowess he had to fear. I guess I was right.

What don't you like about the way I host 'Hardfire'? Jim Fetzer and the Boys could not have been too pleased with the outcomes of their debates, but they didn't complain about their treatment. You are too accustomed to preaching to an extremely dull-witted and uncritical choir.
 
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You can't be serious.

But just in case you are: what I am saying is there is 309,000 plus pounds of force for every square inch to be sheared in the situation of a one pound charge of C-4 and the 56 sq. inch cross section.

no you are not:

realcddeal said:
Dividing 17.3 x 10e6 inch pounds by 56 sq. inches = 309,540 pounds per inch.

A pound per inch and a pound per square inch are not the same, one could represent a distributed load while the other is in the dimensions of stress. Checking units is just a quick and simple way to check your calculations, yours don't jive.

it seems you are either trying to say that and pound inch is a force, or an inch is an area. Does any of that make sense?

Shaped charges don't fire in all directions.

erm, yes they do. They are designed to concentrate the explosion on the desired point, but any explosive, will expand in all directions, and energy losses due to this is unavoidable.

Why don't you tell us what the efficiency is then since you are trying to pour water on the argument? Do you know? I think you are the one guessing.

last time I checked it was you who was proposing the CD scenario, I apologize for pointing out that you should take realistic factors into consideration in your calculation. The heat of the explosion for TNT is 4000 kJ/kg

If I wanted to pour water on the arguement, I would just cite the papers below, then be done with it.






Autocatalytic thermal decomposition kinetics of TNT
Thermochimica Acta, Volume 388, Issues 1-2, 18 June 2002, Pages 175-181
Gregory T. Long, Brittany A. Brems and Charles A. Wight

Abstract

previous termThermalnext term analysis has been employed to determine the kinetics and the energetics of the slow cook-off chemistry of 2,4,6-trinitrotoluene (previous termTNT)next term by isothermal differential scanning calorimetry (DSC) in high-pressure crucibles sealed under air. Model-free isoconversional analysis of the DSC kinetic traces has been used to determine activation energies (Eα) and the functional form of the reaction model (dependence of reaction rate on the extent of conversion, α). While the variation in Eα with α is in qualitative agreement with the literature it is nevertheless constant within the 95% confidence limits at 140±10 kJ mol−1. Hence, no systematic variation in Eα occurs over the course of the reaction. Rather, the reaction model exhibits a large increase in the range 0.1<α<0.25 and a decrease for 0.25<α<0.43. Thus, the observed acceleratory period is caused by an increase in the reaction model, not by a decrease in activation energy, as might be expected for autocatalysis. This kinetic behavior is ascribed to nucleation and growth of reaction centers in liquid state previous termTNT.next term In addition, a heat of reaction, Q=(4.9±1.5)×102 kJ mol−1 during the previous termthermal decomposition of TNTnext term has been shown to be independent of the heating rate and sample size.

Thermal Decomposition of a Melt-Castable High Explosive: Isoconversional Analysis of TNAZ
Long, G. T.; Wight, C. A.
J. Phys. Chem. B.; (Article); 2002; 106(10); 2791-2795.

Abstract:

The thermal decomposition kinetics of the high explosive 1,3,3-trinitroazetidine (TNAZ) have been measured by nonisothermal differential scanning calorimetry (DSC). Samples of TNAZ in open pans and pierced pans undergo mainly melting (Hfus = 27 ± 3 kJ mol-1) and vaporization (Hvap = 74 ± 10 kJ mol-1) during heating. However, when confined in sealed high-pressure crucibles, exothermic thermal decomposition is observed. The activation energy for thermal decomposition has been determined as a function of the extent of reaction by isoconversional analysis. The initial value of 184 kJ mol-1 at the start of the reaction decreases to 38 kJ mol-1 near the end of the reaction. The rates clearly exhibit acceleratory behavior that is ascribed to autocatalysis. The measured heat release of thermal decomposition (Q = 640 ± 150 kJ mol-1) is independent of the heating rate and the sample mass. These results are consistent with proposed mechanisms of TNAZ decomposition in which the initial step is preferential loss of the nitramine NO2 group over loss of a gem-dinitroalkyl NO2 group.

Thermal Activation of the High Explosive NTO: Sublimation, Decomposition, and Autocatalysis
Long, G. T.; Brems, B. A.; Wight, C. A.
J. Phys. Chem. B.; (Article); 2002; 106(15); 4022-4026.

Abstract:

Thermogravimetric analysis (TGA) and differential scanning calorimetry (DSC) show that the heating of 5-nitro-2,4-dihydro-3H-1,2,4-triazol-3-one (NTO) leads to competitive sublimation and condensed-phase exothermic decomposition. Model-free isoconversional analysis has determined activation energies (E) for these processes as a function of the extent of conversion, . Sublimation occurs most readily in an open pan; although more than simple sublimation was observed, a global activation energy of E = 130-140 kJ mol-1 for sublimation was determined. Nonisothermal TGA and DSC traces run on pierced pan samples provide convincing evidence for competitive sublimation and condensed-phase decomposition of NTO. Confining NTO samples in a closed pan results in condensed-phase decomposition that leads to the formation of gaseous reaction products and shows autocatalytic behavior during the latter stages. Isoconversional analysis of DSC traces of closed pan samples yield activation energies for exothermic decomposition that increase from E = 273 kJ mol-1 for = 0.01 to a plateau of 333 kJ mol-1 for 0.17 0.35 prior to decreasing to 184 kJ mol-1 for = 0.99. The decrease in E with during the latter stages of decomposition agrees with previous reports of autocatalytic behavior.

Competitive Vaporization and Decomposition of Liquid RDX
Long, G. T.; Vyazovkin, S.; Brems, B. A.; Wight, C. A.
J. Phys. Chem. B.; (Article); 2000; 104(11); 2570-2574.

Abstract:

The thermal decomposition of hexahydro-1,3,5-trinitro-1,3,5-triazine (RDX) has been studied by thermogravimetric analysis (TGA) and differential scanning calorimetry (DSC). Activation energies as a function of the extent of conversion, , have been determined by model-free isoconversional analysis of these data. In open pans, evaporation is a prevalent process with an activation energy of ~100 kJ mol-1. Confining the system in either a pierced pan or a closed pan promotes liquid state decomposition of RDX that occurs with an activation energy of ~200 kJ mol-1, which suggests scission of an N-N bond as the primary decomposition step. In such a confined environment, gas phase decomposition is a competing channel with an activation energy estimated to be ~140 kJ mol-1. In a closed pan, RDX generates a heat release of ~500 kJ mol-1 that is independent of both the heating rate, , and the mass.
 
Someone on the LC forum brought you to my attention by claiming that there was a fantasist Mark was afraid to debate. I replied that there were no doubt fantasists who Mark regarded as too ridiculous or loathsome to debate, but as the entire evil movement is built on bogus science, distorted quotes, and outright falsehoods, there couldn't possibly be anyone whose forensic prowess he had to fear. I guess I was right.

What don't you like about the way I host 'Hardfire'? Jim Fetzer and the Boys could not have been too pleased with the outcomes of their debates, but they didn't complain about their treatment. You are too accustomed to preaching to an extremely dull-witted and uncritical choir.

I would definitely not say I am preaching to an extremely dull-witted and uncritical choir here and I have been spending a good bit of my time here lately.

As for Mark I have shown that the premise of his video, claiming that those who believe the buildings were taken down via controlled demolition think there needed to be a huge amount of explosives, is simply wrong.

A controlled demolition of the twin towers would have used a minimum amount of explosives. As I have shown here, one to two pound charges were plenty capable of taking out the core columns. All that was necessary to bring those buildings down was to take out a majority of the core columns.

It wouldn't have been anything like the huge charges that Mark shows on his video, and since the charges would have been on the core columns they would have been well inside the perimeter, keeping the demolition charges from being visible, although a few did escape the perimeter. The collapse of the buildings would have masked the audible sounds of these relatively small charges.

For now I think the written word is the appropriate way to debate this topic. Maybe when it is all over, if you want, I will stop by your show.
 
Sounds like you are a bad sport who isn't tired of sour grapes. Get used to it.
Have you fixed the errors in you paper yet?
You have zero clue on any thing CD realcddeal is a real neat name for someone with a paper which proves you have no clue about CD or 9/11.

But Gravy did an outstanding job on his video; makes your attempt at a paper look like junk. Good job Gravy, outstanding job Mark. Free beer if you ever find me. realcddeal needs a lot of free beer too, for being so bad at engineering (as seen in his paper).
 
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Well you seemed to have joined the fray and appear to have been wrong about whether the core columns could have been taken out by 1 to 2 pound C-4 charges.

You're funny, but I'm not laughing. Here's the start genius, go back to first principles. Shear energy is derived from the triple integral of (Txy)*y*dx*dy*dz.

Where Txy = T / dA

Where

T = Shear Force
A = Cross-sectional Area


This ultimately translates to yet another integral of (1/2)*(T^2)/(A^2*G)*dx*DA

Where G is the shear modulus of steel (roughly 0.4*E)


Now, that ******** you did, is exactly that: ********. And I regret that this forum censors that out.
 
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A pound per inch and a pound per square inch are not the same, one could represent a distributed load while the other is in the dimensions of stress. Checking units is just a quick and simple way to check your calculations, yours don't jive.

You'd think he'd at least make sure that his units matched when making up calculations.

Truther-logic at it's best, where in*lb = lb/in!
 
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here is a video of a linear shaped charge cutting a plate, you can decide if 100 percent of the energy goes into the plate:

http://www.youtube.com/watch?v=tZRAbUcUkIc

Where did I say it was 100% efficient?

I did the calculations to show how much energy was available and it is many times what is necessary.

Your problem is that you do not want to believe that one to two pound charges could take out the core columns.

The reality is that they could and the energy calculations show that even a relatively low efficiency still allows far more than enough to do the job.
 
Where did I say it was 100% efficient?

I did the calculations to show how much energy was available and it is many times what is necessary.

Your problem is that you do not want to believe that one to two pound charges could take out the core columns.

The reality is that they could and the energy calculations show that even a relatively low efficiency still allows far more than enough to do the job.

Please elaborate on how in*lb = lb/in. You forgot the first part of his post where he showed you were MAKING **** UP.
 
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Where did I say it was 100% efficient?

I did the calculations to show how much energy was available and it is many times what is necessary.

Your problem is that you do not want to believe that one to two pound charges could take out the core columns.

The reality is that they could and the energy calculations show that even a relatively low efficiency still allows far more than enough to do the job.

Sorry for putting words in your mouth. I am only trying to point out the fact that there are energy losses to this and to heat when you re do your energy calculations, since your current ones do not mean anything. You do realize by now the importance of correct units right?
 
You're funny, but I'm not laughing. Here's the start genius, go back to first principles. Shear energy is derived from the triple integral of (Txy)*y*dx*dy*dz.

Where Txy = T / dA

Where

T = Shear Force
A = Cross-sectional Area


This ultimately translates to yet another integral of (1/2)*(T^2)/(A^2*G)*dx*DA

Where G is the shear modulus of steel (roughly 0.4*E)


Now, that ******** you did, is exactly that: ********. And I regret that this forum censors that out.

deleted due to redundant post
 
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You're funny, but I'm not laughing. Here's the start genius, go back to first principles. Shear energy is derived from the triple integral of (Txy)*y*dx*dy*dz.

Where Txy = T / dA

Where

T = Shear Force
A = Cross-sectional Area


This ultimately translates to yet another integral of (1/2)*(T^2)/(A^2*G)*dx*DA

Where G is the shear modulus of steel (roughly 0.4*E)


Now, that ******** you did, is exactly that: ********. And I regret that this forum censors that out.

Stop trying to impress everyone with the integral BS. You know darn right well that the rule of thumb for shear stress for steel is .5 of tensile stress and the less conservative Von Mises stress for steel is .577 of tensile stress.

Are you trying to say that the shear yield stress for A36 steel is not in the vicinity of 21,000 psi?
 
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You'd think he'd at least make sure that his units matched when making up calculations.

Truther-logic at it's best, where in*lb = lb/in!

I wish I could change units like that when doing an assignment, save time, better grades, it sounds like a much easier system for everyone involved.
 
Have you fixed the errors in you paper yet?
You have zero clue on any thing CD realcddeal is a real neat name for someone with a paper which proves you have no clue about CD or 9/11.

But Gravy did an outstanding job on his video; makes your attempt at a paper look like junk. Good job Gravy, outstanding job Mark. Free beer if you ever find me. realcddeal needs a lot of free beer too, for being so bad at engineering.

Not you again!

Mark's video is very nice and I enjoyed the music but its premise is wrong.

The size of the charges on the core columns in the towers would not need to be large.
 

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