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PE determination of M(16) from Gravity Free Fall

The only Force allowed to act on M(16) is Gravity…. Use basic Free-Fall High School Newtonian Physics to calculate your answer in Energy units of Joules (J).
You might chose to use this Equation: Potential Energy(16) = M(16)∙g∙h(94)
We know the Gravitational Free Fall acceleration constant for earth is (g) = (9.8 m/s^2)

We also can know that h(94) = 355 meters, since h(94) = (94/110)(h(110)) = (0.855)(415 m)

The mass of the alleged free-falling 16 floor block, M(16), needs now to be accurately calculated based on the correct relative steel thicknesses of the steel walls contained in M(16) as compared to M(94), on a % Mass or % Weight ratio basis.
It turns out, after viewing my Steel Thicknesses Power Point Graphics at: http://911scholars.ning.com/photo/zaks-wtc-tt-crosssection-of?context=user, and http://911scholars.ning.com/photo/graphical-mathematical?context=user, and http://911scholars.ning.com/photo/tall-crosscut-wall-of-wtc-tt-1?context=user,
that M(94) is essentially 55 times more massive as M(16) due to the greater thicknesses of the steel in M(94), at 4”, 5”, and 6” reported in the lower floors and the 6 additional basement floors.
M(94) = 98.2% of the total Tower’s Steel.
M(110), the total Tower’s weight is 5 x 10^5 Tons = 500,000 Tons = 1 x 10^9 Lb = 4.6 x 10^8 Kg
M(94) = (98.2%)(4.6 x 10^8 Kg) = 452 x 10^6 Kg = M(94)
M(16) = (1.8%)(4.6 x 10^8 Kg) = 8.3 x 10^6 Kg = M(16)

We now have all 3 of the quantities, in red, required to calculate the Energy of a 355 meter M(16) Free Fall and the necessary Physics Equation, in red, to calculate the final result.
PE(16)FF = (M(16))∙g∙(h(94))
PE(16)FF = (8.3 x 10^6 Kg)(9.8 m/s^2)(355 m) = 2.9 x 10^10 Joules

The Energy released by one Little Boy Atomic Bomb (LBAB) = 6.3 x 10^13 Joules = 15 KiloTons TNT.
Therefore the PE(16)FF divided into the PE(LBAB) gives the following:
[PE(LBAB)/PE(16)FF] = [(6.3 x 10^13 Joules)/(2.9 x 10^10 Joules)] = 2,200:1
Meaning PE(16)FF is (1/2,200) the energy of 1 LBAB = 6.8 Tons of TNT explosive (v. small amt)

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Comment by Mehmet Inan on November 17, 2010 at 6:00pm: Chuck Boldwyn said: "All of those energies will be compared to the energy of 1 LBAB (Little Boy Atomic Bomb) = 6.3 x 10^13 Joules = 15 KiloTons of TNT."

What's the interest to compare these energies to LBAB? There is no reason to make such compare. Any LBAB will have one huge explosion. But the towers collapsed by several successive distributed explosions. That means the LBAB theory is false, definitely false. You can make thousands of calculations during hundreds of years, the initial base is wrong. So all these calculations are meaningless.

The collapse method consistent with all available evidence is explained in my power point on slides 49 to 52. Before you make those calculation, it you should try to debunk my explanation; if you manage to debunk it then I will follow your LBAB theory.

Comment by Chuck Boldwyn on November 16, 2010 at 5:03pm: Hello Mehmet,

This is just an intermediate calculation that will be used for comparison's sake with the energy of M(16) to fall 1 floor, 2 floors, 3 floors, the standing Potential Energy of M(94) = (1/2)M(94)(355 m), the support energy of M(94) using a Safety Factor of "5" and of "20". All of those energies will be compared to the energy of 1 LBAB (Little Boy Atomic Bomb) = 6.3 x 10^13 Joules = 15 KiloTons of TNT.

The graphic is just a tutorial calculation of one of the comparitive energies.

KE = (1/2)M(16)(velocity of M(16) at impact after 355m Free Fall)^2

There are 6 energies needed to be calculated and compared by their energy ratios, once you can accurately calculate the mass of M(16) and the mass of M(94)using their unequally distributed masses associated with their heights, 5 inches at lower levels to (1/4) inch at topmost levels, progressively less massive as one goes from the basement floors to the topmost levels.

1. Energy of M(16) falling one floor = 3.8 m, at collision with top of M(94).

2. Energy of M(16) falling 94 floors = 355 m

3. Potential Energy of a free standing M(94) = (1/2)M(94)(355 m)

4. Support energy (electromagnet/Atomic bonding) of M(94) using Safety Factor of "5", which is not gravity based, but is molecularly bonding of steel atoms based.

5 Support energy (electromagnet/Atomic bonding) of M(94) using Safety Factor of "20". which is not gravity based, but molecularly and atomic forces of attraction based.

6. Energy of 1 LBAB = 6.3 x 10^13 Joules = 15 KiloTons of TNT = 15,000 Tons of TNT

Now divide each of the first 5 energies into the LBAB energy for comparison purposes and calculate the amount of TNT required for each energy.

That is what I will be doing in the very near future.

I have already done this for "equally distributed weight" for the Twin Towers, which is far less that when the weight in unequally distributed.

This work is still in progress, so I can understand why you would ask your question.

I appears that I am the only researcher looking at the required energies from this very basic and most simplistic approach, which all of the physicista and engineers should be doing. Why they have not is a total mystery to me and to Jim Fetzer as voiced often in our internet "Real Deal" online interviews.

Check out some of my older power point graphics on this site.

Chuck Boldwyn

Comment by Mehmet Inan on November 16, 2010 at 2:54pm: What do you want to prove with that?

PE determination of M(16) from Gravity Free Fall

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