Findor
Miembro habitual
Pero si todo el mundo sabe que han sido estos:
Sois unos caras. Cuando se demuestre que han sido unas explosiones termonucleares no vais a entrar aquí pidiendo perdón. Desde luego....
Después de páginas y páginas de himbestigaciones donde se demostraba el uso de thermite para cargarse la estructura de acero de las torres, ahora el amigo Human nos sale con una explosión termonuclear! No, qué digo una, ...tres!!!
Entonces, eso quiere decir que todas aquellas sesudas parrafadas, aquellos "hechos irrefutables" y aquellas "hevidencias científicas" del thermite, ...era una bola? o la bola es lo de las bombas atómicas?
No te das cuenta de que ambas teorías son absolutamente incompatibles, panellet? Porque al final, cuál es la teoría buena?
Vaya, yo diría que te pirras por cualquier cosa que suene a conspiración y las acabas comprando todas, aunque entre ellas se den de ostias.
Imagine que todo el combustible del avión se inyectara en sólo un piso del WTC, que el combustible se quemó con una eficiencia perfecta, que los gases caliente se mantuvieron en el piso y que nada del calor escapó de este piso por conducción. Con estas asunciones ideales nosotros calcularemos la temperatura máxima que este piso podría alcanzar.
"El Boeing 767 es capaz de transportar 23,980 galones de combustible y se estima que, en el momento del impacto, cada avión tenía a bordo, aproximadamente 10,000 galones de combustible sin usar (información de fuentes Gubernamentales). "
Cita del informe de FEMA en el derrumbe de las y torres del WTC Uno y Dos (Capítulo Dos).
Desde que el avión sólo estaba volando de Boston a Los Ángeles, ellos no deberían cargar sus estanques llenos al despegue (el avión tiene un rango máximo de 7,600 millas). Ellos habrían llevado bastante combustible para el viaje junto a una cantidad como factor de seguridad. Recuerde, llevar exceso de combustible significa costos más altos y los pasajes menos provechosos económicamente. El avión también habría quemado un poco de combustible entre Boston y Nueva York.
Lo que nos proponemos, es pretender que se inyectaron los 10,000 galones completos de combustible en sólo un piso de la torre de WTC, que el combustible se quemó con la cantidad perfecta de oxígeno, que no escaparon gases calientes del piso y que tampoco escapó a otros pisos por conducción. Con estas asunciones ideales (que son imposibles en la realidad) nosotros calcularemos la temperatura máxima que este piso podría alcanzar. Por supuesto, ese día las temperaturas reales alcanzadas en cualquier piso debido al combustible quemado, serían considerablemente más bajas que aquellas que hemos calculado, pero esta estimación nos permitirá demostrar qué temperatura máxima se habría alcanzado, repetimos, en condiciones ideales.
Un galón de combustible de avión pesa aproximadamente 3.1 kilogramos, por lo cual 10,000 galones pesan 10,000 x 3.1 = 31,000 kg.
El combustible de avión es incoloro, combustible, es un destilado del petróleo. Sus usos principales son como ingrediente en lámparas de petróleo, fluidos para encender el carbón de leña, combustible de motores a reacción y en insecticidas.
Es también conocido como petróleo #1, parafina, petróleo del rango, petróleo de carbón y combustible de la aviación.
Comprende hidrocarburos con un rango de carbono de C9 - C17. Los hidrocarburos son principalmente los alkanes CnH2n+2, con n que va de 9 a 17.
Tiene un punto de encendido dentro del rango 42° C - 72° C (110° F - 162° F).
Y una temperatura de ignición de 210° C (410° F).
Dependiendo del suministro de oxígeno, el combustible se enciende por una de las tres siguientes reacciones químicas,:
(1) CnH2n+2 + (3n+1)/2 O2 => n CO2 + (n + 1) H2O
(2) CnH2n+2 + (2n+1)/2 O2 => n CO + (n + 1) H2O
(3) CnH2n+2 + (n+1)/2 O2 => n C + (n + 1) H2O
La reacción (1) sólo ocurre cuando el combustible esta bien mezclado con aire antes de encenderse, como por ejemplo, en los motores de reacción.
Las reacciones (2) y (3) ocurren cuando un contenedor de combustible arde. Cuando ocurre la reacción (3) el carbono que se forma muestras humos negros en las llamas. Esto hace el humo muy oscuro.
En la colisión del avión contra el WTC, habría mezclado bastante bien, el combustible con la cantidad limitada de aire disponible dentro del edificio, pero la combustión posterior habría sido principalmente una combinación de reacciones (2) y (3) dado que la cantidad de oxígeno realmente se restringió.
Desde que nosotros no sabemos cuales eran las cantidades exactas de oxígeno disponible al fuego, asumiremos que la combustión era perfecta y eficiente, que es, la totalidad del combustible del avión se quemó mediante la reacción (1), aun cuando sabemos que esto no fue así. Esta asunción generosa dará una temperatura que sabemos será más alta que la temperatura real del fuego atribuible al combustible del avión.
Necesitamos saber que el valor calorífico (neto) del combustible de avión cuando es quemado vía reacción (1) es 42 - 44 MJ/kg. El valor calorífico de un combustible es la cantidad de energía liberada cuando el combustible es quemado. Usaremos el valor más alto, el de 44 MJ/kg esto llevará a una temperatura máxima más alta que con el valor más bajo de 42 (y deseamos continuar siendo ultrajantemente generosos en nuestras asunciones).
Para una presentación más limpia y cálculos más simples, asumiremos también que nuestros hidrocarburos son de la formula CnH2n. El dejar caer los 2 átomos de hidrógeno no representa mucha diferencia en resultado final y el lector interesado podrá recalcular fácilmente las fórmulas para un resultado ligeramente más exacto. Así, nosotros estamos ahora asumiendo la ecuación:
(4) CnH2n + 3n/2 O2 => n CO2 + n H2O
Sin embargo, este modelo, no tiene en cuenta que la reacción se está llevando a cabo en el aire, en cual solo parcialmente es Oxígeno.
El aire seco es 79% nitrógeno y 21% oxígeno (por volumen). Aire Normal tiene un contenido de humedad de 0 a 4%. Nosotros incluiremos el vapor de agua y los otros gases atmosféricos menores con el nitrógeno.
Así la proporción de los principales gases atmosféricos: oxígeno y nitrógeno, es 1: 3.76. en la condición molar:
Aire = O2 + 3.76 N2.
Porque oxígeno viene mixto con el nitrógeno, tenemos que incluirlo en las ecuaciones. Aunque no reaccione, es para " lo largo del paseo" y absorberá el calor, afectando el equilibrio de calor global. Entonces, necesitamos usar la ecuación:
5) CnH2n + 3n/2(O2 + 3.76 N2) => n CO2 + n H2O + 5.64n N2
De esta ecuación nosotros vemos que la proporción molar de CnH2n a aquella de los productos es:
CnH2n : CO2 : H2O : N2 = 1 : n : n : 5.64n moles
= 14n : 44n : 18n : 28 x 5.64n Kg.
= 1 : 3.14286 : 1.28571 : 11.28 Kg.
= 31,000 : 97,429 : 39,857 : 349,680 Kg.
En la conversión mol a kilogramos, hemos asumido los pesos atómicos de hidrógeno, carbono, nitrógeno y oxígeno en 1, 12, 14 y 16 respectivamente.
Ahora cada uno de las torres contenía 96,000 (corto) toneladas de acero. Ése es un promedio de 96,000/117 = 820 toneladas por piso. Permítanos suponer que los pisos inferiores contuvieron aproximadamente dos veces la cantidad de acero de los pisos superiores (ya que los pisos inferiores soportaban más peso). Para lo cual estimamos que los pisos inferiores contuvieron aproximadamente 1,100 toneladas de acero y los pisos superiores aproximadamente 550 toneladas = 550 x 907.2 +/-500,000 Kg. Nosotros asumiremos que los pisos golpeados por el avión contuvo a lo menos la estimación de 500,000 Kg. de acero. Esto infravalora la cantidad de acero generosamente en estos pisos, y lleva una vez, a una estimación más alta de la temperatura máxima.
Cada piso tenía una loza de suelo y una loza de techo. Estas lozas eran de 207 pies de ancho, 207 pies largo y 4 (en partes 5) pulgadas de espesor y se construyó de hormigón ligero. Así que cada loza contuvo 207 x 207 x 1/3 = 14.283 pies cúbicos de hormigón. ¿Ahora un pie cúbico de hormigón ligero pesa 50kg, cada loza pesaba 714,150, +/- 700,000 Kg. Juntos, las lozas del piso y del techo pesaron unos 1,400,000 Kg.
Así que, ahora nosotros tomamos todos los ingredientes y estimamos una temperatura máxima a que ellos podrían alcanzar por los 10,000 galones de combustible del avión. Llamaremos a esta temperatura máxima :T. Desde que el valor calorífico del combustible de avión es 44 MJ/Kg. Sabemos que 10,000 galones = 31,000 Kg. de combustible de avión liberará:
31,000 x 44,000,000 = 1,364,000,000,000 Joules de energía.
Esta es la cantidad total de energía que está disponible para calentar los ingredientes a la temperatura T. Pero ¿Cuál es la temperatura T? Para averiguar esto, tenemos que calcular primero la cantidad de energía absorbida por cada uno de los ingredientes .
Es decir, necesitamos calcular la energía necesitada para elevar:
39,857 kilogramos de vapor de agua a la temperatura T°C
97,429 kilogramos de anhídrido carbónico a la temperatura T° C,
349,680 kilogramos de nitrógeno a la temperatura T° C,
500,000 kilogramos de acero a la temperatura T° C,
1,400,000 kilogramos de hormigón a la temperatura T° C.
Para calcular la energía necesaria para calentar las cantidades antes mencionadas, necesitamos sus calores específicos. El calor específico de una sustancia es la cantidad de energía necesaria para levantar un kilogramo de la sustancia en un grado centígrado.
Sustancia
Calor específico [J/Kg*C]
Concreto 3.300
Acero 450
Nitrógeno 1.038
Vapor de agua 1.690
Dióxido de Carbono 845
Sustituyendo estos valores en los anteriores, obtenemos:
39,857 x 1,690 x (T - 25) Joules se necesitan para calentar el vapor de agua de 25° a T°C,
97,429 x 845 x (T - 25) Joules se necesitan para calentar el anhídrido carbónico de 25° a T°C,
349,680 x 1,038 x (T - 25) Joules se necesitan para calentar el nitrógeno de 25° a T°C,
500,000 x 450 x (T - 25) Joules se necesitan para calentar el acero de 25° a T°C,
1,400,000 x 3,300 x (T - 25) Joules se necesitan para calentar el hormigón de 25° a T° C.
La asunción que los calores específicos son constantes sobre el rango de temperatura de 25°-T°C, es una buena aproximación si T resulta ser relativamente pequeña (como lo es). Para valores más altos de T, esta asunción, lleva una vez más, a una temperatura máxima más alta (Ya que el calor específico para estas substancias aumenta con la temperatura). Hemos asumido que la temperatura ambiente inicial es de 25° C. La cantidad, (T - 25° C), es la elevación de la temperatura.
Entonces que la cantidad de energía necesaria para elevar un piso a la temperatura T°C es :
= (39,857 x 1,690 + 97,429 x 845 + 349,680 x 1,038 + 500,000 x 450 + 1,400,000 x 3,300) x (T - 25)
= (67,358,300 + 82,327,500 + 362,968,000 + 225,000,000 + 4,620,000,000) x (T - 25) los Julios
= 5,357,650,000 x (T - 25) Joules
Desde que la cantidad de energía disponible para calentar este suelo es 1,364,000,000,000 Joules, tenemos entonces:
5,357,650,000 x (T - 25) = 1,364,000,000,000
5,357,650,000 x T - 133,941,000,000 = 1,364,000,000,000
Por consiguiente:
T = (1,364,000,000,000 + 133,941,000,000)/5,357,650,000 = 280° C (536° F).
Entonces, si nosotros asumimos un incendio de una oficina normal del WTC, el combustible del avión podría agregar sólo 280 - 25 = 255° C (máximo) a la temperatura del incendio.
Resumiendo:
Hemos asumido que la cantidad total del combustible del avión fue inyectado en sólo un piso del WTC, que el combustible se quemó con una eficiencia perfecta, que no escaparon gases calientes del piso y que no se perdió calor de este piso a otros por conductividad.
Hemos encontrado que es imposible que el combustible de avión, por si mismo, elevara la temperatura de este piso más allá de 280° C (536° F).
List of Publications
NIELS HARRIT
“Pyrylium Salts and Hydroxylamine in Acid Medium. Synthesis of Pyridine N-Oxides from Pyrylium Salts.” N. Harrit, C.L. Pedersen and O. Buchardt, Acta Chemica Scandinavica 24 (1970) 3435-3443.
“Light-Induced Ring Expansion of Pyridine N-Oxides.” O. Buchardt, C.L. Pedersen and N. Harrit, Journal of Organic Chemistry. 37 (1972) 3592-3595.
“Detection of Transients in Low-temperature Photochemistry of 4-Phenyl-1,3,2-oxathiazolyio-5-oxide by Ultraviolet and Electron Spin Resonance Spectrometry.” A. Holm, N. Harrit, K. Bechgaard, O. Buchardt, and S. Harnung, Chem.Commun. 1972, 1125.
“The Photochemical Behavior of Aromatic 1,2-Diazine-N-Oxides.” K.B. Tomer, N. Harrit, I. Rosenthal, O. Buchardt, P.L. Kumler and D. Creed, .J.Am.Chem.Soc. 95 (1973) 7402-7406.
“Photochemical Rearrangement of 3-Methyl-2-methylthio-5-phenyl-(1,3-thiazol-4-ylio)oxide to 3-Methyl-4-methylthio-5-phenyl-1,3-thiazol-2-one.” O. Buchardt, J. Domanus, N. Harrit, A. Holm, G. Isaksson and J. Sandström, Chem. Commun. 1974, 376-377.
“Photolyse af 4-Phenyl-1,3,2-oxathiazolylio-5-oxid. Mekanisme og metode.” Licentiatafhandling. 215 sider. Kemisk Laboratorium II, H.C.Ørsted Institutet, Københavns Universitet 1975.
“Photochemistry of Thiatriazoles. Benzonitrile Sulphide as Intermediate in the Photolysis of Phenyl-Substituted 1,2,3,4-Thiatriazole, 1,3,2-Oxathiazolylio-5-oxide and 1,3,4-Oxathiazole-2-one.” A. Holm, N. Harrit and N.H. Toubro, J. Am. Chem. Soc. 97 (1975) 6197-6201.
“The Photocycloaddition of Cyclohexene to Carbostyrils.” O. Buchardt, J. J. Christensen and N. Harrit, Acta Chemica Scandinavica B30 (1976) 189-192.
“Oxathiiranes as Intermediates in the Photolysis of Sulfines.” L. Carlsen, N. Harrit and A. Holm, J. Chem. Soc. Perkin Trans. I, 1976, 1404-1408.
“On the mechanism of Photolysis of 4-Phenyl-1,3,2-oxathiazolylio-5-oxide in ethanol. Evidence for Ketene Intermediates.” A. Holm, N. Harrit and N.H. Toubro, Tetrahedron 32 (1976) 2559-2563.
“Enviromental Effects as a Determining Factor in Photochemical reactions: Photolysis of matrix-isolated 4-Phenyl-1,3,2-oxathiazolylio-5-oxide.” I.R. Dunkin, M. Poliakoff, J.J. Turner, N. Harrit and A. Holm, Tetr. Letters 1976, 873-876.
“Photoinduced Ring Opening and COS Elimination of Mesoionic Thiadiazoles.” A. Holm, N.H. Toubro and N. Harrit, Tetr. Lett. 1976, 1909-1912.
“Photolysis of 1,2,5-Selenadiazoles. Formation of Nitrile Selenides.” N. Harrit, C.L. Pedersen, M. Poliakoff and I. Dunkin, Acta Chemica Scandinavica B31 (1977) 848-858.
“Strong Evidence for Thiazirines as Stable Intermediates at Cryogenic Temperature in the Photolytic Formation of Nitrile Sulfides from Aryl Substituted 1,2,3,4-Thiatriazole, Thiatriazole-3-oxide and 1,3,4-Oxathiazol-2-one.” A. Holm, N. Harrit and I. Trabjerg, J. Chem. Soc. Perkin I 1978, 746-750.
“Evidence from the Absorption and Emission Spectra of Trimethylenemethane Derivatives for two Molecular Species in Thermal Equilibrium.” N. Harrit, N. Turro, M.J. Mirbach, J.A. Berson and M. Platz, J. Am. Chem. Soc. 100 (1978) 7653-7658.
“Photoinduced Ring Opening and Fragmentation of Isomeric Mesoinic Anhydro-4(5)-hydroxythiazolium Hydroxides and of Anhydro-5-hydroxy-3-methyl-2-phenyloxazolium Hydroxides.”
N.H.Toubro, B. Hansen, N. Harrit, A. Holm and K.T. Potts, Tetrahedron 35 (1979) 229-231.
“Absence of Triazirine Intermediates in the Photolytic Formation of azides from Mesoionic 3-Substituted 1,2,3,4-Oxathiazolylio-5-oxides.” C. Bjerre, C. Christophersen, B. Hansen, N. Harrit, F.M. Nicolaisen and A. Holm, Tetrahedron 35 (1979) 409-411.
“Photolysis of 1,2,3-Selenadiazole. Formation of Selenirene by Secondary Photolysis of Selenoketene.” N. Harrit, S. Rosenkilde, B. D. Larsen and A. Holm, J. Chem. Soc. Perkin I 1985, 907-911.
“Viscosity-dependent Fluorescence and Low-temperature Photochemistry of Mesoionic 4-Phenyl-1,3,2-oxathiazolylium-5-olate.” N. Harrit, A. Holm, I. Dunkin, M. Poliakoff and J.J. Turner, J. Chem. Soc. Perkin II 1987, 1227-1238.
“On Selenoketenes. Formation of 1,3,4-Selenodiazoline from Carbodiselenide and Diazoalkanes.” R.H. Berg, N. Harrit, E. Larsen and A. Holm, Acta Chemica Scandinavica 43 (1989) 885-887.
“Pregnancy Zone Protein, a Proteinase Binding Macroglobulin. Interactions with Proteinase Binding Macroglobulin. Interactions with Proteinases and Methylamine.” U. Christensen, M. Simonsen, N. Harrit and L. Sottrup-Jensen, Biochemistry, 28 (1989) 9324-9331.
“Mixed-function-oxygenase in Juvenile rainbow Trout Exposed to Hexachlorobenzene or 3,3’,4,4’-tetrachlorobiphenyl.” H. Tyle, M. Egsmose and N. Harrit, Comp. Biochem. Physiol. 100 (1991) 161-164.
“Pregnancy Zone Protein, a Proteinase Binding -Macroglobulin. Stopped-flow Kinetic Studies of its Interaction with Chymotrypsin.” U. Christensen, L. Sottrup-Jensen and N. Harrit, Biochem. Biophys. Acta, 1076 (1991) 91-96.
“The Peroxide Chemistry of triaryl Substituted Imidazoles. Fenflumizole, a Non-steroidal, anti-inflammatory Agent.” P.L. Frandsen, K. Håkansson, A. Holm and N. Harrit, Acta Chemica Scandinavica 45 (1991) 627-631.
“Photolysis of N-2,4,6-Trinitrophenyl Substituted Amino Acids.” J. Frederiksen, B.D. Larsen and N. Harrit, Tetrahedron Lett. 32 (1991) 5823.
“Photolysis of 1,2,3-Thiadiazole. Formation of Thiirene by Secondary Photolysis of Thioketene.” B.D. Larsen, N. Harrit, H. Eggert and A. Holm, Acta Chem. Scand. 46 (1992) 482-486.
“Two Different Ca2+ Ion Binding Sites in Factor VIIa and in Des(1-38) factor VIIa.” J. Schiødt, N. Harrit, U. Christensen and L. Petersen, FEBS Letters, 306 (1992) 265-268.
“Correlations Between the Rate Constant of Singlet Oxygen Quenching by Imidazole Derivatives and Anti-inflammatory Activity in Rats.” R.V. Bensasson, J. Frederiksen, M. Rougée, D. Lexa and N. Harrit, Mol. Pharmacol. 42 (1992) 718-722.
“Synthesis of New Sulfur Heteroaromatics Isoelectronic with Dibenzo[g,p]chrysene by Photocyclization of Thienyl- and Phenyl-Substituted Ethenes.” E. Fischer, J. Larsen, J.B.
Christensen, M. Fourmigué, H.G. Madsen and N. Harrit, J. Org. Chem. 61, (1996) 6997-7005.
“Ordering of the Disk-like 2,3,6,7,10,11-Hexakis(hexylthio)triphenylene in Solution and at Liquid-Solid Interface.” Gabriel, J.-C., Larsen, N.B., Larsen, M., Harrit, N., Pedersen, J.S., Schaumburg, K. and Bechgaard, K. Langmuir 12 (1996) 1690-1692.
“Photoconductivity of Langmuir-Blodgett Films of Corbathiene.” A. Komolov, K. Schaumburg and N. Harrit, Thin Solid Films 293 (1997) 159.
“Photophysical properties of 2,3,6,7,10,11-Hexakis(n-hexylsulfonyl)triphe-nylene and 2,3,6,7,10,11-Hexakis(n-hexylsulfonyl)triphenylene in solution.” D. Baunsgaard, M. Larsen, N. Harrit, J. Frederiksen, R. Wilbrandt, H. Stapelfeldt, J.Chem.Soc., Faraday Trans., 93 (1997) 1893-1901.
“Fluorescein-Conjugated Lysine Monomers for Solid Phase Synthesis of Fluorescent Peptides and PNA Oligomers.” J. Lohse, P.E. Nielsen, N. Harrit and O. Dahl, Bioconjugate Chem. 8 (1997) 503-509.
”Synthesis, Structure, and Fluorescence Properties of 5,17-Distyryl-25,26,27,28-tetrapropoxycalix[4]arenes in the Cone Conformation.” Mogens Larsen, Frederik C. Krebs, Mikkel Jørgensen, og Niels Harrit, J.Org.Chem., 63 (1998) 4420-4424.
”2,6,10-Tris(dialkylamino)trioxatriangulenium Ions. Synthesis, Structure, and Properties of Exceptionally Stable Carbenium Ions.” Bo W. Laursen, Frederik C. Krebs, Merete F. Nielsen, Klaus Bechgaard, Jørn B. Christensen, and Niels Harrit, J. Am. Chem. Soc., 120 (1998) 12255 –12263.
”Vibronic Activity in the Phosphorescence Spectra of Disklike Aromatic Molecules: A Combined Experimental and Theoretical Investigation.” D. Baunsgaard, N. Harrit, M. El Balsami, F. Negri, G. Orlandi, J. Frederiksen and R. Wilbrandt, J. Phys. Chem. A, 102 (1998) 10007 –10016.
”The Phosphorescence Spectra of Triphenylene and Truxene: A Combined Experimental and Theoretical Investigation of The Vibronic Structure.” D. Baunsgaard, M. El Balsami, J. Frederiksen, N. Harrit, F. Negri, G. Orlandi and R. Wilbrandt, Laser Chemistry, 19 (1999) 349-351.
”Synthesis and conformational studies of a series of 5,17-bis-aryl-25,26,27,28-tetrapropoxycalix[4]arenes: The influence of - interactions on the molecular structure.” Mogens Larsen, Frederik C. Krebs, Niels Harrit, and Mikkel Jørgensen. J.Chem.Soc.Perkin Trans. 2, 1999, 1749-1757.
”Vibrational spectroscopic and quantum chemical studies of the trioxatriangulenium carbocation.” Johannes Reynisson, G. Balakrishnan, Niels Harrit, and Robert Wilbrandt. J.Mol.Struct., 520, 2000, 63-73.
“Structural properties of DNO investigated with pyrene excimer formation.” Johannes Reynisson, Lise Vejby-Christensen, Robert Wilbrandt, Niels Harrit, Rolf H. Berg, J. Pept. Sci. 6, 2000, 603-611.
“Synthesis of a Hoechst 32258 Analogue Amino Acid Building Block for Direct Incorporation of a Fluorescent, High-Affinity DNA Binding Motif into Peptides.” Carsten Behrens, Niels Harrit, Peter E. Nielsen, Bioconjugate Chem. 12 (2001) 1021-1027.
„Photophysics of Trioxatriangulenium ion. Electrophilic reactivity in the ground state and excited singlet state.“ Jóhannes Reynisson, Niels Harrit, Robert Wilbrandt, Vibeke Brinck, Bo W. Laursen, Kasper Nørgaard, and Albert M. Brouwer, Photochemical and Photobiological Sciences 1 (2002) 763-773.
“System and Method for the Classification of Biological Samples and their Diagnostic Potential”, Lars Nørgård, Morten Albrechtsen, Ole Olsen, Niels Harrit og Rasmus Bro-Jørgensen, Patent (WO2001092859).
“Redox regulation of gene expression of chemoprotective enzymes and of chemoprotection against inflammation and carcinogenesis.“ Vincent Zoete, John Frederiksen, Niels Harrit, Rodger Scurlock, Michel Rougée and René V. Bensasson, Free Radical Research, 36 (2002) 111-114.
“Intercalation of Trioxatriangulenium Ion (TOTA+) in DNA: Binding, Electron Transfer, X-Ray Crystallography, and Electronic Structure.” Jóhannes Reynisson, Gary B. Schuster, Sheldon B. Howerton, Loren D. Williams, Robert N. Barrnett, Charles L. Cleveland, Uzi Landman, Niels Harrit and Jonathan B. Chaires, J. Am. Chem. Soc. 125 (2003) 2072-2083
“Novel Synthesis of Protected Thiol End-Capped Stilbenes and Oligo(phenylenevinylene)s (OPVs).” Nicolai Stuhr-Hansen, Jørn B. Christensen, Niels Harrit, and Thomas Bjørnholm, J. Org. Chem., 68 (2003) 1275-1282.
“Oligodeoxynucleotides Containing -L-ribo configured LNA-type C-aryl Nucleosides: Synthesis and Properties as a Universal Base.” B. Ravindra Babu Raunak, Mads D. Sørensen, Virinder S. Parmar, Niels H. Harrit and Jesper Wengel, Organic & Biomolecular Chem. (RSC), 2 (2004) 80-89.
“Spectroscopic Properties of Sintered BaMgAl10O17:Eu2+ (BAM)
Translucent Pellets. Comparison with Commercial Powder.” E. Zych, W. Goetz, N. Harrit and H. Spanggaard, J. Alloys and Compounds, 380 (2004) 113-117.
”2,6,10-Tris(dialkylamino)-trioxatriangulenium salts - a new promising fluorophore. Ion-pair formation and aggregation in non-polar solvents.” Bo W. Laursen, Jóhannes Reynisson, Kurt V. Mikkelsen, Klaus Bechgaard and Niels Harrit, Photochemical and Photobiological Sciences (RSC), 4 (2005) 568-576.
“Fluorescent Pyrene-Functionalized 2’Amino-LNAs for Nucleic Acid detection in Homogenous Assays.” Patrick J. Hrdlicka, B. Ravindra Babu, Mads D. Sørensen, Niels Harrit, and Jesper Wengel, J. Am. Chem. Soc., 127 (2005) 13293 – 13299.
“Expanding the Optical Trapping Range of Gold Nanoparticles.” Poul Martin Hansen, Vikram Kjøller Bhatia, Niels Harrit, and Lene Oddershede, Nano Letters, 5 (2005) 1937 – 1942.
" Fluorescence spectroscopy and chemometrics for classification of breast cancer samples-a feasibility study using extended canonical variates analysis." Lars Nørgaard, Gyorgy, Soletormos, Niels Harrit, Morten Albrechtsen, Ole, Olsen, Dorte Nielsen, Kristoffer Kampmann, and Rasmus Bro, Journal of Chemometrics (2007), 21, 451-458..
"Close Columnar Packing of Triangulenium Ions in Langmuir Films." Jens B. Simonsen, Kristian Kjaer, Paul Howes, Kasper Norgaard, Thomas Bjornholm, Niels Harrit, Bo W. Laursen, Langmuir 25 (2009) 3584-3592.
"Time-resolved X-ray scattering of an electronically excited state in solution. Structure of the 3A(2u) state of tetrakis-mu-pyrophosphitodiplatinate(II)." Morten Christensen, Kristoffer Haldrup, Klaus Bechgaard, Robert Feidenhans'l, Qingyu Kong, Marco Cammarata, Manuela Lo Russo, Michael Wulff, Niels Harrit, Martin Meedom Nielsen, Journal of the American Chemical Society, 131 (2009) 502-8.
"Structural Tracking of a Bimolecular Reaction in Solution by Time-Resolved X-ray Scattering." Kristoffer Haldrup, Morten Christensen, Marco Cammarata, Qingyu Kong, Michael Wulff, Simon O. Mariager, Klaus Bechgaard, Robert Feidenhans'l, Niels Harrit*, Martin M. Nielsen*. Angewandte Chemie. 2009. "Hot paper" pre-published on the web: www3.interscience.wiley.com/cgi-bin/fulltext/122369907/PDFSTART
"Active Thermitic Material Discovered in Dust from 9/11 World Trade Center Catastrophe." Niels H. Harrit, Jeffrey Farrer, Steven E. Jones, Kevin R. Ryan, Frank M. Legge, Daniel Farnsworth, Gregg Roberts, James R. Gourley and Bradley R. Larsen, The Open Chemical Physics Journal, 2 (2009) 7 – 31.
Tengo ciencias físicas aprobadas con un notable de media.Quizás seas tu el que tienes que volver a repasar un poquito la Física y explicar por ejemplo esto:
Con este resultado, me gustaría que me explicaras cómo fue posible el colapso de los edificios (y sobre todo del WTC 7 que NO tuvo el impacto de ningún avión) y cómo es posible que se encontrara acero fundido en los cimientos de los tres edificios, tal como atestiguaron tanto los bomberos como las empresas que se encargaron de limpiar la zona cero. Y además, tal como se puede ver en numerosas imágenes.
Si en vez de mostrar tanta prepotencia te dedicaras a analizar un poco los hechos que nunca se quisieron mostrar quizás sí que podríamos hablar. Ahora mismo desde luego que no.
Y si esto fuera verdad, sería una excelente demostración de cómo las fogosas creencias irracionales y subjetivas pueden llegar a desbancar a la inteligencia.Tengo ciencias físicas aprobadas con un notable de media.
Hace tiempo traté de explicarte de manera educada, con datos, con CIENCIA por qué estás equivocado y por qué tus teorías y las de los tipos que las defienden se derrumban como castillos de naipes. Lo siento, pero si realmente sacaste un notable en física deberías de darte cuenta enseguida que las cosas que propone esta gente son basura.
Para ello, imaginemosnos que todo el combustible del avión se inyectara en sólo un piso del WTC, que el combustible se quemó con una eficiencia perfecta, que los gases calientes se mantuvieron en el piso y que nada del calor escapó de este piso por conducción. Con estas asunciones ideales calcularemos la temperatura máxima que este piso podría alcanzar.
"El Boeing 767 es capaz de transportar 23,980 galones de combustible y se estima que, en el momento del impacto, cada avión tenía a bordo, aproximadamente 10,000 galones de combustible sin usar (información de fuentes Gubernamentales). "
Cita del informe de FEMA en el derrumbe de las y torres del WTC Uno y Dos (Capítulo Dos).
Desde que el avión sólo estaba volando de Boston a Los Ángeles, ellos no deberían cargar sus estanques llenos al despegue (el avión tiene un rango máximo de 7,600 millas). Ellos habrían llevado bastante combustible para el viaje junto a una cantidad como factor de seguridad. Recuerde, llevar exceso de combustible significa costos más altos y los pasajes menos provechosos económicamente. El avión también habría quemado un poco de combustible entre Boston y Nueva York.
Lo que nos proponemos, es pretender que se inyectaron los 10,000 galones completos de combustible en sólo un piso de la torre de WTC, que el combustible se quemó con la cantidad perfecta de oxígeno, que no escaparon gases calientes del piso y que tampoco escapó a otros pisos por conducción. Con estas asunciones ideales (que son imposibles en la realidad) nosotros calcularemos la temperatura máxima que este piso podría alcanzar. Por supuesto, ese día las temperaturas reales alcanzadas en cualquier piso debido al combustible quemado, serían considerablemente más bajas que aquellas que hemos calculado, pero esta estimación nos permitirá demostrar qué temperatura máxima se habría alcanzado, repetimos, en condiciones ideales.
Un galón de combustible de avión pesa aproximadamente 3.1 kilogramos, por lo cual 10,000 galones pesan 10,000 x 3.1 = 31,000 kg.
El combustible de avión es incoloro, combustible, es un destilado del petróleo. Sus usos principales son como ingrediente en lámparas de petróleo, fluidos para encender el carbón de leña, combustible de motores a reacción y en insecticidas.
Es también conocido como petróleo #1, parafina, petróleo del rango, petróleo de carbón y combustible de la aviación.
Comprende hidrocarburos con un rango de carbono de C9 - C17. Los hidrocarburos son principalmente los alkanes CnH2n+2, con n que va de 9 a 17.
Tiene un punto de encendido dentro del rango 42° C - 72° C (110° F - 162° F).
Y una temperatura de ignición de 210° C (410° F).
Dependiendo del suministro de oxígeno, el combustible se enciende por una de las tres siguientes reacciones químicas,:
(1) CnH2n+2 + (3n+1)/2 O2 => n CO2 + (n + 1) H2O
(2) CnH2n+2 + (2n+1)/2 O2 => n CO + (n + 1) H2O
(3) CnH2n+2 + (n+1)/2 O2 => n C + (n + 1) H2O
La reacción (1) sólo ocurre cuando el combustible esta bien mezclado con aire antes de encenderse, como por ejemplo, en los motores de reacción.
Las reacciones (2) y (3) ocurren cuando un contenedor de combustible arde. Cuando ocurre la reacción (3) el carbono que se forma muestras humos negros en las llamas. Esto hace el humo muy oscuro.
En la colisión del avión contra el WTC, habría mezclado bastante bien, el combustible con la cantidad limitada de aire disponible dentro del edificio, pero la combustión posterior habría sido principalmente una combinación de reacciones (2) y (3) dado que la cantidad de oxígeno realmente se restringió.
Desde que nosotros no sabemos cuales eran las cantidades exactas de oxígeno disponible al fuego, asumiremos que la combustión era perfecta y eficiente, que es, la totalidad del combustible del avión se quemó mediante la reacción (1), aun cuando sabemos que esto no fue así. Esta asunción generosa dará una temperatura que sabemos será más alta que la temperatura real del fuego atribuible al combustible del avión.
Necesitamos saber que el valor calorífico (neto) del combustible de avión cuando es quemado vía reacción (1) es 42 - 44 MJ/kg. El valor calorífico de un combustible es la cantidad de energía liberada cuando el combustible es quemado. Usaremos el valor más alto, el de 44 MJ/kg esto llevará a una temperatura máxima más alta que con el valor más bajo de 42 (y deseamos continuar siendo ultrajantemente generosos en nuestras asunciones).
Para una presentación más limpia y cálculos más simples, asumiremos también que nuestros hidrocarburos son de la formula CnH2n. El dejar caer los 2 átomos de hidrógeno no representa mucha diferencia en resultado final y el lector interesado podrá recalcular fácilmente las fórmulas para un resultado ligeramente más exacto. Así, nosotros estamos ahora asumiendo la ecuación:
(4) CnH2n + 3n/2 O2 => n CO2 + n H2O
Sin embargo, este modelo, no tiene en cuenta que la reacción se está llevando a cabo en el aire, en cual solo parcialmente es Oxígeno.
El aire seco es 79% nitrógeno y 21% oxígeno (por volumen). Aire Normal tiene un contenido de humedad de 0 a 4%. Nosotros incluiremos el vapor de agua y los otros gases atmosféricos menores con el nitrógeno.
Así la proporción de los principales gases atmosféricos: oxígeno y nitrógeno, es 1: 3.76. en la condición molar:
Aire = O2 + 3.76 N2.
Porque oxígeno viene mixto con el nitrógeno, tenemos que incluirlo en las ecuaciones. Aunque no reaccione, es para " lo largo del paseo" y absorberá el calor, afectando el equilibrio de calor global. Entonces, necesitamos usar la ecuación:
5) CnH2n + 3n/2(O2 + 3.76 N2) => n CO2 + n H2O + 5.64n N2
De esta ecuación nosotros vemos que la proporción molar de CnH2n a aquella de los productos es:
CnH2n : CO2 : H2O : N2 = 1 : n : n : 5.64n moles
= 14n : 44n : 18n : 28 x 5.64n Kg.
= 1 : 3.14286 : 1.28571 : 11.28 Kg.
= 31,000 : 97,429 : 39,857 : 349,680 Kg.
En la conversión mol a kilogramos, hemos asumido los pesos atómicos de hidrógeno, carbono, nitrógeno y oxígeno en 1, 12, 14 y 16 respectivamente.
Ahora cada uno de las torres contenía 96,000 (corto) toneladas de acero. Ése es un promedio de 96,000/117 = 820 toneladas por piso. Permítanos suponer que los pisos inferiores contuvieron aproximadamente dos veces la cantidad de acero de los pisos superiores (ya que los pisos inferiores soportaban más peso). Para lo cual estimamos que los pisos inferiores contuvieron aproximadamente 1,100 toneladas de acero y los pisos superiores aproximadamente 550 toneladas = 550 x 907.2 +/-500,000 Kg. Nosotros asumiremos que los pisos golpeados por el avión contuvo a lo menos la estimación de 500,000 Kg. de acero. Esto infravalora la cantidad de acero generosamente en estos pisos, y lleva una vez, a una estimación más alta de la temperatura máxima.
Cada piso tenía una loza de suelo y una loza de techo. Estas lozas eran de 207 pies de ancho, 207 pies largo y 4 (en partes 5) pulgadas de espesor y se construyó de hormigón ligero. Así que cada loza contuvo 207 x 207 x 1/3 = 14.283 pies cúbicos de hormigón. ¿Ahora un pie cúbico de hormigón ligero pesa 50kg, cada loza pesaba 714,150, +/- 700,000 Kg. Juntos, las lozas del piso y del techo pesaron unos 1,400,000 Kg.
Así que, ahora nosotros tomamos todos los ingredientes y estimamos una temperatura máxima a que ellos podrían alcanzar por los 10,000 galones de combustible del avión. Llamaremos a esta temperatura máxima :T. Desde que el valor calorífico del combustible de avión es 44 MJ/Kg. Sabemos que 10,000 galones = 31,000 Kg. de combustible de avión liberará:
31,000 x 44,000,000 = 1,364,000,000,000 Joules de energía.
Esta es la cantidad total de energía que está disponible para calentar los ingredientes a la temperatura T. Pero ¿Cuál es la temperatura T? Para averiguar esto, tenemos que calcular primero la cantidad de energía absorbida por cada uno de los ingredientes .
Es decir, necesitamos calcular la energía necesitada para elevar:
39,857 kilogramos de vapor de agua a la temperatura T°C
97,429 kilogramos de anhídrido carbónico a la temperatura T° C,
349,680 kilogramos de nitrógeno a la temperatura T° C,
500,000 kilogramos de acero a la temperatura T° C,
1,400,000 kilogramos de hormigón a la temperatura T° C.
Para calcular la energía necesaria para calentar las cantidades antes mencionadas, necesitamos sus calores específicos. El calor específico de una sustancia es la cantidad de energía necesaria para levantar un kilogramo de la sustancia en un grado centígrado.
Sustancia
Calor específico [J/Kg*C]
Concreto 3.300
Acero 450
Nitrógeno 1.038
Vapor de agua 1.690
Dióxido de Carbono 845
Sustituyendo estos valores en los anteriores, obtenemos:
39,857 x 1,690 x (T - 25) Joules se necesitan para calentar el vapor de agua de 25° a T°C,
97,429 x 845 x (T - 25) Joules se necesitan para calentar el anhídrido carbónico de 25° a T°C,
349,680 x 1,038 x (T - 25) Joules se necesitan para calentar el nitrógeno de 25° a T°C,
500,000 x 450 x (T - 25) Joules se necesitan para calentar el acero de 25° a T°C,
1,400,000 x 3,300 x (T - 25) Joules se necesitan para calentar el hormigón de 25° a T° C.
La asunción que los calores específicos son constantes sobre el rango de temperatura de 25°-T°C, es una buena aproximación si T resulta ser relativamente pequeña (como lo es). Para valores más altos de T, esta asunción, lleva una vez más, a una temperatura máxima más alta (Ya que el calor específico para estas substancias aumenta con la temperatura). Hemos asumido que la temperatura ambiente inicial es de 25° C. La cantidad, (T - 25° C), es la elevación de la temperatura.
Entonces que la cantidad de energía necesaria para elevar un piso a la temperatura T°C es :
= (39,857 x 1,690 + 97,429 x 845 + 349,680 x 1,038 + 500,000 x 450 + 1,400,000 x 3,300) x (T - 25)
= (67,358,300 + 82,327,500 + 362,968,000 + 225,000,000 + 4,620,000,000) x (T - 25) los Julios
= 5,357,650,000 x (T - 25) Joules
Desde que la cantidad de energía disponible para calentar este suelo es 1,364,000,000,000 Joules, tenemos entonces:
5,357,650,000 x (T - 25) = 1,364,000,000,000
5,357,650,000 x T - 133,941,000,000 = 1,364,000,000,000
Por consiguiente:
T = (1,364,000,000,000 + 133,941,000,000)/5,357,650,000 = 280° C (536° F).
Entonces, si nosotros asumimos un incendio de una oficina normal del WTC, el combustible del avión podría agregar sólo 280 - 25 = 255° C (máximo) a la temperatura del incendio.
Resumiendo:
Hemos asumido que la cantidad total del combustible del avión fue inyectado en sólo un piso del WTC, que el combustible se quemó con una eficiencia perfecta, que no escaparon gases calientes del piso y que no se perdió calor de este piso a otros por conductividad.
Hemos encontrado que es imposible que el combustible de avión, por si mismo, elevara la temperatura de este piso más allá de 280° C (536° F).
http://www.debunking911.com/fires.htmTowers Collapse: Fires and Fire ProofingThe two major factors in the collapse itself were the fires and the lack of fire proofing around the trusses and some columns. With fireproofing, the floor system was rated for 2 hours of continuous fire. Without it, that time is cut dramatically.
The below test results are used by conspiracy theorists to suggest the collapse couldn't have happened because, if you note the test under "Failure to support load" there are three asterisks (***) which indicates that failure did not occur. What they don't stress is the fact that all four tests have fire proofing on the trusses. Note the fire rating with 1/2 inch spray on fire proofing is 45 minutes. Some trusses and columns in the towers impact zone had none.
Construction.com
http://www.nist.gov/public_affairs/factsheet/wtc_fire_resistance_data.htm
An incredibly detailed account of the fires can be found here
http://wtc.nist.gov/NISTNCSTAR1-5A_chap_9-AppxC.pdf (56 megs)
Below is evidence from an A&E documentary 'Inside the Twin Towers' that some fireproofing may have been off well BEFORE the planes hit.
Photo used in the NIST report below.
Faulty Fireproofing Is Reviewed as Factor in Trade Center Collapsehttp://www.mzaconsulting.com/Faulty%20Fireproofing_WTC.pdf
By JAMES GLANZ with MICHAEL MOSS
Excerpts:
"Large areas of fireproofing are missing from the core columns in some of the photographs, and the architect who took them, Roger G. Morse, a consultant in Troy, N.Y., said his work had shown that the fireproofing did not stick properly. But Mr. Reiss said the problems were caused by the swaying of the buildings in the wind and the impact of elevator cables against the beams. "It was an ongoing maintenance headache," he said. Although measures were repeatedly taken to prevent the problem, he said, "every March and April when you had these windstorms and the building rocked back and forth, you would still knock some of the fireproofing down."
In an interview, Mr. Morse said the problems were far more widespread than that, probably because the fireproofing had been applied improperly to rusty steel. Mr. Morse, who at the time of his inspections was a consultant to the manufacturer of the fireproofing, said his examinations had never reached above the 78th floor in either tower, but that the nature and dimensions of the problem convinced him the failings of the fireproofing would be found on virtually all parts of the buildings. Investigators think the planes struck around the 90th to 94th floors of the north tower and the 78th to 84th floors of the south tower.
Mr. Morse said his inspections on several floors also found problems with the fireproofing of the lightweight, weblike trusses that held up the floors. He said his inspections, which began in 1986 and continued intermittently until June 2000, showed stretches of the tubelike structural steel supporting the trusses without any fireproofing, and other areas of extremely thin fireproofing.
Port Authority officials dismissed those allegations, saying that they doubted the photographs were representative of the entire building and that fireproofing on the trusses was regularly replaced and upgraded whenever there was a major renovation or a change of tenants."
One of the BIG lies in the "truth" movement is that UL certifies steel. An ex-employee named Kevin Ryan, who worked as a water tester at UL, said "The buildings should have easily withstood the thermal stress caused by pools of burning jet fuel." He says he knows this because UL certified steel components of the World Trade Center and that someone from the company who was connected with the UL testing told him this fact.
Now for some facts the "truth" movement doesn't tell you.
UL doesn't certify steel components like a steel truss or column. They certify assemblies. That means they certified the total assembly, all put together. They also didn't replicate the impact levels. They replicated a floor system with fireproofing as it would have been before the impact. They also tested it with various fireproofing thicknesses. (See above UL test results) The test trusses were physically undamaged and had intact fireproofing for the purpose of standard rating. What this means is that Mr. Ryan doesn't even know what his former employer does, much less what it did during the World Trade Center investigation. Maybe that's why they fired him...
The other lie in the "truth" movement is the characterization of what the NIST said was the cause of the collapse.
1) The NIST NEVER said burning jet fuel was the cause of the collapse. Only that it was a factor
2) If the assembly stayed together, it only SUPPORTS the NIST hypothesis that the trusses pulled the columns in.
3) The UL test caused the test trusses to sag even with fireproofingNIST Tests Provide Fire Resistance Data on World Trade Center Floor SystemsKevin Ryan is the editor of the "scholars" Journal of 911 Studies and one of the "peers" who review their so called "peer" reviewed papers like the flying elephant paper. It took the "scholars" months to debunk there own paper which Mr. Ryan should have debunked himself. Something which took me about 2 minutes.
National Institute of Standards & Technology
Aug 27, 2004 10:14 AM
The Commerce Department’s National Institute of Standards and Technology today reported that results from a series of four fire resistance tests conducted this month on composite concrete-steel trussed floor systems typical of those used in the World Trade Center towers showed that the test structures were able to withstand standard fire conditions for between one and two hours. The tests are part of NIST’s building and fire safety investigation of the WTC disaster on Sept. 11, 2001.The 1968 New York City building code – the code that the towers were intended but not required to meet when they were built – required a two-hour fire rating for the floor system.
Shyam Sunder, lead investigator of the NIST WTC investigation, explained that the four laboratory tests provide only a means for evaluating the relative fire resistance rating of the floor systems under standard fire conditions and according to accepted test procedures. Sunder cautioned, “These tests alone cannot be used to determine the actual performance of the floor systems in the collapse of the WTC towers. However, they are already providing valuable insight into the role that the floors may have played in causing the inward bowing of the perimeter columns minutes before both buildings collapsed.”
“The fire conditions in the towers on 9-11 were far more extreme than those to which floor systems in standard U.S. fire rating tests are subjected,” Sunder said to a group that gathered to watch yesterday’s final test at Underwriters Laboratories (UL) in Northbrook, Ill. “Our investigation’s final assessment of how the floor system performed in the WTC fires also must consider factors such as the combustible fuel load of the hijacked jets, the extent and number of floors involved, the rate of the fire spread across and between floors, ventilation conditions, and the impact of the aircraft-damaged towers’ ability to resist the fire,” Sunder said.
All four WTC floor system fire tests used the standard procedure known as ASTM E119 for rating the fire resistance of a building structural unit such as a floor system, column or beam under prescribed conditions. The tests were conducted as part of a NIST contract at two separate UL fire test laboratories to take advantage of the different capabilities available at these facilities.
The first two tests, conducted earlier this month at the UL facility in Toronto, Canada, looked at the fire performance of 11-meter (35-foot) floor systems coated with a near-uniform 19-millimeter-thick (0.75-inch) layer of fireproofing material. This is representative of the span size and as-applied average fireproofing thickness of the floor systems in the WTC towers.
http://firechief.com/news/nist-tests-wtc9835/index.html
Since we are on the subject of fire proofing, let me touch on something which keeps being repeated by the so called "scholars". They keep suggesting that lease owner Larry Silverstein had a very good reason to blow up the towers.Fetzer: My impression has been that there were a couple of problems with the towers and it may have been that they were chronic problems. One of course was that it was laden with asbestos and that any proposal to remove that asbestos which was used as a coating on the steel as I understand it would have been a gargantuan task at incredible expense. Can anyone imagine for example of constructing scaffolding around a 110 story building? And second of all that there were difficulties with occupancy that Larry Silverstein wasn’t getting a full return on his investment from the ordinary use of the buildings, because a tremendous large numbers of offices were unoccupied. Whole floors and sections of the buildings.The fact is that asbestos in the towers was limited to floors only up to the 38th floor of WTC 1 and it was encapsulated. There was no asbestos in WTC2 .
"Several materials were considered for the sprayed thermal insulation. The exterior columns required insulation not only for fire protection but also to control column temperatures under service conditions. Alcoa recommended for the exterior columns the use of a sprayed material produced by U.S. Mineral Products, Co. known as BLAZE-SHIELD Type D. The same material was eventually selected for the floor trusses and core beams and columns. This product, however, contained asbestos fibers. On April 13, 1970, New York City issued restrictions on the application of sprayed thermal insulation containing asbestos. The use of BLAZE-SHIELD Type D was discontinued in 1970 at the 38th floor of WTC 1. The asbestos-containing material was subsequently encapsulated with a sprayed material that provided a hard coating. A green dye was added to the encapsulating material so that the asbestos containing SFRM could be identified. Thermal protection of the remaining floors of WTC 1 and all of WTC 2 was carried out using BLAZE-SHIELD Type DC/F, a product that contained mineral wool (glassy fibers) in place of the crystalline asbestos fibers. On the basis of tests, it was reported that the thermal properties of BLAZE-SHIELD Type DC/F were equal to or "slightly better" than those of BLAZE-SHIELD Type D"So much for asbestos, now what about the occupancy...
NIST NCSTAR 1-6A
May 31, 1998 As the market for office space in midtown has tightened and rental rates increased, tenants have been looking to downtown as a cheaper alternative. Over the last year, those seeking large blocks of space have been finding them at the trade center, which had many vacancies as a result of the 1993 terrorist bombing and the shrinkage of the financial industry in the early part of the decade.So much for the "difficulties with occupancy".
''In January 1997 we had about an 80 percent occupancy rate,'' said Cherrie Nanninga, director of real estate for the Port Authority of New York and New Jersey, which owns the complex. Twenty percent of 10.5 million square feet of space is 2.1 million, which would be a substantial building by itself.
But as a result of the last year's work, Ms. Nanninga, said the complex is over 90 percent occupied and expects to it reach the 95 percent mark by the end of the year. That, she said, would be about as full as the center is likely to get, since there is almost always someone moving in or out. ''Ninety-seven percent occupancy would be full,'' said Ms. Nanninga, whose name is pronounced NAN-in-gay.
Downtown; At the World Trade Center, Things Are Looking Up
February 12, 2001
As Real Estate Director, a position Mrs. Nanninga has held since 1996, the occupancy rate at the trade center has risen from 78 percent to a healthy 98 percent, retail soared in the trade center's mall, and available office space in the Newark Legal Center has nearly been filled.
Today, only about 250,000 of the 10.4 million square feet of office space in the trade center remains vacant. And the legal center has an occupancy rate of over 99 percent.
http://www.panynj.gov/AboutthePortAuthority/PressCenter/
PressReleases/PressRelease/index.php?id=61
http://www.911myths.com/html/windfall.html
This is yet another example of what the "scholars" want to spend millions investigating. I don't want my child paying one cent just because these "scholars" do sloppy research or have hidden political agendas. Because with the 8.4 trillion national debt, the millions they want spent will be paid by our children. It's cheaper to just send me a thank you at debunk911@hotmail.com.
One more thing, if making money off the murder of 3000 people mean you committed the murder, does that also mean losing money off the murder of 3000 people mean you're innocent?
http://www.debunking911.com/freefall.htmThe towers did not fall at or below free fall speeds…In every photo and every video, you can see columns far outpacing the collapse of the building. Not only are the columns falling faster than the building but they are also falling faster than the debris cloud which is ALSO falling faster than the building. This proves the buildings fell well below free fall speed. That is, unless the beams had a rocket pointed to the ground.
Just look at any video you like and watch the perimeter columns.
Deceptive videos stop the timer of the fall at 10:09 when only the perimeter column hits the ground and not the building itself. If you notice, the building just finishes disappearing behind the debris cloud which is still about 40 stories high.
Below is a more accurate graphic using a paper written by Dr. Frank Greening which can be found at: http://www.911myths.com/WTCREPORT.pdf
The paper takes the transfer of momentum into account. Like a billiard ball being hit by another on a pool table, each floor transferred its momentum to the next as represented below. The more weight, the less resistance each floor gave.
The time required to strip off a floor, according to Frank Greening, is a maximum of about 110 milliseconds = 0.110 seconds. It is rather the conservation of momentum that slowed the collapse together with a small additional time for the destruction of each floor.Below are calculations from a physics blogger...
When I did the calculations, what I got for a thousand feet was about nine seconds- let's see,
d = 1/2at^2
so
t = (2d/a)^1/2
a is 9.8m/s^2 (acceleration of gravity at Earth's surface, according to Wikipedia), [He gives this reference so you can double check him.]
d is 417m (height of the World Trade Center towers, same source)
so
t = (834m/9.8m/s^2)^1/2 = 9.23s
OK, so how fast was it going? Easy enough,
v = at
v = (9.8m/s^2 x 9.23s) = 90.4m/s
So in the following second, it would have fallen about another hundred meters. That's almost a quarter of the height it already fell. And we haven't even made it to eleven seconds yet; it could have fallen more than twice its height in that additional four seconds. If the top fell freely, in 13.23 seconds it would have fallen about two and one-half times as far as it actually did fall in that time. So the collapse was at much less than free-fall rates.
Let's see:
KE = 1/2mv^2
The mass of the towers was about 450 million kg, according to this. Four sources, he has. I think that's pretty definitive. So now we can take the KE of the top floor, and divide by two- that will be the average of the top and bottom floors. Then we'll compare that to the KE of a floor in the middle, and if they're comparable, then we're good to go- take the KE of the top floor and divide by two and multiply by 110 stories. We'll also assume that the mass is evenly divided among the floors, and that they were loaded to perhaps half of their load rating of 100lbs/sqft. That would be
208ft x 208ft = 43,264sqft
50lbs/sqft * 43264sqft = 2,163,200lbs = 981,211kg
additional weight per floor. So the top floor would be
450,000,000 kg / 110 floors = 4,090,909 kg/floor
so the total mass would be
4,090,909 kg + 981,211 kg = 5,072,120 kg/floor
Now, the velocity at impact we figured above was
90.4m/s
so our
KE = (5,072,120kg x (90.4m/s)^2)/2 = 20,725,088,521J
So, divide by 2 and we get
10,362,544,260J
OK, now let's try a floor halfway up:
t = (2d/a)^1/2 = (417/9.8)^1/2 = 6.52s
v = at = 9.8*6.52 = 63.93m/s
KE = (mv^2)/2 = (5,072,120kg x (63.93m/s)^2)/2 = 10,363,863,011J
Hey, look at that! They're almost equal! That means we can just multiply that 10 billion Joules of energy by 110 floors and get the total, to a very good approximation. Let's see now, that's
110 floors * 10,362,544,260J (see, I'm being conservative, took the lower value)
= 1,139,879,868,600J
OK, now how much is 1.1 trillion joules in tons of TNT-equivalent? Let's see, now, a ton of TNT is 4,184,000,000J. So how many tons of TNT is 1,139,879,868,600J?
1,139,879,868,600J / 4,184,000,000J/t = 272t
Now, that's 272 tons of TNT, more or less; five hundred forty one-thousand-pound blockbuster bombs, more or less. That's over a quarter kiloton. We're talking about as much energy as a small nuclear weapon- and we've only calculated the kinetic energy of the falling building. We haven't added in the burning fuel, or the burning paper and cloth and wood and plastic, or the kinetic energy of impact of the plane (which, by the way, would have substantially turned to heat, and been put into the tower by the plane debris, that's another small nuclear weapon-equivalent) and we've got enough heat to melt the entire whole thing.
Remember, we haven't added the energy of four floors of burning wood, plastic, cloth and paper, at- let's be conservative, say half the weight is stuff like that and half is metal, so 25lbs/sqft? And then how about as much energy as the total collapse again, from the plane impact? And what about the energy from the burning fuel? You know, I'm betting we have a kiloton to play with here. I bet we have a twentieth of the energy that turned the entire city of Nagasaki into a flat burning plain with a hundred-foot hole surrounded by a mile of firestorm to work with. - Schneibster edited by Debunking 911
Let me make this clear, I don't assume to know what the ACTUAL fall time was. Anyone telling you they know is lying. The above calculation doesn't say that's the fall time. That was not its purpose. It's only a quick calculation which serves its purpose. To show that the buildings could have fallen within the time it did. It's absurd to suggest one can make simple calculations and know the exact fall time. You need a super computer with weeks of calculation to take into account the office debris, plumbing, ceiling tile etc.. etc... Was it 14 or was it 16? It doesn't matter to the point I'm making, which is the fall times are well within the possibility for normal collapse. Also, the collapse wasn't at free fall as conspiracy theorists suggest.
For more analysis of the building fall times, go to 911myths free fall page.
Please refer to Dr Frank Greening's paper for detailed calculations.
http://www.911myths.com/WTCREPORT.pdf
Italian debunker shows us more than 16 seconds to collapse. That's almost twice free fall speed from the 110th floor.
One of the more absurd arguments is the idea that there was a "Pyroclastic flow" during the collapse. This is easily debunked. You will note not one person was poached at ground zero. Pyroclastic flows are a minimum of 100C, or 212F.
http://www.youtube.com/watch?feature=player_embedded&v=omzWzMLOfUI
http://www.youtube.com/watch?feature=player_embedded&v=qLShZOvxVe4
The gas is usually at a temperature of 100-800 degrees Celsius. The flows normally hug the ground and travel downhill under gravity, their speed depending upon the gradient of the slope and the size of the flow.Not ONE person, even the ones trapped INSIDE the towers, complained of dusty air burning their skin. Trees were left green next to the towers. Paper floated around ground zero without being burned.
http://en.wikipedia.org/wiki/Pyroclastic_flow
When I brought this up to one conspiracy theorist, he produced some photos showing burning cars and such. Yet I easily found photos which show their photo was being taken out of context.
Are the cars, papers and trees in this photo made of asbestos except for the ones on fire? If you think there was a pyroclastic flow and photos of fires at ground zero is your proof then that's exactly what you must think.
It's obvious that the collapse rained paper on fire and even hot steel which could easily explain the spotty fires. Unless the pyroclastic flow hopped from one place to another.
Critical thinking skills will tell the average person there was NO pyroclastic flow but since this was brought up by a "scholar," thinking seems to be optional.
What really makes this argument absurd is the amount of explosives needed to turn that much concrete into dust. (We are only talking about 10% of the total concrete in the building anyway. There was a massive amount of gypsum as well, which conspiracy theorists would like you to forget.) The argument is the pyroclastic flow (which there is no evidence of) was created by explosives. (Some have suggested an absurd amount of thermite) If the incredible amount of POTENTIAL ENERGY (Energy the building had just standing there due to the stored energy of lifting the steel into place.) which converted to Kinetic energy (as it collapsed) is not enough to create the dust cloud, then the assumption is explosives must have created it. How much? And why would they overload the building with powerful explosives? Why put more than would be needed to cut the steel? Why put enough to cut the steel AND create a pyro show? As you can see above, the collapse released enough energy to equal 272 TONS of TNT. Why wouldn't this amount of energy be enough to cut the steel connections AND create some dust as the floors impacted each other 110 times per building?
More on the pulverization of concrete
Another absurd straw man is that they say Greening is saying the collapse weakened the steel. Nowhere in Greening's paper does it say the collapse "weakened" the steel. The massive potential energy converted to kinetic energy in the collapse and was MORE than enough to destroy the connections. No "weakening" of steel needed. The only weakening was on the fire floors which had its fireproofing blown off. This has NOTHING to do with Greening's paper.
Reader contribution:
Just a few numbers that make 9/11 conspiracies nearly impossible:
J.L. Hudson’s in Detroit, Michigan, the tallest building ever razed, was 439 ft. (26 stories)
http://www.implosionworld.com/records.htm
WTC 7 was 570 ft. (47 stories) 1.3 times the height of the J.L. Hudson. http://en.wikipedia.org/wiki/7_World_Trade_Center
WTC 1/2 was 1,368 ft. (110 stories) 3.12 times the height of J.L. Hudson.
http://en.wikipedia.org/wiki/1_World_Trade_Center
http://en.wikipedia.org/wiki/2_World_Trade_Center
So, on 9/11, three buildings were razed with perfect precision. One was 131 ft. taller than the record tower and the other two (minus cell phone antennas) were 929 ft. taller than the record holder.
The Hudson Building “It took us 24 days with 12 people doing nothing but loading explosives…” James Santoro – Controlled Demolition Incorporated"
http://www.history.com/media.do?id=most_hudsons_implosion_broadband&action=clip
Even according to the Loose Change guys, the heightened security and bomb-sniffing dogs had only been lifted for 5 days.
Of course, the construction is different and the towers would need less explosives if they were the same height. However, the towers were much taller and had more columns to cut as a result. Even if they did have the same amount of columns it would still take over 72 days with 12 people doing nothing but loading explosives. That's just one building. Add the second tower and WTC7 and you see where this is going. It quickly becomes absurd. As if this absurdly complex plan was the ONLY way to scare Americans.
I'd like to thank Slugman from Political Myths blog for his contribution.
http://politicalmythsdebunked.blogspot.com/
Y como sabes que eso es acero fundido?
Podria ser plastico, aluminio, o cualquier otra cosa.
Ademas, me estas diciendo que ahi arriba hay un tipo con un soplete?
A ver, Human, utilizando el sentido común:
- No me has respondido lo del tipo con un soplete. Si minutos antes de derrumbarse sale acero fundido de una fachada, debes tener alguna explicación lógica para ello, no? Había algun malo cortando vigas de acero ...desde dentro?
- Si lo que de verdad pretendían los malos era derrumbar las torres -con gente dentro-, y ya habían conseguido llenar las torres de explosivos, ...para qué molestarse y arriesgarse en estrellar en ellas dos aviones?
No era mucho más sencillo darle al botón rojo, que todo saltara por los aires y luego darle la culpa a Bin Laden? Estamos de acuerdo en que el efecto hubiera sido el mismo, no?
No te parece que secuestrar dos aviones y asegurarse de que se estrellaran en las torres suponía correr un riesgo absurdo e inútil, que hacía peligrar el objetivo final?