- #1
vteclimey
Hello all,
New member here, I am in need of someone familiar with Physics to explain something to me.
Its 9-11 conspiracy related so apologies to anyone who is sick of that subject but a someone has posted a stack of formula in response to a comment of mine and I would like someone who knows this stuff to explain to me what they refer to and why they might or might not be wrong.
The post is very long so I apologise (again) in advance, for context it is referencing the North (I think) WTC tower, which collapsed first.
The comment I refer to now follows in its entirety ...
-----------------
Dear Mr. Limey,
Your original posting to Mr. McKee was during the period you were being rather obtuse. Now that you brought it to his attention, he writes:
I’m not sure why this comment wasn’t approved at the time it was sent. Given that it addresses the science of the issue and isn’t a personal attack, it seems that it should have been approved at the time. I will be posting a response to this comment when I have the time to go through it again.
I responded to you there.
But here it is again for your loyal readers.
Dear Mr. Limey,
As probably your one and only subscriber, I see that your grasp of physics as it applies to 9/11 has not improved any from December of 2010 until today September 2011.
Without equations, your grasp of physics is all hand-wavy, and even pseudo in nature. As such, I’m not going to waste my time within your internal “fluff”. I’ll go right to your conclusions.
Mr. Limey wrote in his conclusion:
Given that the collapse was slower than free fall speed, its obvious that the floors offered some resistance. After all, no resistance would have meant free fall speed. Once momentum has been established the fall will continue until it meets something strong enough to withstand it. In this case it was the ground.
Had the vertical supports and floors offered continual resistance as expected by an intact structure, the demolition would not have been (slightly) slower than free fall. It would have been significantly slower than free fall, if not arrested outright at some point in the upper floors.
Your entire “momentum” sentence is misleading and ignores significant factors, including ejection of material laterally and pulverization of content, both of which reduce the momentum and left-over energy to affect the lower portion of the towers.
Momentum is defined as mass times velocity and is conserved during both elastic and inelastic collisions:
P = M1 * V1 + M2 * V2 = (M1 + M2) * V3 {inelastic}
P = M1 * V1 + M2 * V2 = M1 * V1′ + M2 * V2′ (elastic; V1V1′, V2V2′}
The theory of the alleged pile driver rests on inelastic collisions, whereby mass is accumulated with each collision and V2=0.
For the sake of understanding and discussion, I’ll step readers through the real physics in a simplified fashion that purposely does not take into consideration all factors observable in the collapse or known to be inherent in the structure. The simplified physics will debunk your assertions, which will be even more in error once those more complex mechanisms are brought into the equation.
Let’s make the following assumptions:
(a) The upper block had a mass M1.
(b) Each individual floor had a mass M2.
(c) The alleged plane impact (or weak) point was 20 floors from the top. Thus, the starting “pile driver” M1=20*M2 (or M2=M1/20).
(d) Each individual floor was hanging in mid-air without any support below it. (e) Distance between each floor (to make the math easy) 9.8 meters. [Actual distance was probably 1/3 of that.]
(f) The over-design factor in the static weight that any given level N can support is 2*Mass(110-N) [although I think is was bigger than this.]
D is distance (between floors)
g is gravity [9.8 m/(s^2)]
V0 is initial velocity (V0=0)
t is time
D = (1/2) * g * t^2 + V0 * t
9.8 m = (1/2) * [9.8 m/(s^2)] * t^2
Solve for t to find out how long it took M1 to free-fall the spacing of one floor into M2.
t=2^(1/2)=sqrt(2)=1.414 s
Derivative of acceleration at a point in time is velocity:
V1 = a * t = 9.8 * 1.414 = 13.8 m/s
Thus, before hitting M2, M1 reached a velocity V1=13.8 m/s.
M2 is stationary, so its velocity is V2=0. Thus momentum P before impact is:
P = M1 * V1 + M2 * V2 = M1 * V1 (because V2=0)
Conservation of momentum in this very ideal example of an inelastic collision (masses combine rather than bounce) says:
P = M1 * V1 = (M1 + M2) * V3
Re-arrange terms and plug in for V1, you get:
V3 = V1 * [M1/(M1 + M2)] = (13.8) * [M1/(M1 + M2)]
Because M1 & M2 are greater than zero, the new velocity V3 of the combined mass has to be less than the initial velocity V1 which was equivalent to free-fall. Similar calculations can be performed for subsequent levels, plugging in a newly calculated velocity of the combined & growing mass at impact with that level.
In this ideal world with floors of mass M2 just hanging in the air, we can further assume that the starting “pile driver” M1=20*M2 (or M2=M1/20) under the premise that one of the alleged airplane impacts was 20 floors from the top.
V3 = V1 * [M1/(M1 + (M1/20))] = V1 * [1/1.05] = V1 * (0.95)
In this ideal world example, we see quantitatively how the speed of the pile driver hitting a floor M2 hanging in mid-air slows down slightly from its previous free-fall velocity.
However, M2 was not just hanging in mid-air. Using the assumed over-design factor of 2, the vertical support offered by the steel shell and the inner core at level M2
Force(M2) = [Over-Design Factor] * M1 = 2 * M1
In order for floor M2 to fail, V1*M1 has to also be greater than 2*M1. In this example, it is. However, the momentum equation needs to be updated to account for the consumption of energy in smashing floor M2 and its supporting structures.
P = M1 * V1 = (M1 + M2) * V3 + [2 * M1]
V3 = [(M1 * V1) - (2 * M1)]/[M1 + M2] = [V1 - 2] * [M1/(M1 + M2)]
V3 = [(13.8) - 2] * [M1/(M1 + M2)] = (11.8) * [M1/(M1 + M2)]
V3 = (11.8) * [M1/(M1 + (M1/20))] = (11.8) * [1/1.05]
V3 = 11.23 m/s < 13.8 m/s = V1
Thus we see after impact with the first M2 floor taking into consideration the supports under M2, that velocity of the pile driver M1 should be measurably less than the velocity of free-fall at that point.
The ejection of content laterally reduces the mass of M1 to M1′. Also the Force of material ejection must be accounted for in the momentum equation P in a manner similar to the over-design factor. More importantly, the force of content pulverization is another massive energy sink that also gets inserted into the momentum equation P in a manner similar to the over-design factor.
P = M1 * V1 = (M1 + M2) * V3 + Force(M2) + Force(ejection) + Force(pulverization)
V3 = [(M1*V1) - Force(M2) - Force(ejection) - Force(Pulverization)]/[M1'+M2]
Revisiting my assumptions, if the assumed static-load over-design factor is low and a higher factor is employed, the resulting V3 will be reduced further. Similarly, using more accurate floor spacing will factor in by reducing the available M1*V1 energy at impact with floor M2.
Anyway you slice the above, the destruction of the towers should have been measurably and significantly slower than free-fall. It wasn’t.
Mr. Limey wrote:
Each floor failed because it was impacted with more weight than it could hold up.
This is actually mal-framed. This assumes that the “accumulating pile driver” went through the sequence: “(gravity) accelerate down the distance of one floor; slow down upon impacting new floor; (gravity) accelerate down the distance of another floor; …”
In reality, the outer shell and inner core offered continual support and resistance to collapse (not incremental at each floor). Thus, the mass in the middle and edges didn’t even have the advantage of repeated periods of (gravity) accelerating down the distance of one floor. Only stuff between the core and edges might do that.
Thus it is pseudo-science to say “each floor failed.” No, the inner core and outer-structure seemingly failed because supposedly:
M1*V1 > Force(M2) + Force(ejection) + Force(pulverization)
Yet, how could we even have a sufficient V1 from falling, unless something removed the continual support within the core and edges? In fact, they did such a good job of support removal, collapse speed is slower than but close to free-fall.
Mr. Limey wrote:
To explain the collapse does not require any magic, imaginary physics or even explosives.
The imaginary physics (or pseudo-science) that you speak of must be your own.
Explaining the collapse does require using real physics and real observation of the event. You state: Given that the collapse was slower than free fall speed. How much slower than free-fall speed was it? Observation proves that it was slightly slower than free-fall, while real physics suggests that it should be significantly and measurably slower.
It wasn’t. Thus, real physics returns to those equations and the observations. It says in order for collapse speed to remain fast (near free-fall), energy would have to be added to negate the energy sinks represented by:
[Force(M2) + Force(ejection) + Force(pulverization)]
How do you add energy to this system, Mr. Limey?
Not just explosives will do it. DEW will do it, and without too many tell-tail audio decibel signatures of detonation that your heroes of NIST confidently stated weren’t present. How do you power DEW? Maybe a mini-cold-fusion or mini-nuclear reactor were deployed like a portable generator. This would then match measurements of radiation signatures that didn’t match known nuclear weapons but still suggest nuclear type activity.
New member here, I am in need of someone familiar with Physics to explain something to me.
Its 9-11 conspiracy related so apologies to anyone who is sick of that subject but a someone has posted a stack of formula in response to a comment of mine and I would like someone who knows this stuff to explain to me what they refer to and why they might or might not be wrong.
The post is very long so I apologise (again) in advance, for context it is referencing the North (I think) WTC tower, which collapsed first.
The comment I refer to now follows in its entirety ...
-----------------
Dear Mr. Limey,
Your original posting to Mr. McKee was during the period you were being rather obtuse. Now that you brought it to his attention, he writes:
I’m not sure why this comment wasn’t approved at the time it was sent. Given that it addresses the science of the issue and isn’t a personal attack, it seems that it should have been approved at the time. I will be posting a response to this comment when I have the time to go through it again.
I responded to you there.
But here it is again for your loyal readers.
Dear Mr. Limey,
As probably your one and only subscriber, I see that your grasp of physics as it applies to 9/11 has not improved any from December of 2010 until today September 2011.
Without equations, your grasp of physics is all hand-wavy, and even pseudo in nature. As such, I’m not going to waste my time within your internal “fluff”. I’ll go right to your conclusions.
Mr. Limey wrote in his conclusion:
Given that the collapse was slower than free fall speed, its obvious that the floors offered some resistance. After all, no resistance would have meant free fall speed. Once momentum has been established the fall will continue until it meets something strong enough to withstand it. In this case it was the ground.
Had the vertical supports and floors offered continual resistance as expected by an intact structure, the demolition would not have been (slightly) slower than free fall. It would have been significantly slower than free fall, if not arrested outright at some point in the upper floors.
Your entire “momentum” sentence is misleading and ignores significant factors, including ejection of material laterally and pulverization of content, both of which reduce the momentum and left-over energy to affect the lower portion of the towers.
Momentum is defined as mass times velocity and is conserved during both elastic and inelastic collisions:
P = M1 * V1 + M2 * V2 = (M1 + M2) * V3 {inelastic}
P = M1 * V1 + M2 * V2 = M1 * V1′ + M2 * V2′ (elastic; V1V1′, V2V2′}
The theory of the alleged pile driver rests on inelastic collisions, whereby mass is accumulated with each collision and V2=0.
For the sake of understanding and discussion, I’ll step readers through the real physics in a simplified fashion that purposely does not take into consideration all factors observable in the collapse or known to be inherent in the structure. The simplified physics will debunk your assertions, which will be even more in error once those more complex mechanisms are brought into the equation.
Let’s make the following assumptions:
(a) The upper block had a mass M1.
(b) Each individual floor had a mass M2.
(c) The alleged plane impact (or weak) point was 20 floors from the top. Thus, the starting “pile driver” M1=20*M2 (or M2=M1/20).
(d) Each individual floor was hanging in mid-air without any support below it. (e) Distance between each floor (to make the math easy) 9.8 meters. [Actual distance was probably 1/3 of that.]
(f) The over-design factor in the static weight that any given level N can support is 2*Mass(110-N) [although I think is was bigger than this.]
D is distance (between floors)
g is gravity [9.8 m/(s^2)]
V0 is initial velocity (V0=0)
t is time
D = (1/2) * g * t^2 + V0 * t
9.8 m = (1/2) * [9.8 m/(s^2)] * t^2
Solve for t to find out how long it took M1 to free-fall the spacing of one floor into M2.
t=2^(1/2)=sqrt(2)=1.414 s
Derivative of acceleration at a point in time is velocity:
V1 = a * t = 9.8 * 1.414 = 13.8 m/s
Thus, before hitting M2, M1 reached a velocity V1=13.8 m/s.
M2 is stationary, so its velocity is V2=0. Thus momentum P before impact is:
P = M1 * V1 + M2 * V2 = M1 * V1 (because V2=0)
Conservation of momentum in this very ideal example of an inelastic collision (masses combine rather than bounce) says:
P = M1 * V1 = (M1 + M2) * V3
Re-arrange terms and plug in for V1, you get:
V3 = V1 * [M1/(M1 + M2)] = (13.8) * [M1/(M1 + M2)]
Because M1 & M2 are greater than zero, the new velocity V3 of the combined mass has to be less than the initial velocity V1 which was equivalent to free-fall. Similar calculations can be performed for subsequent levels, plugging in a newly calculated velocity of the combined & growing mass at impact with that level.
In this ideal world with floors of mass M2 just hanging in the air, we can further assume that the starting “pile driver” M1=20*M2 (or M2=M1/20) under the premise that one of the alleged airplane impacts was 20 floors from the top.
V3 = V1 * [M1/(M1 + (M1/20))] = V1 * [1/1.05] = V1 * (0.95)
In this ideal world example, we see quantitatively how the speed of the pile driver hitting a floor M2 hanging in mid-air slows down slightly from its previous free-fall velocity.
However, M2 was not just hanging in mid-air. Using the assumed over-design factor of 2, the vertical support offered by the steel shell and the inner core at level M2
Force(M2) = [Over-Design Factor] * M1 = 2 * M1
In order for floor M2 to fail, V1*M1 has to also be greater than 2*M1. In this example, it is. However, the momentum equation needs to be updated to account for the consumption of energy in smashing floor M2 and its supporting structures.
P = M1 * V1 = (M1 + M2) * V3 + [2 * M1]
V3 = [(M1 * V1) - (2 * M1)]/[M1 + M2] = [V1 - 2] * [M1/(M1 + M2)]
V3 = [(13.8) - 2] * [M1/(M1 + M2)] = (11.8) * [M1/(M1 + M2)]
V3 = (11.8) * [M1/(M1 + (M1/20))] = (11.8) * [1/1.05]
V3 = 11.23 m/s < 13.8 m/s = V1
Thus we see after impact with the first M2 floor taking into consideration the supports under M2, that velocity of the pile driver M1 should be measurably less than the velocity of free-fall at that point.
The ejection of content laterally reduces the mass of M1 to M1′. Also the Force of material ejection must be accounted for in the momentum equation P in a manner similar to the over-design factor. More importantly, the force of content pulverization is another massive energy sink that also gets inserted into the momentum equation P in a manner similar to the over-design factor.
P = M1 * V1 = (M1 + M2) * V3 + Force(M2) + Force(ejection) + Force(pulverization)
V3 = [(M1*V1) - Force(M2) - Force(ejection) - Force(Pulverization)]/[M1'+M2]
Revisiting my assumptions, if the assumed static-load over-design factor is low and a higher factor is employed, the resulting V3 will be reduced further. Similarly, using more accurate floor spacing will factor in by reducing the available M1*V1 energy at impact with floor M2.
Anyway you slice the above, the destruction of the towers should have been measurably and significantly slower than free-fall. It wasn’t.
Mr. Limey wrote:
Each floor failed because it was impacted with more weight than it could hold up.
This is actually mal-framed. This assumes that the “accumulating pile driver” went through the sequence: “(gravity) accelerate down the distance of one floor; slow down upon impacting new floor; (gravity) accelerate down the distance of another floor; …”
In reality, the outer shell and inner core offered continual support and resistance to collapse (not incremental at each floor). Thus, the mass in the middle and edges didn’t even have the advantage of repeated periods of (gravity) accelerating down the distance of one floor. Only stuff between the core and edges might do that.
Thus it is pseudo-science to say “each floor failed.” No, the inner core and outer-structure seemingly failed because supposedly:
M1*V1 > Force(M2) + Force(ejection) + Force(pulverization)
Yet, how could we even have a sufficient V1 from falling, unless something removed the continual support within the core and edges? In fact, they did such a good job of support removal, collapse speed is slower than but close to free-fall.
Mr. Limey wrote:
To explain the collapse does not require any magic, imaginary physics or even explosives.
The imaginary physics (or pseudo-science) that you speak of must be your own.
Explaining the collapse does require using real physics and real observation of the event. You state: Given that the collapse was slower than free fall speed. How much slower than free-fall speed was it? Observation proves that it was slightly slower than free-fall, while real physics suggests that it should be significantly and measurably slower.
It wasn’t. Thus, real physics returns to those equations and the observations. It says in order for collapse speed to remain fast (near free-fall), energy would have to be added to negate the energy sinks represented by:
[Force(M2) + Force(ejection) + Force(pulverization)]
How do you add energy to this system, Mr. Limey?
Not just explosives will do it. DEW will do it, and without too many tell-tail audio decibel signatures of detonation that your heroes of NIST confidently stated weren’t present. How do you power DEW? Maybe a mini-cold-fusion or mini-nuclear reactor were deployed like a portable generator. This would then match measurements of radiation signatures that didn’t match known nuclear weapons but still suggest nuclear type activity.