David Chandler Proves that Nothing Can Ever Collapse

[tfk]

You are wrong. If you set an item down very slowly it is being decelerated at 1g and will apply a static load. The static load involves the first g.

Wrong.

When you "set an item down very slowly", its velocity is close to zero over the interval that you are setting it down. This velocity doesn't change much at all during this process.

Since, by definition, acceleration is the change in velocity per unit time, its acceleration is approximately zero too.

How many times...?


Tom

PS. Tony, something being "set down at 1 g" is NOT something that is "set down gently". It is something that is "dropped". It does not matter if it is dropped from .1", 10' or 1000'. Each of those objects are being "set down at 1 g".

According to your BS, those poor, desperate people who jumped out of the towers were "set down gently at 1 g".

Your definition sux.

 

[newton3376]

Newt,

Sorry it took awhile to get back. Damn real world...

Yeah, a bad choice of words on my part.

Position, velocity & acceleration are all vectors. And like all vectors, you can add, subtract, multiply & divide them, like all vectors.

But any given rigid object (treated as a single piece) that is translating (i.e., no rotation) in 3D space can have only one instantaneous velocity and acceleration.

[And it can have only one average velocity & average acceleration too. As long as you average over the same interval of time.]

You can "decompose" that velocity or acceleration into components in any number of ways. Typical example: if you're using {x, y, z} coordinates, then you decompose the position, velocity and acceleration into components along those axes. If you're using cylindrical or spherical coordinates, then you decompose into {r, Ø, Ω} or {r, Ø, z} coordinates.

In each case, the components of the vectors will be different, but when you add them all up correctly, you get the identical answer. (As long as you've got your coordinates tied to the same frame of reference.)

My point with BB was that attempting to add up component accelerations can lead to confusion. It confuses causes & effects.

The (multiple) forces are the causes. The (single) acceleration that results from those forces is the effect.

Tony's attempts to add or subtract velocity from acceleration is a sure fired path to massive confusion.

Tom

Tom,

Thanks for the reply....I figured I was likely misunderstanding what you were saying...and thats what it turned out to be.

I follow your explanation above and it makes perfect sense.

 

[tfk]

For anyone confused about the difference between "g" (gravitational acceleration) and "a" (an object's acceleration)...

What "g" really means:

The "g" in Weight = mass x g is NOT an acceleration. It is the proportionality constant between mass & weight for an object that is stationary (i.e., v = 0, a = dv/dt = 0) on the surface of some planetary body.

An object has a mass of 1 kg. This mass is the result of the sum of all the little masses of the atoms that make up the object. This mass does not change for this object anywhere in the universe.

On earth, it takes 9.8 newtons of force to stop this body from accelerating towards the center of the earth. Therefore, to any supporting structure that is trying to prevent its fall, it "weighs" 9.8 newtons. And it weighs this amount while it is stationary (i.e., no velocity, no acceleration).

On the moon, it takes 1.6 newtons of force to stop this body from accelerating towards the center of the moon. Therefore, to any supporting structure that is trying to prevent its fall, it "weighs" 1.6 newtons.

"g" is a direct reflection of the mass & diameter of the attracting body. It is not a reflection of the object's acceleration.

The matter is made muddier by the fact that sometimes it is convenient to express an object's real acceleration is given in proportion to its static weight (i.e., in "g"s). Ah well...


Tom

 

[rwguinn]

You are functionally innumerate. A 2g deceleration of a falling body subject to gravity requires an additional applied force of 3mg, and if this is applied by a struicture below it comes to three times the static load. Go back to high school.

Dave

Not even.
a 2 g acceleration applied opposite the g vector will yield a 32.2 ft/sec velocity reversal for each second that it is applied.
Time is important.
An object on Earth falls for 1 second, unimpeded.
It has fallen 16.1 feet. It is moving at 32.2 feet/second.
Apply 1 g acceleration opposite gravity.
It is now, after 2 seconds, moving at 32.2 feet/second, and has moved an additional 32.2 feet.
after 3 seconds, it is still moving at 32.2 feet/second.

If, at 1 second, you apply a 2g vector opposite gravity, after 2 seconds, it will have stopped moving, and will have traveled an additional 16.1 feet downward.

ETA: The biggest difficulty I have in replying to Tony is sheer incredulity that anyone could possibly work as an engineer, in any field, while possessing a degree of ignorance so profound that he completely misunderstands simple mechanics. It's quite a relief to see that I'm not the only one who sees this, because it's almost easier to believe that I'm delusional than that Tony is saying things this idiotic.

^^

 

[newton3376]

Not even.
a 2 g acceleration applied opposite the g vector will yield a 32.2 ft/sec velocity reversal for each second that it is applied.
Time is important.
An object on Earth falls for 1 second, unimpeded.
It has fallen 16.1 feet. It is moving at 32.2 feet/second.
Apply 1 g acceleration opposite gravity.
It is now, after 2 seconds, moving at 32.2 feet/second, and has moved an additional 32.2 feet.
after 3 seconds, it is still moving at 32.2 feet/second.

If, at 1 second, you apply a 2g vector opposite gravity, after 2 seconds, it will have stopped moving, and will have traveled an additional 16.1 feet downward.

^^

yeah good point......you really do have to take into account the time period that any and all forces are actually applied.....

 

[Dave Rogers]

Not even.
a 2 g acceleration applied opposite the g vector will yield a 32.2 ft/sec velocity reversal for each second that it is applied.

Yes, but that's a constant deceleration.

If an object is falling at some non-zero velocity under 1g of gravity, then an additional force of 3mg is required to decelerate it at 2g (i.e. d2x/dt^2 = -19.62 m s^-2, which is the only correct definition of a 2g deceleration). The total force on the object of mass m is +3mg-mg, where positive force is defined as upwards, for a resultant force of +2mg which gives an upward acceleration of +2g. If the initial motion is downwards then this is a deceleration.

Tony persists in believing that a static object, for which d2x/dt^2=dx/dt=0, is accelerating upwards at 9.81ms^-2. This is nonsense, and a fundamental misunderstanding of simple mechanics.

Dave

 

[rwguinn]

Yes, but that's a constant deceleration.

If an object is falling at some non-zero velocity under 1g of gravity, then an additional force of 3mg is required to decelerate it at 2g (i.e. d2x/dt^2 = -19.62 m s^-2, which is the only correct definition of a 2g deceleration). The total force on the object of mass m is +3mg-mg, where positive force is defined as upwards, for a resultant force of +2mg which gives an upward acceleration of +2g. If the initial motion is downwards then this is a deceleration.

Tony persists in believing that a static object, for which d2x/dt^2=dx/dt=0, is accelerating upwards at 9.81ms^-2. This is nonsense, and a fundamental misunderstanding of simple mechanics.

Dave

Not disagreeing with you on Tony's incompetence. That is why I was very specific in my terminology.

 

[Cl1mh4224rd]

Force is a function of acceleration.

I'm pretty sure you've got that backwards.

 

[Tony Szamboti]

deleted

 

[Tony Szamboti]

You are functionally innumerate. A 2g deceleration of a falling body subject to gravity requires an additional applied force of 3mg, and if this is applied by a struicture below it comes to three times the static load. Go back to high school.

Dave

ETA: The biggest difficulty I have in replying to Tony is sheer incredulity that anyone could possibly work as an engineer, in any field, while possessing a degree of ignorance so profound that he completely misunderstands simple mechanics. It's quite a relief to see that I'm not the only one who sees this, because it's almost easier to believe that I'm delusional than that Tony is saying things this idiotic.

The situation you are describing is for an arrested fall, although you don't mention that. My point was concerning an impact that results in collapse as we see in the towers.

This will end up just like it did with others here who tried to sound knowledgable while saying I was mistaken about the actual in-service factor of safety of the central cores of the towers being 3.00 to 1. They kept saying it was less than 2.00 to 1 at about 1.67 to 1.92, but that was for worst case design loads which were not the actual loads in the twin towers. So it turns out they were wrong.

 

[sylvan8798]

The situation you are describing is for an arrested fall, although you don't mention that. My point was concerning an impact that results in collapse as we see in the towers.

This will end up just like it did with others here who tried to sound knowledgable while saying I was mistaken about the actual in-service factor of safety of the central cores of the towers being 3.00 to 1. They kept saying it was less than 2.00 to 1 at about 1.67 to 1.92, but that was for worst case design loads which were not the actual loads in the twin towers. So it turns out they were wrong.

How interesting that you seem to refuse to use actual equations to illustrate your assertions.

Get your equations here, folks! Get 'em while they're hot! They won't last! Mister, hey mister! How about some equations for the little lady and the gang?

 

[Grizzly Bear]

The situation you are describing is for an arrested fall, although you don't mention that. My point was concerning an impact that results in collapse as we see in the towers.

This will end up just like it did with others here who tried to sound knowledgable while saying I was mistaken about the actual in-service factor of safety of the central cores of the towers being 3.00 to 1. They kept saying it was less than 2.00 to 1 at about 1.67 to 1.92, but that was for worst case design loads which were not the actual loads in the twin towers. So it turns out they were wrong.

I did the math right here if you want to show me anything I got wrong. But it looks to me like it puts the values in Chandlers' graphics into context; and explains pretty well where you guys are faltering, other than the obvious ..

 

[Tony Szamboti]

How interesting that you seem to refuse to use actual equations to illustrate your assertions.

Get your equations here, folks! Get 'em while they're hot! They won't last! Mister, hey mister! How about some equations for the little lady and the gang?

I have given the equations in this thread. Get off your behind there little lady and go look for them instead of falsely complaining that they aren't given.

Dave Rogers is adding in the weight of the mass after impact to get more than I show in the equations I gave.

 

[Dave Rogers]

The situation you are describing is for an arrested fall, although you don't mention that. My point was concerning an impact that results in collapse as we see in the towers.

Wrong. A deceleration of 19.62ms^-2 of an object falling under gravity requires a 3g applied force, whatever the situation. You're confusing force with acceleration.

Dave

 

[Dave Rogers]

Dave Rogers is adding in the weight of the mass after impact to get more than I show in the equations I gave.

It looks to me like you're neglecting the weight of the mass after impact. What makes you think it goes away?

Dave

 

[Tony Szamboti]

It looks to me like you're neglecting the weight of the mass after impact. What makes you think it goes away?

Dave

If the collapse continues there is no structure to provide a reaction for the mass to act on and the deceleration of 2g is indicative of an applied force twice the static weight.

To get your 3mg force being applied by the lower structure you are adding in an extra g for the weight of the mass to be supported, which is not there if the structure below is collapsed, and that should be clear to anyone with at least high school physics training.

 

[alienentity]

Tony, you're being badly pwned by the math here, here, here, here

just on one page of this thread.

You need to respond to those challenges demonstrating with your own math where they are in error or withdraw your contentions. At least that's what a man would do.

You make a lot of accusations Tony, big, big powerful and dangerous accusations. But it looks like you're dead wrong. Perhaps a little humility and and some apologies are in order...just sayin'.

 

[Tony Szamboti]

Tony, you're being badly pwned by the math here, here, here, here

just on one page of this thread.

You need to respond to those challenges demonstrating with your own math where they are in error or withdraw your contentions. At least that's what a man would do.

You make a lot of accusations Tony, big, big powerful and dangerous accusations. But it looks like you're dead wrong. Perhaps a little humility and and some apologies are in order...just sayin'.

I showed the math supporting what I am saying in this thread. To suggest otherwise is simply false. There is no need for me to apologize when what I have said is accurate. Reality is what it is and I am only calling it like it is.

Essentially the point here is that there was no deceleration and velocity loss by the upper section of WTC 1 during its fall and in a natural collapse there would have been significant deceleration and velocity loss during impact. The real danger is in not re-investigating those collapses and finding out who planted demolition devices in them, as it certainly was not the aircraft impacts and fires which brought them down.

Any apologies here should be coming from those who have resorted to the use of ridicule in lieu of honest reality based discussion.

 

[TruthersLie]

So tony...

you have your paper through peer review right?

What will you do when the peer reviewers tell you the same things you are getting here for free? Throw a tantrum, or do what they suggest and fix your errors?

 

[Tony Szamboti]

So tony...

you have your paper through peer review right?

What will you do when the peer reviewers tell you the same things you are getting here for free? Throw a tantrum, or do what they suggest and fix your errors?

Actually, Dr. Frank Greening has admitted that both he and Dr. Zdenek Bazant reviewed and discussed the paper and did not find any errors which would affect the thesis of the paper.

The vast majority of the criticisms I see here are in error and often founded in simple disbelief with no basis in reality.