David Chandler Proves that Nothing Can Ever Collapse

[W.D.Clinger]

The equation for force due to a static load is F = mg. What do you propose the g signifies in that equation?

As many people have tried to explain to you, the g in that equation signifies the gravity-opposing acceleration vector required to offset the acceleration that would be caused by gravity in the absence of the opposing force given by the equation.

That is a consequence of the fact that the sum of the forces acting on an object at rest (or, more generally, moving at any constant velocity) is zero.

You, however, have persisted in thinking that an object at rest near the earth's surface (because some other object is supporting its weight) is being accelerated by gravity (or, possibly, is being accelerated upward by the object supporting its weight). No, Tony, an object at rest is not moving. That means its velocity is constant. That means it is undergoing zero acceleration.

Your inability to understand the previous sentences appears to be one of the root causes of your repeated but grotesquely incorrect statements saying that a 1g deceleration (that is, a 1g upward acceleration) corresponds to the static load, a 2g deceleration (that is, a 2g upward acceleration) to twice the static load, and so forth.

But who am I to diagnose your mistakes? For all I know, you're making mistakes on purpose, just as you persisted in refusing to see a tilt that was obvious to everyone else.

 

[Mel Odious]

Delete

 

[Tony Szamboti]

As many people have tried to explain to you, the g in that equation signifies the gravity-opposing acceleration vector required to offset the acceleration that would be caused by gravity in the absence of the opposing force given by the equation.

That is a consequence of the fact that the sum of the forces acting on an object at rest (or, more generally, moving at any constant velocity) is zero.

You, however, have persisted in thinking that an object at rest near the earth's surface (because some other object is supporting its weight) is being accelerated by gravity (or, possibly, is being accelerated upward by the object supporting its weight). No, Tony, an object at rest is not moving. That means its velocity is constant. That means it is undergoing zero acceleration.

Your inability to understand the previous sentences appears to be one of the root causes of your repeated but grotesquely incorrect statements saying that a 1g deceleration (that is, a 1g upward acceleration) corresponds to the static load, a 2g deceleration (that is, a 2g upward acceleration) to twice the static load, and so forth.

But who am I to diagnose your mistakes? For all I know, you're making mistakes on purpose, just as you persisted in refusing to see a tilt that was obvious to everyone else.

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.

A 2g deceleration will cause the impacting mass to apply a load of F = m x 2g which is essentially twice the static load. You can't even show the math to say it does anything but that.

You are over complicating and confusing yourself.

 

[WildCat]

Irreducible delusion.

QED

Again.

 

[rwguinn]

What's your net acceleration when sitting on the ground, floor, or chair applying a static load? And why do you think this is important to differentiation between a load that is static, and a load that is dynamic?

Well done. It ain't moving (sort of a pre-requisite for the term static)
The correct form of the equation for a static solution is SUM(F)=0.0

There are MANY instances where m*g is incorrect for a static solution. And for side loads, such as wind (Which can be treated as static to a certain extent), "g" doesn't even enter into the equation...

ETA. Damnit. Gome of you guyas gots fast fingers. You sure you're engineers amd not part of the typing pool?

 

[Grizzly Bear]

You, however, have persisted in thinking that an object at rest near the earth's surface (because some other object is supporting its weight) is being accelerated by gravity (or, possibly, is being accelerated upward by the object supporting its weight). No, Tony, an object at rest is not moving. That means its velocity is constant. That means it is undergoing zero acceleration.

Dynamic loads are time-dependent and influenced by the change in velocity over a said time period (this is the deceleration TS is talking about). He knows what it is, but he makes assumptions which render the conclusion he arrives at maddening.

First, like Chandler, Gage, Heiwa, et al, he's assuming A) the building's performing AS-BUILT, and B) that the lower section is acting against the upper mass in unison.

Secondly that the load remains uniform when the upper section mass begins its descent

He makes several others which are much more obvious that I won't get into :\
But the one above are critical; The load capacity changes significantly between a uniformly applied load and a point load. It's akin to the difference between traversing a frozen lake surface; you may be able to spread your weight out over a larger area and avoid breaking the ice, but you can't do this if you're standing with your weight acting on a smaller area.

The second one is even more important. The load isn't uniform, and it isn't axially aligned with the columns any more. Meaning a direct path to the ground is lost where the collapse is impacting the structure at any given point. This means that the structure is dismembered by a large series of localized failures, (this basically what a progressive collapse is).

Or it could be a mix of this and that. I'm sure somebody might be able to expand on what I'm getting at, and that it's already been covered at some point in time. This crap has gone through the meat shredder for several years now :\

 

[newton3376]

Yes, I believe this is just a misunderstanding.

I assumed it was since Tom rarely posts anything that I find even remotely questionable...plus he does a good job of explaining ME concepts to people like myself (an EE not an ME) who are not MEs...

Yes, velocity and acceleration components may indeed be calculated separately and summed. (When Tony S pretends to attempt that, however, he tends to forget about the 1g acceleration due to gravity.)

Nevertheless: The acceleration^WP is defined as the derivative of the velocity^WP with respect to time, or equivalently as the second derivative of position with respect to time. There are infinitely many correct ways to divide the velocity or the acceleration into components for accounting purposes, but there is only one velocity or acceleration function to which those components will add.

Hence Tony's calculations of the expected acceleration or velocity or force are invalid, because they take some of the components into account while ignoring others. For example, he has stated repeatedly that applying twice the static load requires a 2g deceleration:
http://www.internationalskeptics.com/forums/showpost.php?p=5666766&postcount=120
http://www.internationalskeptics.com/forums/showpost.php?p=5667601&postcount=133
http://www.internationalskeptics.com/forums/showpost.php?p=5668405&postcount=144

In reality, zero acceleration (constant velocity) implies the static load.
In reality, 1g deceleration implies twice the static load.
In reality, 2g deceleration implies three times the static load.

Not only has Tony refused to accept those facts, he has maligned everyone who has tried to set him straight.

I can't find anything in what you typed that I disagree with....especially the whole "2g" thing with Tony...I'm not sure why Tony isn't getting this point...

 

[Dave Rogers]

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.

A 2g deceleration will cause the impacting mass to apply a load of F = m x 2g which is essentially twice the static load. You can't even show the math to say it does anything but that.

You are over complicating and confusing yourself.

Complete garbage. If you set an item down very slowly, it is being decelerated at very much less than 1g, and will apply a dynamic load very slightly greater than mg. A 1g deceleration will require a force of 2mg, therefore the dynamic load is twice the static load. You don't understand the difference between force and acceleration. This is what everybody's been trying to tell you for months.

Dave

 

[McHrozni]

A 2g deceleration will cause the impacting mass to apply a load of F = m x 2g which is essentially twice the static load. You can't even show the math to say it does anything but that.

You are over complicating and confusing yourself.

A 2e deceleration will cause the impacting mass to apply a load of Fd = m x 2e.
The weight of the impacting mass will apply a load of Fl = m x g.

The total load, we'll call it F, it F = Fd + Fl.

Now a task for you. Knowing:

F = Fd + Fl
Fd = m x 2e
Fl = m x g
e = g

What is F?

McHrozni

 

[Tony Szamboti]

Complete garbage. If you set an item down very slowly, it is being decelerated at very much less than 1g, and will apply a dynamic load very slightly greater than mg. A 1g deceleration will require a force of 2mg, therefore the dynamic load is twice the static load. You don't understand the difference between force and acceleration. This is what everybody's been trying to tell you for months.

Dave

A 2g deceleration of a freefalling body will cause a load which is twice the static load. There is no way around that.

The first g of deceleration takes the acceleration to zero. The second g provides the dynamic effects and amplification of the load.

Force is a function of acceleration. It isn't hard to understand. Some of you guys can't get your heads unwrapped that a static load isn't moving but g is still involved in the force it applies. You simply won't admit that it is simultaneously being decelerated by the same amount by a lower structure causing equilibrium.

 

[sylvan8798]

A 2g deceleration of a freefalling body will cause a load which is twice the static load. There is no way around that.

The first g of deceleration takes the acceleration to zero. The second g provides the dynamic effects and amplification of the load.

Force is a function of acceleration. It isn't hard to understand. Some of you guys can't get your heads unwrapped that a static load isn't moving but g is still involved in the force it applies. You simply won't admit that it is simultaneously being decelerated by the same amount by a lower structure causing equilibrium.

Can you spell this out as a sum of forces = ma relationship, instead of just in words? Just pretend I'm in high school physics class.

 

[dafydd]

Can you spell this out as a sum of forces = ma relationship, instead of just in words? Just pretend I'm in high school physics class.

Would you want Tony teaching a physics class? By the way,he claims to be an engineer,does anyone know of any structures that he has worked on,I want to keep well away from them.

 

[Grizzly Bear]

You simply won't admit that it is simultaneously being decelerated by the same amount by a lower structure causing equilibrium.

For a one floor drop:

13 stories for the North tower falling on the floor below
d=distance, g=acceleration of gravity, t=time, v=velocity
d = 0.5 g x t²


Solve for time when g=9.81 m/s², d= 3.8 meters:
3.8=0.5(9.81)*t²3.8=4.905*t²3.8/4.905=t²0.775=t²(SQRT)0.775=t
0.88 Seconds = t

Solve for v:
V_F=V_I+gt
Since initial velocity is 0:
V_F=gt

G=9.81 m/s² t=0.88 seconds
V_F=9.81*0.88
V_F= 8.63 m/s

Now assume a floor has the capacity to stop the mass completely. (DeltaV=-8.63 m/s). Depth of the floor is assumed to be 18 inches or ~.46 meters. Find time:

d = 1/2 (Vi+Vf) x t
D= .46 meters Vi=8.63 m/s Vf=0 m/s t=?

.46=1/2(8.63+0)*t
.46= 4.315t
t= .46/4.135
t= 0.11 seconds

Acceleration:
A=deltaV/DeltaT
Since we count the direction of gravity as positive and the force required to stop the mass is in the negative direction dV=-8.63 m/s

dT=.11 seconds
A=-8.63/.11
A= -78.45 m/s²
or ~7.99 g's

But the floor doesn't have nearly the capacity to bring the mass to a stop. Meaning it fails before an 8g amplification is met. Let's find something closer:

p = momentum = m x v
m1 = mass of the top 15 stories
m2 = mass of the top 16 stories = aprox. (16/15) x m1
v1 = velocity before the additional mass is added = 8.63 m/s
v2 = velocity after the mass is added (unknown)

Momentum is conserved, so:
p = m1 x v1 = m2 x v2 = (31/30) x m1 x v2
Solving for v2:
V₂=V₁*(15/16)
V₂=8.63*(15/16)
V₂~ 8.09 m/s

We want to find DeltaT for an accurate calculation of the instantaneous acceleration caused by impact with the floor so:


d = 1/2 (Vi+Vf) x t
d=.46 meters Vi=8.63 m/s Vf=8.09m/s t=?

0.46=1/2(8.63-8.09)t
0.46=1/2(0.54)t
0.46=0.27t
t= 0.46/0.27
t= 1.7 seconds

A=dV/dT
dV=0.54 m/s dT=1.7 seconds

A=0.54/1.7
A= 0.31 m/s²
Since the direction of gravity is considered positive, the upwards direction will be considered negative here so the value for acceleration relative to the direction of gravity in this final value should read:

A=-0.31 m/s²
This means that the floor could never have offered enough resistance to the mass above to make it stop, even from a drop of one floor height. The net acceleration of the upper mass is still positive relative to the direction of gravity (9.81-0.31= 9.49 m/s²), so the mass is still descending.

It means while the maximum load amplification would be almost 8 times the static weight of the mass, significantly less was required for the floors to fail. The 8g figure is unattainable because the floor failed before it could get anywhere near that. Notice that the change in velocity to get 8g's was 8.63 m/s vs this later figure of a 0.54 m/s change (meaning it's still moving at 8.09 m/s after losing a little speed with the first floor impact).

The difference in these values is significant wrt the load amplification as well since acceleration is both time dependent and velocity dependent.

This calculation requires zero explosives or external factors to facilitate BTW.

If I made any errors people can feel free to pick them out. The point to demonstrate is that even after the floor impacts, not only is the mass of the next floor added, but the mass is still traveling with a net positive acceleration in the same direction as gravity.


EDIT: Wow... looks like chandlers' error is worse than I initially realized. He has a problem with conservation of momentum which easily explains the net values derived in his graphics. A physics teacher that doesn't understand conservation of momentum, bloody wonderful...

 

[tfk]

Hey BB,

I've seen Tony talk about deceleration and velocity loss before.

You'd do well not to listen to his interpretations. They tend to be unique.

Its been a while since my high school physics, (and its completely fine if I've gotten this completely wrong), but is it not correct that deceleration does not necessarily equal velocity loss i.e a reduction in speed? Doesnt it just mean that the acceleration is now less than it was?

Nope. It means the velocity is less than it was.

It's simple.

You got position.

Change in position per unit time = velocity.

Change in velocity per unit time = acceleration.

Change in acceleration per unit time = jerk.

Note that all are vectors. That means that they have a magnitude and a direction. So a change in either magnitude or direction counts as a change.

In common terms, a deceleration is considered a decrease in speed.

In technical terms, an acceleration is ANY change in velocity, whether it be increasing speed, decreasing speed or changing direction. Therefore, in physics, a deceleration is just one particular form of acceleration.

It not deceleration a reduction in acceleration, which just means the change in the increase in speed was less.

Nope. As above. A deceleration is a decrease in speed.

Or have I got this incorrect? i.e the upper block continued to accelerate, but less than before the impact event?

In the case of the upper blocks, each impact slows its instantaneous velocity.

During the descent, the blocks have three "phases":
Note: all of these velocities and accelerations are averaged, not instantaneous.
1) velocity increasing, acceleration approximately constant.
2) velocity increasing, acceleration decreasing (approaching terminal velocity).
3) velocity constant, acceleration = zero. (at terminal velocity)

Tony measured the first phase, and came up with an approximation that the acceleration was equal to a constant (=0.7G).

I can then see that, assuming the impact force is enough to collpase the impacted floor, for a small drop in acceleration, we now have the upper block start to accelerate once again, but with the added mass of another collpased floor.

You just wouldnt see the collapse "slow down", just not speed up as much.

If you were to take many, many images around the time of the impact, you actually will see a sudden drop in velocity. And therefore a large instantaneous change in acceleration.

After the impacted part broke, then the acceleration would take over again, starting from the lower (post impact) velocity.

Or, is deceleration actually a negative change in speed. Is that where I've got it wrong?

Yup.

Stick with it.

You don't have anything conceptually wrong. It's just the definition of the term "deceleration". That kind of mistake is easy to make, and easy to fix.

Conceptual mistakes are harder.


Tom

 

[tfk]

What do you mean velocities and accelerations don't add vectorily?

If I give you some velocity vector you can decompose it to its vertical and horizontal components.....adding those two together will yield the original vector....

Maybe I am misunderstanding what your meaning is here...

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

 

[W.D.Clinger]

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.

A 2g deceleration will cause the impacting mass to apply a load of F = m x 2g which is essentially twice the static load. You can't even show the math to say it does anything but that.

You are over complicating and confusing yourself.

Feckless arrogance rocks!

Having confirmed for the umpteenth time that you simply do not understand the concept of acceleration^WP, I'll hazard a guess that your difficulties result from a similar ignorance of calculus^WP. After all, it would be hard to misunderstand velocity^WP and acceleration^WP if you knew the first thing about calculus and its applications to Newton's laws of motion.

Since you challenged me to "show the math", I will illustrate the point by considering the equations of motion for one of the simplest situations in all of Newtonian mechanics: an object at rest.

Choose any Newtonian frame of reference, and consider an object at rest with respect to that frame. Since it is at rest, its position will be some constant vector s₀. Its equations of motion are:

[latex]
\begin{eqnarray}
\hbox{{\bf s}}(t) = \hbox{{\bf s}}_0 \\
\frac{d{{\bf s}}}{dt} = \hbox{{\bf 0}} \\
\frac{d^2{{\bf s}}}{dt^2} = \hbox{{\bf 0}}
\end{eqnarray}
[/latex]

Those three equations express the position, velocity, and acceleration of the object as functions of time. Note well that equation (3) says the object's acceleration is the constant function whose value is the zero vector.

Note also that those equations of motion depend only upon the fact that the object is at rest. They are independent of any forces that may be acting upon the object, so long as those forces add up to zero (as implied by the fact that the object is at rest). In particular, the equations of motion shown above still hold even if the object at rest lies within a gravitational field and its fall is being opposed by a force equal and opposite to the gravitational force. That is true even if the gravitational field is time-varying (which of course implies that the gravity-opposing force is also time-varying, else there would be moments at which the net force is nonzero and the object would move, contrary to our assumption that the object is at rest).

From the equations of motion, it is clear that the acceleration of an object at rest is zero. You, however, have regularly denied that fact, proving beyond all doubt that you do not understand the concept of acceleration^WP.

From the one post you have made in this thread since the post I quoted above, I believe you have been attempting to compensate for your ignorance of calculus by developing your very own idiosyncratic view of physics along with an incorrect terminology that might make sense to you but leads to confusion when you try to explain yourself to others. Your home-brewed theory of physics may also have contributed to some of the substantive errors you have been making in your calculations.

 

[rwguinn]

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.

^^ what he said

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

Tom

A condition not unlike adding airspeed and wingspan together to get fireball diameter...

 

[Newtons Bit]

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.

We need a special award for statements like this. Gravity is a force, not an acceleration.

If you set an object down very, very, VERY slowly, it has an acceleration of zero.

 

[sylvan8798]

I don't understand how Tony Z. can possibly have any credibility if he can't grasp something from Physics I.

But then, we're talking about truthers here.

 

[Dave Rogers]

A 2g deceleration of a freefalling body will cause a load which is twice the static load. There is no way around that.

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.