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David Chandler Proves that Nothing Can Ever Collapse

tom.

I hate to nitpick.. but I think you mean mass instead of weight.

I know that Tony has been nailed for it. Just letting you know

ETA: BTW tom, it is an excellent explaination.
 
tom.

I hate to nitpick.. but I think you mean mass instead of weight.

I know that Tony has been nailed for it. Just letting you know

ETA: BTW tom, it is an excellent explaination.
Weight is a force--it is the force the gravitational acceleration exerts on the mass of the block.
This force is acting on the lower (undamaged) block, also.
 
Seymour,


Is this explanation (of the acceleration & forces) clear?

Tom

Yep.

Although to be more clear, I should have said, instead of "deceleration", used "deceleration vector".??????

And then pointed out that the sum of the whole - acceleration vector minus deceleration vector - can still result in net positive acceleration, until, as you pointed out, equilibrium is reached?????
 
tom.

I hate to nitpick.. but I think you mean mass instead of weight.

I know that Tony has been nailed for it. Just letting you know

ETA: BTW tom, it is an excellent explaination.


Hey TL,

rwguinn got it right.

Whenever anyone is talking about force, then weight is the appropriate term to use.

Do you get the difference between the two?

Note carefully the use of the word "acceleration". (Not velocity.)

Mass is an "inherent property" of matter. Matter can never get rid of its mass. An object in outer space has zero weight but the same mass that it would if it were sitting on the earth.

A gravitational force is applied to matter by the mass of the earth. If that force is unopposed (as in a free falling object), then the object will be accelerated towards the earth.

F = m a. There is only one "F" on the object (gravity). Therefore the "m" has to "a".

Near the surface of the earth, this value for "a" is equivalent to "g", if one falls without air resistance.

Weight is the force that one has to impart onto an object to PREVENT it from accelerating towards the center of the earth. Weight only appears when that downward acceleration is decreased.

If you bring the downward acceleration to zero, as in someone standing or sitting, then the full "weight" appears (as the term is commonly used).

Now watch the difference between acceleration & velocity.

If you were standing on a bathroom scale in an elevator, and it is descending at a very high constant rate of speed, do you weigh less?

The answer is "no". If the elevator is at a constant speed, then your acceleration is zero. It doesn't matter how fast you're going. If your speed is not changing, then your acceleration is zero.
(Little correction, if your speed and direction are not changing...)

Now, during those first gut disturbing seconds while the elevator is ACCELERATING to that speed, that's when you weigh less. And the scale will show it.
___

While standing on a scale, someone weighs W lbs.

If his downward acceleration is 0.0G's, he "weighs" 1.0*W.
If his downward acceleration is 0.2G's, he "weighs" only 0.8*W.
If his downward acceleration is 0.8G's, he "weighs" only 0.2*W.
If his downward acceleration is 1.0G's, he "weighs" only 0.0*W.

Note that in all cases, his velocity can be anything. Using the elevator, if the elevator is going up, stopped or going down at a constant speed, then the velocity will be positive positive, zero and negative respectively. But in all cases, the acceleration is zero. And a person on a scale will show his stationary weight, W.

In a similar fashion, a person's instantaneous velocity can be upwards, zero or downwards. His acceleration is the CHANGE in that velocity. So the acceleration can be positive (velocity increasing), zero (velocity remains the same) or negative (velocity decreasing).

TL, I get the feeling from your posts that you've got this idea pretty well. This is for anyone still struggling with the difference between weight & mass.

Too many words?

Tom
 
Yep.

Although to be more clear, I should have said, instead of "deceleration", used "deceleration vector".??????

And then pointed out that the sum of the whole - acceleration vector minus deceleration vector - can still result in net positive acceleration, until, as you pointed out, equilibrium is reached?????

Well, here's a bit of a correction.

Forces add (vectorily).

Velocity & accelerations don't.

A solid object can only have one velocity & one acceleration. It can have lots of forces.

So the right way to do these analyses is to add up all the various forces, and then calculate the single resultant acceleration and/or velocity.
 
Well, here's a bit of a correction.

Forces add (vectorily).

Velocity & accelerations don't.

A solid object can only have one instantaneous velocity & one instantaneous acceleration. It can have lots of forces.

So the right way to do these analyses is to add up all the various forces, and then calculate the single resultant instantaneous acceleration and/or velocity.
Let's be precise, here. :D
 
Mass is an "inherent property" of matter. Matter can never get rid of its mass. An object in outer space has zero weight but the same mass that it would if it were sitting on the earth.

Tell that to Richard Gage, AIA. One of my favorite quotes from him:

http://screwloosechange.blogspot.com/2008/03/physics-of-richard-gage.html

Richard Gage said:
In the case of the North Tower, the 15 story section above the point of jet plane impact, you see it decreasing by half before it ever begins to move down. In other words, half of that mass, which we're told was so great, is decreased, destroyed, disintegrated by these explosions, and then within two seconds the rest of it is completely gone such that you don't have this 15 story mass crushing the rest of the building and its 80,000 tons of structural steel designed to resist this load for the life of that building.
 
Hey TL,

rwguinn got it right.

Whenever anyone is talking about force, then weight is the appropriate term to use.

Do you get the difference between the two?

Note carefully the use of the word "acceleration". (Not velocity.)

Mass is an "inherent property" of matter. Matter can never get rid of its mass. An object in outer space has zero weight but the same mass that it would if it were sitting on the earth.

A gravitational force is applied to matter by the mass of the earth. If that force is unopposed (as in a free falling object), then the object will be accelerated towards the earth.

F = m a. There is only one "F" on the object (gravity). Therefore the "m" has to "a".

Near the surface of the earth, this value for "a" is equivalent to "g", if one falls without air resistance.

Weight is the force that one has to impart onto an object to PREVENT it from accelerating towards the center of the earth. Weight only appears when that downward acceleration is decreased.

If you bring the downward acceleration to zero, as in someone standing or sitting, then the full "weight" appears (as the term is commonly used).

Now watch the difference between acceleration & velocity.

If you were standing on a bathroom scale in an elevator, and it is descending at a very high constant rate of speed, do you weigh less?

The answer is "no". If the elevator is at a constant speed, then your acceleration is zero. It doesn't matter how fast you're going. If your speed is not changing, then your acceleration is zero.
(Little correction, if your speed and direction are not changing...)

Now, during those first gut disturbing seconds while the elevator is ACCELERATING to that speed, that's when you weigh less. And the scale will show it.
___

While standing on a scale, someone weighs W lbs.

If his downward acceleration is 0.0G's, he "weighs" 1.0*W.
If his downward acceleration is 0.2G's, he "weighs" only 0.8*W.
If his downward acceleration is 0.8G's, he "weighs" only 0.2*W.
If his downward acceleration is 1.0G's, he "weighs" only 0.0*W.

Note that in all cases, his velocity can be anything. Using the elevator, if the elevator is going up, stopped or going down at a constant speed, then the velocity will be positive positive, zero and negative respectively. But in all cases, the acceleration is zero. And a person on a scale will show his stationary weight, W.

In a similar fashion, a person's instantaneous velocity can be upwards, zero or downwards. His acceleration is the CHANGE in that velocity. So the acceleration can be positive (velocity increasing), zero (velocity remains the same) or negative (velocity decreasing).

TL, I get the feeling from your posts that you've got this idea pretty well. This is for anyone still struggling with the difference between weight & mass.

Too many words?

Tom

Tom.

Perfect clarification. It has been years since my engineering/physics classes... I was incorrect. Thank you for providing clarification for me.
 
Tell that to Richard Gage, AIA. One of my favorite quotes from him:

http://screwloosechange.blogspot.com/2008/03/physics-of-richard-gage.html


Originally Posted by Richard Gage said:
In the case of the North Tower, the 15 story section above the point of jet plane impact, you see it decreasing by half before it ever begins to move down. In other words, half of that mass, which we're told was so great, is decreased, destroyed, disintegrated by these explosions, and then within two seconds the rest of it is completely gone such that you don't have this 15 story mass crushing the rest of the building and its 80,000 tons of structural steel designed to resist this load for the life of that building.

There are no words.
 

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In the case you mention the static load of the truck is far more than sufficient to crush the box. It isn't a dynamic event either.

Many of you guys don't seem to understand what a dynamic event actually is.

There is a need for a dynamic event in the collapse of the towers since the structure below was built to withstand several times the static load above it.

The way a load is amplified is when the deceleration is several times that of gravity. That requires velocity loss, yet there is no velocity loss in the fall of the upper section of WTC 1.

The only way it could happen is if the columns are missed, but that has been analyzed and deemed impossible.
Hi Guys.

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

It not deceleration a reduction in acceleration, which just means the change in the increase in speed was less. That doestn mean it actually got slower though. Or have I got this incorrect? i.e the upper block continued to accelerate, but less than before the impact event?

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.

Or, is deceleration actually a negative change in speed. Is that where I've got it wrong?
 
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Part of your delusion is that the aircraft impacts had anything to do with the collapses. As for fire causing the collapses, the NIST analysis of the small amount of steel they did get does not show the steel experienced high enough temperatures to even weaken it.

Sorry Beachnut, but the present official government analysis and explanation for those collapses is not convincing at all. We need a new investigation.

We have a winner here.

Bolding mine.

Ya hear it, beachnut? Reallity is delusion! I didn't know that, you didn't either. But now, straight from Truther University this marvel of human knowledge came to us.
 
Part of your delusion is that the aircraft impacts had anything to do with the collapses. As for fire causing the collapses, the NIST analysis of the small amount of steel they did get does not show the steel experienced high enough temperatures to even weaken it.

Sorry Beachnut, but the present official government analysis and explanation for those collapses is not convincing at all. We need a new investigation.

Regress to type.

Woof!
 
I just love when truthers demand a new investigation because the "present official government analysis and explanation for those collapses is not convincing at all", as if 1) there's such a thing as a "present official government analysis" and 2) as if the fact it doesn't convince THEM actually means something.
 
Well, here's a bit of a correction.

Forces add (vectorily).

Velocity & accelerations don't.

A solid object can only have one velocity & one acceleration. It can have lots of forces.

So the right way to do these analyses is to add up all the various forces, and then calculate the single resultant acceleration and/or velocity.

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...
 
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...
Yes, I believe this is just a misunderstanding.

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 accelerationWP is defined as the derivative of the velocityWP 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.
 
Yes, I believe this is just a misunderstanding.

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 accelerationWP is defined as the derivative of the velocityWP 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.

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

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?
 
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