Heiwa's Pizza Box Experiment

[Heiwa]

Are you familiar with the equation F = ma, Heiwa?

What does it represent?


Now, suppose you have a mass falling downwards (a = g, therefore F = mg), and after falling a certain distance (and thereby gaining speed), it impacts a body which provides a resistance force of R (which, being a resistance, is opposite in direction to F).

To re-iterate, you have a mass (m) falling under gravity (g) onto another object. By the time the falling mass hits the object, it has gained some speed (V, or else it wouldn't be falling, would it?).

So it is moving, and moving downwards. Even you can agree with the assumption that a mass must move downwards to strike an object below it, requiring said mass to be in motion.


Now, you have a mass with speed V. It impacts an object which provide a resistance force of R.

You state that if R is greater than mg, the mass will move upwards.

That acceleration vector will be upwards. That I grant.
But the motion of the object will still be downwards. It's downward acceleration will be slowed, but it will not "fly up into the sky"


Follow me here (and remember, forces and acceleration are vectors)

mg (force of falling mass) = F.

Well, we know that R will impart an acceleration on the mass. R is a force, therefore we can write R = ma.

Let's look at three cases (and note that I take upwards as the "positive" direction):
1) R = -F (forces R and F are equal in magnitude and opposite in direction),
2) R < -F (force R is less in magnitude and opposite in direction to force F),
3) R > -F (force R is greater in magnitude and opposite in direction to force F)

Starting with scenario 1):

R = -F
Therefore ma = -mg.
We cancel out the like term (m), and we wind up with a = -g.

Thus, we see that the acceleration imparted by the resistance force is equal to the acceleration of gravity. There is no net acceleration.
But the mass was already moving! Therefore, in this case, the moving mass continues to move downwards at a constant velocity! Exactly as Dave Rogers said!​

Now on to scenario 2):

R < -F
Therefore ma < -mg.
We cancel out the like term (m), and we wind up with a < -g.

Thus, the acceleration imparted by the resistance force is less than the acceleration of the mass under gravity. There is a net acceleration downwards.
Therefore, in this case, the moving mass continues to accelerate downward! This, by the way, is what was observed on September 11, 2001.​

Finally, scenario 3):

R > -F
Therefore ma > -mg.
We cancel out the like term (m), and we wind up with a > -g.

Thus, the acceleration imparted by the resistance force is greater than the acceleration of the mass under gravity. There is a net acceleration upwards.
Therefore, in this case, the moving mass slows it's descent! It will slow down to a complete stop. It cannot shoot into the air, because that would remove it from the object providing the resistance force, and thus the upwards acceleration would no longer be imparted. This is what you seem to think should have happened. Your errors in this assumption, however, numerous though they are, have been pointed out to you before, and in great detail, and so I'm not going to cover them here.
​

But surely you can no longer deny that for the upper block to slow down and stop moving requires a resistance force greater than the force of the falling block, not equal to it.

Dave Rogers is correct in what he said to you..

It was Dave that suggested that R>mg (statically), so I pointed out that it means that m must then fly away (in direction of R). Dave got it wrong.

It seems you have not read what I have written at all. So I repeat.

When m impacts anything say X, it may

A. bounce (m is solid and elastic, X is strong),

B. get destroyed (m is solid and inelastic, X is strong),

C. slip off X and drop beside X (imperfect impact),

D. damage itself and X locally and then m (damaged) gets stuck in X,

E. punch a hole in X and continue (m is rigid, ++, X is not strong enough) or

F. globally collapse X (m is rigid, ++++, X is like a house of cards).

So there are many things that may happen when m impacts X, none of which is a lie.

That F. is the only possible event, as suggested by NIST and Bazant, is not evident, particularily when m is, in fact, smaller and weaker than X.

So I investigate A - F, and suggest that D is a very likely result. It is called collapse arrest ... and happens very frequently when structures are locally damaged and parts of it drop from it. That's why structures do not globally collapse on a regular basis due to local failures.

NIST and Bazant make a lot of unscientific assumtions about m and ignore basic physical laws to demonstrate F. It is NWO physics and has nothing to do with the real, not very complicated, world.

I suggest you apply your knowledge to A - F and let me know your findings.

 

[Smackety]

You forgot one Heiwa.
G. m destroys top floor of X. X does not globally collapse.
Then m + 1 floor of X destroys the top -1 floor of X. X does not globally collapse.
Then m + 2 floors of X destroys the top -2 floor of X. X does not globally collapse.
Then m + 3 floors of X destroys the top -3 floor of X. X does not globally collapse.
Then m + 4 floors of X destroys the top -4 floor of X. Wash. Rinse. Repeat.

It is actually very similar to your scenario D. so you are close to being right.
At every floor the mass of the top section increases and arrest becomes less and less likely. Welcome to gravity.

You are treating X as a solid mass.

 

[GreyICE]

Homeland Insurgency will not be mounting a defense of Heiwa's zany experiments or assertions.

Well neither will Heiwa since I'm one of the 10,000 people on ignore.

I'll agree that Heiwa is correct if he ships me half the pizza. Hey, he gets his first supporter, I get 50 pizzas, sounds good to me!

 

[Heiwa]

To Smackety. Re your G, what happens to m, when it destroys top floor of X, and how much energy is consumed at this event? Does m remain intact, i.e. is perfectly rigid? If yes, you believe in WHO physics. It seems you assume m not only being rigid but also solid.

And the detroyed top floor of X. In what condition is it after impact? Why would it drop? And why would it glue itself to m, increasing m? The destroyed top floor - was it solid before impact? What is it after impact?

Or is X a house of cards, each floor a card?

In my example X is anything and not a solid mass. But it is very similar in structure as m.

Evidently the top floor of X may go into several pieces at impact and then you have to check what happens to all these pieces.

If you believe each piece of destroyed top floor X is glued to (rigid) m and becomes rigid, you believe in WHO physics.

 

[Smackety]

To Smackety. Re your G, what happens to m, when it destroys top floor of X, and how much energy is consumed at this event? Does m remain intact, i.e. is perfectly rigid? If yes, you believe in WHO physics. It seems you assume m not only being rigid but also solid.

And the detroyed top floor of X. In what condition is it after impact? Why would it drop? And why would it glue itself to m, increasing m? The destroyed top floor - was it solid before impact? What is it after impact?

Or is X a house of cards, each floor a card?

In my example X is anything and not a solid mass. But it is very similar in structure as m.

Evidently the top floor of X may go into several pieces at impact and then you have to check what happens to all these pieces.

If you believe each piece of destroyed top floor X is glued to (rigid) m and becomes rigid, you believe in WHO physics.

It does not have to be rigid. You need to get beyond these crazy assumptions.

 

[Dave Rogers]

So Heiwa is wrong because he doesn't have the resources to rebuild the complex and fly airplanes into it.

Heiwa is wrong because he's treating the collapse of the towers as a problem in statics, not dynamics. This is painfully obvious from his last response to me, where he claimed that the upward friction force on the falling block is equal to the gravitational force downwards on it, and from the way he keeps claiming that the upper block cannot collapse the lower because its weight is less than the yield strength of the support columns. Because he treats the collapse as a statics problem, he starts with the assumption that all forces balance and that no bodies are in motion. When he reaches the conclusion that the towers don't collapse, the only reason for that conclusion is that his starting assumption is that the towers don't collapse. Heiwa is wrong because he's arguing in a circle, and I think he's doing so with deceptive intent because his work shows too much intelligence for him to be unaware he's arguing in a circle. The only other possibility is that he really doesn't understand the difference between statics and dynamics, in which case he must be lying when he says he's an engineer.

Dave

 

[Dave Rogers]

It was Dave that suggested that R>mg (statically), so I pointed out that it means that m must then fly away (in direction of R). Dave got it wrong.

Lying again, Heiwa. I asked you to determine the dynamic force on the falling block due to friction and entanglement, and show that it must be greater than mg. You responded by calculating the static force, assuming that the force due to friction and entanglement is sufficient to arrest the falling block. I made it perfectly clear that R must be greater than mg while the upper block is falling in order for that fall to be arrested. You dodged the question because you know you don't have an answer. Your repeated assertion that friction will bring the falling block to a halt is a lie, because you have no way of calculating the dynamic friction force on the falling block.

Dave

 

[beachnut]

How do you know pomeroo beachnut? Do you both attend the same anger management class? Do you pay for the both of you or does pomeroo have assistance? Does he ever try to bum any change or toast off of you?

Your evidence free stance is noted, and your inability to understand structural engineering was evident from your first post. You are a member in good standing of the biggest anti-intellectual movement since the NAZIs in the 1930s.

If you could support the failed ideas with more fantasy inputs, it would be more interesting, but you choose to support the failed ideas with nothing. Good; you are maximizing the anti-intellectual aspect.

Heiwa should stick to CTs based on boat accidents.

 

[Zipster]

Are you familiar with the equation F = ma, Heiwa?

What does it represent?


Now, suppose you have a mass falling downwards (a = g, therefore F = mg), and after falling a certain distance (and thereby gaining speed), it impacts a body which provides a resistance force of R (which, being a resistance, is opposite in direction to F).

To re-iterate, you have a mass (m) falling under gravity (g) onto another object. By the time the falling mass hits the object, it has gained some speed (V, or else it wouldn't be falling, would it?).

So it is moving, and moving downwards. Even you can agree with the assumption that a mass must move downwards to strike an object below it, requiring said mass to be in motion.


Now, you have a mass with speed V. It impacts an object which provide a resistance force of R.

You state that if R is greater than mg, the mass will move upwards.

That acceleration vector will be upwards. That I grant.
But the motion of the object will still be downwards. It's downward acceleration will be slowed, but it will not "fly up into the sky"


Follow me here (and remember, forces and acceleration are vectors)

mg (force of falling mass) = F.

Well, we know that R will impart an acceleration on the mass. R is a force, therefore we can write R = ma.

Let's look at three cases (and note that I take upwards as the "positive" direction):
1) R = -F (forces R and F are equal in magnitude and opposite in direction),
2) R < -F (force R is less in magnitude and opposite in direction to force F),
3) R > -F (force R is greater in magnitude and opposite in direction to force F)

Starting with scenario 1):

R = -F
Therefore ma = -mg.
We cancel out the like term (m), and we wind up with a = -g.

Thus, we see that the acceleration imparted by the resistance force is equal to the acceleration of gravity. There is no net acceleration.
But the mass was already moving! Therefore, in this case, the moving mass continues to move downwards at a constant velocity! Exactly as Dave Rogers said!​

Now on to scenario 2):

R < -F
Therefore ma < -mg.
We cancel out the like term (m), and we wind up with a < -g.

Thus, the acceleration imparted by the resistance force is less than the acceleration of the mass under gravity. There is a net acceleration downwards.
Therefore, in this case, the moving mass continues to accelerate downward! This, by the way, is what was observed on September 11, 2001.​

Finally, scenario 3):

R > -F
Therefore ma > -mg.
We cancel out the like term (m), and we wind up with a > -g.

Thus, the acceleration imparted by the resistance force is greater than the acceleration of the mass under gravity. There is a net acceleration upwards.
Therefore, in this case, the moving mass slows it's descent! It will slow down to a complete stop. It cannot shoot into the air, because that would remove it from the object providing the resistance force, and thus the upwards acceleration would no longer be imparted. This is what you seem to think should have happened. Your errors in this assumption, however, numerous though they are, have been pointed out to you before, and in great detail, and so I'm not going to cover them here.
​

But surely you can no longer deny that for the upper block to slow down and stop moving requires a resistance force greater than the force of the falling block, not equal to it.

Dave Rogers is correct in what he said to you..

Big thumbs up to you! An explanation even a non-engineer could follow (well, except for Heiwa...).

 

[Heiwa]

Lying again, Heiwa. I asked you to determine the dynamic force on the falling block due to friction and entanglement, and show that it must be greater than mg. You responded by calculating the static force, assuming that the force due to friction and entanglement is sufficient to arrest the falling block. I made it perfectly clear that R must be greater than mg while the upper block is falling in order for that fall to be arrested. You dodged the question because you know you don't have an answer. Your repeated assertion that friction will bring the falling block to a halt is a lie, because you have no way of calculating the dynamic friction force on the falling block.

Dave

No, you didn't. And any forces, including friction, applied by the lower structure (static) on the upper block, still moving, are static and they cannot be greater than mg in neither the dynamic, nor the static phase..

Because these forces (smaller than mg), when applied on the upper moving part, consume/absorb energy ... and when the available energy is consumed/absorbed, the destruction is arrested. It takes some time. And then equilibrium is achieved (static condition) and the resisting forces are equal to mg.

So there are two phases during a collapse arrest: the dynamic one when forces of the lower structure are applied to the downward moving part(s), which according all physical laws decelerate the mowing part(s) (to zero speed) and the static one when the same forces = mg = collapse is arrested.

Pls do not suggest that I lie.

NIST and Bazant of NWO physics fame, suggest that no forces are applied to the upper part(s) and that the upper part(s) is (are) thus accelerating destroying everything below. This is not in accordance with Newtons's third law.

Force are applied at every point of contact, they create, i.a. friction and deceleration, and in 99.99% of the incidents the destruction is arrested. It seems the 0.01% remaining cases would be WTC1,2 but I doubt it.

It seems that I have probability on my side.

Dave, you are putting your money on a losing horse. NIST and Bazant have no credibility. They just invent things because in NWO physics there is no real physics.

 

[Jonnyclueless]

So Heiwa, while pretending to be an engineer, is referring to the top portion of the building as a solid block?

That would explain his use of the term "NWO physics".

And then Heiwa who obviously is NOT an engineer and oviously has NO engineering experience is claiming NIST has no credibility. And at the same time Heiwa, who is obviously not an engineer has yet to publish a paper showing what should have happened?

Heiwa, if you're going to be a con artist, then at least try to be a good one.

 

[X]

It was Dave that suggested that R>mg (statically), so I pointed out that it means that m must then fly away (in direction of R). Dave got it wrong.

It seems you have not read what I have written at all. So I repeat.

You are correct. I did not read the rest of your post.
When I find an error like that in the first line of a post, I see no point in reading further. At least not until the error is addressed. For much the same reason I am not addressing the rest of your most recent post to me until you understand your assertion is wrong.


You claim that Dave Rogers is wrong.
As has been pointed out to you (repeatedly), you are treating this as if it were a static scenario, instead of a dynamic one.

I'm going to go into more depth on this.

I apologize in advance, the formulae I'm going to be using are a little more advance than the elementary school level F = ma I used previously.
Now we are moving into the realm of high-school physics.
I understand that this may be a little too complicated for the children you write your papers for to understand. Accordingly, I'm going to explain everything as weel as I possibly can. If said children still have trouble understanding what is going on, please direct them to talk to a local high-school physics teacher.

And now I shall introduce the equations that shows Dave Rogers is correct, and you are wrong:

Newton's Second Law of Motion: F = ma
where

F = force
m = mass
a = acceleration​

This formula gives the relationship between the force acting on a mass, and the acceleration that the mass undergoes as a result of the force.


Uniformly accelerated rectilinear motion: v = v₀ + at
where: v = instantaneous velocity
v₀ = initial velocity
a = acceleration
t = time

This is the equation for determining the velocity of a mass that is undergoing acceleration. Here we need to note again that acceleration and velocity are vectors. A vector is something that has both magnitude (numerical value) and direction. For comparison, a scalar has only magnitude (time is a scalar)
This formula looks complicated, but don't be put off by appearances, children!
It is very straightforward to derive.

You know that the velocity of an object can be determined from its acceleration, right?
If a car takes off from a stationary starting point, and accelerates at a constant rate (which we shall call "a"), we can find the cars velocity from the length of time the car is accelerating. v = at.
To verify that this is true, look at the dimensions:
Acceleration has dimensions of (length/time/time), or (length/time²).
Time has the dimension of (time).
Thus, av has dimensions (length/time²)*(time). One of the (time) units cancels, leaving you with dimensions (length/time). This is velocity!

(As a side note: position can also be determined, but the formula is not at², as you might guess based on dimension. Instead, position requires integration. Integration is a little too advanced for our discussion today, so you'll just have to take my word on it that position is determined from x = 0.5*at². Towards the end of high school you may get into calculus, and be able to derive this for yourselves.)​

But what happens if the car is already moving when it begins to accelerate?
Uh oh! How do we account for that?

Don't worry, it's easy! This is the great thing about Dynamics. It allows you to perform calculations based on moving objects. Because not everything is stationary, kids.

In fact, accounting for initial motion is ridiculously simple. We just add it into the equation we determined above.
If you recall , we derived the equation v = at for a car accelerating at a constant rate from a stationary start.
If the car is moving before it begins accelerating, it will have some initial velocity. We will call this initial velocity v₀, and add it into out equation.
Thus, we get v = v₀ + at.
See? Easy and straightforward. Nothing complicated!


Now let's apply it to the falling upper section of Heiwa's tower as it impacts the lower section and experiences a resistance force due to the impact.

For simplicity's sake, we will assume the resistance force is constant (in reality, it will not be, since the tower is not of homogeneous (<-- New word! It means the same everywhere. For example, Jello is homogeneous. It is Jello everywhere. Building are not homogeneous. There are floors, trusses, beams, open spaces, etc which are not distributed in an even mixture. Rather, a cross-section of a tower will show steel in some places, concrete in others, open space in still other locations, but none of it is mixed together uniformly))

Now, we know the upper block it falling.
It's not hard to figure out. It must be falling, because the lower block can't leap up to meet it (being fixed tot he ground and all), and it is coming into contact with the lower section. Plainly, it must be moving.

So we know it has some initial velocity, v₀.

Now the fact that velocity and acceleration are vectors come into play. Let us assume that upwards is the positive direction. Thus, any accelerations, forces or velocities acting upwards are positive, and any acting downwards are negative.

Since the upper block is falling downwards, its initial velocity is negative. So we will write -v₀.

As a side note, it is possible to determine this initial velocity. The formula is v₀ = (2ad)^0.5.
where d is the distance traveled. Under gravity, the acceleration will be -g (negative because it acts downwards)
This is another equation requiring calculus, and so the derivation is a little too advanced for our audience today. But feel free to ask a high school math or physics teacher to show you. Oh, and the exponent of 0.5 means the same as a square-root sign.​

So now we have a falling mass whose velocity can be determined by:
v = at - v₀ = at - (2gd)^0.5
Now we need to find the acceleration imparted on the falling mass.

If you recall, we assumed the lower section of the tower would provide a constant resistance force. We will call this force R. If you recall from before, force = mass*acceleration. Thus, since the force is a resistance, we can write:
R = ma.

Being a resistance, this force will act against the motion of the upper block. Since the upper block is moving downwards, the resistance force will act upwards, and thus be positive.
Since R is positive, it follows that the acceleration it imparts on the falling mass is positive as well. This is obvious from the equation (R = ma), since mass cannot be negative, thus R and m must have the same direction.

We can now re-arrange the force equation (R = ma) to find a.
Doing so gives us: a = R/m.

We can now plug the acceleration back into our velocity formula. We get:

v = Rt/m - (2gd)^0.5.

But wait! We are still missing one component! We determined acceleration based on the resistance force alone. But we overlooked the acceleration due to gravity! We accounted for it in determining the initial velocity, but not in the instantaneous velocity calculation!
This is a major point, and I hope you caught onto it without me telling you! You have to account for all the forces acting on an object when you are determining its resultant motion. Otherwise all your calculations are meaningless!
And there are two accelerating forces acting on the falling mass. There is the upward-acting resistance force. but there is still the downward-acting gravity force! We have to account for both. Fortunately, gravity is as easy to account for as R was. In fact, it is the same equation, only instead of being R = ma, it is F = -mg where g is acceleration due to gravity and F is the force exerted by gravity. We can re-arrange this as we did with the resistance force, and write: g = -F/m. Remember that it is negative, since gravity acts downwards.

v = Rt/m -Ft/m - (2gd)^0.5 = (R - F)t/m - (2gd)^0.5
The velocity of the upper block is determined from the initial velocity of the upper block added to the acceleration imparted by the resistance force and the acceleration due to gravity.

Now let's look at the same three cases we look at earlier, this time using the approach of dynamics instead of the more simple Newton's Second Law approach.

For simplicities sake, I'm going to go back and replace (2gd)^0.5 with the simpler to write v₀. Otherwise the equation will look messy and confusing. Since the only reason for including it is to show that there is initial velocity, there's no reason to clutter the equation by including the method of calculation initial velocity.


So we have v = (R - F)t/m - v₀


1) R = -F (forces R and F are equal in magnitude and opposite in direction),
2) R < -F (force R is less in magnitude and opposite in direction to force F),
3) R > -F (force R is greater in magnitude and opposite in direction to force F)

Starting with scenario 1):

R = -F
v = (R - F)t/m - v₀
We can replace R with -F, and we get
v = (-F - F)t/m - v₀
But -F - F is zero, so the equation comes out to:
v = v₀

Therefore if the resistance force is equal to the force due to gravity, the falling block will continue to move downwards at a constant velocity, v₀. This velocity can be determined from (2gd)^0.5, where d is the distance the upper block fell before it impacted the lower section.​

Now on to scenario 2):

R < -F
v = (R - F)t/m - v₀
Since R is less than -F, the function R - F will yield a resultant force equal to R - F. This resultant force will be (magnitude of R) less than (magnitude of F), and negative in direction (eg. if R = 3 and F = -5, R - F = -2). We will call this resultant force B.
v = -Bt/m - v₀

Therefore if the resistance force is less than the force due to gravity, the falling block will continue to accelerate downwards with an acceleration rate of less than that of gravity. This, by the way, is what was observed on September 11, 2001.​

Finally, scenario 3):

R > -F
v = (R - F)t/m - v₀
Since R is greater than -F, the function R - F will yield a resultant force equal to R - F. This resultant force will be (magnitude of F) less than (magnitude of R), and positive in direction (eg. if R = 5 and F = -3, R - F = +2). We will call this resultant force C.
v = +Ct/m - v₀

Therefore if the resistance force is greater than the force due to gravity, the falling block will accelerate upwards at a constant rate.
But now things get interesting.
If the upper block is not moving when this resistance force is applied, it will indeed move upwards. BUT IT IS ALREADY MOVING DOWNWARDS AT SOME INITIAL VELOCITY v₀!.
The acceleration due to the resistance will act opposite to the direction of motion, slowing the falling upper section down.
IT WILL NOT SHOOT INTO THE AIR! At least, not until the downward motion has been slowed to a stop.
And after that, the situation is static, and the lower section would only provide a resistance force equal to the now-static load of the upper section. Enough to counter the weight of the upper section (mg), nothing more. Doing otherwise would contravene physics.
Besides, if the upper block moves away from the lower section, it is no longer in contact and thus the lower section has no way of applying the resisting force to it!​

--- --- --- --- ---


In summary, Dave Rogers is right when he says:

You've demonstrated that A*S > M₁g, which was never in doubt. You've made no attempt to determine F₁, and you've asserted that it's equal to M₁g. If that were the case, the falling block would fall all the way to the ground at constant speed, therefore the collapse wouldn't arrest. Clearly you have no idea what the magnitude of the friction and entanglement force is, so you have no grounds for claiming it would necessarily arrest the collapse.

and you, Heiwa, are wrong in saying:

Sorry, F₁ cannot be bigger than M₁g, because then the latter would fly up into the sky.

I have now proven it using Newton's second Law and[i/] using dynamics of motion!

As I said, I'm not reading the rest of your post when your opening line contains an error so egregious my 10-year-old nephew could spot it.

And shame on you for trying to pass off your incompetent physics on innocent children!
I only hope your target audience is older than 10, so that they at least have the knowledge to see your mistakes.


YOU ARE WRONG!


And if you are wrong on so basic a calculation, why should the rest of us waste time with the rest of your claims?

Fix this error,and then we can move onto the next issue.



No wonder you have no concept of the intricacies of scaling.

 

[Zipster]

Holy ◊◊◊◊ [X] has done it again!

How long did it take for you to type all that up? Good thing that you have the patience to type all that up. I would love to have done the same, but alas, Heiwa has put me on ignore.

It would have been like talking to a brick wall. Wait...

 

[X]

Holy ◊◊◊◊ [X] has done it again!

How long did it take for you to type all that up? Good thing that you have the patience to type all that up. I would love to have done the same, but alas, Heiwa has put me on ignore.

It would have been like talking to a brick wall. Wait...

Too long. And I hate algebra.

But I had the chance to contribute something worthwhile, and I did so.

 

[NobbyNobbs]

The velocity of the upper block is determined from the initial velocity of the upper block added to the acceleration imparted by the resistance force and the acceleration due to gravity.

Wonderful post, X, very clear. The only problem I have is in the above sentence. It implies you are adding velocity to acceleration to get velocity. The units don't allow this. You probably meant to phrase it: "The velocity of the upper block is determined from the initial velocity of the upper block added to the velocity caused by the acceleration imparted by the resistance force and the acceleration due to gravity."

Otherwise, exceedingly well done.

 

[twinstead]

mmmmmmmm. Pizza

 

[X]

Wonderful post, X, very clear. The only problem I have is in the above sentence. It implies you are adding velocity to acceleration to get velocity. The units don't allow this. You probably meant to phrase it: "The velocity of the upper block is determined from the initial velocity of the upper block added to the velocity caused by the acceleration imparted by the resistance force and the acceleration due to gravity."

Otherwise, exceedingly well done.

Yes, that was the intent. Sorry for the oversight.
I realized it was cumbersome and could have been worded better, but that doesn't excuse completely overlooking the point you brought up.
My mistake. I just rambled on, figuring that since I knew what I was talking about everyone else would, too.



I also missed the edit window on the following:

This:

Uniformly accelerated rectilinear motion: v = v0 + at
where: v = instantaneous velocity
v0 = initial velocity
a = acceleration​

Should have been formatted thus:

Uniformly accelerated rectilinear motion: v = v0 + at
where:

v = instantaneous velocity
v0 = initial velocity
a = acceleration
t = time​

Ah well. I just hope I don't wind up on ignore for this.


and "av" should be "at". Bugger.
(re: 16 lines after "Uniformly accelerated rectilinear motion: v = v0 + at)

 

[Zipster]

I'm anticipating Heiwa's response to your post [X]. Probably will still hold onto his "if R > mg, then m will fly way" nonsense.

But you might just end up on ignore. Come join the rest of us that he doesn't have an answer to from his "engineering knowledge"! I hear we're having a pizza party.

 

[Heiwa]

To [X]. Thanks for your long post. It seems you assume that the upper block is one rigid mass m when dropping, which it is not. But you are right that the lower section applies a force on it at contact. In reality there are multiple forces - one at every contact point.

If the upper block would bounce on the lower section (a possibility not investigated by NIST) that force exists only during the short bounce/compression/decompression of upper block and then it disappears when the upper block flies up. It is the energy of the upper block that creates that force, and that force creates the energy that allows the upper block to bounce (laymen's terms). Compare rubber ball bouncing on ground.

If the upper block gets damaged at contact (another possibility that is not investigated by NIST) that force is evidently applied to the upper block and overloads it in particular locations that fail, etc, etc. Compare crystal glass dropping on marble top. The energy of the upper block is actually used to destroy it. According NWO physics, it cannot happen, as the upper block is assumed rigid.

But let's assume the upper block is rigid and is not destroyed at contact. Then only the lower structure is destroyed (locally) and for that you require energy, i.e. a force that displaces parts of the structure below a distance (Energy is simply force times distance).

We know the energy of the upper (moving) block and it is not very big. In one of my papers it corresponds to about 40 litres of petrol. And that energy can only locally destroy some upper parts of the lower structure. When that energy is then consumed or absorbed as structural failures, the destruction is arrested.

NIST does not investigate that. NIST simply assumes that the energy available exceeds the total energy that can be absorbed by the lower structure and that the latter globally collapses. It is evidently absurd as the energy that the lower structure can absorb is 1000 times bigger than the energy applied by the upper block.

But that is assuming the upper block is rigid.

As the upper block is not rigid, it (and parts of the lower structure) should be destroyed at the multiple impacts between parts upper block/lower structure as shown in my papers in clear figures. The upper block would be sliced into 1000's of smaller parts at impacts that can then not cause further damages/destruction to the lower structure.

NIST evidently does not investigate that possibility, i.e. they didn't do a complete analysis. I wonder why?

 

[beachnut]

As the upper block is not rigid, it (and parts of the lower structure) should be destroyed at the multiple impacts between parts upper block/lower structure as shown in my papers in clear figures. The upper block would be sliced into 1000's of smaller parts at impacts that can then not cause further damages/destruction to the lower structure.

NIST evidently does not investigate that possibility, i.e. they didn't do a complete analysis. I wonder why?

Your complete lack of knowledge on this issue is summed up by you! Excellent work.

You debunk yourself so well, who needs to waste more time with your CT ideas best saved for maritime accidents and mishaps. Good luck; I can not understand why you fail to understand physics and claim to be an engineer. How can this be?