Why scale models are ineffective

[BigAl]

To model an impact of a structure C (WTC 1 top part) with a similar structure A (WTC 1 btm part) is quite easy and has nothing to with gravity or scale!

You don't have a clue what "model" and "scale" mean in the context of understanding structures.

 

[Justin39640]

You don't have a clue what "model" and "scale" mean in the context of understanding structures.

that was obvious with his water tank bs
he thought a campfire was a good scaled inferno lol

he also thinks snow cant crush a building lol

 

[sylvan8798]

I find Heiwa's obsession with his C and A "structures" as strange as psikey's fixation with the masses of steel and concrete at every location. I wonder what a psychiatrist would say about this type of obsessive behavior?

 

[Hokulele]

Scale gravity? A few of you need to take basic physics classes again!

If you drop a bowling ball and a golf ball from the same height, at the same time, which will hit the ground first? You can even ignore air resistance for this example as it plays a very minor role.

That is exactly my point, acceleration due to gravity is a constant^*.

To take this back to scaling, now calculate how much force is required to stop the bowling ball? How much force is required to stop the golf ball? There is no way to scale gravity to make these forces equal, so there is no way you can call your golf ball and its behavior a perfect scale model of your bowling ball and its behavior.


* - At least for the purposes of model-building here on Earth, as Dave Rogers points out in his reponse.

 

[psikeyhackr]

That is exactly my point, acceleration due to gravity is a constant^*.

To take this back to scaling, now calculate how much force is required to stop the bowling ball? How much force is required to stop the golf ball? There is no way to scale gravity to make these forces equal, so there is no way you can call your golf ball and its behavior a perfect scale model of your bowling ball and its behavior.

* - At least for the purposes of model-building here on Earth, as Dave Rogers points out in his reponse.

.
But if you have a series of bowling balls spaced in a tube with holes to let the air out and the top ball has to break the supports of the balls below then the mass of all of the balls is relevant to the acceleration that can occur because the top falling ball will try to force what it hits to accelerate faster than gravity. The conservation of momentum becomes a factor. So not knowing the distribution of mass through the tower is absurd.

The strength of the supports will be an additional factor in slowing things down.

http://www.youtube.com/watch?v=LXAerZUw4Wc

psik

 

[Hokulele]

But if you have a series of bowling balls spaced in a tube with holes to let the air out and the top ball has to break the supports of the balls below then the mass of all of the balls is relevant to the acceleration that can occur because the top falling ball will try to force what it hits to accelerate faster than gravity. The conservation of momentum becomes a factor. So not knowing the distribution of mass through the tower is absurd.

I recommend bacon bits and papaya seed dressing. They go with anything, even word salad.

And yes, your bizarre obsession with TONS of STEEL and TONS of CONCRETE is noted.

 

[FineWine]

First you must decide what you are going to model, e.g. a steel tower structure where part C one way crushes part A, when C is dropped on A by gravity; C = 1/10 A, A carried C before.

How to go about it? And does size/scale matter? Let's do it full scale!

First select a suitable building standard to get the design stresses and the redundancy right. Let's also simplify and that only static design stresses due to gravity loads are considered (in this example).

Let's start with a very simple tower 100 floors high. It has a core of 8 vertical columns (100 floors high) in a square; three columns each side; 4 corner core columns and 4 intermediate columns. At every floor the core columns are interconnected with horizontal/slooping beams for redundancy reasons. The core is like a big mast of 8 columns. It is evidently self-supporting due to the beams between the core columns.

The core is surrounded by perimeter wall columns; say 7 each side, thus total 24! At every floor the perimeter core columns are interconnected with horizontal spandrels for redundancy reasons.

All the columns are primary, load bearing elements. At present no loads (except own weights) are applied to the columns. The horizontal beams and spandrels are secondary elements installed for redundancy reasons and act as support to keep critical stresses (buckling) low in the columns (as per the standard).

Now we connect the core columns with the perimeter columns with horizontal floor trusses! 5 perimeter columns at each wall corner are connected to one core corner column (thus 5 trusses) and the intermediate core columns are connected by one truss to the remaining, intermediate perimeter columns.

Thus there are 24 floor trusses between perimeter columns and core; 4 x 5 trusses connected to the core corner columns, 4 just between remaining columns. You follow? The trusses are just bolted to the columns. The trusses are secondary elements.

This can be done in any scale, size.

Now put a thin steel plate over all floor trusses and add your loads on the floor trusses.

These loads are evidently transmitted to the columns that compress the columns. Now ensure that the static stress in each column is according to standard. Evidently the corner core columns carry more load than any other column but ... the stresses are the same. The columns get heavier lower down.

Now check the structure for redundancy! Say that the criteria according to standard is that you can remove any one column between any floor and that the spandrels and the horizontal beams shall be able to transmit the column load around the 'failure' (the removed column bit) to adjacent coumns, that will not fail. It means, e.g. that the intermediate core columns become quite strong.

OK, your structure is ready. In any size, scale. The static stresses and the redundancy are as per standard.

Now we are going to remove all columns, one after the other, between parts C and A (so we can drop C on A).

Let's say we first remove one centre, the 4th middle one, perimeter wall column in one wall. No problem - there is redundancy enough.

Then we remove the two adjacent perimeter wall columns. What happens? Well, it is a possiblity that the 4th middle perimeter wall column above (and its local loads) will drop down to ground, the spandrels above connected to the 4th middle perimeter column shear off, unless the floor trusses (and the hat truss!) above can pull it (and the load!) in position.

Regardless, it seems that part C will be severely locally stressed and possibly damaged when we start removing columns between parts C and A, thus before C is even dropped on A.

This is another reason why part C cannot one way crush down part A. C is simply less strong than A and subject to bigger local loads due to local failures between C and A. There are many others. And none of them has to with scale or modelling. It is just simple structural damage analysis that is required and it is the same for any structure.

Anyone suggesting that you can remove all columns between parts C and A and then drop C on A is a fool. C is locally damaged before that, parts of C and loads should drop off an the remaining parts of C cannot crush down A.

Read my papers to get a better feel for the problem.

Your papers have been shown by real engineers to be incompetent rubbish.

 

[FineWine]

.
Well then why did 1 washer on each toothpick 1 inch apart stop 20 washers that fell 12 inches faster than toothpick without washers? If anything I bet a created a situation more likely to collapse all of the way than the WTC. My toothpicks didn't get stronger all of the way down like EVERY SKYSCRAPER HAS TO. But you people don't insist on knowing the amount of steel on every level of the towers.

You can just BELIEVE things on the basis of inadequate data.

psik

The real engineers, who are much smarter and know vastly more than you, regard your obsessive claims as nonsense. Why are they wrong? How do you know things they don't?

 

[FineWine]

.
But what it hit didn't weigh a lot. It didn't have a lot of inertia to be overcome.

That "chaos ensues" is just like that idiotic"collapse was inevitable" crap from the NIST.

But I am sure you can't get over it.

psik

A thousand engineers, physicists, architects, and fire safety experts state that "collapse was inevitable." One poorly educated, agenda-driven crank says that they are all wrong. What do you know that they don't, and how did you learn it. (You'll get tired before I do.)

 

[Scott_Milner]

No, you need to progress a little beyond basic physics.

The structural strength of a building depends on the cross-sectional area of the support columns. Let's start with a building that has a safety factor of two. Build another, in which you multiply the width, depth and height of the building by six, and the structural strength increases by thirty-six times, but the weight increases by two hundred and sixteen times. Its weight is now three times higher than the structure can support; the building will fall down. However, build the second structure on the moon, where gravity is one-sixth of that on Earth, and the weight is now only thirty-six times that of the original building. Since the structural strength is thirty-six times higher, the safety factor is now back to two, and the building will stand up. That's what Hokulele means by "scale [things such as] gravity"; it can be used (in theory) to compensate for square-cube laws, but in practice there are one or two fairly major difficulties.

Dave

It is a response such as this that keeps me away from this forum.

We are not comparing a building on Earth vs. the moon, and therefore there is no need to 'scale' gravity. The fact that weight = mass x gravity should automatically tell you that a smaller scale building will proportionaly adjust loading on the support columns for a comparitive test.

If you scale down gravity on top of scaling down the mass of an object, then you are reducing the loading by an unfair factor!

 

[R.Mackey]

It is a response such as this that keeps me away from this forum.

We are not comparing a building on Earth vs. the moon, and therefore there is no need to 'scale' gravity. The fact that weight = mass x gravity should automatically tell you that a smaller scale building will proportionaly adjust loading on the support columns for a comparitive test.

If you scale down gravity on top of scaling down the mass of an object, then you are reducing the loading by an unfair factor!

Scott:

Welcome to the Forums.

Dave Rogers is Ph.D. in Physics, so he knows what he's talking about. Try to read his answers in context, they actually do make sense.

I gave my own treatment of scaling in precisely this situation on Hardfire some months ago; see here for video treatment, or read the slides here. As that discussion demonstrates, it would be nice if we could change gravity to help with scaling, but of course we can't.

 

[Scott_Milner]

That is exactly my point, acceleration due to gravity is a constant^*.

To take this back to scaling, now calculate how much force is required to stop the bowling ball? How much force is required to stop the golf ball? There is no way to scale gravity to make these forces equal, so there is no way you can call your golf ball and its behavior a perfect scale model of your bowling ball and its behavior.


* - At least for the purposes of model-building here on Earth, as Dave Rogers points out in his reponse.

The force required to stop the golf ball will be proportional to the force required to stop the bowling ball if scaled properly.

Once again, if you reduce the mass of the objects in the scaled model, the kinetic energy available also decreases. It is the potential enery and kinetic energy totals that are of interest, along with the action of gravity acting on the structure.

If F= m x a you can change either mass, or acceleration to produce the force. In your example if you scale one box column to 1/10th of the original and THEN reduce the acceleration of gravity the ratio is no longer 10:1 with respect to weight and potential energy.

 

[Hokulele]

The force required to stop the golf ball will be proportional to the force required to stop the bowling ball if scaled properly.

Once again, if you reduce the mass of the objects in the scaled model, the kinetic energy available also decreases. It is the potential enery and kinetic energy totals that are of interest, along with the action of gravity acting on the structure.

If F= m x a you can change either mass, or acceleration to produce the force. In your example if you scale one box column to 1/10th of the original and THEN reduce the acceleration of gravity the ratio is no longer 10:1 with respect to weight and potential energy.

Exactly. It would proportional, while the strength of the 1/10 box column is not proportional to the full-scale version (the Square-Cube Law that is the topic of this very thread). Hence the need to scale gravity to make your box columns behave appropriately, or the need to scale gravity to have the effect of a golf ball match the effect of a bowling ball.

 

[Dave Rogers]

It is a response such as this that keeps me away from this forum.

And yet, here you are.

We are not comparing a building on Earth vs. the moon, and therefore there is no need to 'scale' gravity. The fact that weight = mass x gravity should automatically tell you that a smaller scale building will proportionaly adjust loading on the support columns for a comparitive test.

Let me try again. Weight = mass x gravity. Mass varies as volume, which varies as length cubed. Therefore weight varies as length cubed times gravity. Structural strength varies as cross-sectional area, which varies as length squared. Therefore, if you scale down the entire building by a given factor, without adjusting the relative sizes of the support structures, the only way to preserve the strength-to-weight ratio of the original building is to scale gravity inversely with length. As Hokulele pointed out, this is impractical; therefore, results on scale models are not expected to be the same as results on the full sized object.

If you scale down gravity on top of scaling down the mass of an object, then you are reducing the loading by an unfair factor!

Obviously. What I was pointing out is that scaling down gravity is needed if you scale up the dimensions of the object. An understanding of basic physics would allow one to draw the inference that, if you scale down the dimensions of the object, gravity needs to be scaled up. If you didn't get this, then it looks to me like you need to take basic physics classes again.

Dave

 

[Scott_Milner]

Let me try again. Weight = mass x gravity. Mass varies as volume, which varies as length cubed. Therefore weight varies as length cubed times gravity.

Of course.

Structural strength varies as cross-sectional area, which varies as length squared. Therefore, if you scale down the entire building by a given factor, without adjusting the relative sizes of the support structures,

Why would you not scale down the support structures? This is a scale model, correct?

2' x 2' x 10' box columns vs. 2' x 2' x 10' box columns, NOT 2' x 2' x 10' box columns vs. 1" x 1" x 10" tooth picks.

the only way to preserve the strength-to-weight ratio of the original building is to scale gravity inversely with length.

As Hokulele pointed out, this is impractical; therefore, results on scale models are not expected to be the same as results on the full sized object.

What does the 'work': gravity, mass, or the energy/force?

Obviously. What I was pointing out is that scaling down gravity is needed if you scale up the dimensions of the object. An understanding of basic physics would allow one to draw the inference that, if you scale down the dimensions of the object, gravity needs to be scaled up. If you didn't get this, then it looks to me like you need to take basic physics classes again.

Dave

See previous paragraph.

Take a scenario where the vertical component is removed. Use a rolling, or sliding object crashing into a stationary object. If you were to alter either acceleration, OR mass can you effectively "scale" the force? Yes, or no?

 

[Dave Rogers]

Why would you not scale down the support structures? This is a scale model, correct?

I said relative sizes; in other words, scale down everything by the same amount.

2' x 2' x 10' box columns vs. 2' x 2' x 10' box columns, NOT 2' x 2' x 10' box columns vs. 1" x 1" x 10" tooth picks.

No, because if you keep the column dimensions the same then you aren't scaling the columns.

Let's say you build a half-scale model. The support columns are scaled down to half the width and half the depth, giving one quarter the compressive strength. The total structure is scaled down to half the width, half the depth, and half the height, giving one eighth the mass. The strength to mass ratio has therefore doubled. The only ways to maintain the original strength to weight ratio are to change gravity, which we can't do, or not to scale everything the same.

What does the 'work': gravity, mass, or the energy/force?

That's a very strange question. The effect of gravity on the mass exerts a force which, when moving in the appropriate direction, does the work; the work done is equal to the change in gravitational potential energy. Therefore, the answer to your question, as far as I can offer a sensible one, is "all four of the above".

Take a scenario where the vertical component is removed. Use a rolling, or sliding object crashing into a stationary object. If you were to alter either acceleration, OR mass can you effectively "scale" the force? Yes, or no?

Yes, so arbitrarily changing the mass can correct the scaling error. Nobody's disputing that. However, a model with the mass arbitrarily altered to correct the scaling error is not a scale model.

Dave

 

[sylvan8798]

Let's say you build a half-scale model. The support columns are scaled down to half the width and half the depth, giving one quarter the compressive strength. The total structure is scaled down to half the width, half the depth, and half the height, giving one eighth the mass. The strength to mass ratio has therefore doubled. The only ways to maintain the original strength to weight ratio are to change gravity, which we can't do, or not to scale everything the same.

Can't one also potentially correct this by changing materials? Isn't that part of the question of what needs to be "scaled" in order to achieve a representative model? The scaling can be more than just a dimensional thing.

Yes, so arbitrarily changing the mass can correct the scaling error. Nobody's disputing that. However, a model with the mass arbitrarily altered to correct the scaling error is not a scale model.

Dave

Not a scale model if you only define scaling as scaling the dimensions, but that's too narrow of a definition, I think.

 

[Dave Rogers]

Can't one also potentially correct this by changing materials? Isn't that part of the question of what needs to be "scaled" in order to achieve a representative model? The scaling can be more than just a dimensional thing.



Not a scale model if you only define scaling as scaling the dimensions, but that's too narrow of a definition, I think.

Yes, these are both valid points. However, we then go on to what the truthers' response to these modifications would be. Suppose we construct a 1/10 scale model of the WTC, then reduce the strength of the main structural members by a factor of 10, and show that the model collapses in a similar manner to the real structure. Will the truth movement's response be "OK, that was a valid modification to compensate for the effects of scaling on the strength-to-weight ratio of the structure? Or will it be, "This test was invalid because the model was deliberately weakened in order to ensure collapse"?

Dave

 

[deep]

Yes, these are both valid points. However, we then go on to what the truthers' response to these modifications would be. Suppose we construct a 1/10 scale model of the WTC, then reduce the strength of the main structural members by a factor of 10, and show that the model collapses in a similar manner to the real structure. Will the truth movement's response be "OK, that was a valid modification to compensate for the effects of scaling on the strength-to-weight ratio of the structure? Or will it be, "This test was invalid because the model was deliberately weakened in order to ensure collapse"?

If the modifications were defined/documented/published before the construction of the model, and they remained unchanged for the duration of the test, then I would either disagree with the modifications before construction started or accept the results of the test.

In other words, I would have a problem if the modifications were changed after testing began, for whatever reason.

 

[~enigma~]

You may have seen a number of Truthers try to make scale-models of the World Trade Center and use them to "prove" that the towers couldn't possibly have collapsed. Famous examples include Richard Gage with his cardboard boxes, and Heiwa with his pizza boxes and his (fake) $1 mil challenge.

These don't work. Read and learn this article about the Square-Cube Law.

Did you read it? No you didn't. Stop trying to skip ahead. Read it, I say!

Okay now. If you magically double the dimensions of a tower, its weight will increase by a factor of eight, but the structure's load-bearing capacity will only increase by a factor of four.

This works in reverse, too--if you magically shrink the tower's dimensions by half, its weight will be reduced to 1/8th that of the original, while the load-bearing capacity will only be reduced to 1/4th.

Hence, all these 10- foot "scale models" Truthers have produced over the years to attempt to "prove" the towers couldn't have collapsed, are actually capable of holding a much greater percentage of their own weight than the actual, real-life Twin Towers. Even when you build them out of materials much weaker than steel (such as cardboard), they still have a significant advantage, due to the much smaller scale.


Of course, if you're smart, your intuition should've already told you that Gage and Heiwa's little experiments weren't painting the right picture, but I thought you might like to know the science behind it.

And I am interested in seeing Heiwa's rebuttal.

Is density the Rodney Dangerfield of scaling?