Check my math

[nicepants]

I've been going back & forth with Stundie on a thread at LCF, he believes the towers should have arrested collapse. I have tried to explain that a falling object exerts much more energy on objects below it than if it is just sitting. Stundie didn't get it, so I whipped out some equations. I believe I'm correct, but maybe someone here who knows more about physics can tell me if I've done these correctly.

For an object dropping 10 feet, the speed at impact will be 7.7 meters/sec. (Calculation below)

10 feet = 3.048 meters
v = a * t
d = 1/2 * a * t(squared)
v = sqrt(2*a*d)

v = sqrt(2 * 9.81 m/s * 3.048)
v = sqrt(59.80176)
v = 7.73315977
v = Approx 7.7 meters/second

Kinetic Energy equation
KE = 1/2 mv(squared)

where:
m - mass of the object (in kg)
v - velocity (in meters/second)

The impact force (force required to stop an object) which falls 3.048 meters (10ft), with a .0762m (.25 ft) distance, will be approximately 40x the weight of the object.

Object weight used = 1kg
Favg - impact force (average) over distance D (in Newtons)
D - distance required to stop the moving object (in meters)

Favg * d = - 1/2 mv(squared)
Favg * .0762 = 1/2 1kg * 7.7 (squared)
Favg * .0762 = 1/2 59.29kg m/s
Favg * .0762 = 29.645
Favg = 29.645 / .0762
Favg = 389.041N
389.041N = 87.464lb = 39.67 kg

A 1kg object, falling approx 10 feet, will require a 39.67kg force to arrest its fall over a distance of .25 feet.

 

[Arus808]

what's the point? he's just going to ignore it.

 

[R.Mackey]

You've got a stray minus sign in the first Favg * d expression, that you quickly drop. Other than that the math is correct.

However, the "stop in 0.25 feet" assumption is arbitrary, and you may get dinged for that. Typically back-of-envelope problems like this are treated as impulse problems, though even there you usually have to make hand-waving assumptions.

In actual impact dynamics, the instantaneous force can be extremely high.

 

[e^n]

You might want to make clearer that the weight of the object is the force exerted due to gravity and that the force exerted after a fall at 7.7m/s is much larger over the same time period.

 

[nicepants]

You've got a stray minus sign in the first Favg * d expression, that you quickly drop. Other than that the math is correct.

However, the "stop in 0.25 feet" assumption is arbitrary, and you may get dinged for that. Typically back-of-envelope problems like this are treated as impulse problems, though even there you usually have to make hand-waving assumptions.

In actual impact dynamics, the instantaneous force can be extremely high.

I figured with a .25ft stop distance, I was being extremely lenient. In actuality, if the falling mass was to be stopped by steel and concrete, I imagine that the stop distance would be much smaller to avoid breaking the subsequent floor....and of course the smaller the stop distance, the higher the force.

If even a relatively small multiplier (x40) would support the collapse theory, a more accurate factor (i'm guessing much higher) would exceed the load limits of the tower by an even more considerable margin.

I'm sure none of this will matter to Stundie because he doesn't consider logic, reason, or science. But for my own sake I just wanted to be sure I was calculating correctly.

 

[jaydeehess]

I figured with a .25ft stop distance, I was being extremely lenient. In actuality, if the falling mass was to be stopped by steel and concrete, I imagine that the stop distance would be much smaller to avoid breaking the subsequent floor....and of course the smaller the stop distance, the higher the force.

If even a relatively small multiplier (x40) would support the collapse theory, a more accurate factor (i'm guessing much higher) would exceed the load limits of the tower by an even more considerable margin.

I'm sure none of this will matter to Stundie because he doesn't consider logic, reason, or science. But for my own sake I just wanted to be sure I was calculating correctly.

0.25 feet is only 3 inchs. That would be a 3 inch deflection in the floor. In all likelihood it would take more than that to destroy the floor. Try over 1 meter and you still get a force of 30 N which is basically 3 Kg.

 

[Myriad]

I've used a similar argument, and it seems to work pretty well. You actually don't need to calculate kinetic energy, you can get the same result just with forces and accelerations.

I usually allow an even larger displacement, a whole foot or even half a meter, which still results in forces way above designed loads. Though from an engineeering standpoint this way of illustrating the situation has the disadvantages that R.Mackey pointed out, it also has some advantages for intuitive understanding. When calculations are based on energy, truthers will tend to insist that enormous amounts of energy must be "absorbed" in fracturing the concrete, "plastic" or "elastic" deformation of the structure below, and so forth.

The straightforward f=ma approach cuts through this, as one can ask: "Do you think that the floor could be pushed down a foot (or half a meter, or whatever) without breaking the structure below it? Because that's what would have to happen, in order to arrest the collapse at that point." Followed, as needed, by e.g. "Do you think that the floor could resist a force that's an even greater multiple of its designed load, in order to bring the falling mass to a stop over a shorter distance than that?"

Respectfully,
Myriad

ETA: It works out that the resistive force needed to arrest the collapse (bring a free-falling mass to a stop) is the weight of the falling mass * (1 + d1 / d2), where d1 is the distance it falls and d2 is the distance over which it must stop. This is easy to understand, even for people who can't follow the calculations used to get there. The +1 (which your result didn't take into account, by the way) is for the weight of the falling mass, because in addition to bringing the falling mass to a stop, the resistive force must also resist its normal gravitational acceleration -- that is, bear its actual weight.

 

[Newtons Bit]

Or, instead of assuming an arbitrary deflection, you could what Bazant and Zhou did and actually calculate the deflection.

 

[Myriad]

Or, instead of assuming an arbitrary deflection, you could what Bazant and Zhou did and actually calculate the deflection.

Is that comment really as condescending as I'm reading it? If not, I apologize in advance for the following response in kind.

I could do what Bazant and Zhou did, but why on God's green earth would I?

If I doubted Bazant and Zhou's findings or intended to build on their findings and wanted to first confirm them for myself and understand all their subtleties, then yes, I might calculate the deflection as they did.

But it should be pretty clear that the goal we're talking about here is not performing engineering studies, it's explaining the consequences of certain basic principles to a generally hostile audience, as the OP is attempting to do. Toward that end, performing a calculation that's already been done, and that the audience cannot possibly follow, is utterly useless, as is proclaiming a value as correct based on an engineering paper. if the audience accepted the findings of engineering papers at face value, the discussion in question wouldn't be happening in the first place.

Nicepants' explanation is not fundamentally wrong or misleading in any way. If it's not rigorous enough for engineers, well, engineers' papers are full of weird symbols that most people can't read, so maybe rigor isn't the solution to this particular problem.

Engineering is one thing. Explaining things to people is another. Some engineers (including most here) are good at both, but many are decidedly not. I trust, in those cases, that the one they're good at is engineering.

Respectfully,
Myriad

 

[gumboot]

Myriad raises a good point...

Many of those engineering papers, etc. are highly complex. They should taker some advice from my fellow New Zealander:

A theory that you can't explain to a bartender is probably no damn good.
-Earnest Rutherford.

-Gumboot