Mackey's Appendix B

[elkhound]

I have a question about Appendix B of Ryan Mackey’s paper, “On Debunking 9/11 Debunking” (Version 2.1).

On p. 197, Mackey acknowledges that the acceleration will be nonconstant, “but over time will roughly follow an average acceleration”. He then states that "we can estimate the average acceleration" by using the equation h = ½ a t2.

However, my understanding is that this equation is applicable only for cases of constant acceleration; the equation does not give the average acceleration in cases of nonconstant acceleration.

Right?

[Childlike Empress]

Right, but it's good enough for a rough estimate - and Mackey's argument.

[Doctor Evil]

Actually, Mackey tells you precisely what he is doing. He is taking a very complicated scenario, where the structure resistance to collapse may very in time, and approximates it by a system with constant acceleration. Note that the acceleration in this case will not be that of free fall, but some smaller value.

Now, you are correct that this is not an exact solution. However, this is not the issue here. The only question is whether such an approximation is useful. There are two possible scenarios where such an approximation can be problematic. One is when the (averaged, or long time behavior) functional dependence on time is no longer proportional to t2. The other is when the fluctuations around this averaged behaviour are large enough to make the specific point you would want to make with this estimate meaningless.

Basically, this type of estimate can be useful. It does replace a complex behavior with one or two numbers, but these can still tell us something. The problems arise when one tries to use such models beyond their region of validity. Since did not told us what Mackey was using this estimate for, its hard to know if the model is used inappropriately. In fact, I would be surprised if that is the case. From what I have read so far Mackey clearly knows what he is doing.

[beachnut]

Right, but it's good enough for a rough estimate - and Mackey's argument.
Expand on your non-responsive statement. If you can't then why would you speak up when the equation is used to find an average acceleration? You proved in a short post you are not on speaking terms with algebra. Education is the key.

Solve the equation for what you need with what you have given. Algebra - it's what is needed to figure out the equation. Got Algebra?

[twinstead]

Expand on your non-responsive statement. If you can't then why would you speak up when the equation is used to find an average acceleration? You proved in a short post you are not on speaking terms with algebra. Education is the key.

Solve the equation for what you need with what you have given. Algebra - it's what is needed to figure out the equation. Got Algebra?



CE has ideology, not algebra. They're different ;)

[rwguinn]

I have a question about Appendix B of Ryan Mackey’s paper, “On Debunking 9/11 Debunking” (Version 2.1).

On p. 197, Mackey acknowledges that the acceleration will be nonconstant, “but over time will roughly follow an average acceleration”. He then states that "we can estimate the average acceleration" by using the equation h = ½ a t2.

However, my understanding is that this equation is applicable only for cases of constant acceleration; the equation does not give the average acceleration in cases of nonconstant acceleration.

Right?
Nope--
The equation is correct for a constant, or in this case, average acceleration. High school physics. Us the equation for distance, then. since distance and time are known quantities, solve for a.
He is using known distance and measured time to determine the average acceleration, so what he actually uses is a=(2*h/t2

[fourtoe]

Expand on your non-responsive statement. If you can't then why would you speak up when the equation is used to find an average acceleration? You proved in a short post you are not on speaking terms with algebra. Education is the key.

Solve the equation for what you need with what you have given. Algebra - it's what is needed to figure out the equation. Got Algebra?



I think I might have misread CE's post because I thought she was actually answering the OP.

I always hear high school physics and I never took physics until I was in college--and it was calculus based. Is high school physics just trig based or something?

[TruthersLie]

I think I might have misread CE's post because I thought she was actually answering the OP.

I always hear high school physics and I never took physics until I was in college--and it was calculus based. Is high school physics just trig based or something?

KFC. Usually in high school, they give you the equations in simple algebra. You don't need the calculus because they don't tell you that you are doing calculus. That way it doesn't scare the high school students.

:)

[T.A.M.]

I have a question about Appendix B of Ryan Mackey’s paper, “On Debunking 9/11 Debunking” (Version 2.1).

On p. 197, Mackey acknowledges that the acceleration will be nonconstant, “but over time will roughly follow an average acceleration”. He then states that "we can estimate the average acceleration" by using the equation h = ½ a t2.

However, my understanding is that this equation is applicable only for cases of constant acceleration; the equation does not give the average acceleration in cases of nonconstant acceleration.

Right?

You wil be pleased to know that Ryan Mackey has an account here, so you can PM him your question.

TAM:)

[fourtoe]

KFC. Usually in high school, they give you the equations in simple algebra. You don't need the calculus because they don't tell you that you are doing calculus. That way it doesn't scare the high school students.

:)

Ahhh. I wish I took physics in high school, I didn't get interested in science and stuff until I was in college, sadly.

[rwguinn]

to clarify a bit the equation h=1/2*a*t2 defines the relationship between distance, acceleration and time.
The assumption is implicit in the equation that acceleration is unchanging during the time step (duration) used--i.e., average acceleration for that duration is what is used/what you get.

[Dave Rogers]

However, my understanding is that this equation is applicable only for cases of constant acceleration; the equation does not give the average acceleration in cases of nonconstant acceleration.

Right?

Well, if the acceleration is constant, then the average acceleration is equal to the constant acceleration, so the use of the term "average" is redundant, but I don't think that's quite what you meant. But CE's response, whether meant sincerely or not, is perfectly correct; it's used to estimate, not calculate, the average acceleration, and for the purposes of the appendix it gives a close enough result for the line of argument to be valid. Any physicist would recognise this sort of estimate as a perfectly ordinary, and perfectly acceptable, approach.

Dave

[R.Mackey]

Right, but it's good enough for a rough estimate - and Mackey's argument.


Yup. I believe it even says this in the text.

[triforcharity]

I am not a science guy, nor can I do much (read:hardly ANY) Algebra, but even I know when you are doing that kind of math, you have variables. There is also a margin of error that will occur, you just have to account for it.

Right?? Or am I really that dumb when it comes to math???

[dropzone]

You wil be pleased to know that Ryan Mackey has an account here, so you can PM him your question.But it's EVER so much more fun to do it publicly: If it is successful for the challenger, it's fun for him. If it is fun for the guy defending the title, say, R.Mackey, it's a public trashing of another Troother.

[Par]

Expand on your non-responsive statement. If you can't then why would you speak up when the equation is used to find an average acceleration? You proved in a short post you are not on speaking terms with algebra. Education is the key.

Solve the equation for what you need with what you have given. Algebra - it's what is needed to figure out the equation. Got Algebra?
CE has ideology, not algebra. They're different ;)But it's EVER so much more fun to do it publicly: If it is successful for the challenger, it's fun for him. If it is fun for the guy defending the title, say, R.Mackey, it's a public trashing of another Troother.


Unfortunate stuff.

[T.A.M.]

But it's EVER so much more fun to do it publicly: If it is successful for the challenger, it's fun for him. If it is fun for the guy defending the title, say, R.Mackey, it's a public trashing of another Troother.

Yes...I see.

Well I was merely providing him with the information so he might not be called out for "baiting" someone, if that WAS NOT his intent.

TAM;)

[Childlike Empress]

Unfortunate stuff.


Happens when prejudice clouds judgement.

The cheap shots aside, people who work with applied math have to be always cautious not to confuse the map with the territory, i.e. the equation/model with nature. Engineering students who work with huge safety factors, but do their calculations with four decimal places, will not pass the test.

beachnut just made himself look dirt dumb.

[Hokulele]

To try and get back to the OP, Mackey described his reasoning behind using the average acceleration here:

What about the Kurttila paper? Earlier in this paper, on page 117, we remarked that a similar calculation by Dr. Kurttila [182] was invalid. He used a homogeneous approximation for the structure, which we criticized as a poor model. However, in the calculation above, we have assumed an “average” rate of acceleration during the collapse, which also implies homogeneous structure. Why can we use this simplification while Dr. Kurttila cannot?



The reason is in the interpretation. In our calculation, we have estimated the total energy dissipated by the structure. This total energy is an aggregate quantity. We accept that the actual rate of energy dissipation is going to vary wildly from millisecond to millisecond, as each floor is hit by the descending mass, briefly resists, and then breaks leading to another short period where the resisting force is much lower. We do not know the magnitude of the peak force, but it doesn’t matter – we also do not know the duration of the peak force. We do, however, know the product of the two. If the peak force is high, it must only resist for a short period of time, or else we would get a different time of collapse. If the peak force is low, it must resist for a longer period of time. The total amount of work done is equal to the product of peak force times the distance over which the resistance acts – short if it resists for a brief period of time, long if not – which works out to be roughly constant averaged over the entire collapse, for any given collapse time. We never have any reason to calculate the peak force. We are satisfied with the average.