Full Version: Smrekar's Dialog on Lightpoles and Momentum
Smrekar
QUOTE (dMole @ Jun 13 2008, 09:16 PM)
Looking at the momentum (p) in this alleged inelastic collision:

p = mass * velocity

530 mph ~= 777.3333333333 fps [or approx. 460.5574082073434 knots]

Since the amount of fuel or passenger load in AA77 can be disputed, let's use the operating empty weight for a 757-200 for a "light" estimate:

Operating empty with P&W engines 57,840kg (127,520lb), with RB211s 57,975kg (127,810lb). Basic max takeoff 99,790kg (220,000lb), medium range MTOW 108,860kg (240,000lb), extended range MTOW 115,665kg (255,000lb) or 115,895kg (255,550lb).

http://www.airliners.net/aircraft-data/stats.main?id=101

127,810lb * 777.3333333333 fps = 99350973.3333334 ft-lb/sec [of B757-200 momentum]

Let's "guess" a 1% momentum transfer in this inelastic collision (a quite conservative estimate):

99350973.3333334 ft-lb/sec * 0.01 = 993509.733333334 ft-lb/sec [of light pole momentum]

Now dividing by the assumed 200 lb. light pole mass,

993509.733333334 ft-lb/sec / 200 lb = 4967.5486666667 ft /sec [of light pole velocity]

Just for comparison, most rifles fire a bullet at 2000-4000 ft/sec.

Is a broken windshield all that we would expect to see happen to Lloyd's? cab?

I'm sorry, but this analysis is simply impossible from a physical standpoint. It assumes that an object traveling at 777 ft/s hitting a stationary object could propel the stationary object along it's trajectory at over 4000 ft/s. This is, I believe, quite impossible.
You're welcome to source me to an article that would say otherwise, of course ... though I do doubt there would be any.
KP50
QUOTE (Smrekar @ Oct 13 2008, 09:33 PM)
I'm sorry, but this analysis is simply impossible from a physical standpoint. It assumes that an object traveling at 777 ft/s hitting a stationary object could propel the stationary object along it's trajectory at over 4000 ft/s. This is, I believe, quite impossible.
You're welcome to source me to an article that would say otherwise, of course ... though I do doubt there would be any.

Can we just ban this person now and save ourselves time later? All of this polite

"source me to an article"

bullshit. We know exactly what is happening, don't we? So let's not play the game.
Smrekar
QUOTE (KP50 @ Oct 13 2008, 10:53 AM)
Can we just ban this person now and save ourselves time later? All of this polite

"source me to an article"

bullshit. We know exactly what is happening, don't we? So let's not play the game.

Sure you can. But you'll have to ban one of your global administrators, as well - he's the one that has been demanding the same from me all the time.

The assumption made in the calculation is obviously wrong, and that won't change no matter what you do.
KP50
QUOTE (Smrekar @ Oct 13 2008, 09:55 PM)
Sure you can. But you'll have to ban one of your global administrators, as well - he's the one that has been demanding the same from me all the time.

The assumption made in the calculation is obviously wrong, and that won't change no matter what you do.

"We" won't have to do anything - I'm in favour of banning you because your whole purpose is to waste the time of people who have better things to do. Luckily I am watching cricket so really do have nothing better to do. Farewell as I don't think you'll be around in my morning.
dMz
QUOTE (Smrekar @ Oct 13 2008, 02:33 AM)
I'm sorry, but this analysis is simply impossible from a physical standpoint. It assumes that an object traveling at 777 ft/s hitting a stationary object could propel the stationary object along it's trajectory at over 4000 ft/s. This is, I believe, quite impossible.
You're welcome to source me to an article that would say otherwise, of course ... though I do doubt there would be any.

Actually, I don't think I mentioned trajectory anywhere. Which "it's trajectory" again- the 777 fps object or the stationary object? Thanks for agreeing that the hypothetical Lloyd Cab "lightpole 1" "collision" and 4000+ fps would be quite impossible though. I don't even think there was a collision BTW- that was sort of the point of the calculation. Savvy?

I value my privacy, and you likely wouldn't believe my credentials if I told you anyway Smrekar [so I won't], but you might be surprised. I've posted about 2200 clues around here for you though. Why don't you do a little more homework/reading and less unsourced "I believing?" Here are a few good places to start:

http://hyperphysics.phy-astr.gsu.edu/HBASE/colcon.html#c1

http://www.walter-fendt.de/ph11e/collision.htm
"The total momentum of the involved bodies is conserved, regardless whether the collision is elastic or inelastic. The movement of the common center of gravity (indicated by a yellow dot) is not influenced by the collision process."

http://en.wikipedia.org/wiki/Inelastic_collision
"An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision).

In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. ... Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum."
Smrekar
QUOTE (KP50 @ Oct 13 2008, 11:05 AM)
"We" won't have to do anything - I'm in favour of banning you because your whole purpose is to waste the time of people who have better things to do. Luckily I am watching cricket so really do have nothing better to do. Farewell as I don't think you'll be around in my morning.

I was under the impression that this website is about scientifically correct analysis of the 9/11. I have pointed out a flaw in one hypothesis presented on it. How is that wasting time? I'm quite sure working with an unworkable hypothesis would've wasted much, much more time very quickly.

Shunning me for it IS wasting time, however. But that's another story.
Smrekar
QUOTE (dMole @ Oct 13 2008, 11:09 AM)
http://hyperphysics.phy-astr.gsu.edu/HBASE/colcon.html#c1

http://www.walter-fendt.de/ph11e/collision.htm
"The total momentum of the involved bodies is conserved, regardless whether the collision is elastic or inelastic. The movement of the common center of gravity (indicated by a yellow dot) is not influenced by the collision process."

http://en.wikipedia.org/wiki/Inelastic_collision
"An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision).

In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. ... Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum."

Yes. But that wasn't the point I was trying to make. Obviously, the momentum will be conserved in our estimate, but the assumption was that 1% of aircrafts' momentum would carry down to the light poles. Since that would require them to fly away at roughly three times the speed of sound, and four times the speed of the impacting aircraft, the assumption cannot be correct, and the calculation and it's conclusions are meaningless.
KP50
QUOTE (Smrekar @ Oct 13 2008, 10:10 PM)
I was under the impression that this website is about scientifically correct analysis of the 9/11. I have pointed out a flaw in one hypothesis presented on it. How is that wasting time? I'm quite sure working with an unworkable hypothesis would've wasted much, much more time very quickly.

Shunning me for it IS wasting time, however. But that's another story.

Nice try but I don't think enigmatic comments suit you. Farewell once more.

And mods, feel free to delete this pointless exchange whenever you want.
dMz
QUOTE (Smrekar @ Oct 13 2008, 02:33 AM)
I'm sorry, but this analysis is simply impossible from a physical standpoint. It assumes that an object traveling at 777 ft/s hitting a stationary object could propel the stationary object along it's trajectory at over 4000 ft/s. This is, I believe, quite impossible.
You're welcome to source me to an article that would say otherwise, of course ... though I do doubt there would be any.

OK- we've got Smrekar's unsourced opinion and belief of "simply impossible." I provided collision dynamics, momentum transfer, inelastic collisions (all with hyperlinked sources). I even did the math for everyone, showing my steps for independent verification.

The reason that I didn't provide a link to a formula for momentum ( p = m * velocity ) is because I have it memorized from college. Perhaps Smrekar has a problem with that, too (or possibly physics in general).
Smrekar
QUOTE (dMole @ Oct 13 2008, 12:16 PM)
OK- we've got Smrekar's unsourced opinion and belief of "simply impossible." I provided collision dynamics, momentum transfer, inelastic collisions (all with hyperlinked sources). I even did the math for everyone, showing my steps for independent verification.

The problem is that the links you provided do not even address the issue I have pointed out. Can you point out which one examines the case where the impacted object travels faster after the impact than the impacting object before the impact?
None.
It's impossible to prove the negative, so I can't really source you anything, I'm afraid.

You can, of course, source me something that does speak of that option. None of your links thus far do that.

QUOTE
The reason that I didn't provide a link to a formula for momentum ( p = m * velocity ) is because I have it memorized from college. Perhaps Smrekar has a problem with that, too (or possibly physics in general).

No, not really, no. I just have an issue with blatantly wrong assumptions
There is absolutely nothing wrong with that formula or your calculations (aside from using about 10 decimal spaces more than you should). It's that you made a rather preposterous assumption along the way, and drew conclusions from that.

dMz
QUOTE (Smrekar @ Oct 13 2008, 03:15 AM)
Yes. But that wasn't the point I was trying to make. Obviously, the momentum will be conserved in our estimate, but the assumption was that 1% of aircrafts' momentum would carry down to the light poles. Since that would require them to fly away at roughly three times the speed of sound, and four times the speed of the impacting aircraft, the assumption cannot be correct, and the calculation and it's conclusions are meaningless.

So there is a point you are trying to make then here? Pray tell us what is it?

Now it's time for Smrekar to look yet more foolish, as Smrekar demonstrably does NOT understand the scientific definition of "assumption." When I say "assuming" it has a very specific meaning BTW. I've been waiting to see if Smrekar was aware of that meaning.
------------------
http://en.wikipedia.org/wiki/Fallacy_of_qu..._out_of_context

Fallacy of quoting out of context

The practice of "quoting out of context", sometimes referred to as "contextomy," is a logical fallacy and type of false attribution in which a passage is removed from its surrounding matter in such a way as to distort its intended meaning. Quoting out of context is often a means to set up "straw man" arguments. Straw man arguments are arguments against a position which is not held by an opponent, but which may bear superficial similarity to the views of the opponent. [1]
----------------------
http://www.thefreedictionary.com/assumption

Noun 1. assumption - a statement that is assumed to be true and from which a conclusion can be drawn; "on the assumption that he has been injured we can infer that he will not to play"
premise, premiss
posit, postulate - (logic) a proposition that is accepted as true in order to provide a basis for logical reasoning
major premise, major premiss - the premise of a syllogism that contains the major term (which is the predicate of the conclusion)
minor premise, minor premiss, subsumption - the premise of a syllogism that contains the minor term (which is the subject of the conclusion)
thesis - an unproved statement put forward as a premise in an argument
precondition, stipulation, condition - an assumption on which rests the validity or effect of something else
scenario - a postulated sequence of possible events; "planners developed several scenarios in case of an attack"
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Now before Smrekar can claim that "assumption" does not belong in science:

http://en.wikipedia.org/wiki/Philosophy_of_science

"Philosophy of science is the study of assumptions, foundations, and implications of science. The field is defined by an interest in one of a set of "traditional" problems or an interest in central or foundational concerns in science. In addition to these central problems for science as a whole, many philosophers of science consider these problems as they apply to particular sciences (e.g. philosophy of biology or philosophy of physics)."

http://web.utk.edu/~dhasting/Basic_Assumpt..._of_Science.htm

http://neutralscience.org/
--------------------
OK clearly Smrekar has deemed that 1% momentum transfer "cannot be correct." Carve that in stone, kids- some anonymous internet poster named Smrekar said so. So what percent of momentum transfer is acceptable to the omniscient Smrekar's a priori "beliefs"? 0.000%? Based upon what sourced reasoning or principle? I'll wait for a sourced answer here- likely a very long time.

You may want to think carefully before answering Smrekar, but I fear that you may lack that capacity (or perhaps have been retained not to).

Just out of curiosity, which of the illusionist's acolytes' username(s) might you be? Heeeyy- I might start a "sock pool" or poll here. Gentlemen- place your bets.

Now it's time for Smrekar to go play with balls. Something could be learned.

http://www.kingsford.org/khsWeb/rfs/elemsci/tmomen.html

--------------
My calculator posts many digits- deal with it (I've explained this here before). HINT: truncate or round, as applicable. My college years and work experience are "need to know"- you don't I'm afraid Smrekar. I'll let you wonder or do some research. You also might want to take a closer look at the whole Conservation of Momentum/collision thing again based upon your repeated question- I think you missed a few things there.

Ta.

EDIT: Split Smrekar's debate-oriented portion of topic per forum member request from:
What Brought Down The Light Poles?
http://pilotsfor911truth.org/forum//index....showtopic=13034
Smrekar
QUOTE (dMole @ Oct 13 2008, 12:50 PM)
OK clearly Smrekar has deemed that 1% momentum transfer "cannot be correct." Carve that in stone, kids- some anonymous internet poster named Smrekar said so. So what percent of momentum transfer is acceptable to the omniscient Smrekar's a priori "beliefs"? 0.000%? Based upon what sourced reasoning or principle? I'll wait for a sourced answer here- likely a very long time.

Using your own sources, the maximum acceptable transfer of momentum to the impacted object is the same as the weight difference between the two objects.
In this example, this is 247lb/127,810llb = 0,00193 = 0,193% per pole. The true(er) number is actually (1 - 0,00193) * 0,00193, because the aircraft will also slow down by that ratio, but that's way less than other estimates already in the system (e.g. aircraft weight) and should be ignored.

It's not a matter of belief, really. This is readily observable phenomenon.

Sources ... well I guess posting these again can't hurt.
http://hyperphysics.phy-astr.gsu.edu/HBASE/colcon.html#c1
http://www.walter-fendt.de/ph11e/collision.htm (esp. this one, play a little with it)
http://en.wikipedia.org/wiki/Inelastic_collision

QUOTE
My calculator posts many digits- deal with it (I've explained this here before). HINT: truncate or round, as applicable. My college years and work experience are "need to know"- you don't I'm afraid Smrekar. I'll let you wonder or do some research. You also might want to take a closer look at the whole Conservation of Momentum/collision thing again based upon your repeated question- I think you missed a few things there.

Ta.

What was your major and what do you work in?
You estimated the speed of the aircraft to accuracy of the same order of magnitude as the radius of an atom (10^-10m). HINT: you're supposed to truncate or round as applicable Showing that many decimal spaces isn't a sign of quality work, but of poor knowledge of the subject.

Please, point out what it is that I have missed. You just keep hinting at it, so I'm really wondering if you just aren't covering for yourself.
dMz
QUOTE (Smrekar @ Oct 13 2008, 05:05 AM)
You estimated the speed of the aircraft to accuracy of the same order of magnitude as the radius of an atom (10^-10m). HINT: you're supposed to truncate or round as applicable Showing that many decimal spaces isn't a sign of quality work, but of poor knowledge of the subject.

Perhaps you can source where I mentioned significant figures, accuracy, or uncertainty in the OP, Smrekar. Again, methinks you assume too much about my knowledge base and intentions. Again, you cast aspersions with no sources provided to back them up- I've noticed a pattern there.

Dude, this is an internet discussion forum, not an academic textbook- get over yourself already. Have you ever heard of "scope, " "hairsplitting," or "intended audience?" If my calculator's paste function isn't to your liking (it is called expediency BTW)- you could always find a different internet forum. Your opinion is noted again.

EDIT: Take a look at post #11 above again Smrekar.

Also, BTW- I said "guess" 1% (a conservative estimate), not "assume" above- I was testing your reading comprehension. You might want to read a bit closer and post a bit slower.

You are becoming tedious IMHO.
dMz
QUOTE (Smrekar @ Oct 13 2008, 05:05 AM)
Using your own sources, the maximum acceptable transfer of momentum to the impacted object is the same as the weight difference between the two objects.

Please provide a verifiable source for this assertion.
Smrekar
QUOTE (dMole @ Oct 13 2008, 01:25 PM)
Perhaps you can source where I mentioned significant figures, accuracy, or uncertainty in the OP, Smrekar. Again, methinks you assume too much about my knowledge base and intentions. Again, you cast aspersions with no sources provided to back them up- I've noticed a pattern there.

You haven't said anything about what's significant and what isn't, you just wrote the figures down. It is generally accepted that you write down significant figures and not the ones beyond it, since they are meaningless.
If there is a different reason for writing them down, please, say so. It might be legitimate, though I don't see how.

QUOTE
Dude, this is an internet discussion forum, not an academic textbook- get over yourself already. Have you ever heard of "scope, " "hairsplitting," or "intended audience?"

Yes, I have. Splitting the hairs would be, for example, writing down an impossibly accurate figure, then worm your way out that you never said those numbers were accurate.

QUOTE
Also, BTW- I said "guess" 1% (a conservative estimate), not "assume" above- I was testing your reading comprehension. You might want to read a bit closer and post a bit slower.

You guessed 1%, and stated it is a conservative estimate. Since neither is corroborated by any sources, it mets the criteria of an assumption.

QUOTE
Please provide a verifiable source for this assertion.

I gave you some links, I believe? They aren't there just for show, you know.

The proof for it is in the Wikipedia article:

This is actually sufficient, but you have to do this:

Cr = 0 for a perfectly inelastic collision in your example:

v2f = (M1V1 + M2V2) / (M1 + M2)

Note that V2 = 0, light pole was stationary.

Therefore:

v2f = M1V1/(M1 + M2)
v2f = (M1/(M1 + M2)) * V1

Note that M2 > 0, light pole had a mass. M1 > 0, aircraft had a mass as well.

M1 < (M1 + M2)

(M1/(M1 + M2)) < 1

v2f < V1

This is all very elementary and not a subject to opinion.
grizz
QUOTE (Smrekar @ Oct 13 2008, 03:31 AM)

I have no idea what you guys are talking about. But I assure you that dMole definitely does know what he's talking about. He's likely overqualified for this discussion.
dMz
QUOTE (Smrekar @ Oct 13 2008, 04:31 AM)
The problem is that the links you provided do not even address the issue I have pointed out.

So now you're going to use the links that I provided you to "prove" "the maximum acceptable transfer of momentum to the impacted object is the same as the weight difference between the two objects."

Again, please provide a source for the "maximum transfer of momentum" part of your assertion- you failed to do so.

Now going back to the inelastic collision Wiki, your reading comprehension suffers yet again. Although I didn't assume a "perfectly inelastic collision in your example", you have when you set C_R = 0 in your "proof." Had you read more carefully, you would have noted that: "In a perfectly inelastic collision [1], i.e., a zero coefficient of restitution, the colliding particles stick together."

http://en.wikipedia.org/wiki/Inelastic_col...astic_collision

http://en.wikipedia.org/wiki/Inelastic_collision

So are you now claiming that the lightpoles all stuck to a B757-200 then Smrekar? I'm fairly certain that this conclusion is not supported by photographic evidence- see the Pentagon forum (or the OP) here. You could stand to improve your reading comprehension considerably before doing so IMHO. Assume less, understand more.
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QUOTE
This is actually sufficient, but you have to do this:

Cr = 0 for a perfectly inelastic collision in your example:
- Smrekar, 2 posts above this one.

BTW, you've also dodged several of my questions above Smrekar. I suppose I'll need to number them for you in my next reply Smrekar for your convenience in answering them.
dMz
Although I've addressed my calculator twice before (posts #11 and #13 above), our apparently-obsessive "It might be legitimate, though I don't see how" Smrekar might need a picture drawn for him?:

Again, HINT: Truncate or round, as applicable. Have you ever heard of a "back of the napkin" calculation Smrekar? I'm primarily interested in the underlying concepts, trends, and results, not in bean-counting single digits. You can wait for my v2.0 of my internet discussion forum posts, I suppose.

EDIT:
3 User(s) are reading this topic (0 Guests and 1 Anonymous Users)
2 Members: dMole, Turbofan
dMz
Gee, where'd "anonymous" go TF?
dMz
QUOTE (Smrekar @ Oct 13 2008, 06:23 AM)
...
I gave you some links, I believe? They aren't there just for show, you know.

The proof for it is in the Wikipedia article:
...

irony
Definition:

The use of words to convey the opposite of their literal meaning; a statement or situation where the meaning is contradicted by the appearance or presentation of the idea. Three kinds of irony are commonly recognized:

1. Verbal irony is a trope in which the intended meaning of a statement differs from the meaning that the words appear to express.
2. Irony of situation involves an incongruity between what is expected or intended and what actually occurs.
3. Dramatic irony is an effect produced by a narrative in which the audience knows more about present or future circumstances than a character in the story.

* What Is Irony?
* "A Modest Proposal," by Jonathan Swift
* Sarcasm
* Accismus

EDIT:
QUOTE
"Can you point out which one examines the case where the impacted object travels faster after the impact than the impacting object before the impact?
None.
It's impossible to prove the negative, so I can't really source you anything, I'm afraid."- Smrekar

http://www.fearofphysics.com/Collide/collide.html

Dump truck vs. stationary kid's scooter:
http://www.fearofphysics.com/cgi-bin/colli...m&mode=wrap
dMz
QUOTE (Smrekar @ Oct 13 2008, 06:23 AM)
Yes, I have. Splitting the hairs would be, for example, writing down an impossibly accurate figure, then worm your way out that you never said those numbers were accurate.
...
You guessed 1%, and stated it is a conservative estimate. Since neither is corroborated by any sources, it mets the criteria of an assumption.

I'll assume that Smrekar meant "meets the criteria" (at source provided by me again above) then?

A quiz for Smrekar:

1/3 = ?

2/3 = ?

pi() = ?
bjohnson
Where did he go, this topic was just gettin' good?
Even if you do a perfectly inelastic collision, I believe you still end up with the light poles moving about 490MPH, correct? There is no photographic evidence of damage resulting from the alleged plane and light pole impact as a collision that obeys all known laws of physics. That leaves us with three scenarios
1) The laws are wrong and we need to change the laws of physics(similar to how a new phenomenon for building collapses had to be created to explain WTC7)
2) Photographic evidence showing the resting places of the downed light poles are fake
3) The alleged collision never took place and the poles were staged as part of a massive military deception

Sounds like Smrekar's vote is for #1.
dMz
BTW, I never did say anything about precision or accuracy. From my post, which you quoted at #1 above Smrekar, so you must have seen it before:

QUOTE
Looking at the momentum (p) in this alleged inelastic collision:

p = mass * velocity

530 mph ~= 777.3333333333 fps [or approx. 460.5574082073434 knots]

Since the amount of fuel or passenger load in AA77 can be disputed, let's use the operating empty weight for a 757-200 for a "light" estimate:

http://en.wikipedia.org/wiki/Table_of_mathematical_symbols

"(Note that ~ is used for an approximation that is poor, otherwise use ≈ .)" [Sorry, my keyboard doesn't have one of that key, so I used "~=" in the OP. Again, learn to deal with it.]

Your reading comprehension drops again Smrekar, I'm afraid. [EDIT: It's not somehow linked to the DJIA is it?]
---------------------
Back to hairsplitting:

http://dictionary.reference.com/browse/hairsplitting

"hairsplitting
–noun
1. the making of unnecessarily fine distinctions.
2. characterized by such distinctions: the hairsplitting arguments of a political debate.
[Origin: 1820–30; hair + splitting]

—Related forms
hairsplitter, noun

—Synonyms 2. quibbling, niggling, nitpicking, captious.
dMz
One of the other 4 moderators is welcome to take a turn now (I'm betting sometime around 2:30 to 4:30 early tomorrow morning). I'm ready to toss this "fish" back in the drink.

EDIT: I wonder if "Smreckar" has seen this thread yet:

Ryan Mackey, Nasa Scientist Proved Incorrect, or lying if he won't concede
http://pilotsfor911truth.org/forum//index....showtopic=14765

Looks right up "Smreckar's" alley to me, based upon speech patterns above.
dMz
QUOTE (Smrekar @ Oct 13 2008, 03:10 AM)
I was under the impression that this website is about scientifically correct analysis of the 9/11.

Correct.
QUOTE
I have pointed out a flaw in one hypothesis presented on it.

Incorrect.
Smrekar
QUOTE (dMole @ Oct 13 2008, 10:41 PM)
So now you're going to use the links that I provided you to "prove" "the maximum acceptable transfer of momentum to the impacted object is the same as the weight difference between the two objects."

Again, please provide a source for the "maximum transfer of momentum" part of your assertion- you failed to do so.

No, I did, you just don't grasp it
Unless the mass of the object in question changes - which did not happen here - the speed is the only factor left in the formula. Since that can't change by an infinite amount - as shown - there is also a maximum amount of momentum that can be transferred.

If you can't grasp something this basic is not my fault.

QUOTE
Now going back to the inelastic collision Wiki, your reading comprehension suffers yet again. Although I didn't assume a "perfectly inelastic collision in your example",

Then your calcultion is incorrect in any event. You assumed complete conservation of momentum, which is only valid in perfectly inelastic collisions. Sorry
Incidentally, what coefficient of ellasticity did you use?

QUOTE
you have when you set C_R = 0 in your "proof." Had you read more carefully, you would have noted that: "[color="#008000"]In a perfectly inelastic collision [1], i.e., a zero coefficient of restitution, the colliding particles stick together."
http://en.wikipedia.org/wiki/Inelastic_collision

See above

QUOTE
So are you now claiming that the lightpoles all stuck to a B757-200 then Smrekar? I'm fairly certain that this conclusion is not supported by photographic evidence- see the Pentagon forum (or the OP) here. You could stand to improve your reading comprehension considerably before doing so IMHO. Assume less, understand more.
------------
- Smrekar, 2 posts above this one.

Nope, but I didn't persume to use conservation of momentum to futher my calculations. Like, uh, you did

QUOTE
BTW, you've also dodged several of my questions above Smrekar. I suppose I'll need to number them for you in my next reply Smrekar for your convenience in answering them.

Yes, please do, because I can't actually find any
Smrekar
QUOTE (bjohnson @ Oct 14 2008, 01:11 AM)
Where did he go, this topic was just gettin' good?
Even if you do a perfectly inelastic collision, I believe you still end up with the light poles moving about 490MPH, correct? There is no photographic evidence of damage resulting from the alleged plane and light pole impact as a collision that obeys all known laws of physics. That leaves us with three scenarios
1) The laws are wrong and we need to change the laws of physics(similar to how a new phenomenon for building collapses had to be created to explain WTC7)
2) Photographic evidence showing the resting places of the downed light poles are fake
3) The alleged collision never took place and the poles were staged as part of a massive military deception

Sounds like Smrekar's vote is for #1.

There is a fourth option, actually.
4) The light poles were not struck in their center of gravity, and were not free standing, but fixed to the ground Furthermore, they distorted with impact.
Each of the three factors is sufficient to render the initial calcultion incorrect, and it is fair to say there is some evidence for each of them. Photographic evidence shows that light poles were bent, light poles tend to be fixed to the ground and it is not possible for a 757 to strike lightpoles in their center of gravity without carving a hole in the ground with it's fueslage. Certainly there is no evidence they were struck in their center of gravity, especially since they appear bent on top.

As to where I was, I do like to do something else than hang on internet forums where I'm shunned for showing 'aeronautical engineers with years of experience' that their grasp of basic kinemtics is non-existent
Smrekar
QUOTE (dMole @ Oct 14 2008, 02:17 AM)
Incorrect.

Why don't you show me why do you think 1% of momentum of the aircraft should carry down to the light pole?
Or could, for that matter.

Don't reffer me to some link which shows you wrong, please. Show an equation of sorts that shows that it is possible.
Smrekar
QUOTE (Smrekar @ Oct 14 2008, 11:50 AM)
Then your calcultion is incorrect in any event. You assumed complete conservation of momentum, which is only valid in perfectly inelastic collisions. Sorry
Incidentally, what coefficient of ellasticity did you use?

Typo - I should've said what coefficient of restitution did you use, obviously
dMz
QUOTE (Smrekar @ Oct 14 2008, 03:50 AM)
No, I did, you just don't grasp it
Unless the mass of the object in question changes - which did not happen here - the speed is the only factor left in the formula. Since that can't change by an infinite amount - as shown - there is also a maximum amount of momentum that can be transferred.

If you can't grasp something this basic is not my fault.

Neither is it my fault that your reading mis-comprehension won't allow you to grasp my usage of "guess" and estimate from back in June of this year. Now that it is 4 months later, I have found better values for the "main" light pole mass (247 rather than 200 lbs). If one is going to go seeking an "ultra-precise" hypothetical analysis of the light pole in question (as Smrekar appears to be seeking here) [for a hypothetical event that I don't think ever happened BTW ] then shouldn't one also use the transverse light pole section and lamp head masses, too? Mine was after all, only an estimate from back in June there Smrekar. You seem to have forgotten that estimate part since your post #15 on this thread BTW.

Again, unless you can provide a source for your "maximum momentum limit" assertion, I'll need to assume that you do not have one. Your "'derivation" has already been showed to be severely flawed, as 5 light poles apparently did NOT stick with "perfect inelasticity" to a "B757-200" along the way to the Pentagon.

QUOTE
Then your calcultion is incorrect in any event. You assumed complete conservation of momentum, which is only valid in perfectly inelastic collisions. Sorry

Wait a minute, at post #10 you said "[b]There is absolutely nothing wrong with that formula or your calculations[/b] (aside from using about 10 decimal spaces more than you should). It's that you made a rather preposterous assumption along the way, and drew conclusions from that."
So which is it? Or are you unable to even remember what you have already posted here Smrekar? Are there more than one of "you" perhaps?

To clarify for Smrekar- I assumed that momentum would be transferred in any "collision," and that Conservation of Momentum is the proper way to analyze it. Collision physics is actually fairly fleeting and complex, despite what Smrekar has already claimed. Then I "guessed" a 1% momentum transfer in my estimate, which actually does look excessive, 4 months later. If we trust Smrekar's "perfectly inelastic collision" analysis, then my "guess" was 1.00/0.193 ~= 5.18 times too high for an alleged 530mph aircraft collision of unknown duration. FYI, this was the first consideration of light pole "momentum" that I had seen anywhere by anyone, and there wasn't much information out there- had to assume [previously discussed here] in order to proceed. BTW Smrekar, 5.18 << "infinity" FYI.

QUOTE
Incidentally, what coefficient of ellasticity did you use?

I didn't use a coefficient of of restitution in my estimate back in June. I don't think we have enough information about the "V_2f" in this case to determine that exactly, do we? If so, I'd love to look at something sourced there, Smrekar.

BTW, a typo would actually be say "ellasticity" instead of "elasticity." This appears to be something much different, as are modulus of elasticity and coefficient of restitution drastically different things, Smrekar.

http://en.wikipedia.org/wiki/Elastic_modulus

http://en.wikipedia.org/wiki/Coefficient_of_restitution

So what did you obtain for the light pole base shearing force(s) then Smrekar? You appear to be alluding as much in your "light poles tend to be fixed to the ground" 4th option above. What percentage of the B757-200 momentum have you calculated there Smrekar (and how many times)?

QUOTE
Nope, but I didn't persume to use conservation of momentum to futher my calculations. Like, uh, you did

Yeah about that. That's where you really blew it Smrekar- if you'll go back to the original Wiki that I provided , you'll find in the third paragraph that: "Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum."

http://en.wikipedia.org/wiki/Inelastic_collision

QUOTE
Yes, please do, because I can't actually find any

Actually with your demonstrated inability to comprehend written matter, I'm growing quite weary of conversing with you. I actually have done another analysis that is less of a "back of the napkin" estimate, but it involves several mathematical assumptions Smrekar, so I'll spare you the details. Also, there is a more applicable collision concept to use here Smrekar, but I'll let you discover the particulars for yourself.

FYI, there is a board limit on smileys per post. Also, I might be away from the internet to take care of some business Smrekar, so someone else will need to play with you I suppose.

Buh bye, in case you're gone when I get back.

EDIT: Some will place a higher priority on fundamental Laws of Physics than on counting decimal places and on circular, sophist sematics, and again- deal with it Dude.

EDIT: Oh gawwwd... Proofreading, it's the 4th paragraph in the Inelastic Collision Wiki linked above. I'd prefer to use sources other than Wiki, but my books are not at my internet access point. So we'll use that in the interests of availability for the masses.
dMz
QUOTE (Smrekar @ Oct 14 2008, 03:56 AM)
As to where I was, I do like to do something else than hang on internet forums where I'm shunned for showing 'aeronautical engineers with years of experience' that their grasp of basic kinemtics is non-existent

Where are you quoting that from Smrekar? My grasp of kinematics is just fine BTW- I even know how to spell it. Your spin and interpretation(s)? of what I have written are what appear to me to be disconnected. Until today, I hadn't listed the assumptions made in that estimate, if you can read carefully.

It is nice to see that you've found the courage to log in under username today. Kudos for that.
dMz
QUOTE (Smrekar @ Oct 14 2008, 04:05 AM)
Why don't you show me why do you think 1% of momentum of the aircraft should carry down to the light pole?
Or could, for that matter.

Don't reffer me to some link which shows you wrong, please. Show an equation of sorts that shows that it is possible.

While I've showed Smrekar the same Wiki link about 5 times now that shows where Smrekar's "assumptions" went wrong, here's a non-Wiki source. I was truly hoping that Smrekar might have learned something along the way (and he? might actually have, given more reading and less quoting/posting/emoticonning).

Although it's not about 1.0% "guesses" and equations, it is precisely about the fundamental and underlying physical Laws that should have been the topic back at about post #2 or 3. I think you can all guess my grades on Smrekar's reading comprehension and performance on the several "quizzes" above (some were very "pop" and unannounced- I'm "good" like that, and I've still got several dangling unanswered).
-----------

Optional Unit V: Applications of Kinematics and Dynamics
A. Momentum

2. The Law of Conservation of Momentum

Key Concepts

The total momentum of an isolated system does not change. (Law of Conservation of Momentum) Another way of stating this is that the initial momentum of an isolated system is equal to its final momentum.

An isolated system is one in which no net external force acts on the system.

Momentum is conserved regardless of whether the interactions in the system occur in more than one dimension.

--------------------
It's been fun gang. How's your "pimp hand" doing TF?
dMz
QUOTE (dMole @ Oct 13 2008, 03:09 AM)
Why don't you do a little more homework/reading and less unsourced "I believing?" Here are a few good places to start:

http://hyperphysics.phy-astr.gsu.edu/HBASE/colcon.html#c1

http://www.walter-fendt.de/ph11e/collision.htm
"The total momentum of the involved bodies is conserved, regardless whether the collision is elastic or inelastic. The movement of the common center of gravity (indicated by a yellow dot) is not influenced by the collision process."

http://en.wikipedia.org/wiki/Inelastic_collision
"An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision).

In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. ... Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum."

Hey looky here- from my first post (#5 above) on the split, primarily-semantical (for one of us) thread. Him not too B R I T E.
dMz
QUOTE (Smrekar @ Oct 13 2008, 05:05 AM)
Using your own sources, the maximum acceptable transfer of momentum to the impacted object is the same as the weight difference between the two objects.
In this example, this is 247lb/127,810llb = 0,00193 = 0,193% per pole. The true(er) number is actually (1 - 0,00193) * 0,00193, because the aircraft will also slow down by that ratio, but that's way less than other estimates already in the system (e.g. aircraft weight) and should be ignored.

I guess I'll have to attempt to decipher what Smrekar means here, since I haven't seen a source for this assertion.

BTW, difference means subtraction:

http://www.mathsisfun.com/definitions/difference.html

"The result of subtracting one number from another. How much one number differs from another.

Example: The difference between 8 and 3 is 5. "

It appears that Smrekar actually found a mass ratio above though...

http://www.mathsisfun.com/numbers/ratio.html

"Ratios
A ratio shows the relative sizes of two or more values.

Ratios can be shown in different ways:
3 : 1 Using the ":" to separate example values
¾ as a fraction, by dividing one value by the total (3 out of 4 boxes are blue)
0.75 as a decimal
75% as a percentage"

What was that again about not understanding the "basics," Smrekar?? "If you can't grasp something this basic is not my fault." -- Smrekar, post #29 above

EDIT: Just so everyone knows, Smrekar is currently at a 0.00% warn level in the Forum system...
Smrekar
QUOTE (dMole @ Oct 14 2008, 06:40 PM)
Neither is it my fault that your reading mis-comprehension won't allow you to grasp my usage of "guess" and estimate from back in June of this year. Now that it is 4 months later, I have found better values for the "main" light pole mass (247 rather than 200 lbs). If one is going to go seeking an "ultra-precise" hypothetical analysis of the light pole in question (as Smrekar appears to be seeking here) [for a hypothetical event that I don't think ever happened BTW ] then shouldn't one also use the transverse light pole section and lamp head masses, too? Mine was after all, only an estimate from back in June there Smrekar. You seem to have forgotten that estimate part since your post #15 on this thread BTW.

Well, seeing someone or something deleted my answer to this one, I'll just write it down again.
The mass of the light pole was in no way important as to why your calculation was incorrect. Anyone who understands the case could tell you that - but you just went ahead and calculated something impossible from it anyway, and claimed it to be factual.

QUOTE
Again, unless you can provide a source for your "maximum momentum limit" assertion, I'll need to assume that you do not have one. Your "'derivation" has already been showed to be severely flawed, as 5 light poles apparently did NOT stick with "perfect inelasticity" to a "B757-200" along the way to the Pentagon.

You provided it yourself.

Assuming a perfect elastic collision, the maximum possible speed of the impacted object after the collision is twice the speed of the impacting object. Therefore the maximum amount of momentum transfered can be calculated directly from this equation.
p2(max) = m2 * v2f(max) = m2 * 2 * v1 * (m1/(m1+m2))

Where:
p2(max) - maximum momentum of the impacted body after the collision
m2 - mass of the impacted body (lightpole, in our case)
m1 - mass of the impacting body (757, in our case)
v2f(max) - maximum possible speed of the impacted body after the collision
v1 - speed of the impacting body before the impact

For perfectly inelastic collision:
p2(max) = m2 * v2f(max) = m2 * v1 * (m1/(m1+m2))

Or, for any collision:
p2(max) = m2 * v2f(max) = m2 * (1 + Cr) * v1 * (m1/(m1+m2))

This is derived directly from your own source, which you claim to understand intimately. I already showed you why I'm correct with the v2f(max), and I will not do it again. The rest is just the same old equation of p = m * v.
Therefore, I believe I have adequately shown this to be correct. If not, I can write you down how you get to this, but you have to admit you don't know what to do with the provided equations first (incidentally, where did you graduate and in what field?).

By the way:

QUOTE
Looking at the momentum (p) in this alleged inelastic collision:

It's quite safe to assume you assumed Cr = 0, judging from your opening statement alone.

QUOTE
b]There is absolutely nothing wrong with that formula or your calculations[/b] (aside from using about 10 decimal spaces more than you should). It's that you made a rather preposterous assumption along the way, and drew conclusions from that."[/color]
So which is it? Or are you unable to even remember what you have already posted here Smrekar? Are there more than one of "you" perhaps?

Your initial estimate is incorrect (to the point of being absurd). From that estimate you make a correct calculation, which gets to a wrong result. The calculation is correct - as I explicitly stated - but the initial estimate and the final result are not.

QUOTE
To clarify for Smrekar- I assumed that momentum would be transferred in any "collision," and that Conservation of Momentum is the proper way to analyze it. Collision physics is actually fairly fleeting and complex, despite what Smrekar has already claimed. Then I "guessed" a 1% momentum transfer in my estimate, which actually does look excessive, 4 months later. If we trust Smrekar's "perfectly inelastic collision" analysis, then my "guess" was 1.00/0.193 ~= 5.18 times too high for an alleged 530mph aircraft collision of unknown duration. FYI, this was the first consideration of light pole "momentum" that I had seen anywhere by anyone, and there wasn't much information out there- had to assume [previously discussed here] in order to proceed. BTW Smrekar, 5.18 << "infinity" FYI.

Interestingly enough, I didn't have access to any information which you didn't have, yet I was able to make the calculation, but you weren't. Quite interesting.
Where did you say you worked?

QUOTE
I didn't use a coefficient of of restitution in my estimate back in June. I don't think we have enough information about the "V_2f" in this case to determine that exactly, do we? If so, I'd love to look at something sourced there, Smrekar.

We do have sufficient information to determine your "conservative estimate" was approximately three times above being physically possible, and approximately five times above being plausible (seeing that coefficient of restitution will not be very high for the light pole or the plane).

QUOTE
So what did you obtain for the light pole base shearing force(s) then Smrekar? You appear to be alluding as much in your "light poles tend to be fixed to the ground" 4th option above. What percentage of the B757-200 momentum have you calculated there Smrekar (and how many times)?

Quite simply, I didn't. I didn't intend to calculate the speed of the light pole after the impact, nor did I pretend to have the sufficient information to do it. I only pointed out your calculation was way beyond the realm of reality.

QUOTE
Actually with your demonstrated inability to comprehend written matter, I'm growing quite weary of conversing with you. I actually have done another analysis that is less of a "back of the napkin" estimate, but it involves several mathematical assumptions Smrekar, so I'll spare you the details. Also, there is a more applicable collision concept to use here Smrekar, but I'll let you discover the particulars for yourself.

Go ahead, let's see what you can do
Smrekar
QUOTE (dMole @ Oct 14 2008, 08:08 PM)
While I've showed Smrekar the same Wiki link about 5 times now that shows where Smrekar's "assumptions" went wrong, here's a non-Wiki source. I was truly hoping that Smrekar might have learned something along the way (and he? might actually have, given more reading and less quoting/posting/emoticonning).

Although it's not about 1.0% "guesses" and equations, it is precisely about the fundamental and underlying physical Laws that should have been the topic back at about post #2 or 3. I think you can all guess my grades on Smrekar's reading comprehension and performance on the several "quizzes" above (some were very "pop" and unannounced- I'm "good" like that, and I've still got several dangling unanswered).
-----------

Optional Unit V: Applications of Kinematics and Dynamics
A. Momentum

2. The Law of Conservation of Momentum

Key Concepts

The total momentum of an isolated system does not change. (Law of Conservation of Momentum) Another way of stating this is that the initial momentum of an isolated system is equal to its final momentum.

An isolated system is one in which no net external force acts on the system.

Momentum is conserved regardless of whether the interactions in the system occur in more than one dimension.

You don't seem to understand the problem in the slightest

Assuming the momentum remained the same (certainly true in your approximation, not so in reality, because we aren't dealing in an isolated system with no external forces, but let's ignore that for now), the aircraft would retain more than 99% of the momentum, while the light pole would obtain less than 1%.
The numbers are approximately 99.8% and 0,2%, respectively. That's an overly generous estimate, not a conservative one - we assumed a too light 757, no distortion of the light pole and impact directly into the center of gravity, none of which are true, and each of whom would necessiate a lesser amount of momentum transferred to the pole.
I showed you why the light pole can't get that much momentum in the present example twice, I see no need to do so again.

Well, truth to be told, there is one, and one alone, option in which that can happen. It is so absurd, however, that it is also irrelevant to our present debate.