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3rd July 2011, 12:36 AM | #81 |
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on youtube from fkn newz
"another logical fallacy please contain yourself - the straw man argument - A straw man is a component of an argument and is an informal fallacy based on misrepresentation of an opponent's position.[1] To "attack a straw man" is to create the illusion of having refuted a proposition by replacing it with a superficially similar yet unequivalent proposition (the "straw man"), and refuting it, without ever having actually refuted the original position.[1][2]" how did you know ? |
3rd July 2011, 01:25 AM | #82 |
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I have a nagging suspicion he's seen this kind of thing before.
It might almost be worth (but in my heart I doubt it) going back and asking him to confirm that in his original equation he inserted a value for v when he should have used a value for a , and whether he realises yet that this mistake rendered his equation utterly meaningless. But by now he seems to have muddied the waters so thoroughly that he'll always have an escape route. It's the way Truthers like to operate. |
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3rd July 2011, 02:11 AM | #83 |
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This is the problem with arguing on YouTube-it's too hard to figure out the order of who said what when. What "Strawman" is he claiming you/we've created? Also, where did the value of A = 0.5 g suddenly come from? He's trying to do an analysis of a collision by just assuming values for all the really important unknown variables, and that has very limited utility in the real world. |
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3rd July 2011, 02:22 AM | #84 |
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i think it was because on of the posters who claims that the laws of physics were broken is a loner who has lived in the woods for over a decade thinking the world is gonna end and the government is gonna get him and i was just stating that the idea he knew anything about the laws of physics might be stupid. I am willing to concede that being a paranoid loner might not be a barrier to understanding the laws of physics but when you consider he's a truther i think it is. I just won't accept truther science and instead ask people who have studied maths and physics to rebut some of the claims truthers make. That's just me though. If i don't know about something i tend to track down those who might and see what they've got too say, that way i can reduce my ignorance and deal with truther arguments all in one go
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3rd July 2011, 03:52 AM | #85 |
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High speed collisions are mainly governed by momentum, not by material strength.
Collisions are more or less elastic/inelastic. Elastic means that all the kinetic energy is preserved, which in turn means that, when it's over and done with, none of the kinetic energy is converted into other forms of energy, most notably plastic material deformation. In laymen's terms: Totally elastic collisions means no damage at all. On the other extreme, a totally inelastic collision means that the maximum damage has occured to the colliding bodies. Also, in an inelastic collision, the two bodies stick together rather than bouncing off of one another, creating a composite body that contains both their mass. So let's calculate how much damage, in terms of energy lost in deformation, was done - maximum! - to the plane and the pole! Let's assume the numbers your buddy provided initially, even though they appear to be a bit too low. Momentum of the plane is p = m*v = 80,000kg * 154m/s = 12320000kgm/s Kinetic Energy of the plane is E = 1/2 m*v^2 = 0.5 * 80,000kg * 154^2m^2/s^2 = 948640000J Momentum of the pole before collision is 0 (because v=0) Energy of the pole before collision is 0 (because v=0) After a perfectly inelastic collision with the entire pole, the plane+pole still have the same total momentum, but a mass of 80,000kg + 150kg = 80,150kg. So the following holds true: p = 12320000kgm/s = 80,150kg * v from this follows that the speed of plane+pole after collision is v = 12320000kgm/s / 80,150kg = 153,71179m/s That's a speed reduction of less than 0.2%. The kinetic energy of the composite is E = 0.5 * 80,150kg * 153,71179^2m^2/s^2 = 946864629J. That's a loss of 1775371J, or 1.78MJ, which went into deformation of aluminium, steel, and what have you. Again, this is an energy loss of less than 0.2%. How much is 1.78MJ? It is, for example, the kinetic energy of a compact car (1300kg) going top speed (188km/h or 118mph) and crashing into massive concrete. Again, this is an upper bound of the energy that was converted into damage. We would have to deduct a lot because
I think it is pretty fair to assume that only 20% or less of my calculated value actually went into deformation work, which in turn means that velocity decreased not by 0.29m/s, but by 0.06m/s or less. Let's suppose the poles were hit at a height of 15m and were 20m high, and bent around a pivot at 0m. Then the plane and the pole would have stayed in contact for about 13m, which takes about 0.086s, during which the plane was decelerated by 0.058m/s. That's an acceleration of -0.675m/s^2. The net Force then excerted upon the plane (and the pole) would have been F = m*a = 80,000kg * 0.675m/s^2 = 54,000N. That's 0.4% of what your buddy came up with. |
3rd July 2011, 05:08 AM | #86 |
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this just in
"just had a lengthy discussion with a structural engineer -i phoned a friend hahah. He calculates impact forces for the MOD specifically missile and mine blasts as well as shock loading and other proper science - he says f=MA no doubt about it iti simplistic and ignores many minor attributes (including impulse, shear force/bolts etc ) but it will do for our purposes- then Stress (the deforming force) = Force over Area - so how many newtons is 100,000x05g - whenever your ready" any thoughts? |
3rd July 2011, 05:12 AM | #87 |
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Then is he saying that 0 force was involved then since there was no acceleration component? If a = 0, then F = 0 if he is going to insist on using F=ma as his equation. It is a self-defeating argument he has going on there.
ETA: Oystein has laid the problem out about as simply and straight-forward as can be for such a complex problem with so many unknown variable values. Your YouTube poster did not have the problem set up correctly in the first place and injecting F = ma had no meaning what-so-ever in the context he was using it. The character a represents acceleration, the change in velocity over a given time span, not velocity. Yes, F = ma is applicable as Oystein uses it once the a component is known:
Quote:
Update: Never mind, I asked him a simple question myself on the YouTube thread (I'm spcengineer2003 at YouTube). |
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3rd July 2011, 08:10 AM | #88 |
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As I asked before, where did the 0.5g come from? As that's the essential variable we need to calculate the force from F=MA, he can't just pull a number out of his ass. He needs to justify it, with some sort of calculation or measurement. So where did this value come from? ETA: Oh, and if he now wants to start talking about stress, we'll need another A, that is , the Area over which the force is applied. So what was the contact area of the wing and post? |
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3rd July 2011, 08:19 AM | #89 |
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i don't know where it came from and as it's such a lovely day here i feel we may have to wait till this evening to find out the answer.
This feels like an episode of CSI and i'm off to build a giant screen where i can move the angles around and see if i can make sense of the crime scene. |
3rd July 2011, 08:33 AM | #90 |
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BobHaulk,
you are copy&pasting my work from here to the youtube debate without acknowledging me. Neither here nor there. I consider that bad style. |
3rd July 2011, 09:01 AM | #91 |
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sorry pal, i did ask for help in arguing with a friend of mine but if you object then i will refrain from doing so anymore and i'll remove the offending posts. Will that make you feel better? After all there is a huge amount of stuff available here to help debunk truthers and i didn't really think anyone would mind, after all i've told my friend that i use the jref to avail myself of any relevant facts and he should get himself over here to debate with people such as yourself himself but if you don't want anything you write to be copied i won't.
You can fight the truthers all by yourself now without having to worry about me stealing your glory |
3rd July 2011, 09:06 AM | #92 |
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It's not personal, Bob. It's just good form.
(Look at my sig!) |
3rd July 2011, 09:06 AM | #93 |
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3rd July 2011, 09:07 AM | #94 |
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"Is your claim that the level of penetration is only governed by distance and not the material that is being penetrated?" - DGM |
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3rd July 2011, 09:12 AM | #95 |
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well i did ask for help, i never hid from my mate that i was using the jref and he should get himself over here so really what more could i do? But if he's not happy fair enough, won't happen again.
I've had some good pointers and quotes from others and they didn't seem to want me to put their names on the youtube posts but as i've said if some aint happy i'm happy to stop using their contributions and he can get his ass over to youtube and deal with things as he see's fit |
3rd July 2011, 09:15 AM | #96 |
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3rd July 2011, 09:18 AM | #97 |
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3rd July 2011, 09:24 AM | #98 |
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Actually, I was fairly offended to see my posts copy/pasted w/o attribution as well. Of course Oystein did a lot of actual calculations which most likely took him quite a bit of time to do. My stuff was just giving general direction.
In the real world, it is a very big no-no not to attribute the source of a direct quote. It is called plagiarism and definitely would get you kicked out of any reputable school |
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"Is your claim that the level of penetration is only governed by distance and not the material that is being penetrated?" - DGM |
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3rd July 2011, 09:33 AM | #99 |
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By the way, I post as PapaOystein in that YT discussion.
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3rd July 2011, 09:53 AM | #100 |
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i know that plagiarism is a big no no and that wasn't my intention especially when i have stated that i was no good at physics and was looking for pointers to debate with a friend the error of his ways. I was under the impression that the ideas and points raised were available to use and i make it clear to my mate that i use the jref as a source and he should come and debate. No intention was there to claim other peoples work as anything other than rebuttals too his claims.
Thanks for the info anyways |
3rd July 2011, 10:07 AM | #101 |
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They are available for you to use. However, that is different from a direct quote (copy/paste). I'm not upset about it at all, so no biggee. Just trying to help you avoid misunderstands in the future. If you notice my signature, I use a direct quote from a friend of mine and I attribute him as the source. Now, if you used my data (as someone on P4T did) in a study to prove something without attribution and then turn around and tell me I don't know what I'm talking about, then I get upset. Just keep in mind that sometimes people put hours, days, months or years into what might seem like a small post on JREF to you.
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"Is your claim that the level of penetration is only governed by distance and not the material that is being penetrated?" - DGM |
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3rd July 2011, 01:56 PM | #102 |
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I'm to understand that the poles at the Pentagon were FAA compliant, so I thought I would pull up the relevant standards as a reference.
Quote:
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"Is your claim that the level of penetration is only governed by distance and not the material that is being penetrated?" - DGM |
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3rd July 2011, 02:09 PM | #103 |
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Wow - I'm good!
I have no doubt that there is a huge element of good luck and random chance that got me this close. Had this YT-dumbo used the correct figures for mass and speed, my numbers would have been higher (I think). Or lower? Whatever. Differebt anyway. Also, I was only looking at the mass, not the mechanical stiffness of the poles. Another thing: This guy, fknnewz, now claims that an acceleration value of -154m/s^2 comes from the FDR. I say this is nonsense, it's still 15 times g, and the FDR would probably not even be able to record such values. Help!? |
3rd July 2011, 02:27 PM | #104 |
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I already posted the final g values from the FDR for him and the velocity change (4.5 knots/s) for the pole encounter. The guy is just an idiot and I've wasted enough time on him.
However, this ends the debate. By specification, the maximum force transfered to the wing was 58 kN. He can pull whatever number he wants out of his rear, but there is your upper limit for the encounter. |
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"Is your claim that the level of penetration is only governed by distance and not the material that is being penetrated?" - DGM |
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3rd July 2011, 02:37 PM | #105 |
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3rd July 2011, 03:25 PM | #106 |
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The poles are never used in his calculations:
"once again i must insist you show your sums - F=MA in relation to lamposts at pentagon (I could use any no of anomalous events contradicting the laws of physics) F = 80000kg (mass of 767) x 154 ms (official speed of planes) so F = 12320000 N or 2,769,646.178992 lbs So a force of over 2 million pounds occurs 5 times to parts of a flimsy Aluminium air frame and it flies on - no damage. Im sure you ll want to correct me if i have made an error." So why worry about them? |
3rd July 2011, 04:27 PM | #107 |
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First, F = ma is a product, not a sum. Second, the a is for acceleration, not the official speed of planes. So what do you do with that? If the guy does not even get the basics right, how will I ever show him with physics and math that the force exerted by the plane on the pole is really not that important. What is important is the force exerted by the pole to the plane. That is a kinetic energy and momentum problem which is upper-bounded by the break-away design of the pole's base. In this case, it breaks away at >58 kN. Some folks are just stupid. Sometimes we have to accept that and move on with our lives
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3rd July 2011, 04:28 PM | #108 |
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i'm finished too. Way to much stress to be getting on with.
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3rd July 2011, 04:34 PM | #109 |
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So, in layterms, the essential flaw of the youtube guy in the OP was that he assumed that the full "force" of the plane would work on itself while colliding with the pole, while in reality this force is determined by the resistance the weakest element involved - at the base of the pole - gives to the planes movement before failing, which Oystein found out by calculation and you found out by specification? edit: ah, I think I got that right, you just explained it above. |
3rd July 2011, 05:02 PM | #110 |
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Originally Posted by Childlike Empress
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3rd July 2011, 07:59 PM | #111 |
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At any rate, unless impacted at its center of gravity, I would think the pole would rotate around its CG. His incredulity at the short distance the poles traveled.
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3rd July 2011, 09:38 PM | #112 |
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Pretty much. Oystein freely admits he got lucky since the design failure point is a critical part of the equation. He simply used reason and orders of magnitude to come up with a reasoned estimate. As it turns out, the reasoned estimate is pretty darn close to the specification.
What we were unsuccessful in conveying to the YT poster is that he was using the wrong equation to solve the problem at hand. We never got past that point. |
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3rd July 2011, 11:27 PM | #113 |
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So, at the end of all this, is this guy trying to prove that the airplane should have just exploded over the highway leaving all it's wreckage there?
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4th July 2011, 12:16 AM | #114 |
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4th July 2011, 12:52 AM | #115 |
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The light poles were made by Acme. A fine company, as their product placement ads in the Roadrunner and Coyote documentaries have demonstrated...
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4th July 2011, 01:58 AM | #116 |
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The first essential flaw was that he doesn't know the difference between velocity and acceleration. From there on it got even more bizarre. Insofar as it's possible to sift any sense out of the absurdities he used, mathematically his calculation was equivalent to the assumption that the impact with the light poles must have exerted enough force on the plane to bring it to a dead stop in one second, and he correctly estimated that this force would be enough to shear the wings off. He made it worse, though, when his error was pointed out, by puffing himself up and insisting that he was still right.
Basically, everything he said is so error-riddled as to be hardly worth even correcting as long as he maintains his bad attitude. I hope he isn't taking physics in high school, because he's heading for a very bad fail if he is. Dave |
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4th July 2011, 02:21 AM | #117 |
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There is one more flaw in the OP (the referred argument): The reference to "parts of a flimsy Aluminium air frame ".
I assume some people still think planes are like in 1915. That flimsy airframe part is a wing designed to carry a 200 ton aircraft and more specifically a leading edge designed to take the continuous stress of cutting through air, rain and hailstorms at the speed of a pistol bullet. It is anything but flimsy! Hans |
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4th July 2011, 06:09 AM | #118 | ||||
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I would not normally make the effort of teaching high school physics, but as you all know, this topic does come up often in various forums. Also, the seriously misguided individual, fknnewz is responding to just about everyone, "do keep up - F=MA newtons cradle - post a sum of **** off - next". As I've explained to him already, YT is not the proper venue to fully explore the subject at hand and he has rejected invitations to join us here so that we can walk him through it. So, I thought I would develop the subject so that interested parties can see "a sum".
Oystein has already developed "a sum" based on the principles I am going to outline, so I am not going to duplicate his effort. His result agrees with the 58 kN FAA specification for light pole break-away, so his "sum" is a well-reasoned one. But the question is why is Oystein's "sum" valid while fknnewz's "sum" is completely erroneous? The answer of course is Newton's Second Law.
Quote:
Unfortunately, individuals such as fknnewz see that summary equation without understanding exactly what it means. As noted already, the 2nd Law deals with changes in momentum over time, which means that we are interested in is the sum of the forces (change in momentum) or, the first derivative of the momentum. Since the mass is a constant the equation reduces to, which is essentially the same as the common form above. What we have in this case is a body in motion (the plane) encountering a body at rest (the pole). Oystein set up the encounter in its proper context using the 2nd Law and momentum. The pole has no momentum, so the plane has to 'transfer' some of its momentum to the pole in order for the pole to move. However, the pole is not an isolated body sitting at rest. It is anchored to the ground making it part of the Earth. Unless some other factor comes into play, the plane will have to transfer all of its momentum to the pole and come to a dead stop. That is one of the fallacies made by fknnewz. Oystein simply omits the problem of the pole being anchored and solves for the pole only to simplify the problem. At this point, Newton's Third Law comes into play.
Quote:
The force exerted on the plane's wing is limited to the break-away specification, in this case <= 58 kN. As pointed out in Oystein's solution, this force can be applied to the plane's momentum to determine the resultant momentum change. If the mass remains constant, then the loss will be reflected by a reduction in the plane's velocity (or speed). As Oystein has demonstrated, this reduction would have been negligible. So hopefully this will help fknnewz understand the problem better. However, in light of his most recent comment, I doubt it. But at least it might help some of those who choose to continue the debate.
Quote:
Update: The guy has gone into total melt-down over on the YT thread. I've seen temper-tantrums before, but geessshhh |
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4th July 2011, 06:59 AM | #119 |
Penultimate Amazing
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May I nitpick?
I think you are missing something. The 54kN max required to buckle the pole are a static force. I am convinced that FAA specification ignores, for practical purposes, the additional forces needed to accelerate the mass of the pole. My calculation went exclusively after the latter: The force needed to accelerate the pole to match the speed of the wing, over the span of some feet/milliseconds while the two are in contact (until the pole has rotated sufficiently down to the ground to be out of the way). I assumed the force to break the base of the pole to be 0 - a pre-cut pole, if you like. So maybe, in a worst-case scenario, the force suffered by the wing would be my 54kN plus the 58kN from the pole specification. And it is still mainly coincidence that my result was so close to that specification. |
4th July 2011, 07:13 AM | #120 |
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He's also tried to raise this comparison:
Quote:
http://www.youtube.com/comment?lc=k4...P83l19gfZ5rs-M Let's assume for a moment that the poles hit in this case are the same, with the 58 kN specification. Using F=ma, and using a = 10 m/s/s (the acceleration due to gravity), this 58 kN is equivalent to the weight of 5800 kg. http://en.wikipedia.org/wiki/Gulfstr...fstream_III.29
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Compare that to: http://en.wikipedia.org/wiki/Boeing_757#Specifications
Quote:
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