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#1 |
Graduate Poster
Join Date: Sep 2006
Posts: 1,542
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A Critical Analysis of JDX's math (split)
I decided to split this because the thread is full of alot of discussion of the larger issues surrounding this claim. I want to focus, specifically, on the analysis he presents, and mine.
Since I constantly harp on them for being imprecise, not citing their sources, not dealing with error-ranges, and not stating their assumptions, I've decided to repeat JDX's calculation, first, and then do it myself, properly. Purpose To calculate the height of Flight 77 as it passed over each of the 5 light poles, based on all available data. Methodology Since it has been established that absolute elevation according to the Flight Data Recorder's (FDR) altimeter is imprecise, we are going to construct a motion model and constrain it by the final impact point, the relative elevation changes from the altimeter, and the geometry/elevation of the terrain. From this, we are going to extrapolate the plausibility of hitting the light poles on the way to the pentagon. Data Sources: Lightpole distances: http://img137.imageshack.us/my.php?image=pentuy7.jpg FDR Data: http://i47.photobucket.com/albums/f1...ltfpmpaint.jpg I used the USGS elevation site (the same site he sources, and all the same numbers he uses, for elevation). Note: There is an inconsistency in this data (his data). The FDR's last blip appears to be 900-1000 feet away from the pentagon, and it also appears to be traveling at about 800ft/s, maximum. If those two numbers are correct, the FDR should have had another blip. That means one of the two pieces of data I am using is wrong (and/or the errors in each are large). Error Issues: Relative elevation in the altimeter - I have no idea how accurate these are, and did not account for any error whatsoever Distance errors - I didn't check google earth's precision, but barring a major error, these are relatively insiginificant. Eleveation errors - The site sources +- 7m, minimum. I DID NOT, however, account for this error. Good elevation numbers are (probably poorly) assumed. FDR Timing/Location - I didn't check the error of the locations/timing of the sample points in the FDR data source I used (the same one JDX used), I assumed it was accurate. Others: Probably more First, I repeated JDX's calculation and got this: ![]() JDX's analysis has one major assumption, he assumes a linear trajectory. First order motion is only possible if there is no acceleration, or there is no change in velocity. In order to check this, I plotted the downward average velocities according to the FDR: ![]() You will notice, as marker, there is very clearly acceleration present for the last 3 seconds of the flight data recorders results. This, immedietly, makes JDX's analysis suspect. In order to correct his mistake, I decided to figure out a bounded prediction on the acceleration, and try a second-order motion model (one with constant acceleration). I chose his estimate of a=0 to be the "upper" bound on acceleration. I chose the _worst_ of the three previous upward accelerations to be the "lower bound". Here is a graph of the predictions: ![]() Now, in order to compute a second order motion model, I will need to define several variables. Acceleration will be constant (and is defined by the slope of the prediction line). Initial downward velocity is chosen to be -66 ft/s, the amount of the previous second's average. Final position is taken to be at the pentagon impact point. The last parameters is how long this traversal time takes. How long does it take to go from the point where the plane is at -66 ft/s vertical velocity, until it hits the pentagon? This is a hard question is answer. The Mean Value Theorem, applied to the FDR recorder data, says that at at least one point between 1s and 2s before impact, the plane was at -66 downward velocity. Since this is a lower bound, I chose to use 2s as the more extreme scenario (I also ran 1.5 for fairness). Now that we have all our numbers, we use them to solve for y-initial, and we get a quadratic: y(t) = 0.5at^2 + vo * t + y0 where a is acceleration t is time vo is the initial velocity y0 is the initial height This equation should, at T=2.0 (or T=1.5 for the other one), should be hitting the pentagon. I have spare you the tedium of all the calculations. There are repeatable, and I have the excel file if anyone needs it. What follows is my graph showing the results: ![]() Conclusion What have we shown? That a plane hit the pentagon? No. What we've shown is that all that would be required for a plane to hit the pentagon, and all the lightpoles, given all of JDX's data, is for the pilot to have given the plane a minor amount of acceleration. How much? Roughly the same amount he gave the plane between 2 and 3 seconds before the crash. I did not even take into account the lift that ground-effect probably already gives the plane. The conclusion of this entire thing is, simply, that with a provably doable amount of acceleration (as in, he did it quite literally 2 seconds before), it's possible for the pilot to both hit all the lightpoles, and meet all the other data points. (Note, this asumes that there is no major error in the numbers used, see the error section for which errors I am aware of but rejected, and why. In several cases, substantial error may still exist, i.e. the elevation numbers) |
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#2 |
Downsitting Citizen
Join Date: Mar 2006
Posts: 17,078
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I think these JDX claims are beyond silly, but this is a very interesting analysis, A-S. In your list of potential error sources in JDX's calculations you omitted the main one: JDX's brain. Accounting for that error would take some doing.
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#3 |
Illuminator
Join Date: Sep 2006
Posts: 4,812
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I thought it was odd that his descent wasn't parabolic. In fact, I think it'd be hard to graph at all. I'd imagine controlling a plane manually probably isn't going to match any equation plot, perhaps a series of plottable equations or a "best match" line, but I'd imagine the plane's descent and speed is going to fluctuate quite a bit from all of the other forces in play. Going into a project thinking the descent is going to be plotted linearly is setting yourself up for a disappointment.
Good job though, it's good to see somebody can come up with a better description using the same data he is. |
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#4 |
Guest
Join Date: Aug 2006
Posts: 3,843
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Great job.
It is a far more plausible explanation of what happened that some silly theory, of fly over. Just so I fully grasp what you said and get my head around it. The only way Johnny’s theory will hold up is if the plane maintained a constant speed and did not accelerate or decelerate in any way and if the angle of approach did not alter at all. I should imagine if this was the case, it would be impossible for any pilot, let alone a one just about to fly a plane into a building, to do so manually. I just cannot see any human being able to hold a machine at a constant and precise speed manually, under the given circumstances. Cheers stateofgrace |
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#5 |
Penultimate Amazing
Join Date: Apr 2003
Location: 16 miles from 7 lakes
Posts: 11,098
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Even more to the point--
The "pilot" was aiming at something. Anyone who has ever tried this, walking, on a bicycle, in a car--knows that the closer you get, the bigger your errors are, and the more correction required. A plane, at 450kt will appear to drift from the target rather quickly as you approach, so you would be constantly correcting. I wonder what the control surface position records would show... Coming in on a parabolic curve asymtotic to the ground would make a lot more sense than setting up a rate of descent and expecting to hit the target... |
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#6 |
Banned
Join Date: Oct 2006
Posts: 191
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Good work Anti-Sophist.
I have been looking over JDX's work. It seems the vertical acceleration (G Force) is the most that has me puzzled since it doesnt show any leveling out during the last second. It actually shows an increase in the rate of descent. As for being linear descent rate, it was one second of time frame that shows almost 4000 fpm descent rate at 09:37:44. One would think you would have to pull some massive G's to pull out of that dive in less than a second from :44-:45 impact time. The FDR shows the opposite. Still researching. |
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#7 |
Philosopher
Join Date: Apr 2006
Posts: 7,854
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Thank you, Anti-Sophist. To sharpen up your approach, however, could you go a little bit further into the acceleration values?
What values of acceleration did you find for your "minimum" and "maximum" traces? Units of feet per second seem appropriate here. In the FDR recorder, there should be accelerometer data -- I haven't checked but I will. I believe these are sampled at 8 Hz but they can be averaged over a few seconds to get a reasonable approximation of the sensed acceleration. Did you look at this? If so, does the value agree with the acceleration you computed through the Quadratic Equation? All in all, though, you have clearly demonstrated that considering the approach as a straight line is not a safe or credible assumption. I had no idea that Mr. Doe X had made a straight line assumption when I responded to Skeptic4Sure in this post. I was never presented with his method, merely his results, and was hoping Skeptic4Sure would explain the method. It never occurred to me that he'd make such a foolish assumption. This is yet another reason why his results are not credible... I'll add that to the list. At this point, we can conclusively state that his approach, while not entirely wrong, is far too crude to make his case. The data from the FDR are still plausibly consistent with the official theory. |
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#8 |
Graduate Poster
Join Date: Sep 2006
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My worst-case acceleration assumption is 24 ft/s^2. (about 3/4 of a G force). This is based on the average acceleration between -3 and -2 second from impact.
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#9 |
Philosopher
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#10 |
Graduate Poster
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I don't see either of the effects you are talking about. The rate of descent is clearly decreasing during the final 3 seconds, that implies a positive acceleration, not a negative one. If there is acceleramoter or other data that is giving you that information, I'd like to see it.
A 4000 fpm descent is about 66 ft/s. 66 ft/s to 0 ft/s, in 1 second, requires exactly 66 ft/s^2 of acceleration. That is 2 G-forces, which is standard G force for a normal commercial airline with professional pilots, let alone a hijacked one. Not exactly "massive" Gs. |
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#11 |
Banned
Join Date: Oct 2006
Posts: 191
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Something doesnt look right from JDX's Pilots forum.
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It appears the plane needed 1.7 G's to slow the vertical speed roughly 1800 fpm. It looks like JDX used the actual numbers from the FDR to provide a benchmark for vertical acceleration forces and the result. I'm puzzled. By the way, I have also read that the reason JDX did this alternate analysis working backwards from the poles was to show something to a guy named Billzilla? I guess Billzilla did this analysis first? Do you have that analysis? It seems this analysis was a hypothetical based on Billzilla's claims. I'd like to see that. I'm still highly skeptical of JDX's claims of 480MSL. It appears he went to Boeing and a few other companies to get some answers, but was refused. I'm also trying to work some of my channels. Thanks for all the work guys. |
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#12 |
Graduate Poster
Join Date: Sep 2006
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Quote:
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Like I said before, if you are descending at 4000 fpm, you need to pull 2g positive acceleration (or about 3G), to correct -all- of that descent in a single second.
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gravity = 32 feet per second per second gravity = 32 * 60 feet per minute per second gravity = 1920 (feet per minute) / 1 second acceleration = (+3960 feet per minute) / 1 second acceleration = 2.065 * (1920 (feet per minute) / 1 second) acceleration = 2.065g or, about, 3.065 G-Force (give or take a cosine of a small angle (approx 1 for small angles) to correct for the G force being not perfectly orientated with "up")
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#13 |
Banned
Join Date: Oct 2006
Posts: 191
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I looked over the csv file and it looks accurate. The aircraft was descending almost 6000 fpm at one point, then it pulled 1.7 G's and slowed the descent rate to about 4000 fpm. Put the altitude columns side by side with the vertical acceleration columns. It actually makes sense when looking at the trend.
Find out descent rate in between each second for the last minute. You'll see the correlation in G force as the descent rate increases or decreases. Also remember that 1 G is the "zero" point since straight and level flight is always at 1 G. I'm not a math major (and I wouldn't remember all those forumula's anyway haha), but the csv file makes sense when you put it side by side with the columns described above. The only thing that doesn't make sense is the fact that the plane pulled 1.7 G's to slow the descent rate from 5760 fpm to 3960, and then showed less than 1 G thereafter. In other words, the csv file doesnt show any G force for pulling level across the lawn after the :44 second time stamp. It only shows 1.7 G's prior to slow the descent rate from 5760 to 3960 in the 42-44 time stamp. Very weird. edit, typo |
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#14 |
Banned
Join Date: Oct 2006
Posts: 191
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Quote:
Edit, I would say it takes a good 15 seconds or so to level from a 4000 fpm descent as to keep it "smooth" for the passengers. Still feeling a bit over 1 G. Maybe 1.1 or 1.2 |
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#15 |
Graduate Poster
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Quote:
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#16 |
Cold-hearted skeptic
Join Date: Apr 2004
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#17 |
Graduate Poster
Join Date: Sep 2006
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Well, your opinion isn't up against me, it's up against Isaac Newton. Mass and momentum have absolutely nothing to do with acceleration's relationship on velocity. G-force is an acceleration, and 4000fpm is a velocity. More mass means it requires more force, for the same amount of acceleration.
As I said, 4,000 fpm is not vert fast. It's 66 feet per second. It's 45 mile per hour. If you drop a rock from the top of a building, neglecting air resitence, it will take it all of 2 seconds to reach that speed. In an airplane, it would require, therefore +2G (or, a 3G correction, for a full second) to correct that much descent in a single second. (Again, this is assuming a small angle of descent) |
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#18 |
Banned
Join Date: Oct 2006
Posts: 191
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G Force is a Force. F = ma. Basically, by your claims, a 757 would take the same amount of G's to pull out of a 4,000 fpm dive in one second as a Learjet. I highly disagree. Besides the fact that the csv file never shows 3 g's for one full second, anywhere. But, I do admire your work and thanks for the input. What kind of student are you? Physics? |
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#19 |
Banned
Join Date: Oct 2006
Posts: 191
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I might also add, I dont even think its possible to arrest a 4000 fpm descent rate on a 255,000 lb object in one second without hitting a wall or something.
Can you stop your car in one second from 45 mph (66ft/sec)? What would the G's be if you did? |
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#20 |
Graduate Poster
Join Date: Sep 2006
Posts: 1,542
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No, a g-force is not a force. It is an acceleration. It's a misnomer and often causes confusion. You _experience_ a force, when undergoing g-forcing acceleration, (given by the equation above), but 3Gs is 3Gs, no matter how much you weigh.
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Even still, your claim doesn't add up. Duration matters. 1 second at 3Gs is equal to 2 seconds at 2G is equal to 4 seconds at 1.5G is equal to 8 seconds at 1.25G (in a plane, that is, fighting against normal gravity). |
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#21 |
Graduate Poster
Join Date: Sep 2006
Posts: 1,542
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Again, your intuition is failing you. I've proven it to you time and time again (mathematically, using the laws of the physics) that a simple 3G pull-up, for 1 second, will do the trick.
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#22 |
Cold-hearted skeptic
Join Date: Apr 2004
Posts: 2,084
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#23 |
Banned
Join Date: Oct 2006
Posts: 191
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the aircraft shows a 3960 fpm dive at time stamp :44. The G's after that show no correction. It actually shows less than 1 G. Where is the correction to start pulling out of the dive?
The G's shown prior to the :44 second time stamp are pulling level from 5960fpm to slow the descent rate. 09:37:25 AM 1.3 1690 altitude 1.26 1.24 1.16 1.13 1.08 1.03 0.96 09:37:26 AM 0.93 1647 altitude 0.88 0.84 0.84 0.85 0.88 0.92 0.91 09:37:27 AM 0.95 1596 0.97 0.98 0.98 0.98 0.99 0.99 0.99 09:37:28 AM0.971545 0.97 0.96 0.99 0.99 0.98 0.93 0.87 09:37:29 AM0.811494 0.75 0.7 0.65 0.63 0.6 0.6 0.61 09:37:30 AM0.631432 0.63 0.63 0.64 0.62 0.6 0.52 0.46 09:37:31 AM0.391362 0.34 0.31 0.32 0.31 0.36 0.35 0.52 09:37:32 AM0.541263 0.65 0.69 0.74 0.77 0.82 0.88 0.89 09:37:33 AM0.911158 0.92 0.94 0.99 1.01 1.04 1.11 1.26 09:37:34 AM1.391049 1.51 1.6 1.67 1.66 1.57 1.52 1.52 09:37:35 AM1.56955 1.45 1.42 1.27 1.31 1.21 1.15 1.2 09:37:36 AM0.94869 1.01 1.01 1.03 0.86 0.92 0.83 0.72 09:37:37 AM0.72786 0.67 0.7 0.64 0.72 0.56 0.36 0.61 09:37:38 AM0.56685 0.79 0.93 0.97 0.75 1 1.12 1.14 09:37:39 AM1.29592 1.17 1.38 1.21 1.39 1.43 1.35 1.43 09:37:40 AM1.45496 1.52 0.97 0.93 0.86 0.9 0.9 1.06 09:37:41 AM1.27399 1.31 1.28 1.12 1.1 1.25 1.22 1.28 09:37:42 AM 1.4 307 5520fpm here 1.7 1.71 1.75 <pulling G's to slow descent rate 1.72 1.62 1.58 1.55 09:37:43 AM 1.52 239 <4080fpm here 1.13 1.22 1.13 1.23 1.04 1.03 0.88 09:37:44 AM 0.66 173 <3980 fpm here 0.62 0.63 0.69<doesnt show any G's to level the 3980 fpm descent. 0.76 0.61 0.72 0.62 09:37:45 AM |
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#24 |
Graduate Poster
Join Date: Sep 2006
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Quote:
All I've said is it's not _impossible_, I didn't say it happened. |
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#25 |
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#26 |
Banned
Join Date: Oct 2006
Posts: 191
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But I now also understand JDX's 480MSL based on local pressure correction. Wow.
I really have to look into this further. Thanks for all the help everyone. I'll continue to post so observations. Yes, I'm a retired Airline pilot from Independence Air. Haven't thought much about flying since except for buzzing around in a 172. It good to get the ol' noggin working again. I'm going to try to contact this JDX guy and see if he worked for FLYI. |
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#27 |
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#28 |
Graduate Poster
Join Date: Sep 2006
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Oh dear. How exactly can you tell the plane is level from 5 frames of video, when only one of those frames shows the plane? (and it shows it rather poorly). Certainly the 3 frames of explosions aren't helping.
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#29 |
Cold-hearted skeptic
Join Date: Apr 2004
Posts: 2,084
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#30 |
Banned
Join Date: Oct 2006
Posts: 191
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haha, on the internet? Sure, right after you provide your birth certificate.
Do you think I'm a fake? As I said, I can take my posts back over to flightinfo. No sweat off my back. It already seems some of you people attack anyone that comes here and obviously this JDX guy has many of you so riled up. My best advice is to get some whiskey. Go out and see a movie. Enjoy your life. Don't let one person get so under your skin. Especially one that you don't know and one that you obviously disagree with. How many threads are here on this guy alone? You don't even know him? If I seem protective of him, it's because if he did work for FLYI, it was one big family and we all went through alot with that airline. But, that doesn't excuse his claims if they are wrong. Good luck all. Try not to let one person get to you. |
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#31 |
Mad Mod Poet God
Join Date: Aug 2002
Location: St. Louis, MO
Posts: 3,298
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Here here. Let's all take a breather (well, not me, I just started posting in this thread, and I don't know NTSB from NSTB).
WW said he was going to check out some numbers on JDX site; after he does, he can get back to us with what he discovered. In the meantime, we can discuss this and that, of whatnot and happenstance, of cabbages and kings, and where the sea is boiling hot. (BTW, WW, I suspect that JDX likes the band X. If you do know him, tell him to get a new moniker please) |
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#32 |
Downsitting Citizen
Join Date: Mar 2006
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I think the issue of credentials is only relevant if someone cites their experience, rather than evidence we can check (argument to auto-authority?), or if they get things wrong that someone with the claimed experience should get right. JDX has done both of those things at times, so asking him to show his credentials is to be expected. I don't see why Weedwacker should be asked to show credentials unless he makes questionable claims or doesn't seem to know what he's talking about. I've seen no evidence of that behavior.
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#33 |
Cold-hearted skeptic
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#34 |
Philosopher
Join Date: Apr 2006
Posts: 7,854
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I'm coming to this party late, after a pleasant evening sampling the premier oeonphiliac products of the free world with the wife... so forgive me if this doesn't make sense...
An accelerometer reading of zero G's probably doesn't mean that the aircraft is freefalling. The accelerometer doesn't know about Earth's gravity. If the aircraft is flat and level, experiencing no net acceleration, the accelerometer reads zero G's. If the accelerometer reads -1 G, it means the aircraft is diving at freefall. If it reads +1 G, it means the aircraft is climbing. You have to treat the aircraft itself as its own Newtonian reference frame for this to be consistent. If the aircraft doesn't accelerate at all, it means zero G's. That means flat and level flight. Also, for what it's worth, I've found weedwacker's comments to be completely reasonable. I could care less what his background is, it's the quality of his arguments that matter. That goes for every person here. Let's not make the mistake of the CT's, and let everyone have their say. Thanks. |
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#35 |
Guest
Join Date: Sep 2006
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G-force is indeed no force but acceleration. The ratio with the famous factor g is always used (think about roller coasters)
I didn’t know about JDX math stuff, it looks like that parking camera information is used in your analysis, some people claim that that frames are edited but even if that info is genuine you have to know something about the accuracy of the frames. The frames are crappy, crap in crap out. Even my dad has a better camera for his business bought at the local store for a few euros. Normally when you have n points you can estimate the “real” trajectory by a polynomial of order n-1, I think a 2nd order estimation is sufficient for an airplane, because the heavier they are the straighter the trajectory will be. But of course if there is a lot of force available the momentum (hence velocity) can change rapidly. I assume the acceleration of a plane is public and available information that can be found somewhere? Once you have the position of function of time the velocity as function of time is nothing more than the time-derivative and the acceleration is another time-derivative. Be careful because an initial error in position will strongly affect your end result. |
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#36 |
Cold-hearted skeptic
Join Date: Apr 2004
Posts: 2,084
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This is something I would be interested to know for sure.
I would have thought that a stationary accelerometer at MSL would read +1g since it is, after all, experiencing an accelerative force (gravity). By this logic, zero G's would mean free-fall. An accelerometer does not "know" what the source of the acceleration is. It has no way of segregating gravitational acceleration from any other kind because this is simply not possible and is, in fact, one of the corner stones of relativity. The only way to construct an accelerometer which would read zero G's in the above scenario is to artificially compensate for the expected gravitational force. Perhaps the FDR data has been adjusted for this. I don't know.
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#37 |
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Posts: 652
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Correction there. The accelerometer will read +1 in normal, straight & level flight. If it reads zero, then you are actually in freefall. Here's a piccie of one: ![]() Notice how it is reading 1 in 'normal' condition, and that marking is thicker than the other lines? That's why the DFDR data will show '1' vertical acceleration a lot of the time. So the last DFDR data showed 3960 fpm rate of descent, and about 0.6G during the last recorded second. That's to be expected because the the aircraft was descending, thus lower G. Still, DOH! himself has said that 4000 fpm is not "excessive" for an airliner, so he would have to admit that pulling out of that descent is quite possible - especially some 400 feet above the ground (very rough figure, and assuming DOH! is correct with his calculations) when you are not concerned with overstressing the aircraft. |
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#38 |
Muse
Join Date: Jun 2006
Posts: 652
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You know, what this does go to show is that you need to be a professional trained in this field to properly analyse the DFDR data.
Regardless of DOH!s pilot experience, he is not (I assume) qualified to analyse DFDRs and give an accurate conclusion. There are lots of factors of which the layman - and pilot - is unaware, and must be considered with this stuff. There are no doubt a few, highly qualified people who can look at this correctly. Flight test pilots & engineers would have - most likely - the necessary backgrounds to properly evaluate the data. |
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Any time it can be proved that one of my studies is wrong, I am more eager than anyone to acknowledge AND CORRECT IT. Jack White Little White Lies....... |
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#39 |
Penultimate Amazing
Join Date: Aug 2002
Posts: 24,692
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I think an accelerometer will be reading zero in normal gravity. Of course, it is really in 1G, but this will be its reference. So an accelerometer will show the differential G value.
There is a technical reason for this. The accelerometer has to be set to zero, somehow. The transducer types used for this are not capable of holding an initial setting indefinitely, so you would either have to take them for frequent recalibrations or you set them to zero when there is no differential force. What the reading in an FDR is, I don't know. It could have one G added, but I somehow doubt that. It wouldn't be of any use, since what we are interested in is the differential G. Hans |
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#40 |
Muse
Join Date: Jun 2006
Posts: 652
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I know it might seem counter-intuitive, but they read 1 normally:
Quote:
So if the DFDR says +1.5G, you are pulling +1.5G but are experiencing +0.5 differential G. So normal weight is say 100kg, so at 1.5G you would "weigh" 150kg. |
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Any time it can be proved that one of my studies is wrong, I am more eager than anyone to acknowledge AND CORRECT IT. Jack White Little White Lies....... |
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