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Homer
Mountain climber
742 Evergreen Terrace
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Nov 14, 2010 - 10:07pm PT
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Thanks for your post Mike Bolte.
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Port
Trad climber
San Diego
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Nov 14, 2010 - 10:20pm PT
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I knew this would attract the typical debunkers. It is soooo predictable.
Because once again you've substituted the absurdly complex explanation for the rational one. Don't get all huffy and surprised when we come around, you should expect it by now after making so many ridiculous and unsubstantiated claims.
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graniteclimber
Trad climber
The Illuminati
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Nov 14, 2010 - 10:48pm PT
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How many of you ever heard of the regulas 1, not to mention the fact it burnt a conning tower off of a sub outside pt Mugu nearly sinking the sub? Not one of you would be the answer. THE only reason i know is that my Dad was on the project. How many of you knew that icbm missle silos have gone on FULL alert many times in the past? probably not many...THe only reason i know is that my brother was there. do we see a pattern? America for years knew nothing of the VAST underground structure known as the Strategic Air Command.
None of this stuff has been secret for many years. It's common knowledge now among those interested in this stuff--aviation and naval history buffs.
My Dad was assigned there working on a secret guided missile system...A significant part of Nebraska was OFF LIMITS and the SAC COMMAND CENTER was there UNDERGROUND. did I stutter?? I have my Dads SAC id card..I also got to hunt chukars in the most restricted area of nevada at the time, palomino valley-where they built and test fired the LM ascent and descent stage engines for the apollo series. You never saw a picture of area D have you....I have some...Dad was the final safety inspector for those engines an his stamp is on them on the moon... I could go on with all sorts of things you never knew...
Roy, you are denigrating your Dad's honor and integrity if you are implying he told you classified information. He could tell you about that stuff because it isn't secret anymore, and hasn't been for a long time.
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 01:14am PT
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Apparently Ed you think every 3D stereogram image must be mathematically evaluated and proved to show that you are actually seeing it in 3D and with depth of field, and that you are actually seeing what you are indeed seeing. Ed do you go around mathmatically proving that you are seeing in 3D and with depth of field?
Ed that is laughable.
well the short answer is, yes I do think about these things, it is something that I had to work through in various experiments that I did, and yes, photogrammetry is powerful, but it is not absolute and in situations where it is set up to make do a measurement it often fails... anyone who has ventured out into wilderness areas where there has been no "ground truth" verification knows that topos made form just photogrammetry can be very wrong.
There are a number of things to consider in making your claim, you have done nothing but to state some strange absolutes without any verification. In particular, your ability to see how far out something is with a "3D stereogram" depends critically on the parallax of the measurements. Your experience of depth perception is a combination of many different strategies, most of which are not available to view your particular scene. In human vision, most people can resolve the horizontal separation of 2.3 minutes if arc, about 1 mrad... that's a millimeter at a meter, at a kilometer that would be a meter, at 100 kilometers that's 100 meters... displacements less than that cannot be resolved stereoscopically...
For your particular problem we do not know the distance, what we presume to know is the heading of the object, and the position of the camera... let's assume for a moment that the path the helicopter is flying along is perpendicular to the object to simplify things.
At position 1 we make a bearing measurement Θ1 and at the second point a bearing of Θ2, the separation distance between points 1 and 2 we can take as S, then the distance the object is away from us, D, is given by:
D = S/(tanΘ1 - tanΘ2)
Let's say that S is 1 mile, then to get a D of 30 miles the difference in the angles would have to be something like 0.03 radians, or about 2 degrees...
I'm skeptical that you can get that sort of measurement from your pictures (and by the way, your forgot that the earth is globe when you drew the bearings on the satellite view of the contrails, you've got your coordinates badly goofed up there).
We don't know, for instance, the path of the helicopter, this matters because it can change the angles that we measure. Where do your stereogram positions come from? And the resolution of the object, such as the claim that it is a spiral as opposed to something else depends on our ability to resolve the depth of the feature (meters?) at that distance. A 10% error in the measurement of the angles leads to a 0.3 mile error in the position of the object, we're talking 0.2 degrees, and in your picture, there are roughly 60 degrees across (if I believe your labeling) for about 800 pixels, that's 0.075 degrees per pixel, 2 degrees is 27 pixels, and an error of just 3 pixel displacement gives you the 1584' error... pretty big 3D features...
Put the other way, an object 30 miles out on your picture can be resolved to about 500' with these estimates...
Once again, this makes assumptions about the helicopter path... which I actually know nothing about, having guessed some numbers which may be reasonable for the purpose of estimate.
So you have nothing, except perhaps laughter...
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Klimmer
Mountain climber
San Diego
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Nov 15, 2010 - 01:26am PT
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Stzzo,
I'm gonna ignore your ignorance and just let you hang yourself.
For everyone else:
Ed H., Mike B., and Tom C.,
Quick trig. without using the curvature of the Earth, assuming a right triangle only (just a rough estimate but a reasonable one). . .
Using the WNW high point of the exhaust/vapor plume that has drifted just South of the green arrow head plotted, and measuring the flat linear distance to this point from Long Beach Harbor = 177.53 miles.
The angle Theta = 9.5 degrees, from the horizon to the tip of the exhaust/vapor plume.
Knowns:
Adj. = 177.53 miles
Theta = 9.5 degrees
Unknown:
The altitude of the last exhaust/vapor plume is the opposite . . .
Opp. = ?
Tangent (theta) = opposite/ adjacent
opp. = (Adj.) (Tan theta)
opp. = (177.53 miles) (Tan 9.5 degrees)
opp. = 29.708333 miles = 29.7 miles
Convert miles --> feet:
(29.7 miles/1)x (5280 feet/1 mile) = 156, 816 feet
The missile went at least 156, 816 feet high. Even higher considering the curvature of the Earth.
Obviously this is much higher than commercial jets with passengers go or even military jets. This is a higher altitude than a record breaking military jet has flown, as far as I know.
Twas a missile.
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 01:32am PT
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Ed H., Mike B., and Tom C.,
Quick trig. without using the curvature of the Earth, assuming a right triangle only (just a rough estimate but a reasonable one). . .
Using the WNW high point of the exhaust/vapor plume that has drifted just South of the green arrow head plotted, and measuring the flat linear distance to this point from Long Beach Harbor = 177.53 miles.
The angle Theta = 9.5 degrees, from the horizon to the tip of the exhaust/vapor plume.
Knowns:
Adj. = 177.53 miles
Theta = 9.5 degrees
Unknown:
The altitude of the last exhaust/vapor plume is the opposite . . .
Opp. = ?
Tangent (theta) = opposite/ adjacent
opp. = (Adj.) (Tan theta)
opp. = (177.53 miles) (Tan 9.5 degrees)
opp. = 29.708333 miles = 29.7 miles
Convert miles --> feet:
(29.7 miles/1)x (5280 feet/1 mile) = 156, 816 feet
The missile went at least 156, 816 feet high. Even higher considering the curvature of the Earth.
Obviously this is much higher than commercial jets with passengers go or even military jets. This is a higher altitude than a record breaking military jet has flown, as far as I know.
Twas a missile.
you teach science?
you make a very elementary mistake, klimmer, in converting the angular extent to a spatial size without enough information, you know you can't do that without understanding something about the position of the object, which you assume, and your assumption is very incorrect.
a jet could fly a trajectory that ends up tracing out the exact same angle across the sky but at a constant altitude, you can't know the difference in your analysis, which is wrong.
look at this, Half Dome is larger than the Moon!
visual proof!
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Klimmer
Mountain climber
San Diego
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Nov 15, 2010 - 01:47am PT
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Simple trigonometry Ed. Just like the British used to tell the heights of the great Himalaya. Ok they had a very accurate theodolite and got within 20 feet of the known elevation of Mt. Everest.
I know where the exhaust plume is. I know the scale of the GOES satellite image. I can quickly calculate the horizontal distance from Long Beach Harbor to the tip of exhaust plume. Yes, the hypotenuse would be a little longer. Close enough.
I used a visual inclinometer to measure the angle theta from the horizon to the tip of the exhaust/vapor plume, and I got 9.5 degrees.
Solve for the Opposite = Altitude of the missile plume.
Rough estimate but pretty darn close.
We do this in model rocketry all the time. Have you not ever calculated the height of a Testes model rocket?
Might want to rent "October Sky" for inspiration.
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Mike Bolte
Trad climber
Planet Earth
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Nov 15, 2010 - 01:47am PT
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To be more generous, I'd say it differently Ed. If Glenn wants to:
(1) assume this is a rocket
(2) assume a trajectory (basically flying straight up perpendicular to the surface of the Earth) and
(3) assume a distance for the launch point (I'm a little fuzzy on how he picked this particular distance, but perhaps I have not been paying attention), then he is right! He has the right triangle trig down pat.
But of course he has simply made a bunch of assumptions and done some trivial math. This doesn't prove anything about the nature of the event last week.
EDIT: Ah! I see that he is using the clouds from the weather satellites as the point where the rocket was fired. OK - assumption #3 stands, but now I'm not so fuzzy on where it came from.
Glenn - from your response, you have no comprehension of Ed's point. That is a little mind-boggling, but I guess this is hopeless. Signing off on this thread!
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 01:51am PT
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klimmer, you are a joke...
I doubt you could fully understand how the British Survey calculated what they did, it is not "simple trigonometry" and to say so is either a tremendous demonstration of your ignorance or you're being disingenuous.
Your Testes rocket calculation works because you know how far you are from the launch point, and you know the angle of the rocket's trajectory, which is perpendicular to the surface.
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Klimmer
Mountain climber
San Diego
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Nov 15, 2010 - 02:06am PT
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Ed,
I know how they did The Great Triangulation of India. I have read about it, studied it, known the trig behind it, and even have a great DVD on it.
Your Testes rocket calculation works because you know how far you are from the launch point, and you know the angle of the rocket's trajectory.
Bingo Ed. I know how far I am from the highest portion of the exhaust/vapor plume of the rocket. You can see the exact exhaust plume and the high point when it runs out on the GOES image. I know the horizontal distance to this point = 177.53 miles.
I know the angle Theta from the distant ocean/sky horizon to the last tip of the exhaust/vapor plume in the sky in the green marked up image. Yes, it is a rough estimate, but it is darn good one using my visual clinometer.
Yes, it will be higher because I haven't taken into account the curvature of the Earth.
I already admitted it isn't exact, it is a rough estimate. But it also is not widely off.
The same way we do it in model rocketry. No different. Basic Trig.
Edit:
You sight from a flat level horizon to the top of the exhaust/vapor plume. The ocean/sky horizon is that flat horizon to the top of the last known point, where the exhaust runs out. This is Theta.
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 02:10am PT
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no, you still do not understand your model rocket,
you know two of the angles of the triangle with the corners: launch point, your position and the top of the trajectory, the one you measure at the top of the trajectory and the angle that the rocket launches at...
in your LA rocket hypothesis you only know one angle... and you are guessing what the distance is, you don't know. So there is insufficient information to calculate what you claim to have calculated.
that's simple trig...
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Klimmer
Mountain climber
San Diego
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Nov 15, 2010 - 02:18am PT
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Ed,
Do I have to put a point on the GOES image for you?
I do know where the exhaust plume runs out.
I suppose I'm gonna have to do this for you. It is getting late. I'll have to do it later. But I will.
I can see the entire exhaust/vapor plume from near after the launch to where the last really evident wisp is on the the GOES image as it flew to the W - WNW. and then we can see the entire exhaust plume drift to the SSE.
You can disagree all you want, put I have located it and pointed it out. You see it build just as it builds and spreads out in the oblique images. No doubt about it. Does the same thing in the GOES satellite image. The exact same thing and in the exact same direction I said it would. Because it is leaning away from the coast toward the W - WNW, just as I said it does in the stereogram images. And yes the parallax is sufficient to see this clearly. But then you wouldn't know that since you haven't checked it out.
I know the adjacent side of the right triangle. I know theta. Solve for opposite. That is the elevation of the last missile plume seen.
Even my physics students can get this.
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 02:35am PT
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here's what we are talking about:
you don't have any estimate of the angle of the rock launch... so you don't know how high it is...
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Shack
Big Wall climber
Reno NV
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Nov 15, 2010 - 03:02am PT
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That is not an exhaust plume on the satellite image. That is a cloud.
One forms right next to it at the same time, and the scale is totally wrong
for a rocket trail.
If that "exhaust trail" is going from the surface to 150,000 feet or so,
how is it staying straight? Rockets don't fly straight.
Rockets fly on an curved trajectory.
Rocket trails do not remain straight. The wind shear at different altitudes, twist it into all kinds of weird shapes.
Like this...
If it was a Chinese missile it would have been a JL-1 or a JL-2, either way,
a trajectory to the NW would have made the trail appear to actually curve back down toward the horizon. Like this...
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 03:09am PT
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the curvature of the earth is important...
The solid arc is the surface of the Earth drawn to scale, the dashed line is a 20,000' constant altitude trajectory above the surface... from the viewer on the left, it appears to rise out of the horizon, and it's subsequent angular position still has it out at sea when that angle is 9º but it is still 20,000' high.
Klimmer's "flat earth" is shown as the dotty line... a very different story.
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Shack
Big Wall climber
Reno NV
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Nov 15, 2010 - 03:19am PT
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Come on Ed, don't confuse the argument with facts.
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Klimmer
Mountain climber
San Diego
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Nov 15, 2010 - 09:40am PT
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I will discuss these sometime today; I'm just getting them posted for now . . .
The massive missile/vapor plume is right where I said it would be, and pointing and it grows in the W - WNW direction just as suspected. The stereograms I put together easily show this.
Please note the tip of the missile/vapor plume is behind and much, much, much higher than the "alpenglow" illuminated Cirrus cloud in the fore-ground, which are 16,500 - 40,000 feet high at mid latitudes.
We will go over the simple trig again. But you have to realize Ed that a few thousand feet off the deck in a helicopter is not going to make much of a difference over 177 miles. I did say it was a quick and rough calculation, but it is more accurate than the smoke you are blowing. It is a very simple calculation to do regardless of the curvature of the Earth. With that included, it will be even higher.
At 177 miles, you wouldn't even see your thin wispy barely perceptible contrail from a commercial jet. Nor would you see the flashes of light from a fuselage. What we saw glowing and emitting light was from a missile rocket engine over 177 miles away.
When I have the chance I will mark up a right triangle trig illustration of my own.
I'll get back . . .
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monolith
climber
Berkeley, CA
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Nov 15, 2010 - 09:58am PT
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Klimmer, you still have not shown your work for 240 degrees.
Use GE and show the line thru the harbor. Along with a still from the vid.
You will find the 240 degree line does not fit correctly if you put it close to the end of the San Pedro peninsula as is shown in a still
Classically bad science Klimmer. You fell in love with 240 and have been sucked further into lunacy cuz it pointed to some clouds somewhere along its path.
Were two missiles fired, cuz there are two clouds there Klimmer?
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Ed Hartouni
Trad climber
Livermore, CA
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Nov 15, 2010 - 10:38am PT
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look Klimmer, at 3000' the horizon is at about 50 miles, it does not change the curvature issues, but pushes it back a bit further, the Earth is not flat, and on the scale that you are calculating, you cannot ignore that curvature...
...the possibility that it was a missile is very very small, while the likelihood that it was an aircraft is much higher, overwhelmingly higher. Your calculations are not correct and do not lend any credence to the missile hypothesis.
You have a faith based hypothesis. There is not sufficient information from just the images to make a definitive statement on the missile. Given all of the information, the aircraft hypothesis is by far the strongest.
It is why we actually do the calculations, klimmer, to make sure that what we believe to be true is supported by the science. Just waving your arms in the air, in your case very rapidly, will no more cause you to take flight as to be right... you have to do the supporting work. Where you have tried to do it has exposed your set of assumptions, which is part of the process. In your case, that set of assumptions seems rather naive.
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monolith
climber
Berkeley, CA
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Nov 15, 2010 - 11:56am PT
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Klimmer, did your missile go more or less straight up, cuz we don't see much of a downrange plume from the satellite pic.
As Shack's pic points out, missiles go downrange fast. And if we can see the 'wispy' contrail of the Hawaii flight from a satellite, then surely we would see your missile's plume if it went downrange.
And if it went more or less straight up (and very slowly), then why did it lose most of it's plume?
And why is it so narrow at the base?
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