Another Amelia Theory of Disappearance

The fact that there were 30 or 40 odd people living on the island at the time running a copra plantation, including their European overseer, I would have thought that someone somewhere might have noticed the event occurring. And to cap that the commander of the search aircraft flight that arrived a couple of days after her disappearance actually landed in the lagoon and spoke with the overseer who didn’t even understand what all the fuss was about as they hadn’t seen a thing. But then in the world of Earhart theories perhaps I am being a little too sensible.

Why not just dive on it and see ?

Still Pareidolia. With any mass of pixels I’m sure you can see all sorts of things that the brain wants to put into some order. Some see nothing and some see what looks like an aircraft. You are just looking at natural patterns in the sand. If you look at other close ups of a shallow water seabed you will be amazed at what patterns you can see. So in my humble opinion all this refraction reflection stuff is rather a red herring because the plane is not there in any shape or form.
Ian

Amelia? Is that you? I’ve got a bad feeling about this….

And this is allegedly within a lagoon at snorkelling depth 🙂

I understand about line of sight and the virtual image. Everyone does. It does not need another diagram.

You do not understand how refraction works. That is why your virtual image is much too large unless your ‘Electra’ is visible kilometres down.

You have looked for some blobs that, if you squint, might be called aeroplane-shaped. Do that for long enough you will find them.

When someone pointed out that the shape was too large you misinterpreted a piece of physics to explain that away. Your refusal to step back and learn the theory properly suggests that deep down you really do not want to. But I can’t psychoanalyse, I can only repeat. You have the geometry wrong.

Refraction happens at the water surface. Incident light, ie any light not meeting the surface at 90 degrees, does not leave the interface at the same angle that it approached it from. This deflection defines the size of the virtual object. If the electra is 100 metres down, from 500 km vertically above the virtual object would be less than 5 cm larger than the 12 metre real one.

It seems that, if anything, Earhart’s Electra has grown larger over those 80 years. Therefore we can conclude that shallow salt water is good for aircraft. :very_drunk:

Let the scrying continue. Surely staring at Google Earth and seeing patterns in the random patterns of colour in the empty ocean is the 21st C. equivalent.

Putting aside all the stuff about refraction; suppose Amelia Earhart had ditched her aircraft intact on the ocean -and Earhart was the lady to do that- there is then the 80 years of deterioration since to think about. If the Electra is in water shallow enough to be seen from high up then that is pretty shallow water, full of heat and currents and life. There are 8o years of storms churning the water and currents burying things under sand. There is 80 years of saltwater doing wicked things to the airframe. And there is 80 years of life clinging to and growing on the aircraft and turning it into an artificial reef. At this late stage, assuming Earhart’s aircraft is sat underwater in one piece would you even be able to tell its an aircraft unless you get up close?

Well that is confusing. You agree that the virtual fish is larger than the actual fish. You agree that only the virtual fish can be seen when above the surface. But you disagree with the size? In order to determine overall apparent size, a ray tracing diagram must have the eye or camera line of sight touch and end on the extreme of the virtual image. The line of sight line cannot return to the real image or object as it cannot be seen. I’ll prepare some ray diagrams that have larger angles. Maybe that will explain it better. And thanks Meddie and Chris for keeping the thread at the top.

Tom, I am not questioning the maths, it’s the geometry – the application of the maths – that’s off. You are drawing your triangles in the wrong places. Now whether you are being deliberately obtuse (I thank you) or not I don’t know, but please read what we have written and make the most of a fairly benign forum populated by people with esoteric but relevant experience, like Chris Cussen, as otherwise you will look a bit silly taking this further with other audiences,.

Tom, if, according to your calculations, the light is bent by 0.002 degrees then, roughly it will cause a displacement of the image by 2 metres at a depth of 1000 metres (look up far field effect). How deep do you think the sea is at this point?
By your figures it must be somewhere near 2000 metres to give you an factor of 3.5 metres. And you expect to see an aircraft at that depth?

Your maths is total bullocks, and you are grossly misinterpreting the maths to prove a point. Believe me I used to do write, design, develop and test sonar imaging for a living and what you say is total crap.

Note to Otis: you are right we do know exactly the position of the stars, the point I was trying to make is joining some dots and saying it looks like a bear, or a virgin or a Spitfire.

This:

Tom, may I be so impertinent as to ask what your professional area and your academic experience are?

Followed by:

I’m not real smart but I can read..

I’ve been following this thread for a while, and it has kept me amused with the total c**p that is presented at scientific proof the Electra is where claimed. It seems akin to looking at the stars, selecting a few at random and claiming that they are in the shape of an object.

Agreed, except joining the dots on stars is an exact process. They are precisely measured and observable spots in the sky. Tom has imagined his datum points to convenient empty places on his blurred image and pretended they are exact points on the aircraft. Then he tries to tell us how accurately all these imagined points line up with his image of the real aircraft. Like he has not really plotted them on the image to look exactly like the plane outline. Kind of cheap really.

All of this nonsense, and you are still trying to prove that a bit of noise in some Google aerial imagery is Amelia’s lost Lockheed! Time to give it a rest, no?

I have done the math…see my post #263. The angles are tiny indeed. Looking up 89.9985, the bending is .0005. At 400Km height the added size for a 10 meter object is 3.5 meters. The typical distance of the object from the surface is inconsequential considering the 400km height. The added magnification would be 3.3 meters if a right triangle calculator with five digits to the right of the decimal could be found. The best I’ve found is the four digit calculator. Some rounding errors maybe. I’m not real smart but I can read.. All this comes straight out of the Wiki and physics literature. Your turn to do the math.

Apart from the 1.33 thing, Tom, you have it right there!

Now do the math (or just draw the diagram, much like the fish one) on the example of a body a few tens of feet under from the height of a satelite, using Snell’s law and applying the refractive index appropriately. As others have said it’s very close indeed to a 90 degree incidence, and therefore the deflection is tiny and the magnification on something the size of an Electra is a couple of centimetres at most.

People have been charged by The Jockey Club or prosecuted by the RSPCA for less.
This equine is deceased: please stop hitting it.
Tom, may I be so impertinent as to ask what your professional area and your academic experience are?

“The ratio of refractive indices 1 and 1.33 means every virtual image is magnified by 1.33.”
Not to a ray effectively incident at 90 deg to the surface, as would be imaged by a satellite several hundreds of kilometres above a target subtending a mere few metres of resultant incidence angle.

The important item both Beermat and I haven’t discussed yet is the virtual image created by refraction. It is elementary, as Beermat says, to the question. My favorite example is the fisherman who looks down on fish in water below the high pier. He knows from experience that the fish is not quite so big as it appears. And should the fish move away so that the view angle is -say- 45º he knows that in order to spear the fish he must throw under the virtual fish because the real fish is actually displaced below it’s apparent virtual position. All light rays that pass through the water to the air surface is bent so that when looking into water all the objects we might see are virtual images of the real object. We can not see the real object. The ratio of refractive indices 1 and 1.33 means every virtual image is magnified by 1.33. If ray tracing is to be meaningful, the virtual image must be included. The virtual image does not change size as the eye or calibrated camera increases in height. The satellite camera at 400Km measures the size of the virtual image and can not see the actual real object. My stick figure is attached. This is my understanding of how refraction magnification works and makes the satellite measure meaningful with regard to the aircraft size.

[ATTACH=CONFIG]256015[/ATTACH]

Just another note, if the image is taken from directly overhead then the incident light is at 90 degrees and there will be no refraction or magnification.

I’ve been following this thread for a while, and it has kept me amused with the total c**p that is presented at scientific proof the Electra is where claimed. It seems akin to looking at the stars, selecting a few at random and claiming that they are in the shape of an object.