Infra red cameras and curvature?

Vogon

New Member

So, this supposedly shows Three Lakeway Center, in Metairie, Louisiana, a 403-foot (123 m)-tall skyscraper from the other end of Lake Pontchartrain Causeway.

Screen Shot 2016-03-14 at 17.52.24.png

Curvature should show a 116m drop for that distance, and the buildings around it are lower than 116m.

Screen Shot 2016-03-14 at 17.49.58.png (Google Street, from the bridge).

The viewer is standing at an elevation equal to the causeway, which takes a few meters off the drop.
Screen Shot 2016-03-14 at 17.56.17.png

The idea if you haven't seen, is that the IR sensing capability removes the hazing and distortion that prevents us seeing these things normally. Which seems at some levels like BS. We get clear days.
They also claim that you can't get IR images from mirage and atmospheric lensing effects, which again seems like BS. It's radiation, so it's susceptible to certain types of distortion.

How come the "middle" building seems so much higher that you'd think the change in perspective would indicate?

So what's happening here?
 

Mick West

Administrator
Staff member
Their video shows the bottom of the building are obscured, hence the surface of the lake is curved. Case closed.
 

Vogon

New Member
Their video shows the bottom of the building are obscured, hence the surface of the lake is curved. Case closed.
I agree, but the numbers do indicate that the lower buildings shouldn't be visible.

I'm asking why we can see THAT much of the buildings - other than the video being faked.
 

Mick West

Administrator
Staff member
I agree, but the numbers do indicate that the lower buildings shouldn't be visible.

I'm asking why we can see THAT much of the buildings - other than the video being faked.
Refraction and/or the view position being higher.

Here's what it looks like in GE, modeling a ball earth, with the viewer height at 5m above sea level (the lake is at sea level).

20160314-105854-qec9n.jpg
 

Henk001

Active Member
The clearance is 5m so I assume when you stand on the pavement the height of the observer is about 10 m. Then the buildings should be 74 m below the horizon without considering any atmospheric refraction (AR). If you do take AR into account (using the rule of thumb R= 7/6 R,earth in my calculation) the buildings should be 60 m below the horizon. Flat Earthers (FE's) most of the time use the 8 inch/square mile approximation, not realizing that this gives the horizon dip below a tangent at the standpoint of the observer. In other words, when you hold your camera at the surface of the earth/water, or when your face is flat at the surface.
edit: I assumed 42 km distance
 
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Mick West

Administrator
Staff member
It is BS. Longer wavelengths are refracted more. The distance to the radar horizon is about 5% greater than the visible horizon when refraction is accounted for. IR falls in between those wavelengths.
The longer wavelengths (red, infrared) are refracted less. On a prism split spectrum the infrared is right next to the red, which is how it was discovered (the hot region next to red).
infrared-radiation.jpg
Source: http://www.physics-and-radio-electr...ctromagnetic-spectrum/infrared-radiation.html

But if this were actually a factor then objects at a distance would be all rainbow colored around the edges. Since they are not, we can conclude atmospheric refraction does not visibly act different for different wavelengths of light.

You can actually see the splitting with very high magnification on very (visibly) small distinct bright objects, like Venus, when close to the horizon.


Source: http://www.cs.cmu.edu/~zhuxj/astro/html/venusphase.html

Hence there is no benefit (in terms of reducing refraction) for using infrared. Infrared is used for viewing at long distances because it passes through things of low density, like mist and haze - but to see a long way, the camera needs to be high up.
 

Chew

Senior Member
Brain fart on my part confusing the amount of refraction. But yeah, every source I've seen says the radar horizon is farther than the visible horizon understand standard refraction (i.e. multiply Earth radius by 4/3 for radar horizon and by 7/6 for visual horizon).
 

Henk001

Active Member
Brain fart on my part confusing the amount of refraction. But yeah, every source I've seen says the radar horizon is farther than the visible horizon understand standard refraction (i.e. multiply Earth radius by 4/3 for radar horizon and by 7/6 for visual horizon).
Understandable confusion. The refractivity (N: (n - 1)x10^6) decreases with increasing wavelength for UV, optical, IR, but it INcreases again for the radio band. Typical values are N = 284 (UV), 275 (red), 273 (IR) and 346 (mm-wavelengths at 15°C).
 

Gufferdk

New Member
Brain fart on my part confusing the amount of refraction. But yeah, every source I've seen says the radar horizon is farther than the visible horizon understand standard refraction (i.e. multiply Earth radius by 4/3 for radar horizon and by 7/6 for visual horizon).
If the radars they are talking about have relatively low wavelengths (A- and B-band radars (this would especially apply to sources from say around WW2)), surface-wave propagation might also have an effect.
Surface wave is where the conductivity of the surface of the Earth causes electromagnetic radiation to not follow straight paths but instead curve towards the surface of the Earth. The longer the wavelength, and the more conductive the surface, the more powerful the effect gets.
LF broadcast and LORAN primarily propagates beyond the optical horizon this way. In the low end of HF (fx the 80m HAM-band) the radio horison may be up to 80 km away under pristine conditions. On VHF (fx FM-radio) the effect is usually down to single digit percentages. At higher frequencies fx D-band radar (1-2 GHz) the effect is irrelevant.

Source: "Vejen til sendetilladelsen" p. 89-90 ISBN 978-87-85149-08-4 (Danish: The road the the transmission permit) (a book written by HAMs for people studying to become HAMs).
 

Vogon

New Member
Yes... Great stuff guys, thanks.

I used the pythagorean calculation which should allow for someone 6' tall - I think I need to check my formula!
In fact, all the formulae that produce the same result.
 

Henk001

Active Member
Yes... Great stuff guys, thanks.

I used the pythagorean calculation which should allow for someone 6' tall - I think I need to check my formula!
In fact, all the formulae that produce the same result.
Here is an exact calculation with its derivation:
upload_2016-3-16_17-48-52.png
 
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