The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: Bobby Shafto on July 04, 2018, 02:29:05 PM
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I would like to try to apply flat earth theories to analyze a couple of photos I took yesterday. This is from the first photo, taken from Cabrillo Point in San Diego, looking WSW from https://goo.gl/maps/YnzAo9batsT2 (https://goo.gl/maps/YnzAo9batsT2)
(http://oi66.tinypic.com/a5axkz.jpg)
Date/Time
July 3rd, 2018; 10:55 PDT
Weather Conditions
air temperature 70°
water temperature 68°
humidity 66%
dew point 58°
visibility 10+ miles
onshore wind ~5kts
mixed wind/south ground swell 1-3'
Location
N32°40'30.2"
W117°14'46.4"
elevation ~100'
Viewing azimuth WSE ~250°
Camera
Canon PowerShot SX5 HS
1/320 sec; f/8; 215mm
ISO 100
4000x2248 resolution
Cropped to 1000x562
Contrast/color adjusted
Comments: I didn't notice this with the naked eye at first. It was only while panning the horizon at my greatest analog zoom that I saw it. Looked a little like a city skyline peeking over the horizon, but it's actually a loaded container ship some distance off the Southern California coast.
Why does it look like this?
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15 minutes later, it looks different.
(http://oi68.tinypic.com/30lz4eu.jpg)
I didn't change camera or camera settings.
The ship didn't get closer.
The weather conditions didn't change.
What did change was my vantage point.
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https://www.youtube.com/watch?v=ko93J_hXaXQ
https://www.youtube.com/watch?v=rCEOxqne23E
https://www.youtube.com/watch?v=h16PWpLx3tI
(http://i64.tinypic.com/wquc5x.jpg)
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Ocean swells can, and do, alter what you can see on a sea horizon.
I provided wind and ground swell information, and also the viewing elevation, for the first picture. Can that first picture really be explained by obscuring open ocean swells?
San Clemente basin buoy information:
2.4ft @ 13s from 199° (SSW)
2.0ft @ 10s from 288° (WNW)
1.6ft @ 8s from 281° (WNW)
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Can that first picture really be explained by obscuring open ocean swells?
I believe so.
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Can that first picture really be explained by obscuring open ocean swells?
I believe so.
Only if the viewer height is less than the swell height. Otherwise you're looking over it and the swell can't obscure more than it's own height.
(https://image.ibb.co/iL8rC7/waves.jpg)
So if the swell was less than 3 feet then unless Bobby took the image from a lower height than that, the swell can't have obscured more than 3 feet of the boat.
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Can that first picture really be explained by obscuring open ocean swells?
I believe so.
I don't believe so, but I'm asking for a flat earth theory analysis, so okay. This is the "dime can hide an elephant" principle.
A 1-3' ocean swell obscures this amount of vertical height when viewed from 100' of elevation and at X distance:
(http://oi66.tinypic.com/24kwlj9.jpg)
Can we calculate X? We probably need to find the value for the obscured height. What does that look like? 50'? 70'?
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Lrn the difference between waves and swell.
https://www.naturalnavigator.com/news/2017/10/the-difference-between-waves-and-swell
Bobby has given us H0.
Booby has not given us SwH - Swell height.
H0 - "If both swell and wind-waves are present, it should equal the square root of the sum of the squares of the swell and wind-wave heights."
I can't know the swell height unless I have either H0, WwD, SwP and AvP, or am directly given SwH.
https://www.ndbc.noaa.gov/waveobs.shtml
What a bunch of amateurs.
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Lrn the difference between waves and swell.
It actually doesn't matter in terms of my diagram, the key thing is viewer height and whether that could have conceivably been lower than the waves or swell or any other obstacle in between him and the ship. If it could have been lower then you have a case, otherwise not so much.
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Lrn the difference between waves and swell.
It actually doesn't matter in terms of my diagram,
No, because your diagram consists of waves and no swell and bears no relation to reality.
the key thing is viewer height and whether that could have conceivably been lower than the waves or swell or any other obstacle in between him and the ship. If it could have been lower then you have a case, otherwise not so much.
Again wrong. Please learn the difference between waves and swell. You can't progress with this thread until you do.
I provided wind and ground swell information
You did not.
you provided
mixed wind/south ground swell 1-3'
Mixed ... IE H0, not SwH. There are a lot more wind waves than there are swells, dragging the overall max height way down to a lower average. Great if I want an average for my oceanic power plant. Useless if I want to know what I can see or cannot see at a distance. Being as I have neither the period for your wind waves WwP, nor the period of your swell SvP, nor the combined average of the two AvP allowing me to deduce one of them, I can't determine the ratios of wind waves to swell either. This is a mess of an OP.
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Bobby has given us H0.
Booby has not given us SwH - Swell height.
Swell height data (not wave height) provided came from this open ocean buoy (https://www.ndbc.noaa.gov/station_page.php?station=46086).
I'm a naval officer and a surfer. I know the ocean swells. Call me an amateur, will you?
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I provided wind and ground swell information
You did not.
you provided
mixed wind/south ground swell 1-3'
Mixed ... IE H0, not SwH. There are a lot more wind waves than there are swells, dragging the overall max height way down to a lower average. Great if I want an average for my oceanic power plant. Useless if I want to know what I can see or cannot see at a distance. Being as I have neither the period for your wind waves WwP, nor the period of your swell SvP, nor the combined average of the two AvP allowing me to deduce one of them, I can't determine the ratios of wind waves to swell either. This is a mess of an OP.
Opening post was information at the shoreline: a mixed wind/ground swell of 1-3'.
When you claimed ground swell was responsible for the hidden portion of the ship, I provided:
San Clemente basin buoy information:
2.4ft @ 13s from 199° (SSW)
2.0ft @ 10s from 288° (WNW)
1.6ft @ 8s from 281° (WNW)
That's not mixed. That's ground swell. Being the expert you are, I'd expect you to know that.
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Lrn the difference between waves and swell.
It actually doesn't matter in terms of my diagram,
No, because your diagram consists of waves and no swell and bears no relation to reality.
It doesn't matter if it consisted of a hill of beans. This is geometry, it doesn't matter what is in between you and the object you're looking at.
If your viewer height is higher than the highest obstacle in between you and the object then the obstacle will block less of the object than its own height.
My diagram shows why.
It doesn't matter if that obstacle is waves or the swell or a brick wall inexplicably in the middle of the ocean, all that matters it the height of it relative to your viewer height.
If you still maintain I'm wrong then feel free to draw your own diagram showing why I am.
In your diagram you have made the swell artificially high as thought there was a major storm, from the pictures that does not appear to be the case.
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I think I've understood the argument...so the claim is that the ship is bobbing up and down on the swell and Bobby has captured 2 images one where it's at the top, so you can see it and the other when it's in the trough so you can't see the bottom of it.
That makes more sense as an argument but I'm not sure big ships bob up and down on the swell like a small one does, they're generally big enough to be more stable.
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I think I've understood the argument...so the claim is that the ship is bobbing up and down on the swell and Bobby has captured 2 images one where it's at the top, so you can see it and the other when it's in the trough so you can't see the bottom of it.
That makes more sense as an argument but I'm not sure big ships bob up and down on the swell like a small one does, they're generally big enough to be more stable.
That might be the theory, but that's not the observation. The presentation of the ship from vantage point #1 never changed. Everything below the weather deck remained obscured. It wasn't until viewing from vantage point #2 that I could see the ship down to the waterline, and from that point it was never obscured.
Any pitch, roll or yaw was undetectable at that distance. And if there was a ground swell great enough to obscure the ship from a vantage point of 100', I wouldn't have been taking pictures of the horizon. I would have either been in the water or taking pictures of the epic surf at Windansea or Black's.
There IS a hurricane currently that's entered into SoCal's swell window, and we might start picking up some of that soon. That wasn't the case yesterday, and the buoy information confirms that.
But whether it's local wind chop or long train ground swell from the South Pacific, the challenge for the theory is can ocean irregularities be the cause of hiding the lower portions of distant ships at or near the horizon? How does that geometry work? If I can determine the distance to the ship, the height of the observer, and the height hidden portion of the ship, we should be able to calculate the height of the wave or swell or whatever ocean irregularity above the flat plane.
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But whether it's local wind chop or long train ground swell from the South Pacific, the challenge for the theory is can ocean irregularities be the cause of hiding the lower portions of distant ships at or near the horizon? How does that geometry work? If I can determine the distance to the ship, the height of the observer, and the height hidden portion of the ship, we should be able to calculate the height of the wave or swell or whatever ocean irregularity above the flat plane.
Here is the location of the MSC Michela yesterday at the time of the photo. It was in transit, southbound from Long Beach and off the coast of San Diego about 25 statute miles.
I was 100' above the tide pools at Cabrillo Point.
If 50' of ship was being obscured by ocean waves or ground swell, how high would they have had to have been?
(http://oi63.tinypic.com/sdzb4p.jpg)
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Can you upload the full size original version of the photo at the bottom please?
(http://oi68.tinypic.com/30lz4eu.jpg)
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Here is the location of the MSC Michela yesterday at the time of the photo. It was in transit, southbound from Long Beach and off the coast of San Diego about 25 statute miles.
I was 100' above the tide pools at Cabrillo Point.
If 50' of ship was being obscured by ocean waves or ground swell, how high would they have had to have been?
From a point 100' high a point 50' high 25 miles away works out to an angle of 0.022°.
(http://oi66.tinypic.com/2mzjdd1.jpg)
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(http://oi66.tinypic.com/a5axkz.jpg)
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Why does it look like this?
(http://www.sacred-texts.com/earth/za/img/fig85.jpg)
Surface irregularities near the point of H is one flat earth explanation for hull-first disappearance, but it doesn't work as a stand-alone explanation. You need to what and where H is.
H is Rowbotham's "natural law of perspective" explanation for where the earth's flat plane appears to rise to eye level. Geometrically, the convergence to that vanishing point of H occurs for objects or lines of perspective that are the same height above eye level as below. Beyond that, the plane of earth doesn't appear to rise. Instead, objects or lines of perspective above that height equal to eye level converge beyond H, but anything below that line is lost to sight since the earth is not transparent.
Here, I've taken illustrations used to explain this in Earth Not a Globe (http://oi63.tinypic.com/2yum6uv.jpg) and applied the container ship illustration, with eye-level approximately at 50' instead of 100', merely for the ease of illustration. (I couldn't make it work to match the observation if I put the eyeline level at 100')
(http://oi63.tinypic.com/2yum6uv.jpg)
So, the flat earth theory is Perspective (explanation of H and converging sight lines) and Surface Irregularities at H (waves/swell).
Now, how do we calculate distance to H?
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So, the flat earth theory is Perspective (explanation of H and converging sight lines) and Surface Irregularities at H (waves/swell).
Now, how do we calculate distance to H?
I've got an idea for that... let's find that with an experiment!
https://forum.tfes.org/index.php?topic=10016.0
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So, the flat earth theory is Perspective (explanation of H and converging sight lines) and Surface Irregularities at H (waves/swell).
Now, how do we calculate distance to H?
I've got an idea for that... let's find that with an experiment!
https://forum.tfes.org/index.php?topic=10016.0
I don't think that will find the value of H per Flat Earth (ENaG-version) Theory. Though I have yet to get confirmation, my interpretation of Rowbotham's reasoning for Perspective and the horizon is:
(http://oi67.tinypic.com/126d9xf.jpg)
If that's correct, then for a 100' vantage point, H will be around 65 miles away.
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So, the flat earth theory is Perspective (explanation of H and converging sight lines) and Surface Irregularities at H (waves/swell).
Now, how do we calculate distance to H?
I've got an idea for that... let's find that with an experiment!
https://forum.tfes.org/index.php?topic=10016.0
I don't think that will find the value of H per Flat Earth (ENaG-version) Theory. Though I have yet to get confirmation, my interpretation of Rowbotham's reasoning for Perspective and the horizon is:
(http://oi67.tinypic.com/126d9xf.jpg)
If that's correct, then for a 100' vantage point, H will be around 65 miles away.
I don't think I understand this diagram. What are these lines? I thought these lines are the imagined (ie incorrect) projections of perspective lines as seen from a side-view. No?
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these lines are the imagined (ie incorrect) projections of perspective lines as seen from a side-view. No?
Yes, but the key is to see all lines don't converge at a single distance. Depends on origin of perspective line's vertical distance away from the eye, with point of convergence forming a distant angle of 1 arcminute (that's for avg human sight).
But the opaque ground plane creates a limit on convergence distance. Beyond that, lines higher than the eye's level above that ground plane meet at convergent points beyond H, where the ground plane appears to level off.
Vanishing Point is not a single point but variable, with H just being special because the ground is not transparent.
It's mainly described in http://www.sacred-texts.com/earth/za/za32.htm]this chapter of Earth Not a Globe (http://). I just updated the diagram with a formula for calculating distance to H.
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Surface irregularities at the horizon (H) and the Perspective Theory for why there is H at all are just two components of a Flat Earth explanation for what appears in that first picture. There's another one that is evident in the picture from vantage point #1 but not vantage point #2 so much.
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these lines are the imagined (ie incorrect) projections of perspective lines as seen from a side-view. No?
Yes, but the key is to see all lines don't converge at a single distance. Depends on origin of perspective line's vertical distance away from the eye, with point of convergence forming a distant angle of 1 arcminute (that's for avg human sight).
But the opaque ground plane creates a limit on convergence distance. Beyond that, lines higher than the eye's level above that ground plane meet at convergent points beyond H, where the ground plane appears to level off.
Vanishing Point is not a single point but variable, with H just being special because the ground is not transparent.
It's mainly described in http://www.sacred-texts.com/earth/za/za32.htm]this chapter of Earth Not a Globe (http://). I just updated the diagram with a formula for calculating distance to H.
This is all fine, but why are they drawn as lines rather than curves?
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This is all fine, but why are they drawn as lines rather than curves?
Why should the light lines be curved?
The challenge is to explain a visual phenomenon. Is the light curving?
In the Perspective explanation, light is straight, and it's the angle at convergence and the limits of optics that determine what is perceived. vanishing points, like those lying at a horizon, and those beyond it but aligned with the horizon, are perceptual.
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This is all fine, but why are they drawn as lines rather than curves?
Why should the light lines be curved?
The challenge is to explain a visual phenomenon. Is the light curving?
In the Perspective explanation, light is straight, and it's the angle at convergence and the limits of optics that determine what is perceived. vanishing points, like those lying at a horizon, and those beyond it but aligned with the horizon, are perceptual.
In the diagram, the point labeled "H" is a point at eye-level where an object of height "C" above eye level appears to be 1 arcminute above eye-level. Correct? The straight-line side-view path between them should show the line between C and H as horizontal because C is still physically above H by the exact same height it started.
If this diagram is attempting to show how the apparent height of C above eye-level varies with distance, that is a curve and not a straight line. I cannot come up with any reasonable explanation for this diagram. What am I missing? I must be completely misunderstanding the diagram.
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In the diagram, the point labeled "H" is a point at eye-level where an object of height "C" above eye level appears to be 1 arcminute above eye-level. Correct?
I would articulate it that the object (or portion of an object) that is of height "C" above eye level at the location of the observer appears to be less than 1/60 of a degree (1 arcminute) at some distance, called H. So, I think "yes" to your question, but maybe clarified a bit.
The straight-line side-view path between them should show the line between C and H as horizontal because C is still physically above H by the exact same height it started.
No. That would be a separate line, like one parallel to AB in this diagram, but originating at C. (Rowbotham simply didn't draw that it.)
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
And I didn't either on my adaptations. But there is acknowledged the actual line (non-changing height) as distinct from perceptual (diminishing heights).
If this diagram is attempting to show how the apparent height of C above eye-level varies with distance, that is a curve and not a straight line. I cannot come up with any reasonable explanation for this diagram. What am I missing? I must be completely misunderstanding the diagram.
The line represents what we see. Not what we graph. Though the line may curve on a graph, it's because the x axis plot points are equidistant (1x, 2x, 3x, etc.) And that's true in actuality, just as the height doesn't really change. But perceptually, not only is the height appearing to diminish, but the distance between points as they recede into the distant doesn't appear equal but compressing. And if height and distance intervals appear to be decreasing in a direct relation, the line stays straight.
Imagine replacing the (1x, 2x, 3x, ....) intervals in your experiment with measured intervals as they appear on a 2D projection or photograph. The line doesn't curve because the intervals appear to be decreasing too.
H is just a perceptual threshold. Any vanishing point is an apparent one. As Tom put it, it's how the world "presents itself to us" (or something like that). The world doesn't really end at a vanishing point, but vanishing points aren't infinitely away. A horizon isn't an infinite distance away. It's finite. I've only added a calculation to it that I don't believe has been expressed before. It's a simple geometric calculation based on optical resolution, and for the human eye, Rowbotham's source cited put that at 1/60th of a degree.
It's just that instead of all points appearing to compressing at the same vanishing point (or multiple points on a vanishing line), lines that originate higher will merge/compress at a distance further than the one that defines H. And because the earth is not transparent, those points will appear to sink behind it.
Look, you do understand that I don't hold to this myself. I'm only trying to honestly present it as best I understand it. I think I apprehend the rationale, but if I'm botching it, a real FE proponent is always free to take over. My main point is that I put a distance to H and not just leave it vague. It may not be a physical point in space, but from the point of the observer, H must occur at a finite distance away from him or her. I'm putting a value to it, based on the ENaG explanation for phenomena at the horizon.
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In the diagram, the point labeled "H" is a point at eye-level where an object of height "C" above eye level appears to be 1 arcminute above eye-level. Correct?
I would articulate it that the object (or portion of an object) that is of height "C" above eye level at the location of the observer appears to be less than 1/60 of a degree (1 arcminute) at some distance, called H. So, I think "yes" to your question, but maybe clarified a bit.
The straight-line side-view path between them should show the line between C and H as horizontal because C is still physically above H by the exact same height it started.
No. That would be a separate line, like one parallel to AB in this diagram, but originating at C. (Rowbotham simply didn't draw that it.)
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
And I didn't either on my adaptations. But there is acknowledged the actual line (non-changing height) as distinct from perceptual (diminishing heights).
If this diagram is attempting to show how the apparent height of C above eye-level varies with distance, that is a curve and not a straight line. I cannot come up with any reasonable explanation for this diagram. What am I missing? I must be completely misunderstanding the diagram.
The line represents what we see. Not what we graph. Though the line may curve on a graph, it's because the x axis plot points are equidistant (1x, 2x, 3x, etc.) And that's true in actuality, just as the height doesn't really change. But perceptually, not only is the height appearing to diminish, but the distance between points as they recede into the distant doesn't appear equal but compressing. And if height and distance intervals appear to be decreasing in a direct relation, the line stays straight.
Imagine replacing the (1x, 2x, 3x, ....) intervals in your experiment with measured intervals as they appear on a 2D projection or photograph. The line doesn't curve because the intervals appear to be decreasing too.
H is just a perceptual threshold. Any vanishing point is an apparent one. As Tom put it, it's how the world "presents itself to us" (or something like that). The world doesn't really end at a vanishing point, but vanishing points aren't infinitely away. A horizon isn't an infinite distance away. It's finite. I've only added a calculation to it that I don't believe has been expressed before. It's a simple geometric calculation based on optical resolution, and for the human eye, Rowbotham's source cited put that at 1/60th of a degree.
It's just that instead of all points appearing to compressing at the same vanishing point (or multiple points on a vanishing line), lines that originate higher will merge/compress at a distance further than the one that defines H. And because the earth is not transparent, those points will appear to sink behind it.
Look, you do understand that I don't hold to this myself. I'm only trying to honestly present it as best I understand it. I think I apprehend the rationale, but if I'm botching it, a real FE proponent is always free to take over. My main point is that I put a distance to H and not just leave it vague. It may not be a physical point in space, but from the point of the observer, H must occur at a finite distance away from him or her. I'm putting a value to it, based on the ENaG explanation for phenomena at the horizon.
Ok I get you. Perhaps easiest of we don't sweat those lines too much at all and just work the math then. Does that work? It's quite trivial to do this. The angular size of the object is simply arctan(height/distance). Simple as that. So the distance at which an object appears to be 1 arcmin is simply its height/tan(1/60). So that does mean the distance at which an object is 1 arcmin tall varies linearly with the height of the object. (I'm saying "height" only because of the side-view... works for height or width of course.)
Is that what we're talking about? I see you've mentioned that... 100' object vanishes at 65 miles... right! I'm not really certain what the point of the diagram is TBH. It doesn't show anything about objects vanishing bottom-up, and it can't explain the sun going past the horizon as far as I can tell.
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(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
I neither do understand this diagram.
Rowbotham is jumping back and forth from real world to perceptual view/image plane.
- Real world lines A-B and C-D would have to be extended by far (near infinity)
- Only than both lines C-H and A-W in perceptual view/image plane could be a representation of lines A-B and C-D with possible vanishing points at either H or W. (there's only one vanishing point possible)
- Now back in real world Robotham applies eye resolution from real world for the distance to the vanishing point H from perceptual view
- And now comes the ultimate kick: Rowbotham is turning around perception. The distance, defined by the limit of eye resolution is given as real world distance to the vanishing point in perceptual view and is defined by observers hight and not by the size of the observed object.
Think about this again: "Real world distance to the vanishing point in perceptual view"
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Ok I get you. Perhaps easiest of we don't sweat those lines too much at all and just work the math then. Does that work? It's quite trivial to do this. The angular size of the object is simply arctan(height/distance). Simple as that. So the distance at which an object appears to be 1 arcmin is simply its height/tan(1/60). So that does mean the distance at which an object is 1 arcmin tall varies linearly with the height of the object. (I'm saying "height" only because of the side-view... works for height or width of course.)
Yes; height above the observer's eye level.
Is that what we're talking about? I see you've mentioned that... 100' object vanishes at 65 miles... right!
I didn't say it that way, but yes, sort of. (I think?) With eye level at 100', the horizon figures to be 65 miles away. So anything that doesn't exceed that 100' height off the ground will have "vanished" on or before the horizon at 65 miles away or less. And it will have "vanished" as you would expect: to a speck and then gone. No "bottom up" vanishing.
Such "vanishing" you can telescope back into view though because your optical resolution improves to better than 1/60th of a degree, and H is extended (within limits imposed by atmosphere). Those things (or details) aren't beyond the horizon defined by eye level height.
But objects or aspects of objects that are above 100' above a 100' observer don't "vanish" at 65 miles. Vanishing point for them is at a distance from the observer >65 miles, but they do so beyond that apparent horizon, which eclipses the lower portions first before they vanish because of that 65 mile horizon point.
I'm not really certain what the point of the diagram is TBH. It doesn't show anything about objects vanishing bottom-up, and it can't explain the sun going past the horizon as far as I can tell.
I not doing it justice, and that's probably because even just typing it out I see flaws in the explanation. If my 218mm focal length is improving optical acuity to better than 1/60th of a degree, then that should extend H out to some greater distance and 65 miles is no longer the point at which "hull first vanishing" is happening.
I'm also not keen on the reasoning for why anything that is merging with the horizon but beyond the point of H would disappear "bottom first," though I'm trying to include waves/swell or whatever irregularities and undulations are at the flat plane that could impact appearance at H (including one big factor I've been holding for last). But this particular ship was only 25 miles away at the time, so whether with the naked eye (H~65 miles) or camera zoom (H>65 miles), I would not expect it to appear the way Rowbotham/ENaG explain. If the ocean surface is flat, then something else has to be going on, either to supplement or replace what ENaG covers in chapter XIV (http://www.sacred-texts.com/earth/za/).
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(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
I neither do understand this diagram.
Rowbotham is jumping back and forth from real world to perceptual view/image plane.
- Real world lines A-B and C-D would have to be extended by far (near infinity)
- Only than both lines C-H and A-W in perceptual view/image plane could be a representation of lines A-B and C-D with possible vanishing points at either H or W. (there's only one vanishing point possible)
I'm tracking with you, but that last parenthetical isn't true. Right? There are infinite vanishing points. Even if just focusing on the horizontal plane, the Rowbotham explanation of the natural law of perspective is that a single VP at which all things appear to converge is wrong. The horizon is a vanishing point, but vanishing points for other things lie beyond it.
He uses the example of a disc with a smaller circle feature on it. As the disc gets more distant, the circle on the disc will reach 1 arcminute before the disc itself. The VP for the circle is sooner (closer) than the disc as a whole. You will see the disc but not be able to make out its smaller feature. It's all a function of that ability to resolve angular distance of a certain measure.
But the "bottom first" element has to do with proximity to an opaque plane. If you look upward, perspective works the same way, but there's no eclipsing by a non-transparent planar surface. That's not true when looking along a ground plane you can't see through. Then, those elements with angular dimensions smaller than what you can optically resolve "blend" or "merge" (whatever the preferred term is) with the opaque, obscuring plane, and will "vanish" before the elements that are still angularly resolvable -- I'm "word salading" now trying to convey this in stream of thought -- from the opaque plane surface.
I got off track. Point is there isn't one VP. It's this point that Rowbotham makes to explain why not everything vanishes to a point and how an object can become obscured at a horizon line bottom-up.
- Now back in real world Robotham applies eye resolution from real world for the distance to the vanishing point H from perceptual view
- And now comes the ultimate kick: Rowbotham is turning around perception. The distance, defined by the limit of eye resolution is given as real world distance to the vanishing point in perceptual view and is defined by observers hight and not by the size of the observed object.
Think about this again: "Real world distance to the vanishing point in perceptual view"
I have no good response to this. I will say that as a devils' advocate presenting the Rowbotham reasoning, I've applied a distance calculation to H as this eye resolution reasoning goes, but I've never seen actual advocates do that. It's supposed to be a finite point, but one I never see given a finite value. How "the world presents itself" in a perceptual sense vs. a "real world distance" is off-putting to me too. I have to leave that to Tom or someone vested in the idea to defend. I can't.
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Surface irregularities at the horizon (H) and the Perspective Theory for why there is H at all are just two components of a Flat Earth explanation for what appears in that first picture. There's another one that is evident in the picture from vantage point #1 but not vantage point #2 so much.
No one has mentioned "Convergence Zone" yet. That's the 3rd element that I believe works in conjunction with surface undulations + perspective theory to create the appearance that objects sink beyond the horizon.
Look at the picture and notice the distortion in the band right at the apparent horizon.
(http://oi66.tinypic.com/a5axkz.jpg)
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Also "swell" would not magically form a reasonably static line in front of a boat... like the ocean just swells up at one point and stays there? Or like it's continually "swelling" up and down and yet somehow the line in front of the boat doesn't rise up and down i.e. there's a huge swell but also no motion to the swell at all? hmmmmm. Obviously static photos don't show this lack of movement (ironically) but plenty of videos that do.
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Also "swell" would not magically form a reasonably static line in front of a boat... like the ocean just swells up at one point and stays there? Or like it's continually "swelling" up and down and yet somehow the line in front of the boat doesn't rise up and down i.e. there's a huge swell but also no motion to the swell at all? hmmmmm. Obviously static photos don't show this lack of movement (ironically) but plenty of videos that do.
I have 3-4 photos of this view. I'd have to check the timestamps, but at no time did the boat or the horizon change vertically.
Ocean turbulence could only possibly account for 2-3'.
But on to the point of "Convergence Zones," there's a surface inversion layer (common as the morning heats up) that's creating an inferior mirage, evident from 100' but not from 400'.
(http://oi67.tinypic.com/o91xdg.jpg)
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(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
I neither do understand this diagram.
Rowbotham is jumping back and forth from real world to perceptual view/image plane.
- Real world lines A-B and C-D would have to be extended by far (near infinity)
- Only than both lines C-H and A-W in perceptual view/image plane could be a representation of lines A-B and C-D with possible vanishing points at either H or W. (there's only one vanishing point possible)
I'm tracking with you, but that last parenthetical isn't true. Right? There are infinite vanishing points. Even if just focusing on the horizontal plane, the Rowbotham explanation of the natural law of perspective is that a single VP at which all things appear to converge is wrong. The horizon is a vanishing point, but vanishing points for other things lie beyond it.
Rowbotham is wrong for parallel lines. All parallel lines from real world, in perceptual view merge at the one and only vanishing point. The vanishing point in perceptual view represents infinity in real world, no other "point" can be beyond infinity, and no matter how far apart these real world lines are they appear to merge at the vanishing point in perceptual view.
Horizon: What horizon?
The horizon defined by perspective, where viewing plane (eye level) and ground plane of the observer merge? Ok, yes this horizon includes the vanishing point for any bundle of parallel lines, which are parallel to observers viewing plane.
The horizon defined by globe earth curvature. No, this horizon clearly is closer than infinity. That's another type of 'horizon'.
The horizon defined by observers eye resolution. Sorry, that's no horizon, that's something blurring your view. Perspective is pure geometry and mathematics, there's nothing in it for limited eye resolution.
You could instead calculate that point, where in real world eye resolution would find it's limit and than apply perspective.
He uses the example of a disc with a smaller circle feature on it. As the disc gets more distant, the circle on the disc will reach 1 arcminute before the disc itself. The VP for the circle is sooner (closer) than the disc as a whole. You will see the disc but not be able to make out its smaller feature. It's all a function of that ability to resolve angular distance of a certain measure.
The vanishing point represents infinity in real world. The ability to resolve angular distance does not define a 'horizon'. Playing with the zoom of your camera does not change the vanishing point, or? It's just more - or less blur. Or is the vanishing point also 'moved' by atmospheric haze?
But the "bottom first" element has to do with proximity to an opaque plane. If you look upward, perspective works the same way, but there's no eclipsing by a non-transparent planar surface. That's not true when looking along a ground plane you can't see through. Then, those elements with angular dimensions smaller than what you can optically resolve "blend" or "merge" (whatever the preferred term is) with the opaque, obscuring plane, and will "vanish" before the elements that are still angularly resolvable -- I'm "word salading" now trying to convey this in stream of thought -- from the opaque plane surface.
Again this is just a blur. Wouldn't the opaque plane near that - I define a new term for it - 'blurring point' also be blurred in a similar way? That all is merged into one bigger blurred zone, where no one can decide, where the plane ends and the 'vanished' object begins.
I got off track. Point is there isn't one VP. It's this point that Rowbotham makes to explain why not everything vanishes to a point and how an object can become obscured at a horizon line bottom-up.
"Bottom-up": In all examples Rowbotham gives, I see a whole object 'vanish'. It's not the lower part of the ship, it's the hull, clearly distinguishable from the rest of the ship. It's the black boots of the soldiers, in clear contrast to their uniforms. It's the white part of these circles mentioned above, in clear contrast to the bigger black ones.
Another try to explain: Think about a bigger object appearing close above a clearly visible horizontal line. With enough resolution you will see this as described. Now decrease resolution: The horizontal line will blur and appear to get wider: There's no significant change on the bottom side, because the plane below is affected by a similar blur. But on the top side of the blur seems to extend into the object above the horizontal line.
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This is all fine, but why are they drawn as lines rather than curves?
My answer to this above was that though the elevation angle is inversely proportional to distance, so is the perceived interval distance, which offset the "curve" and restores linearity.
But I've been trying to work out a way for the ENaG orthogonal depiction of merging lines to explain a sunset (or appearance of things at a horizon), and something occurred to me that I failed to realize.
I'm going to post it on the Sunset from a Drone discussion topic, but I thought it was apropos to this discussion as well since I've tried to delve into the Perspective Theory here too, though not about the sun specifically. I'll append a cross link after I post it; I just wanted to make note of it here since it's a counter to my reply to you and references the conclusion of your vanishing point experiment.
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Since that other topic was exiled to AR, I'll just amend this post with what I posted there.
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Not to imply that the curved earth explanation is the gold standard, but I've not encountered anyone who objects to rationale of where the horizon is and why objects on the surface of a convex surface beyond it can be hidden.
Whether you believe the surface of earth is convex or not, this makes sense:
(http://oi63.tinypic.com/k9xpqd.jpg)
Increasing "eye level" height above the surface extends the horizon, diminishes geometric drop and changes the amount hidden (or eliminates it entirely). The question is whether or not this is true for the earth, and that depends on whether or not the earth's surface is convex. But I've never heard anyone argue with the geometry or the logic of why it would work this way IF the earth was spherical.
Instead, an alternative explanation is offered for the horizon and objects or features appearing beyond it. Trying to compare and contrast this Perspective alternative, using Earth Not a Globe's diagram as a starting point, we might get something like this:
(http://oi67.tinypic.com/3484pc7.jpg)
Increasing eye level height above surface extends the horizon and changes what's hidden too, but how? The 1/60th degree of resolution is an explanation for the apparent merging of perspective planes/lines that don't actually merge but just appear to. And the opaque flat surface merging to a vanishing point with the transparent plane that is 2x the height of eye level somehow is responsible for the horizon line AT eye level. There is no "dip." There is no calculation for how much is hidden for distances and heights beyond the horizon
It makes sense that increasing eye level height will extend the distance to the horizon, but there is still no alternative explanation to the curved-earth approach to explain why there is a hidden zone beyond the horizon or how to calculate such a thing.
The curved, geometric model has a Drop value from level sight. The flat, perspective model doesn't.
The curved, geometric model can calculate the amount hidden beyond the horizon. The flat, perspective model can't.
The curved, geometric model acknowledges that the horizon doesn't remain level with the eye but dips. The flat, perspective model relies on the horizon always being level with the eye, with no "dip."
I think if you resolve those issues, the flat earth explanation of Perspective gains merit as a competitor to curvature being the reason for horizons and hidden features beyond horizons.
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This is a rendition of how a parallel overhead plane sun can appear to set, according to the Rowbotham "Natural Law of Perspective."
(http://oi67.tinypic.com/141kx8g.jpg)
I've seen this same explanation from the more popular YouTube Flat Earth advocates too using Perspective to explain how the sun appears to descend and merge with a vanishing point horizon.
One of the first critiques of this is that the sun should, according to Perspective, also decrease in size to a point by the time it reaches the vanishing line of the horizon.
(http://oi63.tinypic.com/vymu84.jpg)
Some say it, in fact, does.
This site says it doesn't, but that's because of a magnification effect of the atmosphere as the sun has to pass through greater and greater density of air/moisture, and that effect offsets the expected reduction in sun size.
But also observed with Perspective is that it's not just the apparent size of things that gets smaller with distance away from the viewer. It's also the apparent longitudinal distance compresses. For instance, evenly space things like railroad ties or lamp poles will appear to get closer and they get smaller. To apply this to a setting sun as related to Perspective:
(http://oi66.tinypic.com/15ek421.jpg)
If atmosphere water vapor offsets the diminishing in perceived size, what can account for the offset of diminishing interval distance?
Distance of sun movement along the downward angled line of Perspective is always depicted as being equidistant to its actual straight parallel path position. But if Perspective is at work, the distance should lag too. If not, then why not?