The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: Bobby Shafto on August 03, 2018, 07:40:26 PM
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I just want to comment that Bobby seemed to have spent some time looking at something from the videos, which he deleted. I thought about it and had something to respond with. I am more than happy to look at it with him if he were to repost it in the main forums.
You posted a link to this video:
https://www.youtube.com/watch?v=7-pXWRn_wfk
I posted that that YouTuber was wrong in his analysis of his evidence:
(http://oi68.tinypic.com/34sjpxd.jpg)
He claimed 10,800' Mt San Jacinto shouldn't be visible from a 150' high viewpoint in Malibu, CA, 117 miles away if the earth is curved.
I claim it should be, and his imaging -- despite his accompanying narrative to the contrary -- shows what is expected on a curved earth and not a flat earth.
The graphic I posted was in error. For that reason, and because of the request not to debate in the Media forum, I deleted it. But I'm more than open to discuss it, Tom.
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I was going to respond to the image you created but didn't post here. I think you see where the error was, and so I won't bother with it.
Lets see. I'll go through the items in the video one by one.
Experiment 1.
(https://i.imgur.com/0yY701H.jpg)
Checks out:
(https://i.imgur.com/aAt5jtX.png)
Conclusion: Flat Earth
Experiment 2.
(https://i.imgur.com/KWnpWn3.jpg)
Checks out:
(https://i.imgur.com/xXZ3ql1.png)
Conclusion: Flat Earth
Experiment 3.
(https://i.imgur.com/yUed2hI.jpg)
Checks out:
(https://i.imgur.com/DPFKIYa.png)
Conclusion: Flat Earth
Now he's talking about infrared being able to see through the atmosphere.
(https://i.imgur.com/FgZtVZ4.png)
Good idea!
Experiment 4. Now he gets to the mountain:
(https://i.imgur.com/CQVXM33.png)
Earth Curve Calc says drop should be 7351.2413 feet:
(https://i.imgur.com/DmqOqRV.png)
However at 11:25 he says "If the earth was curved, we shouldn't be able to see Mt. Jacinto. Even though it is 10,600 feet high, the shoreline and the hills on the other side of the bay should be hiding it entirely."
(https://i.imgur.com/iMFUN8o.jpg)
He says the hills in the foreground should cause the mountain to be entirely obscured in the Round Earth version.
I don't have data on the height of those hills in the foreground. I couldn't say whether he is right or wrong.
This is the flat earth version with foreground hills hiding some of the mountain:
(https://i.imgur.com/l3e8S9Q.jpg)
Unless you have all of the data for the hills and can compute what should or should not be seen under RET, I cannot say that your analysis is correct.
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I'm not going to pick apart the entire video. I just want to examine his sighting of San Jacinto from Malibu, which is his capstone.
Work with me on this. If the earth is flat, where would you draw eyelevel on this image?
(http://oi68.tinypic.com/34sjpxd.jpg)
I do believe he was shooting from somewhere on Malibu Bluffs park, below Pepperdine University. The elevation varies, but that's enough location detail and he consistently cites 150' as his viewing elevation.
So, looking across the bay toward Santa Monica and inland, where is the 150' eye-level drawn on that picture?
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Here's my opinion. See if we agree and, if not, let's defend our answers.
If the earth is flat, then I'd draw the 150' eye level line here:
(http://oi67.tinypic.com/313srw0.jpg)
Rationale: For references, I used the 16-story Pacific Plaza on the left that is on Ocean Blvd, elevated about 50' above the beach and rising 180'. And also the La Meridien Delfina hotel on Pico, that's about a half a mile further inland on a 90' elevation. I don't know it's height, but it is a few stories shorter than the Plaza so maybe 140-150'. It's more like 12.5 miles from Malibu Bluffs vice JTolan's note of 12 miles.
So that's how I estimate the FE "eye level" line. Agree or do you have a counter?
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It is indeterminate where the eye level is without knowing the specifics of the lenses in an optical zoom device like that because of collimation.
See: http://www.sacred-texts.com/earth/za/za45.htm
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You don't have to be precise. Is it at the water line? The top of the Pacific Plaza? Is my line in the ballpark?
What makes sense to you? How did JTolan figure the angular distances for his vertical mrad scale? Is it right?
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How much collimation do the lenses in the device make for the zoom seen? Without that information the matter is as erroneous as the theodolite dip. Read Earth Not a Globe.
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I don't agree, but okay. Let me try a different tact then.
Do you agree with JTolan's scale? He measures the height of San Jacinto using his scale for which he placed the 0 point aligned with sea level at Santa Monica and 17.5 millirands (about) at the peak, which equates to about 10,800' at a distance 117 miles away. Is that right? I think not, but do you think that's sound?
How did he derive his scale? He said he used the building he labeled at 12 miles as an index (the La Meridien Delfina). Is that sound given your position on the indeterminacy of eye level?
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You're involved in a lot of discussions right now so let me just jump ahead and show you what I'm getting at.
If we take that condo building on Ocean Blvd and compare it to JTolan's scale, it measures out at around 5.3 millirands, or 0.3°. At 11.9 miles (62,832') away, that angle results in a vertical distance of 329'. But the Pacific Plaza is not 329' feet tall. It's a 15-story building. Using JTolan's method of estimating height, that works out to 180', which I found corroborated on this Web page (https://www.emporis.com/buildings/125191/pacific-plaza-santa-monica-ca-usa).
(http://oi67.tinypic.com/2r73ewx.jpg)
Let's try the La Meridien that he said he used to estimate his scale. This is tougher because I couldn't find any building height data for it, an only roughly estimated it to be about 150' tall. At 10-12 stories, it could be only 75% as tall as the Pacific Plaza, which would make it 135'. Plus, it's tucked away behind some obstacles in the image, so I can't be sure of where it's ground level is.
(http://oi63.tinypic.com/ka4d8z.jpg)
I'll be generous and measure out just what I can see, which is about 2.6 millirands (0.15°). At 12.5 miles away, that's a height of 172'. Even if I use JTolan's distance of 12.0 miles, that's still 166' which is more even than my over-estimate of 150', and that's with skewing everything to try to make it fit.
JTolan's scale isn't right.
Let's use the Pacific Plaza, with a confirmed height of 180' and a measured distance of 11.9 miles to build an alternative scale.
(http://oi63.tinypic.com/1z6tfnb.jpg)
The problem now is that while it appears to rectify the angular dimensions of the buildings in the foreground, San Jacinto is no longer mating up with the 17 millirands that JTolan found as confirmation of a flat earth.
Why?
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Do you have any critique of my proposed reticle/index, Tom?
Or do you have a defense for JTolen's?
Assessing what we're seeing in the distance and discerning flatness or curvature of the earth pretty much depends on it. I'd rather not go through the process if you're going to harbor criticisms of the scale. Can we resolve that first?
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Do you have any critique of my proposed reticle/index, Tom?
Or do you have a defense for JTolen's?
I've given it a few days, so in the absence of any negative assessment, I'll proceed under the assumption my scale is correct and JTolen's is wrong.
That obscuring ridge is 450-480' above mean sea level, 19.9 miles away from the 150' vantage point on Malibu Bluffs and about 8 miles inland from the Santa Monica beach. I can provide supporting arguments for that if you like, but only if you want to contest it. (The closer ridge you can see is a mile closer and about 120' lower, but it's not an obscuring factor. It's just an aid in identification.
The 10,800' peak of San Jacinto lies 97 more miles beyond that ridge.
Without considering atmospheric refraction, I calculated that 1455' of San Jacinto should be visible from 150' Malibu Bluffs on a globe earth if that ridge is 450' high. 1293' if the ridge is 480' high. With standard refraction those figures would increase some, but a no-refraction calculation is the worst case, and even then JTolan would be wrong about San Jacinto not being visible if the earth is curved.
Working it out for a flat earth, 8925' of San Jacinto should be visible over that ridge if it's 450' and 8708' if the ridge is 480'. Again, I can show my work if you want.
How much of San Jacinto is, in fact, visible? By my index, 4.3 milliradians, or 0.25°. At 117 miles, that works out to be 2652' of San Jacinto is visible in that image.
(http://oi66.tinypic.com/23m981u.jpg)
That's around 1200-1350' more than predicted by the no-refraction globe earth calculation.
But it's 6000-6250 less than was a flat earth predicts.
JTolan's scale measures 7.5 milliradians (0.43°) which equates to 4636' at 117 miles, so that's closer but still only a little more than halfway (not to mention it doesn't jive with angular measurements in the foreground.)
I could be guilty of confirmation bias and forced pattern-matching, but if I'm right about these faint features in the IR image aligning with this view of the mountain:
(http://oi68.tinypic.com/4pcm.jpg)
Then that 117-mile perspective captured the area of San Jacinto's elevation seen in the dotted box here:
(http://oi63.tinypic.com/znrbkp.jpg)
Which corresponds to elevations from below 8300' (Fullers Ridge visible) but above 7600' (Black Mountain not visible). That would means we're seeing somewhere around the upper 3000' of the mountain range there. My mrad scale figured 2652'. 348' at 117 miles is 1/2 a millirand. Perhaps my scale is still slightly off.
I had no success getting any images of the upper 4500' (and certainly not the 8700+ feet) of mountain range to match the IR image that might support a flat earth claim. Maybe someone more inclined toward flat earth can.
This still doesn't serve as a "win" for a globe earth since there's still a significant delta between the no-refraction prediction of San Jacinto visibility and what my analysis of the imagery is saying is visible. Even figuring for refraction (which I haven't done yet, but just estimating), I think globe earth only gains an addition 500-600' of predicted visibility, which is still not quite there.
There's a lot of estimating here, but it's not just shots in the dark. Whatever the margin for error, it's much, much closer to validating a globe earth than a flat earth. JTolan's scale is off. He positions it incorrectly. And he misinterprets just how much of San Jacinto is visible. He also incorrectly depicts that the mountain should be entirely obscured if the earth is curved. I argue that his imagery more closely shows what a globe earth predicts than a flat earth.
I don't think his errors are intentional. I think he's just mistaken, caught up in the excitement of making the distant mountains visible through the haze with his IR set up as if he's uncovered something that shouldn't be visible on a curved earth.
Maybe my calculations are wrong. Maybe I'm caught up a round earth premise and making my analysis fit what I already believe.
Maybe. But if so, show me where I've goofed up. I think I'm on the right track, but I welcome reasoned review.
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10,834 foot mountain visible down to around 8,100 feet (approx 2,734 visible). Looks to be about 3,000 feet hidden behind the foreground hill down to the waterline. 5,100 feet are below the waterline. Pretty much fits the globe.
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10,834 foot mountain visible down to around 8,100 feet (approx 2,734 visible). Looks to be about 3,000 feet hidden behind the foreground hill down to the waterline. 5,100 feet are below the waterline. Pretty much fits the globe.
(https://media.giphy.com/media/fnNxDqrUWkEBwZv8p7/giphy.gif)
That 450' ridge of Ladera Hills just 20 miles from the Malibu viewpoint is aligned much too high up the slope of San Jacinto for the earth to be flat. The videographer misinterprets just how much of San Jacinto he is seeing, fooled by the looming of the telephoto image and the incorrect millirand angular index he had created.
Also, if the earth was flat, the Chino Hills breaching 1500' at around 55-60 miles (halfway from Malibu Bluffs to San Jacinto) would be high enough to peek above that 450' ridgeline at 20 miles.
(http://oi67.tinypic.com/2ry57k9.jpg)
On a globe earth, the surface would drop too far by then to be able to see 1500' hills beyond that near ridge. And, as JTolen's video reveals, we don't see the Chino Hills. We don't see any other intermediate distant features beyond that Ladera Hills ridgeline until the land elevation reaches over 8000' MSL. That's not a flat earth signature. That's a spherical earth.
I will say this though. I like the IR filter with spotter scope set up. I want to use something like that to spot the horizon, looking toward San Clemente and Catalina islands, around 65-70 miles off the cost. I know there are times one can see them from San Diego with a little elevation, but since I've been hawking it, the marine horizon to that distance has been cloaked in haze. I accidentally captured Catalina during a sunset last April, when the setting sun provided backlight and cut through the haze.
(http://oi63.tinypic.com/69mh4m.jpg)
But using infrared could help bring the "does the horizon rise to eye level" experiment (https://forum.tfes.org/index.php?topic=9492.380) to a conclusion. (I haven't forgotten about that.)
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I was going to do this. Someone else beat me to it:
https://www.youtube.com/watch?v=doAFr5Cqtb8
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The man in that video seems to be talking about the wrong hills, and trying to estimate the height of that building seems like a fruitless effort since we do not have the architectural height information, do not know how many window panes there are per floor on that building, and we cannot see the entirety of that building from top to bottom. There is a bunch of stuff towards the bottom of the building blocking the view.
Here is my assessment:
Data:
Man saw 10,800 foot tall tall mountain that was 120 miles away, with an eye height elevation of 150 feet.
(https://i.imgur.com/CQVXM33.png)
Tools:
Earth Curve Calculator: https://dizzib.github.io/earth/curve-calc/
Angular diameter calculator: https://rechneronline.de/sehwinkel/angular-diameter.php
Angular Diameter Calc: Full View of Mountain Without Curve
g: 10800 (Full Height of Mountain in Feet)
r: 633600 (120 Mi in Feet)
= a: 0.977 degrees
Angular Diameter Calc: Round Earth Amount of Mountain Hidden Behind Curve
g: 7351.2413 (Amount hidden below horizon according to Earth Curve Calc with elevation of 150 feet over 120 miles)
r: 633600 (120 Mi in Feet)
= a: 0.665 degrees
0.977 degrees - 0.665 degrees = 0.312 Degrees of Mountain Visible Above Curvature of Earth
Now how about the hills in between the observer and the mountain?
Map from the video with red overlay:
(https://i.imgur.com/QZXgwZa.png)
From USGS:
Direct Link to location: https://ngmdb.usgs.gov/topoview/viewer/#15/33.9155/-117.7393
(https://i.imgur.com/17nZP4H.png)
500 meters in Feet = 1640.42 ft
Earth Curve Calc: 1067.0811 ft hidden after 55 miles
1640.42 - 1067.0811 = 573.3389 ft
Angular Diameter Calc:
g: 573.3389 ft
r: 290400 (55 Mi in Feet)
= a: 0.113 degrees
0.312 - 0.113 = 0.199 degrees visible above horizon
Full Mountain: 0.977
Visible Mountain: 0.199
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I was prepping to respond, but I see you've edited. Nothing wrong with that. Just let me know when you're done.
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I'm done editing. But most of this seems like a waste of time since this isn't a water convexity experiment. Most of this experiment takes place over land, not water. We are assuming that land behaves like water.
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It's geometry. We're comparing calculations for a regular flat and convex surface and factoring in irregularities (a.k.a. hills). We're even ignoring effects of atmospheric refraction.
If it's a waste of time, why did you post the link to the video in the first place? It extols the virtues of a flat earth, but now even if you think that land obstruction is Chino Hills 55 miles distant, you're still debunking the videographer who says San Jacinto mountain shouldn't be visible if earth is spherical.
My earlier calculation came to about 13% visible as a globe earth predicts w/o refraction. Using different measures you came up with 11% visible.
Even with errors not favoring a spherical earth you've invalidated a key premise of that video.
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If it's a waste of time, why did you post the link to the video in the first place?
There are a bunch of water convexity experiments in those videos I posted. I posted two other videos in that thread in the Media forum (https://forum.tfes.org/index.php?topic=10295.0). All others are water convexity experiments, which you have ignored.
6+ experiments showing that the earth is flat, and 1 experiment that you think shows that the earth is round?
This mountain experiment isn't even a water convexity experiment. Most of the experiment is taken over land. I'm not sure how accurate we are by assuming that the land behaves like water.
My earlier calculation came to about 13% visible as a globe earth predicts w/o refraction. Using different measures you came up with 11% visible.
Even with errors not favoring a spherical earth you've invalidated a key premise of that video.
Going with your 13%, it seems a little odd that the base of the mountain is way down there in the closer overlay photo you posted:
(https://i.imgur.com/UoUql77.jpg)
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If it's a waste of time, why did you post the link to the video in the first place?
There are a bunch of water convexity experiments in those videos I posted. I posted two other videos in that thread in the Media forum (https://forum.tfes.org/index.php?topic=10295.0). All others are water convexity experiments, which you have ignored.
6+ experiments showing that the earth is flat, and 1 experiment that you think shows that the earth is round?
Yeah. It's the only one I looked at out of your list. I never got past it. The infrared filter appealed to me since I've been frustrated by the summer SoCal atmospheric haze and I found the technique to have potential for the horizon experiments/observations.
If you think that video is not like the others, disavow it then. You should, because despite the narration, it is supporting a convex earth.
I have no idea if the others are valid or not. I got caught up in the JTolen Media one. Sorry, but that's what this thread is about, and I initiated it because you asked and that other board is apparently not the place to comment on it. If you want to address the other videos, by all means, create discussion topics about them. This one is about the Malibu to San Jacinto IR viewing. Let's stick to that on this topic.
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Going with your 13%, it seems a little odd that the base of the mountain is way down there in the closer overlay photo you posted:
(https://i.imgur.com/UoUql77.jpg)
Yeah, that is odd. I wrote about that already. Though I calculated a non-refractive 13% of the mountain to be visible on a globe earth (1455'), my angular analysis of the imagery was that 2000-2500 feet is actually visible, or 19-23%.
That other video found the same thing (23%):
(http://oi68.tinypic.com/kdmmo4.jpg)
The fact that we see more of the mountain than what no-atmosphere, no-refraction geometry figures to isn't proof of earth's flatness. Work out the figures for a flat topology and how much of San Jacinto should be visible if that obstruction is the Chino Hills at 1647' elevation 55 miles away? I came up with 7300' or 68%.
My 13% no-refraction globe earth calculation is closer to the observed 23% than is the 68% flat earth calculation. You need some "bendy light" fudge factor to make up that 45% delta. I only need to make up 10% from the other direction, some of which could easily be accounted for with refraction, bending light the other way.
And yes, if the earth is a globe of radius 3959 miles, then that IS where sea level would appear if you could peer through the earth. On a flat earth sea level would be at eye level, but you can't (or won't) even hazard an estimate as to where "eye level" is on the image.
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We may be looking at the wrong hills. The Chino hills are 55 miles out and he is talking about are 20 miles out. The ones that are 20 miles out seem to be the Baldwin Hills.
Look at the hills JTolan is talking about:
(https://i.imgur.com/VXfLAbm.jpg)
(https://i.imgur.com/R8j85l6.png)
From USGS Topo map zoomed out view:
(https://i.imgur.com/rAa67wV.png)
But if this is like the other USGS map of the Chino Hills from the same system (I have a Topographic map on the previous page), the Baldwin Hills heights are 400+ meters. JTolan may be mistaken that it is 400+ feet.
(https://i.imgur.com/DlGBgF9.png)
Direct USGS Link to above: https://ngmdb.usgs.gov/topoview/viewer/#15/34.0044/-118.3658
The peaks seem able to range from 400 to 500+ meters if the units are the same as the Chino Hills map from the same system (https://i.imgur.com/17nZP4H.png).
400 meters = 1312.34 ft
From the Angular Diameter Calc
https://rechneronline.de/sehwinkel/angular-diameter.php
g: 1312.34 feet
r: 105600 ft (20 miles in feet)
a: 0.712
This combined about what I wrote before:
Earth Curve Calculator: https://dizzib.github.io/earth/curve-calc/
Angular diameter calculator: https://rechneronline.de/sehwinkel/angular-diameter.php
Angular Diameter Calc: Full View of Mountain Without Curve
g: 10800 (Full Height of Mountain in Feet)
r: 633600 (120 Mi in Feet)
= a: 0.977 degrees
Angular Diameter Calc: Round Earth Amount of Mountain Hidden Behind Curve
g: 7351.2413 (Amount hidden below horizon according to Earth Curve Calc with elevation of 150 feet over 120 miles)
r: 633600 (120 Mi in Feet)
= a: 0.665 degrees
0.977 degrees - 0.665 degrees = 0.312 Degrees of Mountain Visible Above Curvature of Earth
If the units are meters, rather than feet, like the Chino Hills map from the same system, it seems to suggest that none of the mountain should be visible.
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We may be looking at the wrong hills. The Chino hills are 55 miles out and he is talking about ones that are 20 miles out.
I'll read the rest of what your wrote after that, but I'd like to first comment that I skipped over your identification of that obscuring ridge at being the elevation 1600+ foot elevation rise 55-60 miles away from Malibu Bluffs because I didn't want to bog down on that since it didn't matter. Claiming that was Chino Hills benefited your flat earth argument, but I let it go because it still didn't work out in FE favor.
I wasn't making that mistake. I know those are the lower hills at 18-20 miles we're seeing, which constitute the area around S. La Brea Ave and the Inglewood oil fields. If you read my previous posts, you can see that. Plus, JTolen also identified that land rise as 400' elevation 20 miles down range. I worked a higher elevation figure (450') than did JTolen, and that's how I arrived at a no-refraction 1455' (13%) visible.
That difference (me using 450' hills at 20 miles; you using 1657' hills at 55 miles) probably accounts for the difference between my 13% visible and your 11% visible conclusions, though you also used 120 miles to San Jacinto whereas I used the correct 117 miles.
On a flat earth, the Chino Hills rise should be visible higher than the Ladera Heights rise. On a globe earth, they should not. I don't see them in the image, so as I said earlier, that's another point in favor of a convex earth.
Now, I'll read the rest of your post.
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I may be mistaken. After checking with other sources on the height of those hills, the USGS map appears to be in feet. It changes units between feet and meters depending on what map you are looking at, without indication that the units have changed.
JTolan is correct that they are 400+ feet in height.
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From USGS Topo map zoomed out view:
(https://i.imgur.com/rAa67wV.png)
But if this is like the other USGS map of the Chino Hills from the same system (I have a Topographic map on the previous page), the Baldwin Hills heights are 400+ meters. JTolan may be mistaken that it is 400+ feet.
You're looking at the wrong spot, Tom. Looks like you do have the right spot (though Baldwin Hills are more to the north). Based on your thinking that the elevation was 400 m vice 400 ft, I thought maybe you were looking at the Hollywood Hills on the topo chart.
Here (https://goo.gl/maps/E4No2JxnaDQ2). This is where the sight line crosses the highest elevation point of those hills.
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I will take a look at your location argument in a minute.
From the Angular Diameter Calc
https://rechneronline.de/sehwinkel/angular-diameter.php
g: 400 feet
r: 105600 ft (20 miles in feet)
a: 0.217
From earlier:
Earth Curve Calculator: https://dizzib.github.io/earth/curve-calc/
Angular diameter calculator: https://rechneronline.de/sehwinkel/angular-diameter.php
Angular Diameter Calc: Full View of Mountain Without Curve
g: 10800 (Full Height of Mountain in Feet)
r: 633600 (120 Mi in Feet)
= a: 0.977 degrees
Angular Diameter Calc: Round Earth Amount of Mountain Hidden Behind Curve
g: 7351.2413 (Amount hidden below horizon in feet according to Earth Curve Calc with elevation of 150 feet over 120 miles)
r: 633600 (120 Mi in Feet)
= a: 0.665 degrees
0.977 degrees - 0.665 degrees = 0.312 Degrees of Mountain Visible Above Curvature of Earth
0.312 - 0.217 = 0.095 degrees visible.
If the units are in feet, it still appears that the entirety of the mountain should essentially be invisible.
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(https://i.imgur.com/rAa67wV.png)
(https://i.imgur.com/TDhvHPW.png)
Yes, the two locations seem to be the same.
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Yes, the two locations seem to be the same.
We're on the same sheet of music now. Yeah, I never noticed that that was also Baldwin Hills on the east side of that Kenneth Hahn State Recreation Area. That ridge line is the western edge of the eastern section of that bedroom community. Baldwin Hills is at the northern edge of those hill. South La Brea Ave runs right along that ridge line.
So, as long as we're agreeing that it's 400' (I'm actually calling it 450' along the point where the Malibu-San Jacinto sight line crosses it) at a distance of 20 miles, we're working from the same data. The only thing you might still want to correct is the distance to San Jacinto. JTolen has it at 120 miles on his graphic, but he cites the correct distance of 117 miles (617,760 feet) in the video. And that's what the distance maps out to be also.
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Without spending a terrible amount of time trying to understand all the numbers being thrown out here, here is where my brain takes me:
If the earth curves 8" per mile, and the viewing distance is 150 miles, that calculates to a drop of 100 feet per 150 miles. (150*8 )/12 = 100
So if you are trying to see a 10,000 foot mountain from 150 miles away, you should still be able to see 9,900 feet of the mountain. Correct?
If I have understood the curvature and done the math correctly, I have no idea why this is proof the earth is flat. You would absolutely be able to see the mountains from that distance on a round earth....
Edit: been a long day, and I just realized it is 8" per mile squared, so when Im not driving I'll recalculate ... Sorry if this is what you have already done.
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Ok, so just using an online calculator it looks like the drop at 150 miles should be around 15,000 feet, so now this is a huge problem for RE.... That is very peculiar to say the least!
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Yes, the two locations seem to be the same.
We're on the same sheet of music now. Yeah, I never noticed that that was also Baldwin Hills on the east side of that Kenneth Hahn State Recreation Area. That ridge line is the western edge of the eastern section of that bedroom community. Baldwin Hills is at the northern edge of those hill. South La Brea Ave runs right along that ridge line.
So, as long as we're agreeing that it's 400' (I'm actually calling it 450' along the point where the Malibu-San Jacinto sight line crosses it) at a distance of 20 miles, we're working from the same data. The only thing you might still want to correct is the distance to San Jacinto. JTolen has it at 120 miles on his graphic, but he cites the correct distance of 117 miles (617,760 feet) in the video. And that's what the distance maps out to be also.
Okay, here it is with your numbers (I haven't checked whether it is 117 miles or 120 miles -- JTolan's map lists the locations at over 120 miles away):
Earth Curve Calculator: https://dizzib.github.io/earth/curve-calc/
Angular diameter calculator: https://rechneronline.de/sehwinkel/angular-diameter.php
Full View of Mountain Without Curve
g: 10800 (Full Height of Mountain in Feet)
r: 617760 (117 Mi in Feet)
= a: 1.002 degrees
Round Earth Amount of Mountain Hidden Behind Curve
g: 6937.2487 feet (Amount hidden below horizon according to Earth Curve Calc with elevation of 150 feet over 117 miles)
r: 617,760 (117 Mi in Feet)
= a: 0.643 degrees
1.002 - 0.643 = 0.359 Mountain Visible Above Earth Curve
Baldwin Hill Height using 450 feet (Your midpoint number -- there are elevations between 400 and 513 feet)
According to earth curve calc, at an elevation of 150 feet, over 20 miles, 16.6866 feet is hidden
450 - 16.6866 = 433.3134
g: 433.3134
r: 105600 ft (20 miles in feet)
a: 0.235
If the mountain is 117 miles away, with a hill height of 450 feet:
0.359 (Visible RET Mountain 117 miles away) - 0.235 (hills) = 0.124 degrees above the horizon
If the mountain is 120 miles away, with a hill height of 450 feet:
0.312 (Visible RET Mountain 120 miles away from last post) - 0.235 (hills) = 0.077 degrees above the horizon
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Ok so the viewing distance is 120 miles, not 150. That makes a huge difference. At that distance about 1/3 of the mountain should still be visible. Which is what it looks like in that video.
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I took my numbers above and broke it down by ratio to see how much would be above the horizon according to RET:
If the mountain is 117 miles away, with a hill height of 450 feet:
0.124 (viewable mountain due to earth curve + hills) / 1.002 (full mountain 117 miles away) = 0.12 or 12% above the horizon
If the mountain is 120 miles away, with a hill height of 450 feet:
0.077 (viewable mountain due to earth curve + hills) / 0.977 (full mountain 120 miles away) = 0.078 or 7.8% above the horizon
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I took my numbers above and broke it down by ratio to see how much would be above the horizon:
If the mountain is 117 miles away, with a hill height of 450 feet: 433.5 feet
0.124 (viewable mountain due to earth curve + hills w/ curve) /1.002 (full mountain) = 0.12 or 12% above the horizon
You got 12% of the mountain visible 117 miles away from an elevated vantage point of 150' and given a terrestrial obstacle of 450' 20 miles away.
I got 13%.
That's close enough for me that I don't think I need to quibble over the small delta.
Work out the geometry for a flat earth topography now.
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Ok so the viewing distance is 120 miles, not 150. That makes a huge difference. At that distance about 1/3 of the mountain should still be visible. Which is what it looks like in that video.
The problem isn't a straightforward earth curve calculation. There's a 450' obstruction 20 miles down range that increases the amount hidden. It, too, is affected by the curvature so not all 450' serves as obstruction.
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Ok so the viewing distance is 120 miles, not 150. That makes a huge difference. At that distance about 1/3 of the mountain should still be visible. Which is what it looks like in that video.
The problem isn't a straightforward earth curve calculation. There's a 450' obstruction 20 miles down range that increases the amount hidden. It, too, is affected by the curvature so not all 450' serves as obstruction.
Ok, I assume that is what the current debate with all the numbers and figs is all about... That was a good summarization of it, thanks!
I'll think about this added variable for a bit and see if I have anything else to add.
Thanks again to you both for putting up with my late and slightly discombobulated entry into this discussion!
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Google Street View Link from a town near that mountain:
https://www.google.com/maps/@33.7190873,-116.8932793,3a,77.1y,62.93h,83.97t/data=!3m6!1e1!3m4!1s9iY3HjUL5TcE0-s1op50IA!2e0!7i13312!8i6656
(https://i.imgur.com/UtN3cdb.jpg)
Does the viewable mountain below looks like it takes up 7.8 - 13% of the total mountain?
(https://i.imgur.com/CQVXM33.png)
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(https://i.imgur.com/UtN3cdb.jpg)
Does the viewable mountain below looks like it takes up 7.8 - 13% of the total mountain?
(https://i.imgur.com/CQVXM33.png)
Not sure if serious.
Before we go onto that aspect of the analysis, are we in agreement that JTolen Media1 was wrong to claim that San Jacinto should not be visible from Malibu Bluffs?
Will you work out what the flat earth predicts the visible/not visible height values should be?
With those in hand, along with our RE calculated numbers, I'm more than eager to dive into what we're seeing in these images.
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Tom, you are talking about perception. I love mountains and pay very close attention to them, and when you are far away from a mountain it always looks like you are viewing the entire mountain. The hint that you are not is that regardless how large the mountain looks from afar, you don't see any trees in the mountain. That means you are looking at the timberline which is the last few thousand feet of a mountain (depending on the mountain).
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If we shrink and overlay the JTolan mountain over the the ground view from the Street View from that nearby town we will get something similar to what we did earlier:
Going with your 13%, it seems a little odd that the base of the mountain is way down there in the closer overlay photo you posted:
(https://i.imgur.com/UoUql77.jpg)
The base of the mountain will be WAY below where it should be, if we are only seeing ~7.8% to 13% of that mountain.
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Street View image from town near mountain with overlay:
(https://i.imgur.com/LbeteVR.png)
Street View Link: https://www.google.com/maps/@33.7190873,-116.8932793,3a,77.1y,62.93h,83.97t/data=!3m6!1e1!3m4!1s9iY3HjUL5TcE0-s1op50IA!2e0!7i13312!8i6656
The Round Earth Theory predicts that the based of the mountain would be at around the area of the red line, or at around the yellow line which is off screen, depending on whether we are using 117 miles or 120 miles for the distance calculations from the last page.
13% and 7.8% is in reference to the amount of mountain the person in the OP should have been seeing of the mountain, according to 117 miles or 120 miles, according to the Round Earth Theory.
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We're past whether or not the mountain is visible from Malibu, then? And now the issue at hand is whether what can be seen of San Jacinto from Malibu Bluffs is more in line with flat earth or convex earth?
Ok.
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OK - sorry for the late response here with my calculations...
So the problem is that there is a 450 ft obstruction at 20 miles out from the vantage point. If I read correctly, the vantage point starts at 150 ft above sea level, so I calculated for a 300 ft obstruction - If this is incorrect, let me know.
If you account for the earth curvature, at 20 miles (105,600ft), the 300 ft obstacle is 33.25 ft higher than your line of sight (since there is a 266.75 ft drop due to the curve).
Using trigonometry, I calculated using a right triangle with tan(x) = 33.25ft/105,600ft. That should give you the angle that you have to look up from 'line of sight' to see over the 300 ft obstacle. This angle is .018 degrees.
So, looking up at a .018 degree angle to see over the hill and to the mountain we can calculate the height that is lost from view due to the obstacle, which is:
tan(.018) = x/633,600ft
x = 199.05 ft
This means the 300 ft obstacle 20 miles away is blocking 199.05 ft of our view at 120 miles.
So, instead of being able to see 1,199.3 ft of the top of the mountain, you can only see 1,000.
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Street View image from town near mountain with overlay:
(http://oi66.tinypic.com/2ymucnd.jpg)
13% of 10,800' is 1400'
Your scale is a little swollen. The lower mound in the foreground is around 5500' and 7 miles away from that viewspot you found in Hemet. It and the lower peak on the western shoulder of the mountain do not appear in JTolan's images. The peak of San Jacinto is to the left of where you've aligned your index, and from that perspective appears slightly lower than the 10,200' mound to the south with which you've aligned.
My counter claim is that the "13%" 1400' vertical elevation we could see in JTolen's Malibu image is within the white segment I've added. As I've re-sized the image, your index is 23 pixels between tick marks. Mine is 13.
Next to consider is the elevation of that Hemet viewpoint. It's at 1900' above sea level, so the mountain peak does not rise 10,800' above that spot but rather 8,900'.
8900'/1400' = 6.35
So, instead of 13% of 10,800, we can evaluate 1400' as a fraction of the remaining 8,900' of elevation rise from Hemet, which is 1/6.35 or 15.75%
If 1400' at 14 miles away is 13 pixels as my index claims, then (6.35)*(13 pixels) = 83 pixels, which I've annotated. That would be the same level as from where that street level photo was taken.
I say mine (white) is a reasonable "eye level" line to the apparent base of 8,900' feet whereas yours (red) is not.
You may continue to offer a 2nd set of calculations (yellow) based on the Malibu-San Jacinto summit distance of 120 miles if you like, but I am going to ignore those since there should be no dispute that the distance is 117 miles.
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We're past whether or not the mountain is visible from Malibu, then? And now the issue at hand is whether what can be seen of San Jacinto from Malibu Bluffs is more in line with flat earth or convex earth?
Ok.
I concur with that statement.
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Bobby, can you overlay the image while keeping scale/aspect ratio as you resize the mountain?
Here is what I get when I resize the mountain and keep the same aspect ratio when I resize:
Your Version
(https://i.imgur.com/2YWouxV.gif)
My Version
(https://i.imgur.com/9NMgqAK.gif)
Which one fits better?
I believe that the shape doesn't exactly line up on mine because the Street View image is taken from a different angle around the mountain.
The obvious width issues with the first image shows that all of that mountain cannot be contained in that small of a space.
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Another view from a better angle around the mountain. Aspect-Ratio is preserved.
(https://i.imgur.com/hbSpEjy.gif)
Google Maps Link: https://www.google.com/maps/@33.9174463,-117.1477034,3a,16.9y,103.01h,89.67t/data=!3m5!1e1!3m3!1svNjcff7JyVpSIeHgC39AvQ!2e0!6s%2F%2Fgeo3.ggpht.com%2Fcbk%3Fpanoid%3DvNjcff7JyVpSIeHgC39AvQ%26output%3Dthumbnail%26cb_client%3Dmaps_sv.tactile.gps%26thumb%3D2%26w%3D203%26h%3D100%26yaw%3D75.10232%26pitch%3D0%26thumbfov%3D100
Context: The Round Earth Theory predicts that the base of the mountain would be at around the area of the red line, or the yellow line which is off screen, depending on whether we are using 117 miles or 120 miles for the distance calculations (discussed on last page).
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Another view from a better angle around the mountain. Aspect-Ratio is preserved.
That works. The angle is steeper than JTolan's, but that's a pretty fair overlay, IMO.
I lost track of what the point of that was, though. That the 1400' round earth calculation (13% of the full 10,800' of mountain elevation) can't be right because that's obviously more than 1455'?
I agree. I had previously agreed (https://forum.tfes.org/index.php?topic=10320.msg162546#msg162546) that, by analysis of the IR image, my estimate was that ~2600' of the mountain was visible from JTolan's Malibu vantage point.
Is the point you're making that 2600' is more reasonable than the no-refraction curvature calculation of 1455'? Or is even that not enough? I've kind of lost the bubble on what issue you're still taking.
What is the flat earth calculation of how much of San Jacinto should be visible from Malibu Bluffs given that 450' ridge at 20 miles? Is it 2600 feet?
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How does this look? I went with 2300' of San Jacinto seen in JTolan's image, and corresponded that to what the segment you were calling 13%.
(http://oi64.tinypic.com/2hh2jw5.jpg)
At 2300' rather than 1455', that's an even 25% of the 9200' of San Jacinto above eye level from that 1600' elevation Morena Valley viewing location. And that pretty convincingly places the "base" of that 9200' at the rise 4 miles across the flats.
Do you want to try to refine that further? Or is 2300' of San Jacinto elevation seen in JTolan's image a fair number?
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Another view from a better angle around the mountain. Aspect-Ratio is preserved.
That works. The angle is steeper than JTolan's, but that's a pretty fair overlay, IMO.
I lost track of what the point of that was, though. That the 1400' round earth calculation (13% of the full 10,800' of mountain elevation) can't be right because that's obviously more than 1455'?
I agree. I had previously agreed (https://forum.tfes.org/index.php?topic=10320.msg162546#msg162546) that, by analysis of the IR image, my estimate was that ~2600' of the mountain was visible from JTolan's Malibu vantage point.
Is the point you're making that 2600' is more reasonable than the no-refraction curvature calculation of 1455'? Or is even that not enough? I've kind of lost the bubble on what issue you're still taking.
What is the flat earth calculation of how much of San Jacinto should be visible from Malibu Bluffs given that 450' ridge at 20 miles? Is it 2600 feet?
By my calculations, for the FE model, the 450 ft ridge (minus 150 ft for the vantage point), blocks 1174 ft of your view, and since we're now talking about a flat plane I don't have to account for a curve, so there should be 9025.92 ft visible on FE.
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By my calculations, for the FE model, the 450 ft ridge (minus 150 ft for the vantage point), blocks 1174 ft of your view, and since we're now talking about a flat plane I don't have to account for a curve, so there should be 9025.92 ft visible on FE.
You're working out the numbers for a no-curve flat earth, right? That's what you figure should be viewable from Malibu Bluffs?
I had previously calculated 8925' so we're close. Are you figuring on 117 miles or 120 miles to the summit?
I have asked Tom to do the flat earth numbers. I believe that if the earth is flat, the Chino Hills at ~1600' 55-60 miles along the line of sight should come into play, and I do think that factored into my 8925' calculated result.
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Do you want to try to refine that further? Or is 2300' of San Jacinto elevation seen in JTolan's image a fair number?
I'm going to refine, and move a little further away still from the earth curvature calculator, bumping my estimate up another 200' from 2300' to 2500' of visible San Jacinto in that JTolen IR image.
(http://oi63.tinypic.com/ix47jk.jpg)
(http://oi66.tinypic.com/n5q35d.jpg)
After adjusting contrast and brightness, I'm fairly convinced we can faintly see San Jacinto's 8500' Fuller Ridge just above the much nearer 450' Baldwin Hills ridge line.
That ridge has a dip to 8300' but mostly runs toward the image's viewpoint at an elevation of 8500'. I figure, based on the angular dimension, at that distance, an extra 200' of San Jacinto is visible.
So, final answer: we can see 2500' of San Jacinto from a 150' high point in Malibu.
Without figuring for refraction, the globe predicts only 1455' should be visible. How much can refraction contribute? 500'? If so, that still leaves 500' unexplained.
But that's just a globe earth discrepancy. What about the flat earth numbers? Do they hit closer to the estimated, observed 2500'? My (and timteroo's) calculations suggest not. For flat earth, over 6000' of San Jacinto elevation is missing. Isn't it?
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By my calculations, for the FE model, the 450 ft ridge (minus 150 ft for the vantage point), blocks 1174 ft of your view, and since we're now talking about a flat plane I don't have to account for a curve, so there should be 9025.92 ft visible on FE.
You're working out the numbers for a no-curve flat earth, right? That's what you figure should be viewable from Malibu Bluffs?
I had previously calculated 8925' so we're close. Are you figuring on 117 miles or 120 miles to the summit?
I have asked Tom to do the flat earth numbers. I believe that if the earth is flat, the Chino Hills at ~1600' 55-60 miles along the line of sight should come into play, and I do think that factored into my 8925' calculated result.
I actually calculated for 120 miles and assumed the line of sight begins at 150' above sea level and that the obstacle is 450' above sea level.
Edit: now that I think of it, I do need to subtract 150' from my mountain calculation, since that is the vantage point. Been too long since I've done math!
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So, final answer: we can see 2500' of San Jacinto from a 150' high point in Malibu.
Without figuring for refraction, the globe predicts only 1455' should be visible. How much can refraction contribute? 500'? If so, that still leaves 500' unexplained.
But that's just a globe earth discrepancy. What about the flat earth numbers? Do they hit closer to the estimated, observed 2500'? My (and timteroo's) calculations suggest not. For flat earth, over 6000' of San Jacinto elevation is missing. Isn't it?
(http://oi64.tinypic.com/2d7u2hx.jpg)
Best I can come up with for a flat earth is that the highest hill obstruction seen in the image is the 1600' hills 60 miles inland, resulting in hiding 3277' of San Jacinto's elevation. Should leave 7573' visible. (9075' visible if obstruction is the 450' hills at 20 miles.)
Globe earth calculation (w/o refraction): 1455' visible
Globe earth guesstimation (w/refraction): maybe +500' bringing it to 2000' visible?
Flat earth calculation (if 1600' hills at 60 miles are blocking): 7573' visible
Flat earth calculation (if 450' hills at 20 miles are blocking): 9075' visible
Photo analysis: 2500' visible
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Bobby, it looks like we pretty much agree.
Tom, sorry I do not have diagrams, and my math is sluggish, but if you like I can create some graphics, and I can recalculate for 117 mile vantage point and adjust my mountain value to accommodate the 150' vantage point that I previously forgot to account for (but just subtract 150 from my visibility calculation and that should do it).
My math was based on tan(x) = opposite/adjacent with opposite being the height of the obstruction and the height of the mountain and 'adjacent' being the 'line-of-sight' distance to the obstruction and mountain, respectively.
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Tom, it looks like we have been able to 'prove' RE by trying to 'prove' FE, arriving at a contradiction - visible mountain does not add up with FE.
Goot debate. How do the judges rule?
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Another view from a better angle around the mountain. Aspect-Ratio is preserved.
That works. The angle is steeper than JTolan's, but that's a pretty fair overlay, IMO.
I lost track of what the point of that was, though. That the 1400' round earth calculation (13% of the full 10,800' of mountain elevation) can't be right because that's obviously more than 1455'?
I agree. I had previously agreed (https://forum.tfes.org/index.php?topic=10320.msg162546#msg162546) that, by analysis of the IR image, my estimate was that ~2600' of the mountain was visible from JTolan's Malibu vantage point.
Is the point you're making that 2600' is more reasonable than the no-refraction curvature calculation of 1455'? Or is even that not enough? I've kind of lost the bubble on what issue you're still taking.
What is the flat earth calculation of how much of San Jacinto should be visible from Malibu Bluffs given that 450' ridge at 20 miles? Is it 2600 feet?
I don't agree that the 2600 feet figure has anything to do with RET.
Lets go over how it was created again. From that link you posted (https://forum.tfes.org/index.php?topic=10320.msg162546#msg162546), which goes back to your earlier post about it:
"How much of San Jacinto is, in fact, visible? By my index, 4.3 milliradians, or 0.25°. At 117 miles, that works out to be 2652' of San Jacinto is visible in that image."
You are using the milliradian method from the beginning of the thread, not a method that uses the RET geometry. Not a method that uses a round earth. You created the index based on the (questionable) building sizes. That is not a Round Earth method. As I see this, you are assuming a Round Earth for the usage of the 2600 foot figure.
The math that dealt with the RET geometry got the values ranging between 13% to 7.8% for the portion of the mountain seen.
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I don't agree that the 2600 feet figure has anything to do with RET.
It's doesn't. It's RET/FET agnostic. It's just mathematics/trig.
Lets go over how it was created again. From that link you posted (https://forum.tfes.org/index.php?topic=10320.msg162546#msg162546), which goes back to your earlier post about it:
"How much of San Jacinto is, in fact, visible? By my index, 4.3 milliradians, or 0.25°. At 117 miles, that works out to be 2652' of San Jacinto is visible in that image."
You are using the milliradian method from the beginning of the thread, not a method that uses the RET geometry. Not a method that uses a round earth. You created the index based on the (questionable) building sizes. That is not a Round Earth method. As I see this, you are assuming a Round Earth for the usage of the 2600 foot figure.
No, Tom. Come on! Even the video author is using that method. It isn't "RET geometry" or "FET geometry." It's using angular ratio to assess dimensions in the background based on objects in the foreground. You know, like "dime hiding an elephant" stuff. You understand that right? It works whether the earth is flat or spherical.
And really, I am quite offended that you're only NOW calling it "questionable." I posted that, invited you to critique it, you never replied so I moved on and applied that scale in my subsequent post. Now you call it questionable, and you offer nothing to back that up? Is JTolen's also "questionable?" If you have a critique, by all means, state your case. But it seems to me you've taken awhile to utter it, and only after you don't like the results.
And milliradian isn't a method. It's just an unit of angular measure. We can talk in degrees or arcseconds. I don't care. I only stuck with mrads because that's what the videographer used. I'm not partial to that unit. But that "method" is valid, and it's not biased toward spherical or planar geometry. (In fact, I'm going to use it again shortly in a post about that other video you found about the IR photography of Clark Island in Washington to assess that guy's flat earth claim.)
The math that dealt with the RET geometry got the values ranging between 13% to 7.8% for the portion of the mountain seen.
That's right. And it didn't jive with the RET/FET agnostic calculation.
But neither does the geometry for FET. FET doesn't "win" by default just because the RET calculation doesn't jive. The RET calculation of 1455' (13%) didn't take into account atmospheric effects. If we do that, we'll get closer to the agnostic 2500-2600' feet but not all the way. But then, I shouldn't have to try to justify getting all the way to 2600' using RET explanations/methods/calculations while you make zero effort to explain the much larger gap between that value and what is predicted if the earth is flat. Please, do that for me, because until then, those IR images of San Jacinto are much more indicative of a convex earth than a flat one.
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Bobby I saw your comment at youtube and came here.
Here's three images to help you. (following this post, in separate posts) Take a look at these images, read carefully what I've annotated and the rationale. Here's some main points I want to impress upon you two:
1) my scale is quite accurate considering the atmospheric distortion, see person on balcony analysis.
2), angular scales based on objects closer to observer are less prone to refraction effects which accumulate over long distances. So I chose the one based on the Le Meridien Delphina in the vertical direction. (one can do a horizontal scale as well, see later comment)
3) the steeper the observation angle the less prone to refraction. peaks of mountains are the best target for observations, as I said in my video.
4) due to the nonlinear behavior of refraction, which changes with elevation angle, the image of distant mountains get's stretched and also it appears as if the "curvature" is hiding objects, but it's not the curvature its the non linear bending, because the peaks come out at the correct height !
5) measuring horizontal distances between buildings produce yet a third scale, which is more accurate, but I choose the vertical one based on the Le Meridien Delphina hotel because the propagation over water (approx 12 miles) introduces yet more compression to the image, and the horizontal scale will lead you astray!
6) I have yet a forth scale based on my image sensor size, pixel count, focal length, etc.. but that's reserved for my book :)
7) don't use the curvature calculator to figure out how much the hills in the foreground are hiding, Tom seems to be making that mistake as well as you Bobby. It's simple trigonometry. Now obviously the analysis is based on straight line propagation which we know apriory is not the case. But for the foreground scenery it is quite accurate. I apologize for my crude image in the video, I said the hills across the "bay" hide the mountain but the graphic was misleading. The tangential point is to the msl level at a distance of about 15 miles, just 3 more miles inland then the shoreline on the other side, so we can't say the water curvature hides any of the land mass in the foreground. Above this tangential line the mountain would still be visible if there were no obstructions above this line (red dotted line tangent to MSL), but obviously there's hills in the foreground! Now use simple trigonometry and see how much 0.01 rad hide at a distance of 117 miles. (since the angle is so small we don't need trigonometry, just multiply the angle in radians times the distance to get the arc length which is pretty damn close to the height calculated based on trig.)
Bottom line folks, the mountain should not be visible, but not only do we see it, it's visible at the correct angular elevation!
What are the chances refraction just happens to do that? none really, its peoples erroneous understanding of refraction that confuses them. The refraction downward assumed by almost everybody who defends the globe is ASTRONOMICAL refraction, astronomers observe stars and planets SPEED up as they get closer to horizon and can calculate this accurately. At zero elevation it's about 0.5 degrees, the diameter of the sun. But ATMOSPHERIC refraction is a whole new ball game. There's temperature gradients, various water vapor distributions, etc., not the same thing as astronomical refraction.
So atmospheric distortions get in the way of the flat earth phenomena, but when the atmosphere effects are understood and removed, we are left with the naked truth, that there is a damn strong flat earth phenomena! It is real, observable, and measurable! I encourage everybody to subscribe to my youtube channel, "JTolan Media1" I have some dynamite information coming, you don't want to miss it.
Anyway, hope this helps you in your quest for TRUTH!
v/r
-JT
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this is a street view picture of the Le Meridien Delphi Hotel from the north east, and a person is standing on the balcony for height reference. I gave this person about 5' 6", its an approximation obviously. seems the floors are spaced at about 12' like I said in my video. People comment how high the ceilings are at this hotel.
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I redid the graphic I had in my video, to clearly show the foreground hills blocking the view of the mountain, in case that wasn't clear in my video. I did say the hills across the water would be blocking it.
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analysis showing why scales don't correspond.
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Bottom line folks, the mountain should not be visible, but not only do we see it, it's visible at the correct angular elevation!
Welcome, JT. Appreciate you following the link and joining the discussion.
I'm a systems engineer with practical experience in military applications of airborne early warning radar, communications/data links, and electronic warfare. I currently work on a satcom program.
It's late for me so I'm just going to pick out the item you bolded for now and ask how you calculated that San Jacinto's summit of 10,800' 117 miles away should not be visible from an elevation of 150', given a 450' obstruction 19.9 miles away and given an earth curvature of radius 3959 miles. Straight geometry, without accounting for a refractive atmosphere, I calculated 1455' of summit visibility. I'd like to know how you came up with your result.
Next morning edit: having read your posts more careful, don't feel you need to answer this since I think I understand your technique. I consider it flawed, but I'll wait to see if you affirm my re-stating of it.
Secondly, how much of San Jacinto do you believe you are seeing in your image? I estimated around 2500' of elevation from the summit, based on the faint detection of Fuller Ridge. Using geometry of a flat earth, that 450' Baldwin Hills ridge should leave about 9000' visible. (Chino Hills another 40 miles in the distance should also contribute to the obstructing height but I don't see them in the image.)
Next morning edit: Again, you can ignore this since you address it in one of your follow-on graphics, showing about a 2600-2800' elevation rise is visible.
Your premise is that San Jacinto should not be visible at all if the earth is convex. I disagree, but even if so, why does seeing 2500' vice 9000' favor a flat earth rather than a convex earth? Being able to see more with an atmosphere than would be predicted in an atmosphere-less calculation doesn't render the conclusion that the earth is flat. The earth could be less convex than the geometric calculation using a greater radius input, but convex nonetheless. The method of accounting for standard atmospheric refraction is done by that very means, "flattening" the level of convexity by an indexed amount such that the earth looks a little less spherical than it actually is. But it's still far from what a flat earth would predict.
Similar to Tom here (and on other discussion threads here), I see you citing atmospheric refraction when it favors a flat earth observation but ignoring it when assessing curved earth RF/light propagation. As a fellow radar engineer, I find that to be curious; biasing, actually.
(I also note that you describe atmospheric refraction as bending light upwards. That would be anomalous propagation: sub-refraction. Standard atmospheric refraction bends light downwards, toward increasing density. What you describe requires an inversion, which is possible but on what basis do you assume it? EDIT: seeing your last post above, I believe I understand you to be deducing "towering is exhibited by San Jacinto.)
Next morning edit: Having read your full post I now see you have rather radically different understanding on standard refraction and anomalous propagation of light, which I find curious given both of our claimed backgrounds. If you're willing to stick around, this may be the crux of a continuing discussion.
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analysis showing why scales don't correspond.
I'll try to describe what you're saying with this to see if I have it correct.
You deduced that the difference in elevation between San Jacinto summit and Saddle Junction was about 2600'. At 117 miles, a rise of 2600' produces angle of about 0.004 radians (0.23°). Saddle Junction is in-line with the ridge line in the foreground, which would make the vertical angular displacement between that ridge and San Jacinto summit 0.004 radians.
So far, I'm in close agreement. I estimated 2500' of visible elevation of San Jacinto and the scale I annotated on your image shows that vertical displacement to be about 4 miillradians too.
But where we depart is you are saying that light at a shallower angle through the atmosphere from the lower elevations of the mountains is bending upwards at an greater rate than the steeper angle of the light coming from the summit. As a result, the expanse of that 2500-2600' distant elevation is being optical stretched by propagation through an anomalous atmosphere, and what would otherwise only appear as a 4 milliradian vertical displacement is "towering" to an 8 milliradian displacement. (You wrote 10, but your scale suggests something slightly less than 8 mrads.)
You feel that your angular gauge, based on vertical estimates of the Le Meridien Defina, is correct and thus this vertical stretching of the distant landscape is not contradicted by your scale. Did I capture that correctly?
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I redid the graphic I had in my video, to clearly show the foreground hills blocking the view of the mountain, in case that wasn't clear in my video. I did say the hills across the water would be blocking it.
And this updated graphic: how did you derive the 10 milliradian angle that the foreground hills would produce above the line tangent to a curved earth surface. It looks like that is based off of your angular gauge as well, measured from where you placed its origin: at the shoreline of Santa Monica beach. Is that correct? If so, then it hinges on the accuracy of the gauge based on the Le Meridien Delfina as well. Yes?
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Bobby,
great to see another RF engineer searching for truth, or are you here just to debate? (which would be ok as a pastime I guess)
You may relate to this: I analyzed data from radar warning receiver payloads a while back and that's when we noticed detections out to twice the distance to horizon calculated from our altitude, then one day a mind blowing 5x to 6x the distance. Highly Anomalous all right, almost out to 1500 miles if my memory serves me right. Simply astounding! I feel like RWRs are the best tools for flat earth research, because the signals only incur a 2 way path loss, so high power radar sets out in the distance can be picked up at incredible distances.
After that a fellow engineer sent me a paper from Sandia Labs written by Armin W Doerry that is excellent at highlighting our confusion, the first sentence in the abstract caught my attention. see image and link below: I thought, somebody knows something here, just like I do!
That's when I realized what were doing wrong as scientists, physicists, engineers, and mathematicians, nobody in their right mind would say the earth is flat, we settled that long ago, so we assume it must be curved electromagnetic wave propagation, so we create elaborate mathematical models and curve fit to the observed data.
But wait a second, what if that is a wrong assumption? Just because we can curve fit to some concocted mathematical model, it doesn't make our model's assumptions necessarily correct. I can assume a different curvature and a different model and curve fit data just fine.
anyway, getting back to the infrared images, it appears after a few reviews you're now understanding my analysis images and what I'm saying. Keep pondering and thinking and it will become even more obvious. My understanding of atmospheric refraction might seem "curious" to you at first, but it's right there in the image, starring us in the face.
v/r
-JT
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You may relate to this:
I relate to it, but in quite a different way from you, apparently.
20-25 years ago, we were not only NOT astounded by greatly extended detection ranges of UHF long range airborne surveillance radar, but knew why it occurred and devised ways to exploit it to our advantage. Same with our passive detection systems (akin to RWR), we could use atmospheric conditions to our advantage that negated the limits that would otherwise be imposed by earth curvature.
I actually find it perplexing that you would lean away from orthodoxy and toward a flat earth model on account of this kind of extended range phenomenon. It's not a mystery. What's still a challenge (apparently, by virtue of the paper you attached having a date of 2013) is finding better and improved ways to model and predict the conditions of a dynamic atmosphere for the purposes of designing and employing such systems to their optimum. But you're the first person I've ever encountered who has experience with such systems who would give credence to the notion that the earth is not of the radius we think we know and that such anomalies are of such an astounding nature as to consider that the earth surface may, in fact, be flat and not convex.
But anyway, back to your IR imagery, though I apprehend your analysis, I'm still confused by it. It seems your starting point for geometric calculations is the gauging of the le Meridien Delfina. It'd odd to me that you would base the rest of your angular calculations based solely on that one datum. The 10 milliradian angle you predict should be formed by the foreground hills is based on the accuracy of that, but it doesn't match with observation. It doesn't match with the topographical data on the elevation of those hills. Your gauge doesn't match with the known height of the nearby Pacific Plaza. It doesn't match with where eye level should be assuming a flat earth. And it certainly doesn't match with the visible amount of San Jacinto, which you've reasoned is just an apparent discrepancy due to assumed sub-refractive atmospheric conditions.
I will freely admit that I have a very strong bias toward a spherical earth. It's not a philosophical or a brain-washed predisposition. It's a consequence of a career in fields in which a flat earth would make zero sense. Right now, I work for a communications satellite program that wouldn't work if the earth were not of the dimensions we believe it to be. I'm here (and probably not for much longer) because I wanted to understand how a flat earth could possible be rationalized. I expected to find more of the YouTube personalities who broadcast the case for a flat earth but don't seem to invite critique. Sadly, that hasn't proved to be the case, with Flat Earth advocates seeming to be sectarian, with different camps or groups or individuals suspicious of each other. This forum is more populated with flat earth critics than flat earth advocates. It's actually quite a treat to have you stop by. Some of your fellow Flat Earth video producers praising you in your comments section will never stop by and seem to block any critical response to their postings. (I won't name names.)
So my hat's off to you for facing up to the critique. And I do believe I am following what you are saying in defense of your analysis. I just think you are analyzing it through a lens that is expecting to find a flat earth (maybe?) In your video, when you grandly exclaim "behold, your flat earth" as the San Jacinto mountain range comes into IR view through the typical southern california surface haze, I was thinking "that's not a flat earth! He's showing us what a curved earth surface looks like." It's frustrating to me to continually see flat earth advocates use a no-atmosphere calculation of earth curvature, identify that it does not match an observation and conclude, therefore, that the globe is busted and the earth flat. That is not logical.
I hope this isn't coming off as a rant. I'm swerving around the point since we're kind of introducing ourselves to each other. I'll try to stay focused here on the topic of the analysis of your video.
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I redid the graphic I had in my video, to clearly show the foreground hills blocking the view of the mountain, in case that wasn't clear in my video. I did say the hills across the water would be blocking it.
And this updated graphic: how did you derive the 10 milliradian angle that the foreground hills would produce above the line tangent to a curved earth surface. It looks like that is based off of your angular gauge as well, measured from where you placed its origin: at the shoreline of Santa Monica beach. Is that correct? If so, then it hinges on the accuracy of the gauge based on the Le Meridien Delfina as well. Yes?
I think the answer is "yes." The reason it seems why you claim none of Mount San Jacinto should be visible from Malibu Bluffs on a spherical earth is because you've concluded that those foreground hills form an obstructing angle of 10 mrad (0.57°); a value which it appears you derived from gauging angular dimensions in the imagery of hotel Le Meridien Delfina in San Monica.
Is that a sensible basis on which to form such a presupposition?
If we were to try to assess whether San Jacinto was visible from Malibu Bluffs before you ever set out to record it, how would we do? I would say we'd using topological data and geometry to predict whether or not the summit of San Jacinto should be visible over, or hidden by, that ridge in the foreground, ignoring atmospheric predictions in the interim. I question whether your measured/calculated for theta Θ in your diagram is accurate:
(http://oi67.tinypic.com/d8u4p.jpg)
First, some flat and spherical earth geometry diagrams, just to make sure we're on the same sheet of music. Here is a diagram of flat earth lines and angles:
(http://oi66.tinypic.com/2e4yj5k.jpg)
P = Position of observer
O = Obstructing object
T = Target
Straight line G is the flat earth ground surface (sea level)
Straight red line EL is the level eye sight from P, parallel to G.
Straight red line SL is the sight line, angled from P to the top of O.
Θ is the measure of that angle.
Compare this with the geometry of a spherical earth:
(http://oi63.tinypic.com/1554cj6.jpg)
P, O and T are all defined the same but they no longer parallel. They are tilted away from each other.
Earth ground surface G is no longer a straight line but is curved. It still represents sea level.
Red Line EL is still straight and perpendicular to a vertical line at P, but it is not parallel to G as it was for flat earth.
There is a new essential line and point that are not applicable for flat earth.
Straight blue line TL drawn from P that is tangent to G at point H (horizon).
This forms an angle that's not part of the flat earth diagram.
Alpha (α) is the "dip" angle between tangent line TL and eye level line EL. On a flat earth, there is no "dip" angle and what is perceived as a horizon is believed to be coincident with eye level.
Straight yellow line SL is defined as it was for flat earth, but the angle theta (Θ) is not formed by it and Eye level line EL but rather with tangent line TL.
All clear?
Assuming there are no other limiting obstacles and ignoring for now atmospheric effects, we can plug in known values for elevations and distances to calculate the angles and the amount of T that is obscured by O in both flat and spherical earth configurations.
Flat Earth Geometry
P elevation = 150 ft
O elevation = 450 ft
Elevation delta between P and O = 300 ft
P-O distance = 105,072 ft (19.9 miles)
Flat Earth Θ = arctan(300/105072) = 0.164° (2.86 milliradians)
P-T distance = 617,760 ft (117 miles)
T elevation = 10,800 ft
Flat Earth T elevation obscured = 617,760*sin(0.164°) = 1768 ft
Flat Earth T elevation visible = 10,800-1768= 9032 ft
(http://oi64.tinypic.com/2mgpfeq.jpg)
Spherical Earth Geometry
Earth radius ≈ 20,903,500 ft (3959 miles)
P elevation = 150 ft
H distance = 79,190 ft (15 miles)
Horizon "Dip" angle α = 0.217° (3.8 milliradians)
P-O Distance = 105,072 ft (19.9 miles)
0 elevation = 450 ft
Spherical Earth O elevation obscured = 16 ft
Spherical Earth O elevation visible = 434 ft
Spherical Earth Θ = arctan(434/105072) = 0.237° (4.1 milliradians)
P-T distance = 617,760 ft (117 miles)
T elevation = 10,800 ft
Spherical Earth T elevation hidden by curve (below TL) = 6937 ft
Spherical Earth T elevation obscured by O (above TL) = 617760*sin(0.237°)= 2555 ft
Spherical Earth T total elevation obscured = 9492 ft
Spherical Earth T elevation visible = 10,800-9492= 1308 ft
(http://oi67.tinypic.com/1zx16it.jpg)
In order to present a 10 milliradian obstructing angle above the spherical earth tangent line, the hill elevation at 19.9 miles would have to be 1051' feet above that tangent line (1067' - 16' hidden). But they are not:
(http://oi63.tinypic.com/2nb7k1k.jpg)
Your angular gauge needs to be reconciled with the vertical elevations of other features (Baldwin Hills, Pacific Plaza) before it can be used to declare what should or shouldn't be visible over a spherical earth topography.
When I did this previously, I came up with 1450' of San Jacinto that should be visible even without atmospheric refraction. Doing the math now that number is reduced to 1308'. I don't know why the difference. If I find my earlier notes, I'll try to find the source. I may have simply used different values for some of the parameters or rounded differently. Or maybe I've made an error here. Whatever the reason, the Baldwin Hills do not rise high enough to obstruct the view of San Jacinto from Malibu if the earth is curved.
In fact, if the earth is flat, much more of San Jacinto should be visible than the 2500-2600' that appears in the captured images. The curved earth geometric calculation is closer to what is observed than the flat earth calculation.
Of course, atmosphere (atmoplane) plays a significant role in that. The question is, does it make more of the mountain range appear or less? If light is being refracted downward, then it should be more, as is the case if the earth is spherical. If light is being refracted upward as you say, then it should be less.
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JT,
If you return to read this topic, here's another question about your video.
At the 2:15 mark, you present a slide from some excursion you took to the Salton Sea where you presented a case for seeing a 130' cell phone tower from a distance of 17 miles across the sea with a camera height of 6'. Is there more to this? I couldn't find anything in your other videos, but this is a pretty extraordinary claim. I'd like to examine it more. Can you provide any additional detail? Tower location? Sighting location?
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I missed this whole thread, which is a shame because I do love these mountain photo analyses. I have done this photo, however, as well as some of JTolan's more recent ones, and my findings were:
Predicted visible amount on sphere earth: ~2350 feet
Predicted visible amount on flat earth: ~8000 feet (generous) or ~9000 feet (more accurate to the FE model)
Actual visible amount: ~2300 feet
It's not too difficult to verify these figures: the main thing was creating a calculator that takes the obscuring ridge into account.
Now I'll check the thread and see how my calculations match up with what others have done. :)
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It was determined that on a Round Earth the amount of mountain that should have been visible to the observer was somewhere between 7.8% to 13% of the mountain, depending on how the math was done.
Not sure how you calculated those figures, Tom. My calculator returns 21.9% of the mountain predicted as visible on a sphere earth. Meanwhile, on the flat earth, it would 74.6-83.8%.
Actual amount visible I make to be around 21.2%.
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snip...
That's around 1200-1350' more than predicted by the no-refraction globe earth calculation.
But it's 6000-6250 less than was a flat earth predicts.
snip...
For what its worth, WP gives a complex formula for terrestrial atmospheric refraction, but also gives a simplified ballpark of one degree for every 932 miles.
For 117 miles away, the mountain would appear 1446 feet higher than it really is.
That brings your calculations to within a couple hundred feet, or about 2% of total height - not too shabby.
(My apologies if this is poorly timed or etc., just catching up..)
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For those interested, here's a download link to the calculator I made in order to calculate predicted hidden amounts for distant landmarks that are obscured by nearer landmarks:
https://www.metabunk.org/attachments/obstruction-calculator-xls.36004/
And this is what it outputs when the figures for JTolan's Malibu to San Jacinto shot are inputted:
(https://i.imgur.com/j7azdmB.png)
The sphere earth predictions are very close to that which was actually observed. The flat earth predictions are more than 300% wrong.