Catholic Info
Traditional Catholic Faith => Fighting Errors in the Modern World => The Earth God Made - Flat Earth, Geocentrism => Topic started by: Matthew on February 05, 2026, 10:35:45 AM
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https://www.youtube.com/watch?v=dMm5jABcqn0
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https://www.youtube.com/watch?v=dMm5jABcqn0
Those are clouds. But believe what you want to believe.
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Those are clouds. But believe what you want to believe.
Wow. Talk about what you WANT to believe. Those mountains at Canagal are a known phenomenon, and it only happens on a few days of the year but it's consistent, and evidently just to mess with us God will create an outline with clouds (just during those times of the year) that happen to match the exact contour known contour of the mountains.
This here is all the proof you need that some people just refuse to shake the programming from their cold dead brains, absolutely refusing to listen to reason and examine the evidence.
You'd be much better off attempting to play the refraction card ...
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Have you ever watched a plane that was flying overhead at cruising altitude (about 36000 ft)? Did you watch it until it disappeared? Did it get close to the horizon or did it disappear before that? If the sun is about 3000 miles over the earth? Do you think you would still see it? How far would it have to go to close the gap between the sun and the earth? These are serious questions to think about.
I did a little geometry.
If the sun is 3000 miles above me and travels in a line parallel to the earth then in 3000 miles I could see as a 45 degree angle over the earth. That means that there would be full sun in California and you would still see the sun on the East Coast. Maybe my math is wrong.
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Another thought. How far are those mountains from where the person is standing? If it is less than 3000 miles there is no way the sun would be lighting them up from behind, since the sun travels parallel to a flat earth.
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Have you ever watched a plane that was flying overhead at cruising altitude (about 36000 ft)? Did you watch it until it disappeared? Did it get close to the horizon or did it disappear before that? If the sun is about 3000 miles over the earth? Do you think you would still see it? How far would it have to go to close the gap between the sun and the earth? These are serious questions to think about.
I did a little geometry.
If the sun is 3000 miles above me and travels in a line parallel to the earth then in 3000 miles I could see as a 45 degree angle over the earth. That means that there would be full sun in California and you would still see the sun on the East Coast. Maybe my math is wrong.
I am correcting this a little. During the 12 hour days the sun is rising in China as it sets on the East Coast. We will choose a point 9638 miles away. For it to be rising in one place and setting in the other the sun would be located at the half way mark between those two point. Which is about 4800 miles.
Right Scalene Triangle
Side a = 3,000
Side b = 4,800
Side c = 5,660.38868
Angle ∠A = 32.005° = 32°0'19" = 0.5586 rad
Angle ∠B = 57.995° = 57°59'41" = 1.0122 rad
Angle ∠C = 90° = 1.5708 rad = π/2
This means that the Sun is 32 degrees in the air still and would still be visible in both places and not hit the horizon.
For the sun to look like it is hitting the horizon then the sun would have to be about 250 miles over the earth. Maybe that is possible?
Right Scalene Triangle
Side a = 250
Side b = 4,800
Side c = 4,806.50601
Angle ∠A = 2.981° = 2°58'53" = 0.052036 rad
Angle ∠B = 87.019° = 87°1'7" = 1.51876 rad
Angle ∠C = 90° = 1.5708 rad = π/2
Just food for thought.
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So ... you've made it abundantly clear that you simply don't want to accept that the earth might be flat.
1) that's just clouds (just blurted out without any evidence or analysis)
-- yes, the exact same cloud formation that just happens to re-appear during about the same time window every year and each year just magically happens to follow the exact contour of the mountain range that just so happens to be in that same place. You'd be more ready to believe that than that the earth might be flat?
2) I did some geometry ... and scrawling out some numbers that are not relevant to the specific example in the video "have you ever seen a plane?" ... reminding me of another poster's "have you ever seen a sunset?" nonsense
-- yes, the geometry of earth curvature is well known, and there are calculators online (not created by FEs) that will calculate the "hidden height" of an object based on the elevation (above sea level) of the observer, the height (above sea level) of the target object, and the distance between the observer and the target object. In the video, the numbers were run, and the entire mountain range should be hidden by at least 3,000 feet above the heighest peak, and it's clear that all but about 1,000 feet are visible (for which there can be many explanations, such as the topography between you and the target, ocean waves, atmospheric conditions, etc.). What's interesting about this phenomenon is that the angle of the sun cuts through those atmospheric conditions that would normally block your ability to see the mountains from that far away.
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There's one example in particular which stands out. There's a world-record long distance photograph (not sure if it still holds the record) of a light-house that stands about 250 feet above sea level (at the very top), and it was photographed from well over 200 miles away, when it should have been hidden by many miles of curvature on a globe. Now, the photo was NOT taken by an FE, and the organization that certified it for the record were not FE either ... but neither the photographer nor the organization were even considering the implications of the photo vis-a-vis the FE question. So this is independent evidence from objective third parties, which makes it special and impervious to debunking that it was faked by some FE. Now given that it was from over 200 miles away and that the picture was extremely clear and detailed, this makes refraction absolutely impossible. From 200 miles away, if the index of refraction due to changing (increasing) atmospheric density between miles 200 and mile 199 wasn't exactly the same as the index between miles 199 and 198, and then 198 and 197, etc. ... for the entire distance, if you had even slight variations in refraction rates at any point long the entire way, you'd have different images refracting into one another, causing the distortion you nearly always see in every true example of refraction, even from much shorter differences. It's simply impossible to get a clear image via refraction from those distances.
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-- yes, the exact same cloud formation that just happens to re-appear during about the same time window every year and each year just magically happens to follow the exact contour of the mountain range that just so happens to be in that same place. You'd be more ready to believe that than that the earth might be flat?
Yes, it's just clouds. :jester:
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Every time I see one of these videos claiming that the curvature of the Earth would hide a distant object, I go back and do the simple math that shows that the object would not be covered by the curvature of the Earth. In the case of the video starting this thread, the curvature of the earth would only be 5100 ft. (cord height) in 175 miles, so it would not hide 9,000 foot mountains. In the case Lad mentioned, the curvature of the earth would be 6,661 feet across 200 miles. I'd like to see more on that case.
Here are the results for curvature of the Earth ("Height" in the calculator) for 175 miles and 200 miles, respectively. I did do the math by hand to verify. Said math is high school sophmore geometry class stuff, BTW.
(https://i.imgur.com/fKNONkf.png)
(https://i.imgur.com/tlGEuMi.png)
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So ... you've made it abundantly clear that you simply don't want to accept that the earth might be flat.
1) that's just clouds (just blurted out without any evidence or analysis)
-- yes, the exact same cloud formation that just happens to re-appear during about the same time window every year and each year just magically happens to follow the exact contour of the mountain range that just so happens to be in that same place. You'd be more ready to believe that than that the earth might be flat?
If the sun is parallel to a flat earth, then the sun would never drop behind those mountains. That is what the geometry shows.
You have made it abundantly clear that you no longer want to think about this topic. I just ask a lot of questions. I use what I know, to understand things. I have been watching the moon and stars for years and from a perspective stand point there are certain things that do not make sense. I try to explain them to you and ask for feedback and you just get mad and project what you think I am doing.
And you don't read all my posts. I said that for the sun to show up behind those mountains it would have to be only 250 miles in the air. If it is only 250 miles in the air, then we have to think about how hot that would be to cause the weather patterns we see today. That requires physics that I don't understand.
If the sun is only 250 miles from earth, then the moon is less than that, so maybe we did go to the moon.
I am trying to understand the science. I am using geometry and physics to ask more questions. Just because you have already formed a biased opinion about me you automatically disregard anything I say. That is not logical, that is just having an emotional reaction.
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Yes, it's just clouds. :jester:
Ok maybe it is the mountain range, but Clovis explained why you can see it. :cowboy:
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Every time I see one of these videos claiming that the curvature of the Earth would hide a distant object, I go back and do the simple math that shows that the object would not be covered by the curvature of the Earth. In the case of the video starting this thread, the curvature of the earth would only be 5100 ft. (cord height) in 175 miles, so it would not hide 9,000 foot mountains. In the case Lad mentioned, the curvature of the earth would be 6,661 feet across 200 miles. I'd like to see more on that case.
Here are the results for curvature of the Earth ("Height" in the calculator) for 175 miles and 200 miles, respectively. I did do the math by hand to verify. Said math is high school sophmore geometry class stuff, BTW.
(https://i.imgur.com/fKNONkf.png)
(https://i.imgur.com/tlGEuMi.png)
You math is way wrong. There’s no need for you to do anything by hand. There’s many different curvature calculators out there.
The mountains would be hidden at 175 miles. It would be over 20,000 ft of curvature.
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You math is way wrong. There’s no need for you to do anything by hand. There’s many different curvature calculators out there.
The mountains would be hidden at 175 miles. It would be over 20,000 ft of curvature.
Can you screenshot your calculator result? My math and the calculator results are absulutely correct: 5100 feet curvature height at 175 mile distance. I'd stake my engineer's license on it.
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Can you screenshot your calculator result? My math and the calculator results are absulutely correct: 5100 feet curvature height at 175 mile distance. I'd stake my engineer's license on it.
https://earthcurvature.com/
https://dizzib.github.io/earth/curve-calc/index.html?d0=175&h0=10&unit=imperial
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https://earthcurvature.com/
https://dizzib.github.io/earth/curve-calc/index.html?d0=175&h0=10&unit=imperial
What does this note mean at the bottom of the second calculator you attached?
Note: Using the formula 8 times the distance in miles squared is not accurate for long distances but is fine for practical use.
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So there's an easy rule of thumb that people sometimes use, where you square the number of miles and multiply that by 8 inches. While some have pooh-poohed that, I've seen charts to where it's accurate to a few feet out to over 100 miles, but didn't see it farther than that.
Using that formula for 175 miles yields 20,416 feet, whereas the real trigonometric calculation is somewhere in the 19,000+ range. Not enough to make a difference regarding the point being made here, and it's easy for anyone to scribble down or even do in their head.
Now, one does have to take into account the elevation of the observer, which the FE folks in these videos always do, since it's critical. Given the elevation of the obsever, in the video, of 1,000 feet, that results in 12,000+ feet of hidden height, and the elevation of the highest peak of the Canagu range is only 9,000. That would mean the entire then should be hidden with about 3,000 feet to spare, and yet we see that most of it is visible, with the exception of maybe 1,000 - 2,000 at the bottom due to various factors. I don't care how you crunch numbers, unless the globe is 10x bigger than the claim, there's no way that physically only the bottom portion of the range would be obscured, and you'd need some massive refraction to pull it off.
So, another reason that Canigou is critical as an FE proof ... it happens every year. Refraction is inherently inconsistent, where some years it wouldn't happen at all, other years it would be badly distorted, etc. etc. ... and yet this same exact phenomenon is consistenlty see every year.
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That sounds like the voice of Jeran, one of the flat earthers who went to Antarctica and, for the record, he is now a glober so it would be interesting to hear his thoughts on this now.
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https://earthcurvature.com/
https://dizzib.github.io/earth/curve-calc/index.html?d0=175&h0=10&unit=imperial
So, the first calculator you provided, earthcurve.com, is correct and does work and proves my calculation as well as the calculator I'd used. The problem is that you used the wrong distance: the distance in earthcurve.com is from the observer to the halfway point to the object (mountain) in the distance, or in this case, 87.5 miles. The curvature of the Earth (cord height) between two points is measured halfway between them. At 87.5 miles, the result is nearly the same as in my first post: (https://i.imgur.com/K1Axx7Y.png)
That other calculator you sent makes no sense and apparently does not apply to what we're talking about. Look at the results from entering 175 miles and 0 ft. observer elevation. It comes back with "0" for d1. What does that even mean :facepalm:?:
(https://i.imgur.com/P1I7VA8.png)
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The distance is the full amount. There’s nothing on that site that says to use 1/2 the distance. :facepalm:
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The distance is the full amount. There’s nothing on that site that says to use 1/2 the distance. :facepalm:
No, and the 8 inches times miles squared gets you very close to the same numbers. I'm sure that if these calculators were doubling the distance they would have been called out long ago by the Globers. I can do the math again, but I have done it for degrees of latitude and longitude for various parts of the alleged globe. It's accurate.
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It's not the "halfway distance" or "distance to the (fictional) horizon". The whole POINT is
1. The mountain is X miles away.
2. The earth curves at a rate of Y.
3. Therefore the mountain should be hidden by Z miles of curvature.
The whole POINT of these calculators is to calculate how tall something would have to be, to still be seen despite the "dropping off" due to earth curvature.
And the calculators are extremely straightforward, easy to use, and criticized by no globe-believers. Except this guy, apparently -- who is showing MASSIVE amounts of cope here.