1. Rory

    Rory Senior Member

    YouTube user JTolan Media 1 has uploaded a video which shows images of San Jacinto Peak in California taken from Malibu, 120 miles away. He believes this is proof that the Earth is flat:

    Source: www.youtube.com/watch?v=7-pXWRn_wfk (start at around 9:48 - before that is preamble)

    His main piece of evidence is this image here (contrast adjusted):

    mt jacinto.JPG

    He gives his observation location as 34.032204, -118.702984 and his altitude as 150 feet. This seems accurate enough. So where has he gone wrong in believing he's seeing something that shouldn't be seen on a spherical Earth?

    The curve calculator predicts a hidden amount for those figures of 6,158 feet. San Jacinto Peak is 10,834 feet above sea level. This leaves a predicted visible amount of 4,676 feet.

    Likewise, peakfinder has no problem with San Jacinto being visible from that spot:

    Source: peakfinder.org

    And this is a fuller image of that mountain range, taken from an elevated location much closer to San Jacinto, to give a picture of what the mountain actually looks like, free from obscuration:

    Source: peakfinder.org

    Indian Mountain, to the right of San Jacinto, summits at 5,790 feet, so what we're seeing in JTolan's video should be the portion of the mountain range a few hundred feet above that. Basically, this:


    As there are hills in the way, though, what we're actually seeing is merely the very top portion of the range, from around Castle Rocks, at 8600 feet, upwards:

    san jacinto.

    Working out which hills are obscuring the range and calculating the predicted visible amount would be the final step here. It's probably about 2,200 feet for the sphere Earth, but substantially more for the hypothetical flat.
    Last edited: Dec 17, 2018
    • Like Like x 3
  2. Rory

    Rory Senior Member

    I was struggling to find a photo of what San Jacinto actually looks like in full from that direction, but I just came across this one:

    Source: https://imgur.com/a/HWZZLoM

    The red boxed part equates to the area JTolan shows in his photo.

    There's no information about a location here, but it claims to be taken from roughly the same angle.

    On another note, JTolan is the chap we've seen before, who shoots images with 'an infrared camera'. He believes it shows him something more than a normal camera can see. I've even heard it claimed, by a different flat earther, that "the infrared camera filter takes out all refraction, atmospheric lensing, and any mirage effects you might normally see on a hot day or where there's high humidity."

    Now, I don't know much about infrared cameras, but I know they certainly can't "remove refraction".

    Also, I'm used to seeing them shooting footage where people's heads' glow white and cold things show dark; JTolan's images look more like simple black and white photography to me.

    In the video above, from about 8:00, he says a little about his set-up. It's made up of:
    • Monochrome astronomy camera
    • Celestron spotting scope
    • 850nm infrared filter
    I'll bet we have some on here who understand what that filter actually does. But, in the meantime, if I go by a quick peruse of the wikipedia article on infrared photography I can learn that: this type of photography is termed "near-infrared", as opposed to thermal imaging, which is called "far-infrared"; that it's good for penetrating atmospheric haze; and that it's mainly just a way to create a nice, dreamlike photo effect - though it can, as JTolan says, sometimes reveal things that are otherwise difficult to see:

    Visible vs. Infrared (900 nm LP) Aerial Photography of Old Hickory Lake, Tennessee, taken seconds apart.
    Last edited: Dec 17, 2018
    • Like Like x 3
  3. Inti

    Inti Active Member

    Nice, Rory. This is the same JTolan who posted an earlier infrared video from a plane, which clearly showed the the amount of curvature that agrees with Walter Brislin's calculator, as already shown here;

    The absurd thing was the number of FE belivers crowing about "no curve" and "believe your eyes" in the comments, until too many debunkers had shown the curve in screenshots.
  4. Rory

    Rory Senior Member

    I've come across a new tool similar to peakfinder which is great at identifying the obscuring ridges. This is the image it gives us for that shot to San Jacinto:

    san jacinto from panorama maker.
    https://www.udeuschle.de/panoramas/panqueryfull.aspx?mode=newstandard&data=lon:-118.702984$$$lat:34.032204$$$alt:45$$$altcam:1$$$hialt:false$$$resolution:600$$$azimut:97.3$$$sweep:6$$$leftbound:94.3$$$rightbound:100.3$$$split:2$$$splitnr:3$$$tilt:0.111111111111111$$$tiltsplit:false$$$elexagg:1.2$$$range:300$$$colorcoding:false$$$colorcodinglimit:215$$$title:Malibu to San Jacinto$$$description:$$$email:$$$language:en$$$screenwidth:1366$$$screenheight:768

    This shows the obscuring ridge is 19.7 to 20 miles away, at an elevation of 390 to 476 feet.

    Using Google Earth, meanwhile, shows that the obscuring ridge on a flat earth model would be one either 56 or 87.9 miles away:

    san jac profile.

    Entering these figures into my obscuration calculator, I get the following:


    Some notes about the above:
    • The predicted visible amount for the sphere earth is based on the height of the ridge directly beneath the peak. 2350 feet tallies pretty nicely with what we see on peakfinder and in reality.
    • There are different figures for the target landmark because the line of sight hits the main body of the mountain in different places depending on the model, as can be seen in the elevation profile above.
    • I noted two different ridges that could be obscuring the flat earth view. The other ridge returns 8085 feet; basically the same.
    • I haven't done the more complex trig which factors in for the tilt. The reason I can't just use the same formula for the Pikes Peak shot is because that one had an obstruction and target below horizontal, and in this one they're both above horizontal. Slightly different formula, and it would take a quite a bit of work to rejig the equation - it's nearly four lines long! Given that the last one returned only a 0.3% difference, I think I can be happy enough with the more basic calculation.
    I also found another nice shot of much more of San Jacinto, taken from Lake Mathews, which is about 40 miles from the mountain, and 3 miles south of the line of sight:

    mt jacinto5.
    Source: https://www.flickr.com/photos/clarence_w/2442230154/

    The black box shows the portion photographed in JTolan's image.
    Last edited: Dec 31, 2018
  5. Mark5300

    Mark5300 New Member

    Rory, the dam at Lake Matthews faces northwest. Mt. San Jacinto is straight east. That is a view of Mt. San Antonio, not San Jacinto.

    Ah! That's Mt. San Jacinto! Another great shot of our local mountains.
    Last edited: Dec 31, 2018
    • Useful Useful x 1
  6. Rory

    Rory Senior Member

    Fixed! Thanks for pointing that out. :)
  7. Rory

    Rory Senior Member

    JTolan's at it again, providing terrific images to demonstrate that the Earth is a sphere:

    Source: https://www.youtube.com/watch?v=KxLwaaU1aNk&t=3m38s

    The measurements in the picture are his own, and are mostly accurate for A-D, but between points E and F he gives an elevation difference of 2800 (or 2700) feet, due to a hitherto undiscovered "compression" effect. (Point F is actually at around 7300 feet)

    The photo is taken from Point Dume at about 140 feet above sea level - somewhere close to 34.001988, -118.806015 (he doesn't give an exact location) - and it's 122.6 miles to the peak.

    Though he's further away this time, and very slightly lower in elevation, he does actually capture more of the mountain - this is due to the obstructing ridge being at a lower elevation than the one in his previous shot.

    This is the elevation profile showing the ridge that would be in the way for the flat earth model (San Juan Hills, at 62 miles from the observer):


    Drawing a line of sight from the observer across the top of this ridge shows that around 8000 feet of the mountain should be visible.

    The view of the mountain from Lake Mathews is showing in the region of 7700 feet, and a simple overlay reveals that we are seeing nowhere near this amount in JTolan's photograph:

    san jacinto  2.

    Based on that, as well as identifying other peaks, it appears that he's seeing somewhere in the region of 3600 feet.

    Here's what the obscuration calculator predicts for the two models:


    • I've entered the sphere earth obstruction as the ridge at 23 miles away. It's possible that it's a slightly taller one about 28 miles away, but the tool to figure this out precisely isn't currently working. The result, however, is more or less the same.
    • Given that the obscuring ridge is clearly one around 25 miles away, we could also use that for the flat earth figure. This would mean that over 10,000 feet of the mountain should be visible.
    • I could probably be around 5% more accurate in calculating the visible amount of the mountain. But given how clearly his shot supports the spherical earth, the extra time to do so doesn't seem worth it.
    • Like Like x 3
  8. Mick West

    Mick West Administrator Staff Member

    Metabunk 2019-01-09 22-24-23.

    This can give some useful perspective. I've added five polygons, each 200m apart, from 2400m to 3200m (use copy and paste and then change the color and height)

    Metabunk 2019-01-09 22-26-49.

    View from pt dume
    Metabunk 2019-01-09 22-28-42.

    Notice the yellow (2400m) is barely visible

    KMZ attached.

    Attached Files:

    • Like Like x 1
  9. Rory

    Rory Senior Member

    It's kind of working now: it really is such a cool tool when looking at these kinds of pictures, and especially for identifying ridges and landmarks.

    Screenshot (152).
    Source: https://tinyurl.com/yddmmpw2

    Using this tells me that the ridge is largely the one at about 23.3 miles, 180 or so feet in elevation - basically in agreement with Google Earth.

    Usefully, it also shows that San Juan Hill only obstructs more or less on the line of sight to the peak, so I think we can use that 23-mile ridge for both models (which would mean a flat earth line of sight hitting the foothills 78 miles away). This results in:


    No prizes for figuring out which one's closer to reality.
    Last edited: Jan 10, 2019
  10. Rory

    Rory Senior Member

    We've a slight discrepancy somewhere, Mick. If the yellow plane is at 2400m and marks the lowest visible point, that means about 2960 feet of the mountain is visible. My calculator says about 3700 feet, as does the panorama maker above, as well as estimating it using known points on the mountain:

    san jacinto2 copy.
  11. Mick West

    Mick West Administrator Staff Member

    I don't think it's the "lowest visible point." There's a number of issues with the way Google Earth displays polygons, but it does give a good indication of the part of the mountain you are looking at.
  12. Rory

    Rory Senior Member

    The more I get to use that German panorama maker, the more I'm loving it. I was looking a bit closer at the point at Castle Rocks, and wondering why the line pointing to it went lower than the ridge. There's a really useful 10x zoom tool that let's you check in even closer:

    crosshair on ridge.

    This is showing that the ridge we see is actually a higher one, but unnamed, approximately 2.6km behind Castle Rocks.

    Sure enough:

    2c. the ridge behind.

    It changes the figures above a tad, but not enough to alter the general estimate of how much we're seeing.
    Last edited: Jan 11, 2019
  13. Ivo Busko

    Ivo Busko New Member

    As for the infrared camera, the 850 nm filter isolates what is called 'near infrared', a band of the electromagnetic spectrum in between 700 and 2500 nm. Not to be confused wit the thermal infared detected by the FLIR cameras (Forward-looking infrared), which extends much further into larger wavelengths, up to 14000 nm. Digital cameras have silicon-based sensors that naturally respond better in the near-IR than in visible light (400-700 nm) so they usually have a coating applied over the sensor surface to absorb this near-IR radiation and enable better contrast and dynamic range for the remaining visible radiation that makes thru the sensor coating. After all, they have to emulate the older chemical emulsions they are meant to replace.

    Thus if you use a 850 nm filter on a regular DSLR, the combination filer+sensor coating is going to block most of the light that strikes the sensor. It becomes a very inneficient system. To overcome this, the author of these videos used an astronomical DSLR camera. By that, we mean a regular DSLR in which the IR-absorbing coating *wasn't applied*. Thus the combination filter+bare sensor will be optimal for registering images in the near-IR band.

    As for the 'haze penetration power' of this combination, indeed it provides better contrast than images made in the visual spectral band, for one reason: the absorption coeffcieints of both Rayliegh scattering and molecular and nano-particulate haze drop dramatically as the wavelength increase. Going from 500-600 nm to 850 nm causes Rayleigh scattering to drop by a factor larger than 2, and absorption by a similar factor. The atmosphere essentially is more transparent and with better contrast in the near-IR.

    The effects on refraction are negligible. The coefficient of refraction decreases by about 1% going from the visible to the near IR bands. And of course, air humidity doesn't play any significant role in atmospheric refraction. This is just a lies purposedly inserted into the flat Earth discourse by people that know it's a lie.
    Last edited by a moderator: Jan 16, 2019
    • Informative Informative x 1
  14. Rory

    Rory Senior Member

    I've been having a further look at some of JTolan's footage, since he also provides some really nice shots of the San Bernardino and San Gorgonio ranges in his video (from 3:04).

    I made this panorama from three screenshots:

    gorgonio panorama.

    I think even at first glance it's a beautiful flat earth debunk, given that we're clearly only seeing down to a little under 7000 feet above sea level for any of the peaks over a hundred miles away. There are so many analyses that could be done here, but here's just a couple:

    1. The ridge in front of Little Green Valley is 16.1 miles away, with an elevation of 312 feet. This predicts a visible amount on the globe of 861 feet, and a visible amount on the flat earth of 5816 feet. Clearly, we're seeing nowhere near the latter, while the former can be quickly confirmed by measuring the difference in apparent height between Little Green Valley and Keller Peak - 840 feet - and noticing that it's pretty much the same measurement down to the ridge.

    2. Looking at the unnamed peak to the right of Cucamonga, where the ridge in front is 17 miles away, at 436 feet above sea level. This gives predicted visible amounts of 3130 feet (sphere) and 5372 feet (flat). Again, taking the difference in height between Cucamonga and the unnamed peak of 2100 feet, we find that the difference on the photo down to the ridge is about one and a half times that - 3150 feet - again matching the globe.

    I also looked again at his main San Jacinto shot and noticed that a point called One Horse Ridge is just below what he can see:

    one horse ridge.

    One Horse Ridge is 3,872 feet below the peak of San Jacinto (and a little bit closer) which again tallies with all the other methods showing that we're only seeing about 3,700 feet of the mountain, which is exactly what we would expect on a sphere.

    In a nutshell:

    gorgonio panorama missing.
    Last edited: Jan 16, 2019
    • Like Like x 2
  15. Nth

    Nth Member

    There's something humorous about all this. I think we often find a popular perception that the curvature of the earth is somehow hard to notice, which I suppose is not entirely untrue depending on your reference frame. That said, the funny part is how applying some known distances and elevations even to landscape photography pretty rapidly makes earth's sphericity clear.

    Nice work on these posts, @Rory, and your recent YouTube videos have been fun to watch. :)
    • Like Like x 1
  16. Inti

    Inti Active Member

    I've just seen this video by Okrelos, who has written his own 3D landscape modelling software as part of his work. He says he became aware of flat earth belief through "Behind the Curve" on Netflix, and then saw JTolan 's San Jacinto image. He has modelled the whole area, and has a slider for the degree of curvature, between the actual value and flat.

    Source: https://youtu.be/RK93TfSYeQU

    The effect is interesting, to say that least. It animated the shift between viewing the full height of the mountain on a flat surface and the globe view, which precisely matches JTolan's view.

    He is careful not to claim it as conclusive proof of the globe, since as he says there migbt be some effect that made it appear to be curved, but tgat it presents a large problem for flat earth claims.

    Perhaps @Rory and @deirdre should try to interpret Okreylos in modelling the Mount Rainier view, too.

    I found his video through an extract in this critique of JTolan's ability to identify and interpret features by Bob the Science Guy;

    Source: https://youtu.be/-TEqAYEa_M4
  17. Mick West

    Mick West Administrator Staff Member


    Excellent illustration. I've fudged this in Google Earth before by raising the camera, but it's not a perfect flat earth view like this.

    But it's given me an idea - I can generate a "tour" for a given camera that will raise and lower the view while keeping the camera pointing at a distant point.
  18. Mick West

    Mick West Administrator Staff Member


    Source: https://www.youtube.com/watch?v=QxjUgCEvGwk

    Okreylos' animation is far better. But this still gives a sense of how big the mountain actually it. It works better with Ocean mountains.
    • Like Like x 2
  19. Steve Funk

    Steve Funk Active Member

    As a nitpick, the radius of the earth in (in #18) is about the same at any latitude. You have the right radius, but by bringing in latitude it implies that you are measuring the radius of the smaller latitudinal circle.
  20. Rory

    Rory Senior Member

    Check out this lovely extended panorama I made, taking all JTolan's images of the mountains behind LA:


    Figures at the bottom show what's missing from what we would expect to see on a flat earth.
    Last edited: Apr 2, 2019
    • Like Like x 1