In this article I would like to check out a video^{1} that has repeatedly been touted as absolute proof that the earth is flat. The video was taken from Miramar Beach, Santa Barbara, California looking towards the Anacapa Islands. The argument goes somewhere along the lines of “if the earth was curved we should not be able to see as much of the islands as we clearly see in this video, therefore the earth is flat.”
Bofore diving in, let me first point out the elephant in the room. We are here not arguing about flat vs globe, but about globe vs globe, in other words, how much curvature we see. It is glaringly obvious that there is at least some curvature drop visible close to sea level. It is therefore beyond me how any sane person could claim that this video proves the earth is flat.
Some flat earthers decided to change the punch line to “we are not seeing the exact amount of curvature we should be seeing according to globe earthers, therefore.. the earth is flat”. As if that is any improvement on their first attempt to be logical.
Never mind, let's have a closer look anyway.
I will go through this video in chronological order first, making a few remarks along the way, and at the end I will gather all the facts and do my own calculations.
0:05
We are immediately shown the date and time at which the video was supposedly taken: 02/07/2016 at 2:08 PM.
Excellent, so the first thing I do is find the actual see level (or tide) at that particular date and time, as we'll need that later on for our calculations. A simple Google search^{2} will tell us that on that particular day there was a high tide of 6.3 ft at 7:55 AM and a low tide of 1.1 ft at 2:52 PM. Given that the first shot in the video was taken at the top of the hill, before descending to the beach, and assuming that it would take at least a few minutes to descend to the beach and to set up the camera, I believe it is fair to say that the essential part of the video was taken at or very close to low tide. In other words, at that time the see level was 7.4 ft below high tide.
0:22
We are given the heights of the three Anacapa islands, West Anacapa being 930 ft, Middle Anacapa 325 ft and East Anacapa 250 ft. Checking on Google Earth this appears to be correct.
0:32
We are given an approximate location of where the video was supposedly taken from. Going to the exact same spot on Google earth we find out that the elevation at that point is at least 9 ft, possible even more depending on where exactly the camera was placed. I hope we'll get another clue later on in the video.
0:41
We are given the distance to the highest point on Middle Anacapa Island: 31.61 miles.
0:48
We are given the height (not the elevation!) of the camera tripod as approximately 3 ft.
0:52
Here we see the first error creep in. We are shown a picture of the Earth Curve Calculator^{3} with some measurements already filled out for us. The “Eye height” value is given as 3 ft, which is incorrect. We have already seen that the camera is located at an elevation of at least 9ft, plus the 3 ft height of the tripod, making the “Eye level” at least 12 ft.
0:56
Now we are shown what the author would expect based on his figures so far, i.e. that islands should be hidden “by 580.68 ft of curvature drop”. As already pointed out above, that figure is out (by 80ft at least) because of the incorrect elevation used in the calculations. Also note that there are still other factors which have not yet been included in the calculations. More on this soon.
1:09
The author tells us that there was a 5 ft ocean swell that day. I find this very hard to believe judging by the small waves we see at the beach. I would guestimate 23 ft maybe. Anyway, even if we go with 5 ft the author still makes yet another mistake. The swell is calculated as the difference in height between the crest and the trough of waves. If there is a 5ft swell, then that would mean the crest is 2.5 ft above sea level, and the trough is 2.5 ft below see level. For every amount of water rising above the sea level (the crest of a wave) there must be an equal amount of water being taken from below the sea level (the trough of the wave). Hence, a 5 ft swell would raise the “visual see level” by only half that amount, 2.5 ft.
In this frame the author also gives us some figures about what “should” happen, but without giving us any clue how he came to these figures. Since his input variables are already skewed at this point I decided to simply make my own calculations.
Instead of assuming an ocean swell has the same effect as an elevation change at the camera (an assumption the author seems to make), I believe the correct way is to account for the rise in horizon and calculate how much more or less of the target will be hidden. As promised, I will give you the calculations at the end. First, let's move on and see what else we're dished up.
1:19
The author now gives us a picture that is supposed to explain how he accounts for the ocean swell. No explanation is given, no calculations, nothing at all, except the figure mentioned before. How is any of these “critical” flat earthers supposed to verify the experiment for himself? Could it be that no one ever bothered doing this, and they are more than happy to go with what “the flat earth expert” tells them?
1:22
Here we finally get to the real pictures, the proof of the pudding!
1:56
Now we get a better idea of where exactly on the beach the camera was located: at the bottom of the steps, between the rocks, almost against the wall. Going back to Google earth, we find that this location has an elevation of 11ft. Which makes absolute sense, common sense that is. If the tripod was really located at zero elevation it would be washed away by the waves crashing onto the beach. Clearly, that is not what we see in this beautiful video.
2:48
We are given a picture taken from supposedly 560 ft elevation. This picture is obviously taken from a location close to the shore. However, since no location is given I went to Google Earth to find the nearest point with this elevation. Turns out that we have to go quite a long way to find such an elevation, either to the left or to the right. Problem is though that the coast line there is NOT perpendicular to the direction we are looking in. Which means, that by going either left or right, we are decreasing or increasing our distance to the Anacapa Islands. And this in turn means that the curvature drop also decreases or increases, and that we are therefore comparing apples with oranges.
If you look closely at the shape of West Anacapa Islands, you will also notice some rather significant discrepancies, which cannot be accounted for by the relatively small difference in elevation. Only a substantial move to the left or to the right can account for this different silhouette.
And a third clue to suspect quite a different location in both images being compared is the fact that in one of them we see some offshore rigs, right in front of our target, while in the other picture we don't. No, they were not hidden by the curvature of the earth :)
3:42
The author indirectly admits that his camera was not really located at sea level, as now he walks towards (and down to) the sea! Now it gets really messy. Notice the funny shape in the frames around 4:18? I'll leave you to figure that one out by yourself for now.
But notice how the camera is supposed to be at sea level now? How come then we are looking down onto the sea? How come we are not IN the water? How come the waves never reach the camera? Never mind, let's move on.
OK, so let's do our own homework now.
We'll start by taking a snapshot of both views, the first one taken at “sea level” and the second one at 560 ft elevation. Then we'll put them underneath each other and analyze what we see.
Those Anacapa Islands are really nice, because the elevation figures are very cooperative. The highest peak in West Anacapa is 930 ft high. To the far left of that peak is a secondary peak, which according to Google Earth is 620 ft high. From these two elevations we can mark on the picture where the islands meet the see, i.e. zero elevation. Doing this on both snapshots gives us the following:
The green line is the horizon, and the bottom yellow line is sea level at a distance of about 31 miles (which is how far these islands are from the camera).
Does anyone else notice how these mountains in the top snapshot appear to be sinking into the sea? Really, this is all the proof a sane person needs to conclude that .. the earth is NOT flat. Still, some people would like the whole works, including the pudding, so let's carry on.
Observation #1 : when standing close to sea level, there is about 300 ft of elevation hidden from view, while the rest of the island appears in perfectly normal proportions.
Observation #2 : when looking at the same object from 560 ft elevation, there is only a few feet hidden from view.
Let's start by calculating the amount of curvature drop we should see at 560 ft. According to the Earth Curvature Calculator^{4} it should be 4.6881 ft, which looks pretty close to what wee see in the bottom snapshot.
Now let's now calculate the theoretical amount of curvature drop, close to sea level, based on the true figures mentioned above. Except, given that we are basing our three elevation lines in the above snapshots on the highest peak of West Anacapa Island, I will use that as the distance to our target, being 30.43 miles.
With an elevation of 11 ft (between the rocks, close to the wall) plus 3 ft tripod height, totaling 14 ft elevation, and a distance of 30.43 miles, the Earth Curve Calculator^{5} tells us that we should have a target hidden height of 445.5465 ft.
Now we have to account for the ocean swell and the low tide. The ocean swell adds 2.5 ft to the level of the horizon (assuming that the 5 ft swell was an accurate assumption, which I very much doubt), while the low tide will subtract about half of the 7.4 ft difference between low and high tide (assuming that the sea level is the mean between high and low tide). Therefore, The horizon is altered by 2.5  7.4/2 = 1.2 ft, in other words, lowered by 1.2 ft relative to the mean sea level.
Using simple geometry, dropping the horizon by a certain amount (h), will cause more of our target to be visible. How much more? To calculate this, we need to know the distance to the horizon (d) as well as the distance to our target (D). The amount of drop in curvature drop is : h * D / d = 1.2 * 31.63 / 4.58 = 8.3 ft. That's not much, but a 8 ft decrease in curvature drop is quite different from a 110 ft increase in curvature drop (as claimed in the frame at 1:09).
So we're down to about 437 feet of curvature drop.
Observation #3: the snapshot taken at 560 ft is nice and crisp, but the snapshot taken close to see level is hazy. Especially when looking closely at the smaller Anacapa Island, we see some strange anomalies. For example, look closely at the shape of the “mountains” in the following snapshots:
Obviously, the Anacapa Islands are not shaped like a mushroom or like Tintin's cowlick. What we see is not what we know exists in that shape. Which brings us to the much dreaded subject of .. mirages.
Check out the following article^{6} on the subject, especially this excerpt:
Towering is quite common in polar regions and during the summer near large bodies of relatively cold water when compared to the overlying air temperatures. Such situations are common along North America's Pacific Northwest coastline during the summer. Towering can make coastal mountains appear to rise and fall in height throughout the course of the day when seen from across cold ocean waters. The illusion formed is of the peaks looming higher, and thus the mountains appear closer than they actually are. This illusion can be quite hazardous to sailors navigating by sight alone by causing them to believe they are closer to shore than they actually are.
Given that the video was taken mid afternoon on a hot sunny day, along North America's Pacific Northwest coastline where mirages are common, I am quite satisfied that the remaining 137 ft in discrepancy is due to a mirage, combined with inaccuracies in “guestimating” the actual sea level, elevation, distances, tides, swells and calculations. No doubt some people will try to claim that this is yet another conspiracy, i.e. that mirages do not exist. Or as some have tried to claim, that mirages always appear upside down. Well, for starters, how else could we see a mushroom shaped mountain, if not because of an upside down mirage? And even so, some mirages (like the towering mirage mentioned above) are not inverted.
Still, even if we discard what we clearly see, even then all that we are arguing about is the exact rate of curvature of the earth, and not whether this curvature exists or not.
I know there will always be naysayers with itchy ears, preferring to believe in fables, but those who have ears to hear and eyes to see, let them hear and see.
Stay tuned for the next episode!

http://tides.mobilegeographics.com/locations/3447.html?y=2016&m=2&d=7 ↩︎

https://dizzib.github.io/earth/curvecalc/?d0=31.63&h0=3&unit=imperial ↩︎

https://dizzib.github.io/earth/curvecalc/?d0=31.63&h0=560&unit=imperial ↩︎

https://dizzib.github.io/earth/curvecalc/?d0=30.43&h0=14&unit=imperialp ↩︎

http://www.islandnet.com/~see/weather/elements/supmrge.htm ↩︎