# Folsom Lake Photographs Demonstrating the Curvature of the Earth

Discussion in 'Flat Earth' started by Mick West, Aug 1, 2016.

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This is Folsom Lake, California. I visited it today (Aug 1 2016) to try to determine the curvature of the Earth.

Most people know the Earth is round, but they know it mostly because they just accept it, as it's long established science. But you can actually go out and test this for yourself, which can be either a fun science experiment, or a life altering paradigm shift, depending on your view of the "Flat Earth" theory.

So that viewpoint is at Brown's Ravine, a boat ramp area run by the California State Parks. Over in the distance is the dam, which is exactly 3 miles away on the right side (2.9 on the left).

I used a Canon 7D with a 500mm zoom lens, and from about 20 feet above the waterline the dam looks like this:

(Full Sized Original: IMG_0150)

Zoom in a bit more, you can see what looks like a pontoon walkway at the base of the dam.

Now as we are 20 feet above the waterline, and only 3 miles from the dam, then nothing is obscured by the curvature of the earth, as the horizon is 5.5 miles away. So we are seeing everything that is there.

So next I went down to the water, and took photos from about one foot above the surface of the water. I did not have to worry about waves, as there was just some 1" ripples. Here's the same shot from there:

(Full Sized Original: IMG_0168)

Here's a close up

Notice the walkway has gone. The water line is now above the floating pontoons, and the walkway itself. It seems like most of the hanging chains are still there though, so it's just over the walkway itself. . It's hard to see how high the walkway is, but based on the height of the safety rails, I'd say around 1 foot is it is obscured by the curvature of the Earth.

So, 3 mile distance, camera at about one foot, about one foot obscured. Is this in the ballpark? Well, if we plug in 3 miles and 1 foot camera to the calculator, we get 2.1 feet hidden, or 1.6 with "standard" refraction.

https://www.metabunk.org/curve/?d=3&h=1&r=3959&u=i
Given the very rough estimates I have for the level of my camera and the height of the pontoon walkway, this is very much in the ballpark. If my "about 1 foot" for both estimates were actually about 1.5 feet, then it would be exactly correct.

Hence the world is not flat.

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I took a LOT of photos, and I'll try to find some more good examples.

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3. ### AuldySenior Member

Did you manage to get one of the dam and the pontoon walkway from up close at all? Just for perspective.

No, on the side of the lake I was on there's nothing closer to the dam. I'll have to go around to the other side, Beals Point, to get a closer view. You can't actually get closer than a mile without a boat.

I found this pic of the back of the dam showing the walkway in a little more detail
So I'd say that's well over a foot from the water surface to the walkway surface

6. ### Sandor SzekelyBannedBanned

Hey Mick that is GREAT to make OWN experiemnts

Mick:
I have problems with this statement LOL

"Given the very rough estimates" "were actually about" ---- "would be exactly correct" "Hence the world is not flat"

WOW! don't mean that seriously right?

SO let me give you MY opinion:

Your picture from the lower height was not at the same zoom, so I had to enlarge it - and I flipped it horizontally for better comparison and put it beside the higher eyeheight picture:

?
what I see there that on the lower height picture the distance between the walkway and the sea level is BLURRED probably because it's hidden by wave or water vapour and I see light diffraction on the waves.

So your picture does not prove that the target hidden height is exactly the same amount of the supposed curvature target hidden height. Your definite proof is missing the EXCLUSION of any other possibilities - like I mentioned here.

THIS IS WHY we need to take measurements from much farther away!

[Off topic material removed]

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7. ### Sandor SzekelyBannedBanned

Well this raises an other debunk subject: HOW do we know that the water level is the same on both pictures? This is a dam and it can change water level

The walkway is floating, so it's always the same height above the water.

Let's keep this tread on topic please. Discussion of the Folsom photos only.

The original photos (linked in the post) are at the same zoom level.

The "light diffraction" is an artifact of the rotation in photoshop. Look at the original.

There were no waves, the lake surface had ripples on it of barely an inch. The walkway is invisible for the entire length of the dam, so it's not a wave. It's a sharp delineation.

I also took multiple photos of multiple points on the dam and the lake shore. They all show the same thing.

Last edited: Aug 2, 2016
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Things that are farther away do show the curvature a lot better, however they are also more prone to refraction problems. I like the 3 mile distance here (besides it being the only convenient lake nearby) as the air is relatively clear over that distance, and there's no apparent bending.

I might pop over to Lake Tahoe sometime though

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In the above I simply checked my rough measurements against the expected value, but how can we take the three values (h = height of camera, d = distance to object, x = amount of object hidden) and get the radius of the earth.

Well, we just reverse the equations. We have:

Distance to horizon, a = sqrt((r+h)*(r+h) - r*r)
Amount obscured x = sqrt(a*a - 2*a*d + d*d + r*r)-r

Substitute a.
x = sqrt(((r+h)*(r+h) - r*r) - 2*sqrt(((r+h)*(r+h) - r*r))*d + d*d + r*r)-r

Solve for r:
r = (d^2*h+d^2*x +/- 2 * sqrt(d^4*h*x-d^2*h^3*x+2*d^2*h^2-d^2*h*x^3) - h^3 +h^2*x+h*x^2-x^3)/(2*(h^2-2*h*x+x^2))))

The negative solution is the correct one for short distances, so, in Javascript:
r = (d*d*h+d*d*x - 2 * Math.sqrt(d*d*d*d*h*x-d*d*h*h*h*x+2*d*d*h*h-d*d*h*x*x*x) - h*h*h +h*h*x+h*x*x-x*x*x)/(2*(h*h-2*h*x+x*x))

Put it in a calculator:

Put in my rough values of 1.5 feet.

Close enough!

http://www-rohan.sdsu.edu/~aty/explain/atmos_refr/horizon.html

Last edited: Aug 2, 2016
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13. ### tinkertailorActive Member

Quick question:
If I wanted to replicate this, should I try to find a relatively still body of water? I'm going to be headed to the SF Bay in a few weeks and think that there are definitely areas where I could try this. I'm assuming waves wouldn't make a huge difference over longer distances like, say, looking at a wharf in Monterey from the beach at San Diego, but over shorter distances couldn't even small waves less than a foot make a difference?

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Yes, a still body of water is better. For short distances you want to be as close to the water line as possible, waves mess this up, especially as they can be different heights in different locations. You want to be above the waves.

San Diego?

6 miles is maybe a good distance. You can have the camera up at six feet above the water, and still have six feet hidden on the opposite side. Then you need to get up to 25 feet (or more, it does not matter much) to be able to see it unobstructed. So a good place is whee there's a pier or walkway or hill 20 feet of the ground, and then a beach where you can stand at the waterline.

This beach might work, with the hills at the east end.

15. ### solreySenior Member

The Farallon Islands would be a nice target when the visibility is good. There is a lighthouse on the highest point at 358 feet in elevation, from that height the distance to the horizon is 25.47 miles. Ocean Beach at the end of Golden Gate Park is 27.25 miles from the Farallons and there is a 30 plus foot elevation gain immediately above the beach. Point Reyes is over 20 miles away from the Farallons and has some good views with beaches below high cliffs. If nothing else it's a good excuse to visit Point Reyes.

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16. ### tinkertailorActive Member

Whoops, Santa Cruz...

I'll definitely try this out if I get the time while I'm down there!

17. ### jeranismMember

I hate to say I don't agree with your evidence but I do not think your camera was one foot from the water. Not even close. Also, I've been there and that isn't Brown's Ravine is it? I think your angle is off. From that shooting location I don't think you'd get the images you got. Can you double check and make sure your google image is correct. Maybe we can meet up Mick- I'm not too far

18. ### tinkertailorActive Member

What about the camera angle makes you think it 'wasn't even close' to being one foot?

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This is the precise location I took the upper photo from, note the GPS coords 38 43 9.18, -121 6 16.03. It's Brown's ravine.

When I took the lower photos I think I went a few yards to the right, to avoid the swimmers.

I think 1 to 1.5 feet is about right. I basically went down to the waterline, then squatted and leaned. Recreating it here gives me the middle of the lens being 8" off the ground.

Well, 11", but still in the ballpark.

(at the lake the straps were around my neck)

Last edited: Aug 9, 2016
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21. ### ThorGoLuckyMember

Fun! And memories of when I lived in that area and visited Folsom Lake often.

22. ### tinkertailorActive Member

@jeranism --
Depending on how far away you live and what kind of camera you have, you could easily recreate this experiment yourself and show us here on MB what mistakes, if any, you think Mick might have made. Or you could just go to Brown's Ravine and try to see the dam from there to check Mick's experiment. There's a clear line of sight from where Mick was standing/squatting to the dam per Google Earth's imagery - what possible error could there be?

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23. ### jeranismMember

Yeah, I'm probably wrong, the pictures just looked like they were from higher than 11" and I've only been there twice but I thought his pictures looked more like they came from the other boat ramp but I was just asking. I also didn't click the larger images.

Isn't this easily explained with horizon dip? As 20 feet off the water would leave you 5 seconds of arc and 1 foot would leave you only 1 minute of arc dip?

24. ### jeranismMember

Thanks Mick-

Hey we should chat because there is some rivers I was looking into close to Sac that are straight shots 7-10 miles. Know anyone who has a boat that navigates the channels?

(oh see above re: horizon dip)

Would you agree that (ignoring refraction) light travels in straight lines?

So the only way something can be obscured is if something is in front of it, in line with it.

If the surface of the water is flat, then it can't obscure anything.

The only way the water can be obscuring the the pontoons is if it is higher than them, and higher than the camera. Hence the lake must curve up more than a foot.

So the surface of the lake is curved.

If you think otherwise, draw a diagram with a flat lake surface, the pontoons, and the camera, and explain what you think is going on.

26. ### jeranismMember

So, see below here, first of all.... were these things in the water? How tall are they?

Second, if you were to be at an angle, say like below. Would you see a different amount of box A vs. Box B based on the angle? I would say they would not be the same, you will see more of red box B. Would you agree?

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27. ### mm1145Member

I am sorry I do not understand what you are saying here can you explaine more?

are you saying that the 2 red boxes are on the dam (it is a dam?) and because the picture is not parralle to the dam one box is further away than the other one and so we should see less of it?

if so then techinley yes it is further away but by so little compared to the disstince to the dam form the photographer that the extra drop would be not detectable on the photos resulation

28. ### TrailblazerSenior Member

Could you explain what you mean here, because it doesn't make any sense to me at all. "Horizon dip" is a phenomenon caused by the curvature of the Earth: it means that the line of the horizon is lower than the true horizontal. You can't attribute the effect to "horizon dip" on a flat earth model.

I will leave it to Nathaniel Bowditch to explain it, in his 1833 tome The New American Practical Navigator:

[...]

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No, it was not there. It's a barrier to keep boats away when the water level rises up to the spill gates, so they can open them. You can see quite clearly from my photos that it is not there, as from the higher angle photos the entire surface of the water is visible.

It would be visually bigger because it is closer (assuming A is the box on the the left). From the angle in your photo the waterline is visible, and so you can see all of both boxes.

Note that the angular difference there is hundreds of times greater than the difference between 1 foot and 20 foot in altitude, so I'm not sure how you think it pertains to my photos.

The remainder of your post was off topic, and has been excised to here:
https://www.metabunk.org/jeranisms-discussion-on-earth-shape-dogmatism.t7823/

30. ### cloudspotterSenior Member

Been trying to come up with my own version of this experiment and it came to me when we were having fish and chips on the beach last night.

We have this wind farm met mast offshore

And this WWII battery observation post overlooking the beach

I think they're about 6 miles apart but I'll have to confirm that

31. ### TrailblazerSenior Member

Is the mast the object in the thick red circle here? (map from http://www.edf-er.com/Portals/edfrenewables/Documents/OurProjects/Blyth/Blyth Offshore Site Boundary.pdf)

I have also circled Blyth Battery on the coast.

If it is that mast then I make the difference approximately 4.1 miles (the co-ordinates on that map don't seem to quite agree with those on Google Maps, despite both supposedly being WGS84).

Last edited: Aug 23, 2016
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33. ### cloudspotterSenior Member

Yeah that's probably it. I was going off the coordinates given and measuring on Google Earth

One thing I did not really address here was the effect of refraction. As seen in the Canandaigua Lake experiment, cool water on a warm day can create a shallow temperature inversion just above the surface of the water. A gradient of 0.11°C per meter ( 0.06°F/foot) is sufficient to bend light around enough to follow the surface of the lake.

And yet here (at Folsom) I have a camera 1 foot off the water, and the obscuration is more or less what you would expect without refraction.

I took the obscured photo at 1:25PM on Aug 1 2016. It was, as I remember, quite hot. Probably in the 90°F range. The water temperature is reported generally as being 75°F in summer. However the temperature is not simple, there's a warm surface layer and a colder bottom layer, giving a profile within the water itself. But the only really important number is the surface temp.

Here's a detailed analysis of those temperatures.

That seems to be 27C (81F) at the surface at a similar time of year.

But it seems quite certain that the late water wall still quite a bit cooler than the air, so it seems quite plausible there was some refraction going on.

Looking at the photos in the other thread taught me to look at the big picture, as this type of local looming compresses the bottom part of the image. So there's the larger view:

The dam is three miles away so the change in height of 20 feet would have a negligible effect of the view.

What we can see here is that there IS a compression of the lower part of the image. In fact it almost looks like you can see parts of the pontoons that are under the bridge.

You can see the compression if you compare the height of the chain railings. These are at identical zoom levels.

So I think that what we are seeing here is a combination of the curvature hiding most of the pontoons, and the refraction compressing the top part of the pontoons, the bridge, and the chain railings (and compressing more as you get lower down)

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And this encouraged me to revisit my Folsom photo collection to see if there are similar examples. Here's a good one:

This is the "high" shot from Folsom point. There's a sloping area on the far shore, with a sun canopy and some water toys down by the shoreline:

Now compare wth the "low" shot.

Significant refraction. This was taken at 2:09 PM, 48 minutes after the Dam shots. You can possibly see a glimpse of the pink and blue inflatable that's near the waterline.

I think the lesson here is that you need to check for refraction. In particular you can't use objects right next to the horizon to set the scale of the image. You need objects that have a line of sight significantly above the refraction.

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