From Hamilton ontario beach to Toronto beach is 33 miles aprox .....you can see the beach in Toronto from Hamilton this is a good view ...not my video but .....i don't see a curve when he pans right to left ....i can't believe this is an issue any more ..... Source: https://www.youtube.com/watch?v=FDdzfWyTME0

NO it wasn't ...but i have lived here for 45 years ..its the same view of tTronto from the beach .......nothing changes nothing is hidden ....zero curvature ...i live here right so I'm not just making this up ..also did you see him pan from right to left.....far right is rochestor NY far left is vaughn there is zero curve .its about 87 miles

maybe tmrw ill head down to the beach or tonight its only 20 min from here ....and 87 miles is about 5000 feet or a mile of curve give or take ....is that right ? there should be a mile of curve from right to left ?

He's viewing it from what looks like a building over Mountain Face park (which he says in the video), which would be well over 600 feet. With some triangulation, he's about here. And no curvature of the horizon is visible below 60,000.

"curve" is irrelevant; only the amount of the object you are looking at that is obscured is relevant. How much of an object is obscured depends on your height of eye. No, it is not right. The usual "drop" formula flat earthers use is only for the how much the curve should drop in the direction who are referencing. Left-right curvature is an entirely different trigonometric function.

This would actually make a great example if someone were to take video or photos from up there, then halfway down the escarpment, and then at sea level. You would see the obscuration very clearly.

View from the Harbor, not sure how high. http://www.torontograndprixtourist.com/2011/02/toronto-ghost-city.html From a boat, probably near Peel Source: https://twitter.com/peelpolicemedia/status/617748662354079744

At the location of the video, ground level is about 184 metres, or 604 feet: http://en-ca.topographic-map.com/places/Hamilton-812832/ So even assuming the cameraman is standing on the ground, call it 610 feet. Central Toronto is 36 miles away from this location. Plugging the numbers into the calculator at https://www.metabunk.org/curve/ we get a hidden height of a little over 20 feet: Edit: Toronto is not at sea level, of course, and neither is the intervening lake. I'd need to think a little more to work out how this affects the calculation. Why would you expect to see a curve from only 600 feet above sea level?

Apples and oranges. You won't see the curve of the horizon until you are high enough to be "looking down on" the plane of the horizon sufficiently. That doesn't prevent things from disappearing over the horizon. At the extreme where you place your eye right at sea level, the horizon will be infinitely close - all you will see is the straight line of the surface of the water intersecting your eyeball. Think about it. The horizon doesn't "dip" from one place to the other. The horizon forms a perfectly level circle, formed by taking a slice through the Earth. The higher your eyeline, the larger the slice, and the lower the horizon is below you. But the horizon is always the same distance below your eye, all the way round the circle. The higher you are, the larger the circle will be, but the more you will be "looking down on" it, until, at an extreme distance, the perimeter of the circle becomes effectively the circumference of the globe.

The visible curvature of the horizon from a vantage point h kilometres above the Earth is given by: k = sqrt [(1 + (h/6371))^2 - 1] https://en.wikipedia.org/wiki/Horizon#Curvature_of_the_horizon For a height of 600ft, or 0.2km, that gives a curvature of 0.0079. That would be the same equivalent curvature as the rim of a circle 10 metres across, viewed from a point less than 8cm above the centre. (As an aside, if the Earth was a flat disc, the horizon would still appear more curved the higher above it you got. The horizon itself is FLAT, whether you are on a globe or a flat disc!)

Radians^{-1} https://en.wikipedia.org/wiki/Horizon#Curvature_of_the_horizon i.e. 1/(angle subtended at the eye by the radius, when looking straight at the center of the circle) I think. [edit] although I don't understand it. [edit2] suspect that wikipedia definition is wrong. Or at least not very clear.

0.0079 radians = 0.4526°. Ok. That is close to the standard dip (height of eye) correction used in celestial navigation for the depression of the horizon. Dip in arcminutes = sqrt(height of eye in feet)*-0.97' sqrt(600')*0.96' = 23.8' = 0.40°

ah it seems to be the angle of the dip, as you mention, which would be arccos(r/(r+h)) arccos(6371/(6371+0.2)) = 0.00792356295 rad sqrt [(1 + (0.2/6371))^2 - 1] = 0.00792372877 Which suggests the units are radians. I'm a little confused, is this a non-trig approximation function? What do they mean by "is the reciprocal of the curvature angular radius"? I seems like it's 1-(curvature angular radius), but reciprocal implies 1/(curvature angular radius)

The formula given on Wikipedia implies that the horizon will have a curvature of 1 at a height of (√2 -1) x the Earth's radius, in other words about 2639 kilometres. At this altitude, the distance to the horizon is equal to the Earth's radius, and thus the angular diameter of the horizon is 90 degrees:

So if i go down to the beach tmrw and take a pic of Toronto from hamilton or Stoney Creek .....how much of Toronto should be hidden ? 35 miles 6-10 feet i get almost 700 feet https://dizzib.github.io/earth/curve-calc/?d0=35&h0=10&unit=imperial

Correct. The exact amount will depend on refraction etc, but the difference between the view from the beach and the view in the video you posted from the top of the escarpment should be striking.

When people talk about 'curvature' they're generally talking about the curve between you (the observer) and the horizon, or something beyond the horizon. The curve, of course, cannot be detected with the naked eye, but its effects can, such as the obscuration of the lower parts of distant buildings and ships. The 'left to right' curve is something different, and cannot be seen until you're way up there. At least 60,000 feet, and more like 100,000 feet for it to be significantly pronounced. To demonstrate: The line ab is the horizon. The observer is at o. The object you're looking at is at v. When people talk about curvature, and use the curvature calculator, it's the line ov that they're measuring: the "straight ahead" curve. Determining the curvature for the horizon - for ab - is an altogether different calculation, as outlined above. For "viewer height in feet" just put in the difference between the viewer and the lake (ie, the elevation of the horizon). Yep, that's right. Will be interesting to see how much of the SkyDome you can see, and how it compares to this photo:

Heres a pic i found taken from van wagers beach ......ill go down today and take my own .....but this looks about right today is much clearer

I don't know where you got this photo ....but its not real ....your missing about 200 buildings in it as you can see in the above photo

The actual skydiome when closed is 282 feet at centre field .....otherwise the building itself is only 50 - 60 feet high ..and the stadium is built under ground ...

It is real, but it is taken from a different direction, so of course the relative positions of the buildings will differ. It is from Olcott NY. Looking from the southeast, from Olcott, the CN Tower is some way to the left of the other tallest buildings: Looking from the southwest, from Hamilton, it is more or less in front of them: --------------------------------- Now, compare the photo you posted, from near the beach in Hamilton: With the photo from up on the hill. Can you spot the difference? Why are all the lower buildings missing? Edit: perhaps this will make it clearer? (screenshot from video contrast-enhanced and resized to roughly match the buildings). Why is more of the skyline hidden from lower down towards the lake if the Earth is flat?

Ijust found that pic this morning ..im going to actually go myself because i want to figure this out .....toronto had the most condo starts in the world the past few years i think 180 or so a year so if the pics old ur going to be missing a lot of things......some time today i will fly down the hill to the beach ....im curious now very curious

That's good. Experimenting for yourself rather than taking people's word for it is something people should be doing more of, so thumbs up for that!

These two photos are excellent examples that perfectly illustrate the curvature of the Earth, even better than the Chicago photos. We can verify the obscuration with Google Earth. Let me make a quick video explaining it, but here's the Google Earth file with both photos overlaid from their approximate viewpoints. Note the red line is required to get the correct water surface height, and GE renders lake water surfaces at sea level.

Wait ..i thought that photo shows all of Toronto from 35 miles away ?...ill go down to the beach in one hour

No, that's the photo from the beach. It shows several hundred feet obscured. I'll upload an explanatory video, probably about 30 minutes, as I have slow internet.

Here's the Google Earth demonstration, showing that what is obscured is what is expected. Source: https://www.youtube.com/watch?v=ogzAufGmBNM

If by chance you've not gone yet. Try taking some photos from: A) as close to the water as possible, like 1 foot above it. (661 feet hidden) B) eye level while standing near the water, so 6 feet (600 feet hidden) C) back in the car park, up a bit, maybe 12 feet (551 feet hidden) You should see differences between all of these (depending on haze, zoom, etc)

I Just got back from the beach ..there is zero visible today i can walk to the water and sit on a rock so ill be around 4 feet from the water .....now I'm obsessed with getting the right pic so sometime over the weekend i will get down there again when its nice..im gonna go by bestbuy and get a camera what should i buy ? what model etc IM positive you can see all of Toronto from the beach i could see to Mississauga to the left right to the ground . its about half way but my phone took nothing you could recognize ... i will get the shot and we will see ...

You want something with 50x zoom or better (the longer zoom superzoom camera) Any of these: https://www.dpreview.com/reviews/2015-superzoom-camera-roundup P900 is ideal. Slightly older models should be fine, for example the Canon SX50 While you can just point and shoot, it's better if you use of a tripod (can be a very cheap one) and take some of the photos with a timer to minimize camera shake.

Minor tip, if you've got an iPhone, etc, then after getting a good picture then take an iPhone picture of the picture displayed on your camera (while still at the same viewpoint). That way you get a GPS record of exactly where the picture was taken, including altitude.

DJC: These three videos were taken by the same person. This is from Ryerson Park, Niagara-on-the-Lake, NY. Google Earth says that park is 251 feet above sea level, while the lake level is 246 feet above sea level (which varies of course). So the park is 5 feet above lake level. Approximate values, but close I think. Also from Ryerson Park on a different day with rougher water. Interesting that the ship is just on the horizon. There's a word for that part of the sea (or lake) that's closest to the horizon - "offing." It's important to know what the horizon is. It's the line at which your line of sight intersects the water. I'm borrowing David Ridlen's illustration from page 3 of the Earth curvature refraction experiments thread. https://www.metabunk.org/earth-curv...nts-debunking-flat-concave-earth.t6042/page-3 The horizon line is marked "C." This is a view from Queenston Heights Park. Google Earth says the elevation is about 580 feet above sea level. (Approximate, because I don't know what part of the park he was in.) That's 329 feet higher than Ryerson Park. Again approximate, but the important fact here is that it's hundreds of feet higher. Queenston Heights Park is inland and about 7 miles farther away from Toronto than Ryerson Park is, yet much more of Toronto is visible. Why? Well, why did lookouts in the age of sail go aloft to the main topgallant yard (or "crow's nest," ye lubber)? Because the higher you are, the farther you can see on a spherical earth. I think you should look at this interactive illustration. It's worth a thousand words. https://www.metabunk.org/earth-curv...g-flat-concave-earth.t6042/page-3#post-161317 Think of the "target" as Toronto, and the "camera" as you. The higher you get, the more of Toronto you can see. The relationship of Queenston Heights Park to Ryerson Park to Toronto.