Explained: The View of Toronto from Hamilton [600+ feet up]

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:

image.jpg

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:



Hi Guys,

Tell you what I find interesting about the above GIF image,

I measured Olcott to Toronto on Google Maps, 65km or 40mls

Now according to curvature maths,

ie H(c) = 8" x D^2 (mls),

8" x 40mls x 40mls = 12,800 ft / 12" = 1,070 Feet or 325m (approx)

so at 40mls away, the curve should hide // 1,070 Feet or approx 325m of the CN Tower,

Ok good .... we're all on the same page ....

Now what I find surprising,

Is that the First Bank Tower (second highest building) after the CN Tower is (according to Wiki) 290m tall minus spires,

and it is clearly visible?

In fact, strangely enough, I'd say near on a good half is clearly visible,

When, according to curvature math, it should all be under the horizon?,

so, but, and

what about the parts that are obscured ie the parts of the CN Tower, First Bank Tower that we cannot see...

Well that's explained by FE's as the effects of waves & atmospheric density on perspective at sea level over distance,

countered by the RE's as light refraction,

makes you wonder where the truth lies doesn't it :)
 
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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.
20160804-103340-lbquq.jpg

And no curvature of the horizon is visible below 60,000.

Why is that?

I'm not being facetious,

but according to Rowbotham Experiment No 7,

Example Here: http://www.sacred-texts.com/earth/za/za12.htm

You should quite clearly be able to detect a curve?
 
ie H(c) = 8" x D^2 (mls),

8" x 40mls x 40mls = 12,800 ft / 12" = 1,070 Feet or 325m (approx)

so at 40mls away, the curve should hide // 1,070 Feet or approx 325m of the CN Tower,

Ok good .... we're all on the same page ....

No, the 8" per mile square is only accurate if your eye is at water level. It's the "drop" amount, not the "hidden" amount. You need to account for the viewer height. See:
https://www.metabunk.org/curve/?d=40&h=20&r=3959&u=i&a=n&fd=60&fp=3264
 
Why is that?

I'm not being facetious,

but according to Rowbotham Experiment No 7,

Example Here: http://www.sacred-texts.com/earth/za/za12.htm

You should quite clearly be able to detect a curve?

You can detect it, it's just hard to see with the naked eye. When you get higher than a few hundred feet the horizon is to indistinct to see the curve until you get really high. However you can detect it at just a few hundred feet with a good camera and a level. See:
https://www.metabunk.org/measuring-the-curvature-of-the-horizon-with-a-level.t7832/
 
Hi Guys,

Tell you what I find interesting about the above GIF image,

I measured Olcott to Toronto on Google Maps, 65km or 40mls

Now according to curvature maths,

ie H(c) = 8" x D^2 (mls),

8" x 40mls x 40mls = 12,800 ft / 12" = 1,070 Feet or 325m (approx)

so at 40mls away, the curve should hide // 1,070 Feet or approx 325m of the CN Tower,

Ok good .... we're all on the same page ....

Now what I find surprising,

Is that the First Bank Tower (second highest building) after the CN Tower is (according to Wiki) 290m tall minus spires,

and it is clearly visible?

In fact, strangely enough, I'd say near on a good half is clearly visible,

When, according to curvature math, it should all be under the horizon?
Ditto to Mick's answer.

Your math is spot on but what you also need to do is add in the height of the viewer above the lake, and also to look at "obscured amount" rather than "drop".

In this case, predicted hidden amount is something like 600 feet.
 
Why is that?

I'm not being facetious,

but according to Rowbotham Experiment No 7,

Example Here: http://www.sacred-texts.com/earth/za/za12.htm

You should quite clearly be able to detect a curve?
The curvature of the horizon (left to right) is an entirely separate thing from the curvature of the Earth. The horizon is flat.

Let me repeat that: the horizon is FLAT even though the Earth is round. When I say "flat" I mean that the horizon is the same vertical distance below you, all the way round. No part of it is higher than any other part (assuming a clear sea-level horizon in all directions, of course).

That's because the horizon represents the set of points where straight lines from your eyes intersect with the globe of the Earth. If you project straight lines from your eyes down to the horizon all the way round, through 360 degrees, you will get a shallow cone shape with your eyeballs at the apex. And if you rest a cone over a sphere, the points at which they intersect form a circle.

So, the horizon is flat, but it is a flat circle. And the reason it only looks curved from higher up is the same reason that if you were standing in a huge flat circular field a few miles across, you would not be able to tell it was a circle unless you were able to hover some distance above the ground: you are looking at it more or less edge-on. It doesn't curve "up and down", it curves "towards you", so the higher you go, the more obvious the curve.
 
I ust 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 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 ...
Did you get the shot?
 
I read this with great anticipation that the OP would produce a photo or two. Quite a disappointment it never materialized.
For some people, alas, their religious vows preclude them from coming too close to evidence that clashes with their beliefs. Don't understand it meself but one must have respect for these things.
 
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