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Saturday, July 16, 2016

Toronto CN Tower from Olcott, NY - explained

I think that the fantastic view of Toronto and the CN Tower from Olcott, NY is one of the best examples of Earth curvature you can find. It's a beautiful view over 39.5 miles of Lake Ontario so there are ample images of the phenomena.

So it is amusing that so many Flat Earthers try to use this to 'debunk' curvature by appealing to some magical version of "perspective" - not surprisingly, one for which they cannot give any formula or explanation of how we would measure such a magical effect or how it would work.

Meanwhile Refraction and Curvature are WELL understood in the surveying and scientific worlds.

Here is what the Flat Earther expects you to accept, that ~359 meters of the CN Tower are perfectly visible and in proper proportion to the actual tower - but suddenly, below that mark, 184 remaining meters just "disappear" to nothing due to their incorrect magical "perspective".  This is simply not how perspective works - if it were going to shrink the tower it would shrink it all in the same proportion.  There is no mechanism where 184 meters of a distant object would shrink to 0 pixels while the upper 359 meters remains fully visible and in the EXACT proportion we expect.

There is simply no way to excuse this as 'Perspective'.

(image credit - click to see animation)

Unfortunately, I do not have enough details for the above image -- it appears to have been taken from more than just a few meters above the water and I don't see a dock so I cannot really calculate anything specific from this image.   But what is very clear is that we can only see the tops of the highest buildings and very significant portion of the city is missing.  The image estimates it at 184m -- which seems in the right ballpark.

Here is another view where we can clearly identify the location which gives us a better estimate for the observer elevation and distance so we can do some calculations and see if this matches what is predicted by an Oblate Spheroid model.

You can clearly see here that we see just a little bit less of the Toronto skyline under these viewing conditions.

(image credit)

First of all, you have to rid yourself of the notion that 8"/mile² has ANYTHING to do with this view.

The observer is plainly NOT at "sea-level" here, so we must account for their elevation in our equation.

To that end I wrote a post: Derivation for Height of Distant Objects Obscured by Earth Curvature

That is a very long and moderately complex post so take your time.

Once you agree that the correct equation to use to estimate how much of a distant object is expected to be obscured by the curvature of the Earth is given by:

h₁ = √[(d₀ - [√h₀ √[h₀ + 2 R]])² + R²] - R

Then you can continue here to evaluate this evidence for yourself.  If you cannot figure out the math here then you really cannot even understand the issues or the claims so why are you even trying?

It's OK not to understand the math -- it's NOT OK to then make false statements when you have been shown, in very straight-forward and fairly simple terms.

Solving The Math

So now all we have to do is make some measurements, plug them into our formula and see if everything makes sense here.

I used Google Earth Pro to find the fairly precise coordinates for the CN Tower and the docks at Olcott, NY, those values are:

CN Tower Base (from Google Earth Pro, 2016)

(image: Google Earth Pro)

Olcott, NY Dock (from Google Earth Pro, 2016)

(image: Google Earth Pro)

Gives us a distance of 63634 meters.

And we look at the elevation profile because we need to know the "low point" and make sure we have a clear line of sight.  Our low point here is 73 meters so we will use that elevation as the BASIS for our radius.

(image: Google Earth Pro)

Next we find the Radius of the Earth, I used 43.340372 latitude and using 73 m ground-level since that is our lowest point in between:

Gives radius at our ground level (actually water in this case) of 6368.183 km
Giving us ~3.5 m for our observation elevation ([75m-73m] = 2m ground + 1.5m observation height = ~3.5m elevation) assuming they are standing on the OTHER dock.

We will first calculate the height obscured ASSUMING 0% refraction as this gives us our worst case (the greatest value that would be hidden behind the curvature) we find:

[(d - (h [h + 2 R]))² + R²] - R, d=63634, h=3.5, R=6368183 = ~254.7 m obscured

And if we assume just 7% refraction we get (this means that the refraction is 7% of Earth radius):

[(d - (h [h + 2 R]))² + R²] - R, d=63634, h=3.5, R=6368183*1.07 = ~236 m obscured

So somewhere between about 236 m and 255 m of Toronto should be obscured given the Oblate Spheroid model, assuming actual refraction is somewhere between 0 and 7% (NOTES: VLBL surveyors measure it using two frequencies of laser because frequency affects refraction and by measuring the differential they can get a very good estimate of the actual refraction along that sightline - EVERY sightline has a slightly different refraction, under normal conditions it will be more severe over the water/land and decrease as you angle up).

We also see that the top of the ICE East building is MISSING from the image, at 234m tall -- so a little bit greater refraction or observer elevation from here and we would expect to start seeing the top of that building -- and that's exactly what we see in the first image where the observer is either a little bit higher or perhaps refraction was a little greater that day.

EITHER WAY we are in VERY CLOSE agreement to the Oblate Spheroid model here (we are down to measurement errors) and NOWHERE EVEN CLOSE to the Flat Earth model.  On the Flat Earth model we should be seeing much, much more of Toronto below that horizon.

The distant view of Toronto only works if that water is curved, no appeal to perspective is going to excuse it, and it MATCHES our expectations when we use the correct mathematics.

(image credit)

About CN Tower:


This soundly refutes and falsifies the Flat Earth model just on the face of it and it supports the Oblate Spheroid model to within our measurement error (get a professional surveyor to get you more precise data and we can run the math and get a more precise predicted value).

Bonus Footage

In the YouTube video "Fakeologist's Flat Earth inquiry - Port Dalhousie to Toronto, Ontario", Flat Earther Fakeologist demonstrates both curvature and that zoom doesn't bring back things over the horizon with his P900!  Thanks mate!

Of course, Flat Earthers stop their analysis at "I can see a distant city, therefore Earth is Flat" but the rest of see an observer that is likely some 40+ feet over the water looking through extremely refractive atmosphere that has stretched the city like a Giraffe's neck and we're still missing a massive amount of the lower city.  Roger Centre still missing, etc.

I would estimate that BEFORE refraction is taken into account this view would be expected to be missing only about 300 feet of the city rather than their claimed "600 feet".  We MIGHT just be catching the wings of the Rogers Centre and stretching them up so with refraction we might be hitting down around 250 feet hidden.


  1. nice try

    1. And by 'nice try' you mean you are going to post a link to a photo without ANY analysis at all and make yourself look like a moron right?

      Did you happen to notice that 49 kilometers is NOT the same as 64 kilometers? Did you check the OBSERVERS elevation as that photo is from very high over the water? Did you fail to notice that you STILL cannot see the bottoms of the buildings?

      The actual skyline of Toronto, also shown IN MY POST already:

      Note how in your photo you can ONLY barely see the white dome on Rogers Centre in your photo? About 60 meters of the bottom part is missing in your picture. Plus another 10 meters from the elevation differences from the two points and you are missing the EXACT amount we would expect from Earth curvature at this distance. Imagine that.

      Na... you would rather assume nonsense than do the work.

      So yeah, nice try.


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