Flat Earth Follies: The Incredible Shrinking Mountain

Here is a picture of the majestic Mt. Rainier which rises 14,410 feet above the nearby Pacific ocean, taken from Lake Washington, about 55 miles / 88513.9 meters away (circa 47.604669°N 122.235966°W)

Associated Press, 2006

The next image is also of Mt. Rainier but taken from about 130 miles / 209215 meters away over the Strait of Juan de Fuca by djboptics from Cadboro Bay (circa 48.454580°N 123.286889°W).

Mt. Rainier, by @djboptics

By scaling the images and matching them up as best we can (the Cadboro Bay image is about 9° further counter-clockwise around Rainier than the Lake Washington image) we can see that a substantial fraction of the mountain is missing! Is the alignment exactly perfect? No -- but the major features are well aligned vertically and horizontally. And this matches the scale expected from the angular size calculations below.

Mount Rainier viewed from 55 miles vs 130 miles - scaled overlay

It is very plain to see the effect of Earth curvature here on the apparent height of the mountain.

You don't need any math or training to see that half the mountain has gone missing and it's not from the effects of perspective or else the top of the mountain would be smaller in proportion as well.

But let's do a little math anyway and see if we can make sense of these observations!

To the Math!

In order to size up these two views of the mountain I'll need to know a few things.

First, what does MSL mean when we say the mountain is 14,410 feet "Above Mean Sea Level"? And what is our basis for MSL at each location. Mean Sea Level is simply the elevation which we believe a static ocean would form by carefully removing the effects of tides, winds, and other external factors. This is the elevation which we agree to call zero, and since common elevations are given in relation to this value we need to understand what this zero value means before we use it improperly.

The basis we're going to start with here is the Earth Ellipsoid model as defined by WGS84. The WGS 84 datum surface is an oblate spheroid (ellipsoid) with major equatorial radius a = 6378137 m and flattening f = 1/298.257223563. From these two values we can get the ellipsoidal radius of curvature for each of our latitude points, this is where our zero value lies. When we are looking over larger distances we have to take into account the fact that our zero point is NOT uniform over that distance. Indeed, we will find there is a significant change in the zero elevation over this distance, so this ellipsoidal shape must be taken into account to get accurate measurements of distant objects.

We can use the equation below to find our approx radius of curvature given our latitude \( \theta \).
\[ a(1 - f + sin^2 \theta) \]
Mt. Rainier is located at 46.852305°N which gives a radius of 6366753.797 m.
Lake Washington view was located around 47.604669°N which gives us a radius of 6366473.847 m.
And our last observation was around 48.454580°N which gives us a radius of 6366158.416 m.

So the radius at Lake Washington was 279.95 meters lower than at Mt. Rainier itself, this is because Mt. Rainier lies South of our location and the Earth is getting wider and wider as we move towards the Equator. And the radius at our Northern-most point was 595.38 meters lower.

So we will need to take these into account when we estimate our view of the mountain.

The next equation that is needed is to find the angular size of a distant object. This is pretty straight-forward but might seem a little confusing at first - this is based on finding the angle of a right-triangle since we know (roughly) the distance and the height of the distant object. The equation for angular size is:

\[ \text{AngularSize} = 2 \arctan \left( \frac{1}{2} \frac{\text{ObjectSize}}{\text{Distance}} \right) \]
So we take one-half the size of the object and divide that by the distance, and then double that one-half angle result to get our full angle. The inner part \(\frac{ObjectSize}{Distance}\) just gives us a Slope and then the \(\arctan\) function converts that into an angle. That's all that is going on. For example, a slope of 1 equals 45°.

Wait, we're on a Globe and the distance is the Great Circle Distance

One possible challenge here is that our ground distance is not over a flat surface. Since we're only talking about 130 miles here let's take a quick estimate based on a circle using the half-sin rule, where R is the radius at our midpoint, arc length s is our 130 miles and we want to find a, chord length:

\[ s = R \times \theta \label{l1} \] \[ a = R \times 2 \times \sin \left(\frac{\theta}{2}\right) \]
Solve equation \(\eqref{l1}\) for \(\theta = \frac{s}{R}\) and substitute that in:

\[ a = R \times 2 \times \sin \left(\frac{1}{2} \frac{s}{R}\right) \]
That gives us:

\[ a = 6366473.847 \times 2 \times \sin \left(\frac{1}{2}\frac{209215}{6366473.847}\right) = 209205.586\,\text{meters} = 129.994\,\text{miles} \]
Well, that was pointless. At least we know that 130 miles is a VERY TINY chord on the total curvature of the Earth and our estimated location is probably wrong by more than .006 miles so we can say that our measurement error here is greater than the actual difference so we can ignore this. However, as you move further away this would increasingly become more significant.

How tall should 14,410' / 4392.168m appear to us?

Ok, so we can now estimate the total angular size of the mountain, let's work in meters, and we will add in our Ellipsoidal delta (because that raises the mountain up a bit) and subtract out the apparent elevation shift expected from Earth's curvature (because that would be lowering the mountain over our horizon). Curvature is approximately 615.37 m and 3438.53 m respectively for these distances.

\[ 2 \times \arctan \left( \frac{1}{2} \frac{\text{RainierHeight}+\text{EllipsoidalDelta}-\text{Curvature}}{\text{Distance}} \right) \]
We do this twice, since we have views from two different distances.

Lake Washington:
\[ 2 \times \arctan \left( \frac{1}{2} \frac{4392.168+279.95-615.37}{88513.9} \right) = 1.2° \]
Cadboro Bay (approx):
\[ 2 \times \arctan \left( \frac{1}{2} \frac{4392.168+595.38-3438.53}{209215} \right) = 0.42° \]
These are pretty small angles so both images appear to have been taken a long focal length (zoomed in) which gives you a fairly narrow field of view. For comparison, an 800mm lens in 35mm format gives about 1.7° Field of View in the vertical so we seem to be in the right ballpark.

The ratio between these two is ~2.82

Looking at our two pictures we find the height of the mountain in pixels in each one. This is difficult to do because we can't see the horizon so I've just made a rough swag (and compared to some other photos with wider views), but we're just doing a rough estimate here (there is only so much we can tell without knowing all the exact details of the observers height, the properties of the lens, and other variables anyway). But this gives us:

Lake Washington pixels: ~425

Cadboro Bay pixels (scaled to match Lake Washington Image): ~148

452/179 gives us a ratio of ~2.87

That's pretty good agreement with observation. At ~418 pixels it would be the same ratio. I do think maybe the horizon could be higher up than I marked it in the below image:

Anyway, thanks to djboptics for the original image and NbZh_ for the original mashup. I just wanted to take it a step further and see if these views aligned well with the Earth model and they seem to align pretty well.

UPDATE

Walter Bislin has this amazing horizon view rendering tool that includes a more complex model of refraction as well so I've loaded up this mountain in that tool as well (NOTE: only the mountain distance and height of main peak is modeled to scale here-- the second peak is NOT to Rainier scale, it's just part of Walter's generic 'mountain' graphic.

Link to Walter Bislin Tool

The relevant settings here are:

Model: Globe+FE, 10m observation height (you can vary this all you want - it doesn't fix the Flat Earth mountain), 400mm focal length, Object 1 is a mountain 209,215m away and 4,392m high, Object 2 is a "bridge" 63,000m away and 230m high (simulating the highest terrain between observer and Mt Rainier).

The "bridge" is just to show where we should see the distant 230m hills fall against the mountain. If you add just a smidge of Refraction to the view the hills and Rainier both rise slightly giving almost a perfect match to the photo from Cadboro Bay.

I've also left refraction as default because the effects of refraction in this view appear to be minimal, at least for our purposes. You can play with the refraction all you want, no amount of refraction is going to hide 2/3rds of a mountain behind a little tiny hill.

Another common Flat Earth lie (it's a lie because I give the positions and they pretend like they debunked it without even bothering to check the views -- that's a LIE because they know they didn't check anything) I get about this image is that the trees in the view from Lake Washington are the same trees blocking our view. Well no, that's impossible because the view of the mountain from Cadboro Bay is off to the right so they could not appear under Rainier from this position -- you're actually looking over all of Seattle in that view but you can't see any of it -- not even the 295 meter top of Columbia Center which should be well above the 230 meter high hills.

This just goes to show how desperate they are to throw up any lie to protect themselves from the facts that they don't even check anything -- they just vomit up the next lie.