|Associated Press, 2006|
The next image is also of Mt. Rainer but taken from about 130 miles / 209215 meters away over the Strait of Juan de Fuca by djboptics. The exact location isn't known but I estimate somewhere around Cadboro Bay based on the view of the mountain (circa 48.454580°N 123.286889°W).
|Mt. Rainer, by @djboptics|
|Mashup done by @NbZh_|
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 θ.
Mt. Rainer 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. Rainer itself, this is because Mt. Rainer 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. Our equation is:
AngularSize = 2 * arctan( ObjectSize/2 / Distance )
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. This inner part ObjectSize/2 / Distance just gives us a Slope and then 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*θ
a = R*2*sin(θ/2)
Solve the first equation for θ = s/R and substitute that in:
a = R*2*sin(s/R/2)
That gives us:
a = 6366473.847*2*sin(209215/6366473.847/2) = 209205.586 meters = 129.994 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 * arctan ( (RainerHeight+EllipsoidalDelta-Curvature)/2 / Distance)
We do this twice, since we have views from two different distances.
2 * arctan ( (4392.168+279.95-615.37)/2 / 88513.9) = 1.2°
Cadboro Bay (approx):
2 * arctan ( (4392.168+595.38-3438.53)/2 / 209215) = 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 NbZh_ for the mashup and djboptics for the original image. 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.