To answer this I grabbed one of the DISCOVR/EPIC frames epic_1b_20170726023808_02 and I found that the Earth was approximately 1506 pixels wide -- this is the semi-major (a) axis:

Next, we need to calculate how tall we would expect the Earth Ellipsoid to be IF it wasn't a perfect circle. To do this we get the model value from what is called WGS-84 (World Geodetic System from 1984), which is the current measured values for Earth. It defines the measured semi-major axis (a) and something called 'flattening', using this you can calculate the semi-minor axis (b) using a simple equation that defines this relationship for an ellipse:

ƒ = (a − b)/a

Therefore, b = a - aƒ

From WGS-84 we have:

1/ƒ = 298.257223563

Therefore, ƒ = 1/298.257223563

So given our semi-major axis (a) is 1506 pixels wide we would expect our semi-minor axis (b) to be approximately:

b = 1506 - (1506 × (1/298.257223563)) ~ 1501 pixels

And that is what we find, to within 1 pixel, in the image from DISCOVR. Since the edges are fuzzy and the atmosphere isn't helping so we are within our margin of error. But this also clearly is not a perfect sphere, we're 4 pixels from it being a perfect sphere.

We can also estimate that since the Earth equatorial radius is 6378137.0 meters radius that we're seeing about (6378137.0*2)/1506 = 8470.3 meters PER pixel here (or 5 1/4 miles per pixel).

So this image comports extremely well with what we expect from an ellipsoidal Earth.

Excellent work! I am now awaiting delicious tears from flat earthers to harvest.

ReplyDeleteGood work, once again. It boggles the mind how Flerfers have let themselves be deceived.

ReplyDelete