At the limb, your line of sight is nearly tangent to the sun's surface. Among other things, that means each pixel on or near the limb represents a much larger area of the sun's surface than pixels nearer the center of the image.
Ok Mr Spock, give us a number per pixel.
Each pixel at the extreme edge covers about 60 times the area covered by a pixel at the center.
Ok Mr. Spock, (FYI, I personally love Spock.) I've looked at the images again carefully and although I personally see no signs of more than perhaps a single pixel of distortion due to the jpg compression, I'll give you two pixels in each direction, both at the limb line and at the photosphere/chromosphere boundary (which looks pretty darn smooth as far as I can tell).
As we have seen, what you personally can see is irrelevant.
The factor-of-60 distortion has nothing to do with JPEG compression. The factor of 60 results from projecting a 3-dimensional sphere onto a 2-dimensional image.
Consider the equations for a sphere of radius 2000 pixel widths, centered at the origin, observed by a camera situated along the z-axis at infinity:
[latex]
\begin{eqnarray}
x^2 + y^2 + z^2 & = & 2000^2 \\
z(x, y) & = & \sqrt{2000^2 - x^2 - y^2} \\
z(0,0) & = & \sqrt{2000^2 - 0^2 - 0^2} = 2000 \\
z(1,0) & = & \sqrt{2000^2 - 1^2 - 0^2} \doteq 1999.99975 \\
z(0,1) & = & \sqrt{2000^2 - 0^2 - 1^2} \doteq 1999.99975 \\
z(1,1) & = & \sqrt{2000^2 - 1^2 - 1^2} \doteq 1999.9995 \\
z(1999,0) & = & \sqrt{2000^2 - 1999^2 - 0^2} \doteq 63.24 \\
z(2000,0) & = & \sqrt{2000^2 - 2000^2 - 0^2} = 0 \\
z(1999,1) & = & \sqrt{2000^2 - 1999^2 - 1^2} \doteq 63.23 \\
z(1999.99974, 1) & = & \sqrt{2000^2 - 1999.99974^2 - 1^2} \doteq 0.2
\end{eqnarray}
[/latex]
Near the center (where x and y are both near 0), the value of z is almost the same at every corner of the pixel, so the area of the sphere covered by a central pixel is almost exactly 1 (in units of square pixels).
At the limb (where x is near 2000), the value of z varies by more than 63, so the area of the sphere covered by a pixel at the limb is more than 63 (in units of square pixels).
The above is analytic geometry, as taught in high school. To have any hope of interpreting the limb of a solar image, you're going to have to understand it.
) and predicts absolutley nothing just makes it a joke. See the over 50 questions that Michael Mozina is incapable of answering.
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