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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Height by sextant angle
From: Peter Monta
Date: 2014 Dec 13, 03:02 -0800
From: Peter Monta
Date: 2014 Dec 13, 03:02 -0800
I tried looking at pixel intensities in an effort to get the size of the penumbra more objectively.
Here's a quick plot of the image values near (x,y)=(157,275). For each x coordinate from 152 to 165, the value of the red channel is plotted from y=250 to y=300, after raising the image to the power 2.2 to convert from JPEG to linear light.
This is a region of the shadow on the upper edge near the rear of the craft. It happens to fall on a fairly uniform dirt-colored background. It's also conveniently horizontal.
I make the slope of the shadow at midpoint to be around -0.048 intensity units per pixel; if we take the light-colored region to have intensity 0.4 (it's a little noisy) and the shadow region zero, then normalizing gets us -0.048/0.4 = -0.12. So if the illumination source were a square ("Minecraft sun") it would subtend 8.3 pixels. Since it's a circle, we multiply by (4/pi) to get a solar diameter of 10.6 pixels.
Google Earth (which has a nice 3D rendering of the building) says the image scale in the image-vertical direction is 0.31 meters per pixel, so with the 57-degree altitude, that's effectively 0.26 meters per pixel normal to the camera axis on the ground. So 10.6 pixels represents a slant distance of 295 meters or 970 feet.
Some or all of these numbers may be wrong. YMMV. :-)
Cheers,
Peter