Hi all,
I came across a most impressive and rare photo taken from the
spacestation MIR during the solar eclipse on Aug. 11 1999:
http://solarsystem.nasa.gov/multimedia/display.cfm?IM_ID=189
The relatively well concentrated region of moon's
core shadow is nicely apparent!
Now I felt it was worth comparing the actual shadow size with the one
in Celestia as an important check!
What I did is this:
I replaced the earth texture by a white globe and just left the clouds and the
nightlights on. This way, the size of the shadow patch in Celestia can
be clearly judged. Moreover I put myself precisely into MIR's orbit
and adjusted the FOV such that the earth curvature of the original photo just
matched the one in Celestia!
Then we can compare the patch sizes with good accuracy. The result is
that Celestia's shadow is way too large and too diffuse
compared to the actual photograph.
Chris, I am afraid, there is some respective work ahead...
Bye Fridger
Celestia's solar eclipse shadow size vs. great MIR photo
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Celestia's solar eclipse shadow size vs. great MIR photo
Last edited by t00fri on 05.06.2004, 18:14, edited 1 time in total.
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Topic authort00fri
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alphap1us wrote:Hello Fridger,
i am just curious: How did you place yourself exaclty in Mir's orbit? Did you just use one of the standard Mir add-ons or did you find some data elsewhere? it looks like the shadow is off by a significant amount, though the fact that it is there at all still amazes me...
Joe
Joe,
You can see part of the MIR hardware on the right of my image above. My altitude was 391km, using the official MIR orbit data from the CVS. I rarely use add-ons...(except my own;-))
Here I reproduce the shadow patch from a much higher altitude so you can see better what it looks like in Celestia. There was no real displacement relative to the shadow, I just fiddled the view carefully in my previous 'montage' to have everything of relevance on...In other words, I displaced the photo on purpose in order not to hide the Celestia shadow patch entirely. This is fine, since all that matters is to adjust the FOV from matching the earth's apparent radius of curvature, no matter where...
Bye Fridger
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The radii of the area of totality and the penumbra are correct, I believe. But the calculation of shadow darkness as a function of the distance from the center of the eclipse is not correct. For each point in the penumbra, Celestia should compute the fraction of the solar disc that is visible. This is a fairly complicated expression (ultimately a circle-circle intersection--see http://mathworld.wolfram.com/Circle-Cir ... ction.html ), but that's not a problem since the set of eclipse shadows (for various sun to occluding disc size rations) are computed just once at initialization time.
I've known about this shortcoming of eclipse rendering for some time, but until you showed me the Mir photo, I never realized that Celestia was so far off from reality. I'd seen images of eclipses on Jupiter, and Celestia matches those reasonably well. Since Jupiter is five times further from the Sun than Earth, the apparent diameter of the solar disc in a Jovian eclipse is one fifth what it is in a terrestrial eclipse. The apparent sizes of the Galilean satellites ranges from larger than the Moon view from Earth (~36 minutes for Io) to one-third the Moon's average apparent size. So, I guess the weakness of the current approximation is more evident when the apparent size of the occluding object and the solar disc are similar.
Time for some experimentation with the penumbra function . . .
--Chris
I've known about this shortcoming of eclipse rendering for some time, but until you showed me the Mir photo, I never realized that Celestia was so far off from reality. I'd seen images of eclipses on Jupiter, and Celestia matches those reasonably well. Since Jupiter is five times further from the Sun than Earth, the apparent diameter of the solar disc in a Jovian eclipse is one fifth what it is in a terrestrial eclipse. The apparent sizes of the Galilean satellites ranges from larger than the Moon view from Earth (~36 minutes for Io) to one-third the Moon's average apparent size. So, I guess the weakness of the current approximation is more evident when the apparent size of the occluding object and the solar disc are similar.
Time for some experimentation with the penumbra function . . .
--Chris