Where will you be..., When Phoenix lands |
Where will you be..., When Phoenix lands |
May 21 2008, 06:46 PM
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#1
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Member Group: Members Posts: 428 Joined: 21-August 06 From: Northern Virginia Member No.: 1062 |
Okay, so there's only a few days till the landing, and I'm just curious, where will you all be? Just trying to keep a similar sounding thread to it's original purpose, that's all.
I am planning as of now to be at the LPL event at the University of Arizona. It should be interesting. I can only be there for about the hour around the landing time, but that's alright. |
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May 24 2008, 12:47 AM
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#2
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Member Group: Members Posts: 340 Joined: 11-April 08 From: Sydney, Australia Member No.: 4093 |
I haven't heard anyone mention NASA's solar system simulator. You can configure it to show Mars from the viewpoint of Phoenix. The view of Mars as seen from Phoenix gets very blurry once you hit around 23:00 UTC on 25 May. Remember that the Solar System Simulator uses spacecraft event time, and please do not use it beyond 23:32:00 on landing day as Phoenix is shown to bounce off (or fly through Mars) and back into outer space! This is at 23:45 spacecraft event time: http://space.jpl.nasa.gov/cgi-bin/wspace?t...=1&showsc=1 The current angular diameter of Mars would also be a nice addition to dmuller's simulation. Unfortunately it's too late now to add, but I'll gladly include it for other missions once I get around coding them. BTW, what's the formula? With my monitor/seating distance, the 20 degree field of view in this image matches what I actually see, so this is how big Mars actually appears to Phoenix! I always wondered what settings to use to make it appear as seen by someone sitting on Phoenix. My knowledge of optics is really zero, I might just as well be theoretically blind Daniel -------------------- |
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May 24 2008, 10:07 PM
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#3
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Senior Member Group: Members Posts: 4252 Joined: 17-January 05 Member No.: 152 |
Unfortunately it's too late now to add, but I'll gladly include it for other missions once I get around coding them. BTW, what's the formula? If r is the radius of Mars, and d our current distance to the centre of Mars (not the surface), then the full disc of Mars will subtend an angle 2*arcsin(r/d). Here arcsin is the inverse sine, also written sin^(-1). Doing the calculation with the current distance, I get that Mars subtends 1.5 degrees right now. That agrees with the solar system simulator: The disc will rapidly increase in size now. Fasten your seatbelts!! |
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May 24 2008, 11:39 PM
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#4
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Member Group: Members Posts: 340 Joined: 11-April 08 From: Sydney, Australia Member No.: 4093 |
If r is the radius of Mars, and d our current distance to the centre of Mars (not the surface), then the full disc of Mars will subtend an angle 2*arcsin(r/d). Here arcsin is the inverse sine, also written sin^(-1). Arrghh silly me ... that's just first grade trigonometry, now isnt it. Brain to mission control - need more coffee! Thanks Fred. Daniel -------------------- |
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