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The Family Portrait, or Portrait of the Planets is an image of the Solar System, created from 60 individual frames, spread over a length of six metres [...]

Why six meters? Photos are usually not measured by length, especially when they come from digital sources. --Abdull (talk) 10:15, 7 December 2007 (UTC)[reply]

Agreed. So edited. Mikaey (talk) 08:09, 8 October 2008 (UTC)[reply]

Pluto?

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Pluto was not yet classified as a dwarf planet at the time these photos were taken, and yet the article makes no mention of any attempt to include it in the portrait. If anyone knows why, I think it might deserve mention in the article --Jleon (talk) 15:49, 22 April 2010 (UTC)[reply]

It was too faint to be observable. Ruslik_Zero 18:50, 22 April 2010 (UTC)[reply]
I removed the reference to Pluto as being a "dwarf planet" as that is an anachronism given that in 1990 it was considered a full planet. 70.72.215.252 (talk) 23:07, 16 June 2012 (UTC)[reply]

60 individual frames?

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With a careful count I can see 45 separate square images. 6 of planets, 5 in the arc over the sun, 4 in the area of the sun, 5 running parallel to the Jupiter-Earth-Venus line, 6 in the line at right angles to that (ending with frame S), 10 in the arc from S to U (not including already-counted S), and 8 in the arc out to frame N. And looking very closely int the large-resolution image I can make out 1 more almost overlapping with frame S. Total: 45. Can anyone explain where the other 15 frames are? Sean Martin (talk) 16:00, 2 August 2013 (UTC)[reply]

I suppose that for some targets, several frames of the same target were taken, each one using a different colored filter, as it was the case for Pale Blue Dot, for which three frames were shot. --Deeday-UK (talk) 23:25, 5 January 2015 (UTC)[reply]

Mars?

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From the phrasing, it appears that an attempt was made to get a photo of Mars, but it just wasn't visible. If this is correct, could modern (c 2014) photo enhancement techniques obtain Mars from the picture/-s taken? CFLeon (talk) 18:55, 30 September 2014 (UTC)[reply]

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