Transit of Venus 2012 user-friendly model

[Mito111]

Hi,

We've just finished for you a brand new model displaying transit as you will see it from your location:

http://www.sunaeon.com/venustransit/

of course itīs free and there are more features waiting too:
- all important event times, trajectory and sun's current position according to horizont
- for admins we are preparing the embedding utility

Enjoy

Mito and SunAeon team, creator of solarsystemscope.com and sunaeon.com

[mapguy]

Wow - informative graphics and impressive programming - thanks!

Can anyone give a simple explanation of the factors that result in the apparent motion of Venus forming an "S"-shaped path across the sky that day?

[ToSeek]

Wow - informative graphics and impressive programming - thanks!

Can anyone give a simple explanation of the factors that result in the apparent motion of Venus forming an "S"-shaped path across the sky that day?

That seems bogus to me. NASA's chart shows it as a straight line: http://eclipse.gsfc.nasa.gov/OH/transit12.html

[grapes]

Can anyone give a simple explanation of the factors that result in the apparent motion of Venus forming an "S"-shaped path across the sky that day?

I haven't looked at it, but if it is S shaped, it is surely wrong, as ToSeek says.

[mapguy]

NASA's chart shows it as a straight line: http://eclipse.gsfc.nasa.gov/OH/transit12.html

Sure enough. But our own Centaur's diagrams show it on a curve (example).

[ngc3314]

I suspect that curved path happens because that display is in azimuth-elevation coordinates, and the view of the Sun in an inertial frame (or the usual RA-dec frame) rotates significantly during the transit as the Earth rotates. In that case, the view it shows would be what you'd see in a magnified view with the zenith up.

[grapes]

I finally looked at the illustrations. I couldn't see the one in the OP (iPhone, flash, etc), but the NASA one was straight (with the ecliptic shown fixed straight across the sun) and Centaur's is curved (but not really S shaped) against the disk (with Zenith indicated--which would change relative to the sun, as ngc3314 says--and since the diagram is in Denver, after noon, the north pole of the sun would be to the right of zenith).

The whole thing happens in under seven hours, right? As Venus moves across the disk of the sun, the earth advances about a quarter of a degree in its orbit.

[Tog]

But the sun also "moves" nearly 1/3 of the way across the sky. Orion stands tall on the meridian, but lays over on it's side by the time it sets. I took the curved path to be the ALT-AZ plotting as well.

If you move to the north pole, it's nearly a straight line.

As you move the location around the map, you do get some really odd looking paths. Some do full loops, like a retrograde path, but those are in areas where the entire event is below the horizon. In some parts of South America, the path is made up of three arcs, like the path of a bouncing ball with a really big middle hop.

[grapes]

Next time I'll actually look at what is being discussed before I make pronouncements!

My apologies. 'Course, I still havent seen it yet.

[ToSeek]

I finally looked at the illustrations. I couldn't see the one in the OP (iPhone, flash, etc), but the NASA one was straight (with the ecliptic shown fixed straight across the sun) and Centaur's is curved (but not really S shaped) against the disk

That looks similar to what the OP's link is showing - the "S" part only figures in if you look at Venus's trajectory before and after the transit, which the Centaur rendition doesn't show.

[Solfe]

Nice music. I like the explorer on the top level of the site.