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It said one spacecraft was active, one was not, but it didn't explicitly say which was active? I was curious if they'll be able to reconnect with the Indian spa
by phyller 10y ago
It said one spacecraft was active, one was not, but it didn't explicitly say which was active? I was curious if they'll be able to reconnect with the Indian spacecraft.
I assume that the LRO is active, since they mentioned people that were still on the team, so they must have just been using it for target practice.
- woliveirajr 10y ago> "Finding LRO was relatively easy, as we were working with the mission's navigators and had precise orbit data where it was located." > The Chandrayaan-1 was more of a challenge because the last contact with the spacecraft was in August 2009. I assume that LRO was active, as you said, because they knew where it was located. Chandrayaan-1 lost contact, so was the inactive one.
- dwc 10y agoYes, LRO is still active. Also, you too can know where it is right now[1]! 1. http://lroc.sese.asu.edu/about/whereislro http://lroc.sese.asu.edu/about/whereislro
- ge96 10y agoIs that on the ground or is that an orbit map? It looks like that cliche 'bell curve flattened earth shadow weather' looking thing in space movies. Also by context seems like its orbit. Also its nuts interfacing hardware with software like that, like a button or sensor attached to a web server, cool stuff.
- dwc 10y agoIt's an orbit map (showing nadir of the orbiter).
- ge96 10y agoI see thanks for the clarification. I guess I don't get why it would be an S shape even if it's a perfect-circle orbit or an ellipse flattened out (side view). Yeah I'll have to read up on that. edit: not sure if this is related, I remember doing this thing when I was a kid, you'd spin a globe and trace a pencil straight down (a lonitude line) and it spiral out (the drawn line). To simulate Coriolis effect edit: Oh I see depends on what it's aligned to, (equator) to create the sinusoidal look... reminds me in calc or engineering creating diagrams/charts (maybe beams and moment/force diagrams) ahhh... sucks when you don't use something and you forget. In calc you did something with max/mins and this determined the concave/convex of a curve on a graph. Ahhh rambling sorry.
- planteen 10y agoThere's also lots of distortion due to the equirectangular projection. When you are at +/- 90 degrees, 1 pixel is being stretched across the entire width. It would look more natural if it was mapped on a sphere. Or if the 2D image dropped off width by cosine of latitude.