Starship Asterisk*

ISS Double Transit

Explanation: Not once, but twice the International Space Station transits the Sun on consecutive orbits of planet Earth in this video frame composite. The scene was captured on August 22 from a single well-chosen location in Schmalenbeck, Germany where the ISS created intersecting shadow paths only around 7 kilometers wide. Crossing the solar disk in a second or less, the transits themselves were separated in time by about 90 minutes, corresponding to the space station's orbital period. while the large, flare-producing sunspot group below center, AR 2043, remained a comfortable 150 million kilometers away, the distance between camera and orbiting station was 656 kilometers for its first (upper) transit and 915 kilometers for the second more central transit. In remarkably sharp silhouette, the ISS is noticeably larger in angular size during the closer, first pass. Of course, tomorrow the Moon will transit the Sun. But even at well-chosen locations, its dark, central shadow just misses the Earth's surface creating a partial solar eclipse.

<< Previous APOD

This Day in APOD

Next APOD >>

[/b]

Why is the first track curved?

Elmar wrote:Why is the first track curved?

It isn’t. You’re experiencing what Neil deGrasse Tyson calls a “brain failure,” and the rest of us call an optical illusion.

I just have to say... This is awesome! Good job, Hartwig Luethen.

The link "partial solar eclipse" goes to Sunspots, not eclipse.

Mikado wrote:The link "partial solar eclipse" goes to Sunspots, not eclipse.

Here, try this one.

It reminds me of a baseball for some reason. Sort of lopsided though.

Elmar wrote:Why is the first track curved?

I "saw" the same thing. I had to put a straight edge (piece of paper) up to my screen to verify that it was a straight line.

I don't know why our brain works that way, but it does.

Forgive my preoccupation, but does anyone else see TIE fighters?
(And yes, that is a really awesome photo!)

If I'm not mistaken, the sun's angular diameter is roughly 0.5 degrees. So if the ISS crossed the disk in one second, wouldn't that imply an orbital period of about 720 seconds, or 12 minutes instead of the actual orbital period of 90 minutes? With a 90 minute orbital period, it seems like it should take the ISS more like 7 or 8 seconds to cross the sun's disk. Or am I making a wrong assumption somewhere?

I am really sorry that I am joining this thread so late that I might not get any responses to my question.

Let us suppose the upper track is close to half way between apogee and perigee, so that the rate of change of altitude is maximized.

Next let us suppose that the orientation of the apogee and perigee does not involve precession (a flawed presumption, but one which can be tolerated for a mere 90 minute experiment).

Ninety minutes after the first track was recorded, the Earth will have rotated in its day so that the sun is at a different angle.

May I assume the huge difference in distance of the ISS from the camera was the change in angle of viewing, rather than rotating under a position where the ISS itself had gained that much altitude in a mere 90 minutes?

My gosh, the mathematics for calculating where to be to capture those images is daunting!

GlenM wrote:If I'm not mistaken, the sun's angular diameter is roughly 0.5 degrees. So if the ISS crossed the disk in one second, wouldn't that imply an orbital period of about 720 seconds, or 12 minutes instead of the actual orbital period of 90 minutes? With a 90 minute orbital period, it seems like it should take the ISS more like 7 or 8 seconds to cross the sun's disk. Or am I making a wrong assumption somewhere?

If you could observe the ISS transiting the Sun from the centre of the Earth, it would take about 7 or 8 seconds.

Craig Willford wrote:I am really sorry that I am joining this thread so late that I might not get any responses to my question.

Let us suppose the upper track is close to half way between apogee and perigee, so that the rate of change of altitude is maximized.

Next let us suppose that the orientation of the apogee and perigee does not involve precession (a flawed presumption, but one which can be tolerated for a mere 90 minute experiment).

Ninety minutes after the first track was recorded, the Earth will have rotated in its day so that the sun is at a different angle.

May I assume the huge difference in distance of the ISS from the camera was the change in angle of viewing, rather than rotating under a position where the ISS itself had gained that much altitude in a mere 90 minutes?

My gosh, the mathematics for calculating where to be to capture those images is daunting!

According to http://www.heavens-above.com/IssHeight.aspx, the mean height of the ISS on Aug 22 was about 401 km, and I think its orbit has a fairly low eccentricity. The large difference in the distances to the ISS, is due primarily to different sun elevation angles, which assuming the quoted distances are correct, must have been about asin(401/656)=38° and asin(401/915)=26°, respectively. These are both a few degrees larger than what I would have calculated from the times and location provided by the photographer, but I'm assuming a flat Earth.

I decided I didn't like the flat Earth assumption in my last, so I've instead drawn a diagram to scale:

What a great photo, what a great discussion.

I wonder what the time-lapse rate of exposures was . . . .

flight of the bumble bee

First at all thanks to the positive responses here

I wonder what the time-lapse rate of exposures was . . . .starobin

The camera was a DMK 41 running at 15 FPS. Scope was a Pentax 75/500 refractor, equipped with a TAL glass frontfilter.
Several people commented on the optical illusion leading to the impression of a curved track for the 1st transit. I was in fact fooled by this and was looking for the error in my stacking technique when I first saw the result until I realised what was going on .

Greetings
Hartwig