Fraction of ISS time spent flying over each country or ocean_

[gdelottle]

Do you know of any estimate of the average time spent over each country or ocean by the ISS?

I don't believe Rogozin's rantings in the slightest. But his nonsense got me thinking out of curiosity about what country/ocean the ISS flies over most of the time. I guess, oceans apart (and in particular the Pacific, where it probably spends ˜50% of the time), the US, or maybe Brazil, but after that? And in which fraction? Curiously enough, I couldn't find any answer online. It would be interesting to know.

[K-P]

Look at the world map.
Look all the land areas between 51,6 degrees north and 51,6 degrees south.
There's your answer.

And yes, it will fly over Russia too.
And China.

[gdelottle]

I know, but this doesn't give a quantitative answer to my question. And I do not care about Russia or any other country in particular.

[IANARS]

https://twitter.com/planet4589/status/1501315098580770822?s=21

Saw this cross by a couple days ago on Twitter and found the percentages interesting myself. Hope it helps with what you’re looking for.

[gdelottle]

[snip]

Saw this cross by a couple days ago on Twitter and found the percentages interesting myself. Hope it helps with what you’re looking for.

Interesting. And counterintuitive. I was actually expecting the "latitude effect" dominates the result, but not up to this level.

[jcm]

[snip]

Saw this cross by a couple days ago on Twitter and found the percentages interesting myself. Hope it helps with what you’re looking for.

Interesting. And counterintuitive. I was actually expecting the "latitude effect" dominates the result, but not up to this level.

I screwed up, the above numbers are wrong.
New numbers :

Ocean/seas  72.8%
China 2.5%
USA   2.3%
Canada 2.1%
Russia 1.9%
Australia 1.5%
Kazakhstan 1.4%
Brazil 1.4%
Argentina 0.75%
Mongolia 0.74%
India 0.58 %
Algeria 0.47%
Ukraine 0.42%

[gdelottle]

Meanwhile, I computed the probability density per deg vs. latitude, and the result is that near 51.6 deg the PD is about 20 times more than at the equator. Does it match your results?

[gdelottle]

Meanwhile, I computed the probability density per deg vs. latitude, and the result is that near 51.6 deg the PD is about 20 times more than at the equator. Does it match your results?

Actually I also have to say I was wrong. The formula I used looks correct, but that ratio depends strictly by the way you compute it. For example: if one compute it at exactly the inclination value the formula diverges, but this is only because one is multiplying for a discreet delta-angle rather than an infinitesimal one.

So, if one calculates the ratio between the time spent near the inclination angle and the equator, the results will strictly depend from the actual delta lat value considered and that "20" doesn't make very much sense.

The formula I derived, by the way, is:

PD(lat)= cos(lat) / sqrt( sin(incl)ˆ2 - sin(lat)ˆ2 )

where:
PD = probability density al deg
lat = latitude
incl = orbit inclination

note the divergence for lat=incl, but PD(lat) * delta-lat where delta-lat is any latitude interval is still working well. In other words, while PD(incl) diverges, the area subtended by the curve is still limited. Note also that the formula must be renormalized to work well (total probability for each orbit must be equal to 1).

[jcm]

Meanwhile, I computed the probability density per deg vs. latitude, and the result is that near 51.6 deg the PD is about 20 times more than at the equator. Does it match your results?

Actually I also have to say I was wrong. The formula I used looks correct, but that ratio depends strictly by the way you compute it. For example: if one compute it at exactly the inclination value the formula diverges, but this is only because one is multiplying for a discreet delta-angle rather than an infinitesimal one.

So, if one calculates the ratio between the time spent near the inclination angle and the equator, the results will strictly depend from the actual delta lat value considered and that "20" doesn't make very much sense.

The formula I derived, by the way, is:

PD(lat)= cos(lat) / sqrt( sin(incl)ˆ2 - sin(lat)ˆ2 )

where:
PD = probability density al deg
lat = latitude
incl = orbit inclination

note the divergence for lat=incl, but PD(lat) * delta-lat where delta-lat is any latitude interval is still working well. In other words, while PD(incl) diverges, the area subtended by the curve is still limited. Note also that the formula must be renormalized to work well (total probability for each orbit must be equal to 1).

That's right, you have to select a delta-lat.