Talk:Orbit darkness time

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New section: It might be useful to add a table.

It might be useful to add a table for all the major celestial bodies in the game. Not everyone feels like doing the math themselves. (They still have to do the math to determine how much power the craft will consume in that time, but that is a lot simpler.) --Ruedii (talk) 04:07, 20 September 2013 (CDT)

What kind table do you mean? I only skimmed through the formulas but it appears, that the ODT depends on the orbit of the satellite itself. The only properties you need from the celestial body are the radius and gravitational parameter. And Kerbol System/Table have it for you (okay µ not directly). Of course you could use some example orbits, but which one? — xZise [talk] 04:59, 20 September 2013 (CDT)
Okay nevermind what I said: I added a table with an orbit of 20% the radius of the body. I don't know if this is any use. With a circular orbit the formula becomes somewhat handy. — xZise [talk] 12:58, 21 September 2013 (CDT)

Possibilities to improve the page

Okay I don't want to screw you over on that article, but there are several things I would change:

  • Remove the upper case from words like “apoapsis”. English is not my native tongue but afaik only names and the first word of a sentence are begin with a upper cased letter. Wikipedia handles it like that too. This would include moving the article to Orbit darkness time.
  • Redo the image as a vector graphic: Scalable and without these jpg artefacts.
  • Instead of only defining the formula for each variable (like ) describe what it does: “, the apoapsis measured from the body's center”. Because some values like eccentricity may be given by the orbit you choose so you don't have to calculate them.
  • Make this page either a tutorial (so like Tutorial:Advanced Rocket Design) or as an article (like Specific impulse). Currently it looks like both: I personally think only in tutorials the reader should be addressed directly via “you” and it is in the Category:Tutorials. On the other hand it looks like Specific impulse or thrust-to-weight ratio.

Now this is only my personal opinion and especially if this is a tutorial it's your choice to do with it whatever you want. — xZise [talk] 04:59, 20 September 2013 (CDT)

Okay I uploaded File:Orbit darkness.svg which is basically the jpeg file vectorized. I didn't add the chart around it as the unit is missing. I also didn't add the time, because it isn't clear which planet or moon this is. Also the new formula looks more clearer and you now define what the variables in the formula mean, nice work. — xZise [talk] 07:39, 21 September 2013 (CDT)

Changes made

Thanks xZise for the suggestions and helping fix this page up. I am just learning Wiki markup and wouldn't have come up with a table that elegant on my own. Anyway, I made a few changes and hopefully the page is getting better. Can we move it to "Orbital darkness time" and make it just a page (since it isn't really a tutorial anyway). I just wanted it to be easily located on a page if people didn't know what to search for. Also, I just read about the capitalization format so didn't realize I named the page wrong until now, I'm just too used to capitalizing almost everything in a title.

Also, what is the length of the shadow? It sounds like at some distance from the body you can't be in its shadow anymore? I'm not sure how KSP calculates the shadow, but it seems like if it's a simple point source, the shadow edges should diverge behind the body. — skaterzero807 [talk] 6:35, 27 September 2013 (EDT)

I have to test it with the shadow, but the length determines at which distance from the other planet/moon, the Sun apparent size is larger that the planet's/moon's apparent size. If your orbit is larger than the shadow of the planet/moon you should (in theory) have always power supply. But I'm not sure how KSP does it (if it even calculate that the shadow gets narrower?). I'll have to do some tests. — xZise [talk] 04:09, 28 September 2013 (CDT)

I did some testing (in the demo version 0.18) and the sun is blocked even outside the shadow zone. You can see the sun model still, but the game does not render the light beams. This leads me to believe the light calculations are done using the center of the sun as the source, but I can't prove definitively that solar panels would not work in this case. I will be buying the full version soon (once my qualifier exams are done, ugh) and should be able to test it more conclusively. — skaterzero807 [talk] 6:28, 3 October 2013 (EDT)

Yep I made a similar discovery and added some of the new knowledge to light. I plan to do additional test to determine if a craft in all three shadowtimes (umbra, penumbra and antumbra) get eclipsed. This of course raises the question if a shadow extend outside the SOI, because it does at least for the Mun. But as the antumbra is as wide as Kerbol at about twice the distance from the planet/moon as the distance of the planet/moon from Kerbol, shouldn't be there more eclipses/transits from planets affecting the solar panels? So my current guess is, that shadows don't leave the planetary SOI or that the body need a certain percentage coverage of Kerbol. — xZise [talk] 02:00, 4 October 2013 (CDT)
I did a quick test with the Mun, and at an altitude of 35,000 km from Kerbin the Mun was able to eclipse my solar panels (KSP even tells you what body is eclipsing the panels, kinda neat). If the Mun's shadow length is approx 10,000 km, I would have definitely been outside this shadow, which made sense because I could still see the sun model, just no rendered light beams. So I guess for determining shadow or not, the body only has to cover the center of the sun. — skaterzero807 [talk] 13:34, 13 October 2013 (EDT)
But the question is, how much does the body need to cover the sun? Afaik all celestial bodies are “there” at all times: You can see Jool from Kerbin when you know where to look. So does a body only need to be in between the sun and your craft? If so, the cones where this is possible must be huge, especially as the penumbra is expanding after the object. — xZise [talk] 13:49, 13 October 2013 (CDT)
Based on what I saw, only the center of the sun needs to be covered by a body so you would be right that the shadows form truncated cones and expand. However, due to the very small ratio of planet diameter to SMA, the cones expand only very little. For example, the edges of Jool's shadow have an angle between them approximately ~0.01 degrees (atan(2*R/SMA)). This means there should be no altitude where the shadow stops, I'll have to try orbiting the sun in the plane of Eve and see if I can get Eve to block it. — skaterzero807 [talk] 14:39, 15 October 2013 (EDT)
So I just checked and it is possible to have Eve block the sun even when orbiting at Kerbin's altitude, see the screencap here. My guess is that this means KSP just checks if a body is covering the center of the sun and uses that to calculate light/shadow state. — skaterzero807 [talk] 15:21, 15 October 2013 (EDT)
A great :), I was doing something similar but with Kerbin and being slightly outside of Kerbin's SOI. But do you know when the eclipse begun? As soon as the planet was in front of the sun? Or only when it is covering the center? And btw, your signature looks crippled, there is only a link to your talk page. Is that intended? — xZise [talk] 15:03, 15 October 2013 (CDT)
I'm not sure when/how it eclipsed since the model is obscured by the sun's ray effect, but my logical guess is that for simplicity the body just needs to cover the center point of the sun. This is what I was able to observe at shorter distances with the Mun. Also, I just have been copying the wrong signature for a while. — skaterzero807 [talk] 22:36 October 17 2013 (EDT)

What about eccentric polar orbits?

The calculation on the page seems to assume that the eclipse occurs at the apoapsis of an eccentric orbit, which would be the worst case for an equatorial orbit. But what about a polar orbit with the periapsis directly over one of the poles? The worst case eclipse would occur just after (or just before) periapsis, when the satellite was moving very quickly, so it doesn't seem likely that it would be very long at all. Say you had a 70kmx50,000km orbit with the periapsis over the south pole and the apoapsis over the north pole. Wouldn't you spend less time in the shadow than you would in a 70x70 orbit, since your orbital velocity would be much higher, but the distance passing through the shadow wouldn't be much longer? Anyone know how to calculate the time in shadow for this sort of situation?

--Khearn (talk) 23:10, 2 March 2015 (CST)

This is absolutely correct. In more precise terms it assumes that the argument of periapsis is 0 degrees. The more general case is.... considerably messier. There are some simplifications you can do in the other useful case, where the argument of periapsis is 90 degrees, but you still lose all your symmetry. I just went ahead and made a spreadsheet to do the general calculation, you can play with it here if you want. I might do a write up explaining it in the future. But again, it's messy. --Hamsterjam (talk) 13:41, 1 June 2016 (UTC)