How would you calculate Transfer windows manually?

[Cato]

Took a bit of a break from this game but decided to come back, and as I was looking at the Kerbol system map I wondered how you would go about doing this, I know there are websites that are set up to calculate windows- but I wanted to be able to do it myself.

Im guessing this is going to be fairly advanced math but nothing ventured nothing gained :)

[tourist]

Checkout Part III & IV (BASICS OF SPACE FLIGHT)
http://www.braeunig.us/space/

This was cited as the source for equations in Alexmoon's github calculator.

The math is doable though what you want is to plot two or more orbits such that you can see the type of transfer that you are looking for as something such as an intersection of these orbital potential transfer plots on that graph.

[Cato]

Originally posted by RoofCatA:

not advanced at all.

First get an idea where is the best (cheapest) interception point for the target. No math needed, just logic.
For bodies on circular orbit it doesn't matter. For bodies with eccentric orbits general rule is target Ap going up and Pe going down. If you think a while about why Minmus transfer is so little more expensive than Mun, while being on a 4 times higher orbit than Mun you will understand the reasons for that approach. Ap is the best point to increase Pe and vice versa. Interception is aligning with the body.

Then you have to find angular difference between Kerbin and the other body based on their orbital velocities and periods. Again, it is more tricky on bodies with eccentric orbits. Use approximate approach test nodes there to see how far you are.
To explain the angular stuff let me give you simplified example for Duna:
1. make optimal (prograde escape, Ap being 180° across the Sun) test node from Kerbin to Duna altitude. Check time needed. It will be close to 300 days.
2. Kerbin and Duna both move on their orbits, but with different velocities. Now all you have to find out is the distance Kerbin and Duna will travel during the time you found in (1). Or more accurately - the DIFFERENCE of the distance traveled by them both.
You could calculate the lenght of their orbit as the altitude is visible in map mode (you would have to add radius of the Sun - check info icon). Then 2 x Radius x 3.14159 gives you the length of the orbit. Orbital velocity is visible in map. So you know the distance and know the velocity. Divide one by another and get the time. Doesn't sound like advanced math to me? And because both bodies move the same direction you would have to use the difference of their velocities at some point. Like 9284.5 (Kerbin) - 7500(~average Duna) = 1784.5m/s. which means each second they get ~1.8km closer. It is not complicated math at all, just pure logic and 5th(?) grade math.


but... it actually goes much easier than that. Use predicted flight time only.

You know you will travel "300" days to intercept Duna at optimal spot. Now the only question is how much days Duna needs to travel the same angular distance (half of it's orbit or 180°) - is it more or less than your 300 days. You could do the math for Duna year or just use info. As it turns out, Duna year is 801 day 3 h 50 minutes. Half of that would be roughly 401 days. Which means Duna needs more time to get to the intercept point and thus must be closer to it (ahead of Kerbin).
300 / 401 = 3/4; from 180° => 135°
Which means Duna will travel just 135° while your ship travels 180° on solar orbit. Or put simply Duna must be 180-135=45° ahead of Kerbin when you launch your ship. Quite simple, no math beyond simplest arithmetics at all.

in fact Duna has very slightly eccentric orbit and your flight time will be somewhere around 260..290 days, but you get the idea.
Few days sooner or later are more important for faster bodies. Mostly Moho. For Duna one week late won't change much.

Also you should know you can make not just one test node, but multiple within chain when you have to deal with inclinations. To see delta v needed for other nodes after the first one, just target them with mouse.

Creating inclined ejection right at Kerbin is trickier. But still nothing you couldn't handle with a bit of logic and a few test nodes adjusted and moved around until you find the sweet spot.

For all test nodes I just use whatever is orbiting Kerbin at the time. Most often a science station at 72-72km orbit. You don't need fuel to make test nodes nor do you have to keep them.



Optimal transfer planers use more advanced math and find best spot within the next 10 orbits, compare them and give you the best one. It is nice to have feature and NASA rather does that similar with just slightly less orbits to fit in limited human time frames. But it is really not needed in your KSP. Spaceflight is easy here. You can wiki-google all math and have your high school level challenges or just have fun with the game in a simple and logic way.

Thank you :)

I have quoted this and saved Nostras source so I will try and understand this later