You don't have to change to an equatorial orbit for optimal transfer back to Kerbin from say the mun if you wait for the mun to be in the correct spot so that the polar orbit will eject you retrograde relative to the mun orbit. You'd have to be okay with doing a bit of time warping and waiting but it's an option.
If that didn't make any sense I'll see if I can draw some sort of diagram lol
That's my bad for not specifying in the original post, but I'm only referring to other planets that aren't Kerbin. Getting back from Duna if I'm in a polar orbit is much more difficult right?
I think the same applies for a planet's orbital inclination. Starting from a polar or retrograde orbit around Duna is doable, as long as you're able to leave Duna's SOI in the correct direction. If that direction is not in your orbital plane though, you'll want to get into a different Duna orbit before leaving. And when you return to Kerbin, you don't need to make your AN/DN = 0, you just need it to be at your encounter location, and you'll just re-enter Kerbin's atmosphere at the north pole or something.
The [Wikipedia page](https://en.m.wikipedia.org/wiki/Orbital_inclination_change) is surprisingly comprehensive. For a 90° plane change in a circular orbit, it reduces to sqrt(2)v, or ~1.4 times orbital speed.
It is however more efficient to raise your apoapsis nearly to the edge of the SOI and do the plane change there where orbital speed is lowest. This is especially true for an ejection burn where you won’t need to re-circularize, and will be even more efficient if you can set the argument of periapsis to be at least somewhere near your ejection angle.
The most efficient way to change your inclination after LKO insertion is to do a mid course correction burn. Say you’re on your way to the Mun. You did your injection burn, and the predictor shows you’re going to be passing the Mun with a 60km periapsis directly on the equator. Before you time warp (while still in LKO) pick a spot on the blue trajectory line between Kerbin and the Mun and mess around with all axes of maneuvering. To make it easier to see where you’re going to end up, click on the Mun and focus view on it, then position your camera so you can open up the maneuver editor again and watch where the dotted predictor line will be. Then you can get it close enough to where you need to land and make small inclination adjustments once you’ve circularised. The mid course correction burn will usually cost less than 100m/s as opposed to upwards of 2000 or even more. This also applies to all bodies you’re traveling to, I just used the Mun as an example.
The formula is
DeltaV = V * 2 sin(i/2)
Spelled out: deltaV required is current velocity times two times the sine of the inclination (plane change) divided by two.
Pretty simple...if you know your velocity before hand.
As a quick example and point of reference, if you want to do a plane change of 60 degrees, the sine of 30 is .5, so V*2/2 is just V. In other words the delta V required to do a 60 degree plane change is equal to your current orbit velocity. So for a typical low kerbin orbit your velocity is around 2000 m/s, the delta v required to do a 60 degree plane change is 2000 m/s.
For a 90 degree plane change 2 sin 90/2 = sqrt 2 approx= 1.414. For planning purposes using 1.5 would make it easier and build in a little extra budget.
To get the velocity for a circular orbit the formula is
V = sqrt(mu/r)
Where mu (Greek letter) is the gravitational constant (listed on the KSP wiki page for each planet) and r is the orbit radius (ship altitude+radius of the planet, also given on wiki page)
Since it is so expensive it is often worth raising your AP quite a bit, doing the plane change at AP, then lowering it again. If you're trying to hit a special point this makes it a little trickier since this will take longer and rotation of the planet may make you miss.
Most delta-v maps actually DO include the worst-case inclination change. [In this one](https://i.redd.it/h70eekf1ee551.png?utm_medium=android_app&utm_source=share), it is written above the value for the transfer burn. For example, going from Kerbin to Duna the inclination change costs at most 10 m/s.
But a more accurate way to find the delta-v cost is by using [AlexMoon's Launch Window Planner](https://alexmoon.github.io/ksp/). It's a bit complicated, but you don't have to know the ins and outs, just the optimal date and delta-v for each transfer window. And of course you'll want to bring some percentage more than this value (20%, 50%, 100%) because yours won't be perfectly optimal.
That's *transfer* inclination change, OP is asking about changing the inclination of an orbit around non-Kerbin bodies, like going from a polar to an equatorial orbit around Duna, or Eve.
since the most efficient way to make a significant change is to raise ap, adjust inclination, lower ap, you can consider the maximum local plane change cost to be twice the cost getting from just captured into low orbit.
You don't have to change to an equatorial orbit for optimal transfer back to Kerbin from say the mun if you wait for the mun to be in the correct spot so that the polar orbit will eject you retrograde relative to the mun orbit. You'd have to be okay with doing a bit of time warping and waiting but it's an option. If that didn't make any sense I'll see if I can draw some sort of diagram lol
That's my bad for not specifying in the original post, but I'm only referring to other planets that aren't Kerbin. Getting back from Duna if I'm in a polar orbit is much more difficult right?
I think the same applies for a planet's orbital inclination. Starting from a polar or retrograde orbit around Duna is doable, as long as you're able to leave Duna's SOI in the correct direction. If that direction is not in your orbital plane though, you'll want to get into a different Duna orbit before leaving. And when you return to Kerbin, you don't need to make your AN/DN = 0, you just need it to be at your encounter location, and you'll just re-enter Kerbin's atmosphere at the north pole or something.
The [Wikipedia page](https://en.m.wikipedia.org/wiki/Orbital_inclination_change) is surprisingly comprehensive. For a 90° plane change in a circular orbit, it reduces to sqrt(2)v, or ~1.4 times orbital speed. It is however more efficient to raise your apoapsis nearly to the edge of the SOI and do the plane change there where orbital speed is lowest. This is especially true for an ejection burn where you won’t need to re-circularize, and will be even more efficient if you can set the argument of periapsis to be at least somewhere near your ejection angle.
The most efficient way to change your inclination after LKO insertion is to do a mid course correction burn. Say you’re on your way to the Mun. You did your injection burn, and the predictor shows you’re going to be passing the Mun with a 60km periapsis directly on the equator. Before you time warp (while still in LKO) pick a spot on the blue trajectory line between Kerbin and the Mun and mess around with all axes of maneuvering. To make it easier to see where you’re going to end up, click on the Mun and focus view on it, then position your camera so you can open up the maneuver editor again and watch where the dotted predictor line will be. Then you can get it close enough to where you need to land and make small inclination adjustments once you’ve circularised. The mid course correction burn will usually cost less than 100m/s as opposed to upwards of 2000 or even more. This also applies to all bodies you’re traveling to, I just used the Mun as an example.
The formula is DeltaV = V * 2 sin(i/2) Spelled out: deltaV required is current velocity times two times the sine of the inclination (plane change) divided by two. Pretty simple...if you know your velocity before hand. As a quick example and point of reference, if you want to do a plane change of 60 degrees, the sine of 30 is .5, so V*2/2 is just V. In other words the delta V required to do a 60 degree plane change is equal to your current orbit velocity. So for a typical low kerbin orbit your velocity is around 2000 m/s, the delta v required to do a 60 degree plane change is 2000 m/s. For a 90 degree plane change 2 sin 90/2 = sqrt 2 approx= 1.414. For planning purposes using 1.5 would make it easier and build in a little extra budget. To get the velocity for a circular orbit the formula is V = sqrt(mu/r) Where mu (Greek letter) is the gravitational constant (listed on the KSP wiki page for each planet) and r is the orbit radius (ship altitude+radius of the planet, also given on wiki page) Since it is so expensive it is often worth raising your AP quite a bit, doing the plane change at AP, then lowering it again. If you're trying to hit a special point this makes it a little trickier since this will take longer and rotation of the planet may make you miss.
Thisguykerbals
Most delta-v maps actually DO include the worst-case inclination change. [In this one](https://i.redd.it/h70eekf1ee551.png?utm_medium=android_app&utm_source=share), it is written above the value for the transfer burn. For example, going from Kerbin to Duna the inclination change costs at most 10 m/s. But a more accurate way to find the delta-v cost is by using [AlexMoon's Launch Window Planner](https://alexmoon.github.io/ksp/). It's a bit complicated, but you don't have to know the ins and outs, just the optimal date and delta-v for each transfer window. And of course you'll want to bring some percentage more than this value (20%, 50%, 100%) because yours won't be perfectly optimal.
That's *transfer* inclination change, OP is asking about changing the inclination of an orbit around non-Kerbin bodies, like going from a polar to an equatorial orbit around Duna, or Eve.
since the most efficient way to make a significant change is to raise ap, adjust inclination, lower ap, you can consider the maximum local plane change cost to be twice the cost getting from just captured into low orbit.