Confused by delta V for plane change?

[ravensierra]

I'm playing around with calculating dV for all the maneuvers as part of mission planning - and found myself stumped by what is going on with this plane change?:

 

 

It's a 45.4 degree simple plane change and velocity at the DN where the maneuver is done is 724.6m/s - Δv I thought was equal to 2.v.sin(θ/2) in this case, or 559m/s

Can any of you orbital mechanics experts explain how I can get away with 373m/s for the maneuver here? I shouldn't complain! But would like to understand

[Gargamel]

Moved to gameplay questions. 

[king of nowhere]

11 hours ago, ravensierra said:

I'm playing around with calculating dV for all the maneuvers as part of mission planning - and found myself stumped by what is going on with this plane change?:

 

 

It's a 45.4 degree simple plane change and velocity at the DN where the maneuver is done is 724.6m/s - Δv I thought was equal to 2.v.sin(θ/2) in this case, or 559m/s

Can any of you orbital mechanics experts explain how I can get away with 373m/s for the maneuver here? I shouldn't complain! But would like to understand

it's not that simple.

actually, if you made a purely normal maneuver (90° inclination over your direction) to get a 45° plane change you'd need to have as much normal velocity as prograde velocity, so you'd need 724 m/s.

but this is not what happens, becase you'll notice that your maneuver also includes a retrograde and radial component.

If you use the plane of Mun as a reference, you'll see that your ship has a complex velocity with a component in all the 3 x, y and z axis. and you want to end up with a different velocity, including a 0 net component on the y axis.

And your maneuver node uses a different frame of reference too - it uses your current orbit.

Now, if somebody held my family at gunpoint and forced me to make the involved calculations, I probably could do it - if I could figure out a way to calculate the exact direction of your ship at the DN to work out the velocity in the three axis. But unless I have that kind of incentive, I'm really not going to try

[Leganeski]

On 1/29/2023 at 7:00 PM, ravensierra said:

It's a 45.4 degree simple plane change and velocity at the DN where the maneuver is done is 724.6m/s - Δv I thought was equal to 2.v.sin(θ/2) in this case

Δv = 2v sin(θ/2) is the correct formula. However, the angle change of the direction of velocity, θ, can be less than the angle change of the orbital plane.

Consider the extreme case where you're in a very elliptic orbit, moving almost directly inwards. No matter how much plane changing you do, your new velocity vector is still pointed "almost directly inwards", so the angle between the old and new vectors remains small. In the example you provided, the eccentricity is not as extreme, but the same idea still applies, and the turn angle θ is substantially less than 45.4°.

The values v = 724.6 m/s and Δv = 372.9 m/s let us calculate θ, which turns out to be 29.82°. If you look at the maneuver from the direction perpendicular to the old and new velocities (rather than looking towards Kerbin), you'll see that the actual angle between the velocity vectors really is about 30 degrees.

[ravensierra]

Thanks very much for the answers - really helped my understanding!

Wow the maths to work out the state vector from the classical elements is quite complicated..