Orbital Mechanics Visual "Hack"?

[mpk10]

Hey nerds,

This is my first time posting in this area of the forums, but I'm hopeful that some of the more math-inclined of you guys can help me with an idea I just had.

I've been playing KSP for years and every time I start a new career mode, my designs and methods get more efficient.  Specifically, I was just docking in LKO to pick up some fuel for a mission to the Jool system and as I was matching orbits with my fuel depot, something occurred to me:

Is it possible that the amount of delta V required for matching orbits is somehow proportional to the area of the space between the two orbits??

That is, if you had two almost entirely overlapping circles (like a venn diagram with almost ALL of it in common) that the delta-V to match the orbits is somehow proportional to the area of the two thin "crescent" shapes carved out on the outside of that "venn diagram"?  

 

There was something tucked away in my memory about Kepler's law when I thought of this.  not sure if his "swept out area" idea is similar to what I'm talking about.

 

Anyway, I'd be grateful for any thoughts.

 

-mpk

[Bill Phil]

There's a minimum Dv, but theoretically no maximum.

You have to basically move your orbit to the target's orbit.

Technically, if you want to rendezvous faster, then it costs more Dv, and if you want less Dv, it usually takes longer.

[mpk10]

25 minutes ago, Bill Phil said:

There's a minimum Dv, but theoretically no maximum.

You have to basically move your orbit to the target's orbit.

Technically, if you want to rendezvous faster, then it costs more Dv, and if you want less Dv, it usually takes longer.

Thanks.  I understand that there's a minimum, i guess i was just wondering that if all the math for calculating that minimum works out to be relatively proportional to the area of the non-overlapping areas of the two orbits (in sq. kms or whatever)?

[goncaloeaguiar]

I guess the only way to find out is doing the math. But what happens when you take inclination into consideration? 

[rudi1291]

The dV needed to do an inclination change depends on your orbital velocity when you execute the burn

 

Edit: That only works that easy for circular orbits. Its more complicated for elliptical orbits. Wikipedia tells you how to calculate it: https://en.wikipedia.org/wiki/Orbital_inclination_change

Edited March 3, 2016 by rudi1291

[Bill Phil]

3 hours ago, rudi1291 said:

The dV needed to do an inclination change depends on your orbital velocity when you execute the burn

 

Edit: That only works that easy for circular orbits. Its more complicated for elliptical orbits. Wikipedia tells you how to calculate it: https://en.wikipedia.org/wiki/Orbital_inclination_change

Wouldn't some simple vector math be sufficient for finding the Dv of a plane change?

[YNM]

There's probably some point here. Cross products are known to be used to measure area. Maaybee it have something to do with angular momentum ? I don't know...

[mpk10]

13 hours ago, goncaloeaguiar said:

I guess the only way to find out is doing the math. But what happens when you take inclination into consideration? 

Interesting!

 I think if you take inclination into account, then it's probably a function of the Volume of the non-overlapping "wedges", rather than the Area of the non-overlapping "crescents". A good visual representation of why plane changes are so dV expensive. 

 

It's not super useful but I find it helpful to think of things visually like that so I understand a little better how to be efficient when matching up orbits at other planets/moons. 

 

Thanks for all the thoughts!

[GoSlash27]

I don't believe that's the case, due to the Oberth effect. Transferring from a circular orbit to a highly- elliptical orbit doesn't change the area disparity linearly with DV. 

IAC, it's such a simple thing to calculate using vis- viva that calculating it by delta area would probably be more difficult anyway.

Best,
-Slashy

[sevenperforce]

13 minutes ago, GoSlash27 said:

 

IAC, it's such a simple thing to calculate using vis- viva that calculating it by delta area would probably be more difficult anyway.

It would still be a neat relationship...like how gas mileage, distance/volume, works out to units of inverse area, and happens to be the instantaneous fuel consumption cross-section along the length of your trip. 

[Shpaget]

I don't see how.

Even without calculation, the answer is an obvious "no", or at least there is a much more significant factor you're not thinking of. Just imagine doing a prograde burn of, let's say, 200 m/s dv from a LEO or LKO and mentally compare it with a 200 m/s dv burn in a much higher orbit. The difference in the change of areas is definitely not equal, not even remotely.

[Yourself]

Even simpler, imagine a parabolic orbit, that encloses essentially an infinite area, but it takes a finite amount of ΔV to change it to a circular orbit, which has a finite area.