Can someone calculate this delta V for me (or show me how)

[Callmedave]

So i have this contract to put a probe in orbit of the sun,

 

Apoapsis: 111,782,138,716m

Periapsis: 108,006,024,920m

Inclination 62.9° This is my problem, I dont know how much Detla v i should pack because of this plane change? 

Do you think 5000 delta V will be enough? Any help please,

 

 

arrows are to show direction of orbit, I know that doesn't change the answer.

[Theysen]

On mobile right now but use Jool and Sun flybys and use free delta v to get it somewhere in the right shape, then work from there

[Tex_NL]

Don't know if 5000 m/s will be enough. I do know you will need quite a lot.

Instead of doing the maths you can find your own answer two ways:

1. Try to duplicate the entire mission in a separate sandbox.
2. Plot multiple manoeuvre nodes and add them up.

[Callmedave]

Just realised, I dont think the picture shows the inclination well enough, 

See here: 

I don't think Jool Flybys will help because of the extreme angle.

[Kobymaru]

30 minutes ago, Callmedave said:

Just realised, I dont think the picture shows the inclination well enough, 

See here: 

I don't think Jool Flybys will help because of the extreme angle.

Oh, but it will. Jool is very powerful!

[LiquidAir]

The orbital velocity at the apoapsis of a Holmann transfer from Kerbin (if I got my math right) is about 1600m/s. The plane change done there will run you around 2200dV. Without gravity assists, I'd budget at least 7500 dV from LKO to get to that orbit.

From solar orbit at Kerbin:

Burn Ap to intersect the orbit: 2750 dV

Plane change: 2200 dV

Burn up Pe: 1606 dV

Note: this doesn't include escape from the Kerbin Gravity well.

Edited December 7, 2015 by LiquidAir

[Snark]

The easy way to know how much dV you'll need from LKO:

Just put a small, cheap satellite in LKO, then give it maneuver nodes to try to put it in the designated orbit.  Once you've got the nodes set up to put you reasonably close to the target orbit, you can just add 'em up for the total dV.

[tomf]

That is a pretty nasty orbit to have to aim for.

You could save a fair amount of dV by making sure you launch while Kerbin is alligned with either the ascending or descending nodes of the orbit. That way you can compine your burn to leave Kerbin with largish component to change plane at teh same time. Then you can combine the burn to raise your PE with more plane changing.

A gravity assist could save a fair bit more, but only if you are able to ensure you arrive at Jool as it passes the nodes of the target orbit.

[Gaarst]

If you're not afraid of doing a little maths, Wikipedia gives a nice formula for orbital plane change dV, but that requires a lot of values KSP doesn't give you explicitely:

With:

• Δi the inclination change
• e the orbital eccentricity
• w the argument of periapsis
• f the true anomaly
• n the mean motion
• a the semi-major axis

While I don't remember what all these values are individually, I'm pretty sure KER can give you most of them (MJ probably as well but I don't use it).

Edited December 7, 2015 by Gaarst

[ElWanderer]

Link to a real-life example of a high-inclination solar probe that used a Jupiter slingshot.

Without using a Jool assist:

I use the formula sqrt( 2 x v^2 x (1 - Cos(delta_inc)) ) to work out the delta-v of an inclination change that leaves your apoapsis and periapsis untouched. The hard part then is working out what your orbital velocity will be at the time of the burn. It's relatively easy to calculate the velocity at apoapsis periapsis, much harder for any arbitrary point.

Your orbital velocity is much lower when far from the object you're orbiting, which is why any inclination burn is best performed far from the sun.

Note that there are some easy-to-remember results:

A 180 degree inclination change requires double your velocity.

A 90 degree change requires 1.4 (sqrt(2)) times your velocity. Note that doing this as two separate burns to kill your orbital velocity and add it in the new direction will cost double your v - it's a massive saving to burn once at 45 degrees. Combining burns saves fuel!

A 60 degree change requires 1 times your velocity. Your 62.9 degree requirement is close to this.

A 30 degree change requires a little over 0.5 times your velocity.

If you can combine your inclination change with your circularisation burn at Apoapsis, that would save you delta-v. That means leaving Kerbin when it's at (or close to) one of the descending/ascending nodes of the target orbit.

[PLAD]

I was thinking about a problem like this recently. Though I wonder if it will really be necessary to set the periapsis and apoapsis values to the nearest meter per second. I see the contract gives it to that accuracy, but that's less than one part in 100 billion and I do not see how you can ever set those values that accurately. If you manage to get your periapsis to 108,006,024,920 meters then you would have to go 3322.77184373 m/s to have an apoapsis of 111,782,138,716 meters. If you go, for instance,  3322.77184372 m/s your apoapsis will be 111,782,138,715 meters.  Maybe if you use one little RCS thruster on a thousand-ton ship?

Anyway, here's how to ballpark the numbers. Pardon me while I do some mental calisthenics. Let's work backwards.
Ra= apoapsis radius, the furthest distance from the sun of the desired orbit.
Rp=periapsis radius, the closest distance from the sun of the desired orbit.
a=the semimajor axis of the desired orbit= (Ra+Rp)/2, in this case 109,894,081,817 m.
The further from the sun you are the slower you go, so we will do the do the big plane change and periapsis change at the apoapsis of the orbit.  How fast do we need to go there?
u=the gravitational constant of Kerbol, from the KSP wiki that is 1.1723328E+18 m^3/s^2.

For all points in an elliptical orbit the speed, v, = sqrt(u*((2/r)-(1/a))). I'll call this "Formula 1".

So the v of your target orbit at r=Ra is 3210.525247 m/s.
Let's round to the nearest meter per second from here on, though you can see you might need to go to the nearest nanometer per second in the end...
Now we need to find the transfer orbit from Kerbin to an apoapsis of 111782138716 m,= Ra. Kerbin's orbit is a circle of semimajor axis 13,599,840,256 meters, so we will use this as our transfer orbit Rp. The semimajor axis for this is (Ra+Rp)/2=62,690,989,488m. Using the Formula 1 from above we find that the v at Rp is 12398 m/s, and the v at Ra is 1508 m/s. So when we reach the target apoapsis from the sun we need to increase the speed from 1508m/s to 3211m/s and change the direction by 62.9 degrees. (Kerbin's orbital plane is inclined 0 degrees, and the Hohmann orbit we will transfer from Kerbin in will also be inclined 0 degrees, so the plane change will equal the inclination of the desired final orbit.) It is much more efficient to make both changes at once, so now we have a problem in trigonometry. Draw a triangle with on leg 1508 units long, and a second leg 3211 units long  that is at an angle of 62.9 degrees to the first leg. We need to figure out the length and direction of the third leg, since that leg is the burn we need to do. From the law of cosines we find the mystery burn will be 2859m/s.  Pythagoras can tell you what the prograde and normal components of that burn will be.

  Now the final problem is the departure from Low Kerbin Orbit. Kerbin's orbital speed around the sun is 9284.5m/s, so you need to leave Kerbin's SOI at 12398-9284=3114 m/s. If we can find the semimajor axis of the hyperbolic departure orbit from Kerbin we can find the velocity at any distance from Kerbin of that orbit. We'll need another formula:
a=(r*u)/(2*u-r*v^2)    
  in this case r is the radius of Kerbin's Sphere Of Influence, which from the wiki we find to be 84,159,286 meters. u is Kerbin's gravitational parameter, from the wiki it is 3.5316E+12.  v is the 3114 m/s of speed we need to have at that SOI. The a therefore = -367,375 meters. I don't know if you have used these formulas before, if not don't worry about the negative value, that is how the semimajor axis for hyperbolas are given. In any case we can now use this value for a in Formula 1 and the value of 'r' will be whatever orbit you like to start from. I usually leave Kerbin from a 75x75km altitude orbit, but you must remember that Formula 1 needs the distance from the center of the planet, not the altitude above it's surface! Therefore I would use 675000 meters for 'r' in Formula 1. I then find my v in the 75x75km orbit needs to be 4481m/s, since the circular speed at that altitude is 2287m/s I will need a 2193 m/s burn to get to the plane-change burn point.
In summary, starting from a 75x75km orbit around Kerbin, you will need to make two burns- a 2193m/s burn that gets you out to 111,782,138,716 meters above Kerbol, and then a burn at that point of 2859m/s that puts you right into the desired final orbit. Total = 5052m/s. In practice I'd allow for a few hundred m/s of correction burns, because there is no way you will be able to do those burns precisely enough! Note that that second burn must be made at the exact Longitude of Ascending Node of 63.7 degrees. How do you find this? Well, in our Hohmann transfer orbit from Kerbin to the plane change we will travel exactly 180 degrees around the sun. This means you must  leave Kerbin when it is at 243.7 degrees. To figure out when this is we note from the wiki that Kerbin starts at  a Mean Anomoly of 3.14 radians at 0 UT. 3.14 radians= 179.9 degrees. Kerbin's orbit is a perfect circle, so it always moves at the same speed. It makes one orbit of Kerbol (=360 degrees)  in 9,203,545 seconds, so we can see it will move 243.7-179.9= 63.8 degrees in (63.8/360)*9203545= 1631072 seconds. So leave Kerbin at 1,631,072 seconds after the game starts, or every 9,203,545 seconds after that.

     Note that you could use a flyby of Jool to throw you into the right plane and out to the target apoapsis. This would save you about 100m/s on the leaving-Kerbin burn and about 2km/s on the 2nd burn. But that would be much more complicated to set up. You would need to flyby Jool when Jool is at 63.7 degrees, but determining what energy you need to arrive at it with is tricky. Considering that I think it's impossible to get into that one-part-in-100-billion orbit I wouldn't spend the time on it...

[DomiKamu]

On 7/12/2015 00:18:39, LiquidAir said:

The orbital velocity at the apoapsis of a Holmann transfer from Kerbin (if I got my math right) is about 1600m/s. The plane change done there will run you around 2200dV. Without gravity assists, I'd budget at least 7500 dV from LKO to get to that orbit.

From solar orbit at Kerbin:

Burn Ap to intersect the orbit: 2750 dV

Plane change: 2200 dV

Burn up Pe: 1606 dV

Note: this doesn't include escape from the Kerbin Gravity well.

...Hohmann transfert orbit (not Holmann) is badly used expression - by definition, it's a transfert from circular orbit to another circular orbit, both on same plane and in same sphere of influence (SOI), by using only two maneuvers.

In real life, Hohmann transfert orbit is used for transfers between LEO (parking) to geostationary orbit (GEO), this Hohmann transfert is also called GTO.