Need some rough estimates of Delta V

[GoSlash27]

Snark,

After running the math both ways, it looks like the single burn may be slightly better in this case. 

Your math was considerably off in both cases by my reckoning, but I'd definitely suggest having OhioBob look it over. I have a lot more confidence in his orbital mechanics skillz.

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 Here's my numbers for the 2 burn solution:

Assuming a LKO velocity of 2300 and therefore escape velocity of 3250...

Kerbin's orbital velocity is 9290 and we need our apoapsis velocity to be 7590. Therefore we need to reduce that by 1700.

sqrt(1700²+3250²)-2300= 1370 m/sec. That's our first burn.

When we exit Kerbin's SoI we will have a velocity of 7590.  Our DV for the inclination change is 2sin(45°/2)*7590= 5,810 m/sec. That's our second burn.

Total DV is 7,180 m/sec.

*edit* I have confirmed this one in- game. Total DV 1307+5766= 7,073.

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The single burn solution is a bit stickier....

You need to create a solution that will result in an x component of 5,370 m/sec and a y component of 5,370.

x plane: Vexc=3920 m/sec Vesc=3250. We cannot subtract our orbital velocity because it's not oriented in a direction that will help us. Our x- plane DV is therefore 5090 m/sec.

y plane: Here we have the opposite problem: Our orbital velocity is aligned to help us, but Kerbin's orbital velocity is perpendicular. Vexc=5,370 Vesc=3250 Vorb=2300 Our y-plane DV is therefore 3,980 m/sec.

Since they are combined in a single burn, the total DV is 6,460 m/sec

*edit* This one is "plausible", but not confirmed. I was able to get a broadly similar orbit for 6,521 m/sec, but it wasn't an exact fit. Finding the launch window that gives a precise fit with no wasted energy is beyond my ability.

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So it's 7,180 (really 7,000 flat due to sidereal assistance during launch) vs 6,460.

The single burn could save about 500 m/sec in that instance unless my math is off. Like you, I'll have to test that and see if it holds up. *edit* Tested and it looks good from my house.

Best,

-Slashy

 

Edited January 1, 2016 by GoSlash27

[Snark]

4 hours ago, GoSlash27 said:

Your math was considerably off in both cases by my reckoning, but I'd definitely suggest having OhioBob look it over. I have a lot more confidence in his orbital mechanics skillz.

I don't think my math was off.  I'm willing to grant that my answer was off for the two-burn case , because from your latest post I realize I was assuming a different burn than you were actually suggesting.  So I'll take your number of 7000 m/s for the two burn solution.  Especially since you've verified it in-game!

However, my answer for the single-burn solution is correct as it stands.  It's 5105 m/s, not the 6460 m/s you suggest.

I've verified this in-game:

Put a ship in LKO, in a circular 90 km orbit with an inclination of 58.3 degrees relative to the Mun, whose AN/DN (relative to the Mun) are on a line that passes thru the Sun.  If you set a maneuver node of 5105 m/s at the appropriate point in the orbit (basically, right near the AN/DN, offset slightly to account for bending trajectory while in Kerbin's SoI), then it will take you to the desired orbit of half Kerbin's Pe with an inclination of 45 degrees.

Therefore, taking your number of 7000 m/s for the two-burn solution, and my number of 5100 m/s for the single-burn solution, that's a pretty significant savings in dV for single-burn.

And I concur, would love to hear @OhioBob chime in on this. 

Bob, if you're just now joining the discussion, I'll give you the executive summary here:  Slashy and I have been having a lively orbital-mechanics discussion about the best way to achieve a maneuver that requires both lowering the periapsis and doing a significant inclination change at the same time.

Simply stated, the problem is this:  You're in LKO, 90 km orbit, can be any inclination you like.  Your goal is to get into a solar orbit whose Ap is the same as Kerbin's orbit, whose Pe is half of Kerbin's orbit, and which is inclined 45 degrees from Kerbin's orbit.  What's the most efficient (least dV) way to do that, and how much dV is required?

 

Edited January 1, 2016 by Snark

[GoSlash27]

Snark,

 Likewise, I'll take your word for the single burn solution. You've already verified it.

 Just out of curiosity, how were you able to derive the correct parking orbit for the burn?

Happy New Year,

-Slashy

Edited January 1, 2016 by GoSlash27

[Snark]

57 minutes ago, GoSlash27 said:

 Likewise, I'll take your word for the single burn solution. You've already verified it.

 Just out of curiosity, how were you able to derive the correct parking orbit for the burn?

Per the earlier post, the ejection from Kerbin's SoI needs 5360 m/s of dV on the Y axis, and 3924 of dV on the X axis (both numbers relative to Kerbin).  So I took 5360 / 3924 = 1.366, the slope of the line.  Arctangent of that gives 53.8 degrees, the inclination of the orbit relative to Kerbin's equator.

As for the longitude of ascending node:  Since the ejection burn needs to include no solar radial in/out component, the AN/DN need to be on a line passing through the sun so that there will be an optimal ejection point at the AN or DN.

And Happy New Year to you, too!

(Edit:  I think I see now where you got your earlier 6460 m/s number for the single-burn solution-- that's pretty close to my 6443 number for ejection speed from Kerbin's SoI.  The dV needed is less than that because you're starting from orbit and can add your LKO orbital speed to the dV you do in LKO.  You only need 5105 m/s dV in LKO to get an ejection of 6443 m/s from the SoI.)

Edited January 1, 2016 by Snark

[OhioBob]

2 hours ago, Snark said:

And I concur, would love to hear @OhioBob chime in on this. 

Bob, if you're just now joining the discussion, I'll give you the executive summary here:  Slashy and I have been having a lively orbital-mechanics discussion about the best way to achieve a maneuver that requires both lowering the periapsis and doing a significant inclination change at the same time.

Simply stated, the problem is this:  You're in LKO, 90 km orbit, can be any inclination you like.  Your goal is to get into a solar orbit whose Ap is the same as Kerbin's orbit, whose Pe is half of Kerbin's orbit, and which is inclined 45 degrees from Kerbin's orbit.  What's the most efficient (least dV) way to do that, and how much dV is required?

 

Thanks for summarizing, that's most helpful.  I've purposely not read any other posts because I don't want to be influenced by anybody else's solution.

We know that Kerbin's orbital velocity is (subscript p for "planet"),

V_p = 9284.5x

For an orbit as you describe, the speed at apoapsis is 7580.8 m/s, but this is inclined 45 degrees, so the spacecraft’s velocity at apogee must be (subscript s of “spacecraft”),

V_s = 5360.4x + 5360.4y

Therefore the velocity the spacecraft must achieve relative to Kerbin is,

V_s/p = (5360.4x + 5360.4y) – 9284.5x = -3924.1x + 5360.4y

The magnitude of which is,

V_s/p = (-3924.1² + 5360.4²)^1/2 = 6643.2 m/s

In this case, the relative velocity represents the hyperbolic excess velocity that the spacecraft must achieve, thus

V_∞ = V_s/p = 6643.2 m/s

We know that

V_∞² = V_bo² – V_esc²

where V_bo is burnout velocity and V_esc is escape velocity.  For a 90 km LKO, V_esc = 3199.5 m/s, therefore

V_bo = (6643.2² + 3199.5²)^1/2 = 7373.5 m/s

The Δv is simply V_bo less the initial orbital velocity, where V_orb = 2262.4 m/s.  Therefore,

Δv = V_bo – V_orb = 7373.5 – 2262.4 = 5111.1 m/s

 

[Doodle]

Ok i decided to cancel the contract and take the cash/rep penaltys for it.

I have never done any gravity Assist transfers or similar stuff and only visited duna once in all my hours of playing.

This is deffintily out of my league. But thanks alot for those calculations was quite interesting to see what would have been needed.