Need some rough estimates of Delta V

[Doodle]

Hi,

I found a nice Satelite Contract and I am curious if anyone could give me a rough estimate for the amount of Delta V that I would need to do it.

 

Not sure if you can see it in the Picture clearly...

Stats are as followed:

AP: 51.421.xxx.xxx meter

Pe: 28.326.xxx.xxx meter

Inclination: 54 Degree!!!!!!!

Since this is a Career Game I would love to know a rough estimate of the Delta V that i need to accomplish this and if it is worth the 400k $

 

Greetings

Doodle

 

[GoSlash27]

That's about a .8t payload (give or take). My rough estimate is "no".

I can work the math on it and give you a figure if you just really want to know.

Best,

-Slashy

[GoSlash27]

I ran the numbers for you.

Transfer from LKO to your desired apoapsis is 1,740 m/sec.

Inclination change at apoapsis is 2,916 m/sec

Circularization to raise periapsis is 1,688 m/sec

This brings the DV budget up to 6,344 m/sec in orbit.

But of course you have to get it to orbit, so you're lookin' at another 3,500 m/sec

This brings the grand total to 9,844 m/sec DV from the pad with no margin.

Figure you pad that out to 10 km/sec DV from the pad, maybe 2km/sec per stage, so 5 stages. And .8t payload or so. That's going to be a big rocket. About a hundred ton launch vehicle with a Mainsail. You could probably get it up there for $50,000 or so.

Best,

-Slashy

Edited December 30, 2015 by GoSlash27

[WuphonsReach]

Personally, I'm not sure it's worth it -- but that's a "balance" complaint with a lot of contracts (not counting opportunity cost of using up the contract slot).

Let's assume that I can take a contract which rewards 10k funds and complete it in a single day.  That means a contract which will probably take 300 days (guessing) should pay out 3 million.  Or at least double that 400k that they are offering.

[Snark]

8 minutes ago, WuphonsReach said:

Personally, I'm not sure it's worth it -- but that's a "balance" complaint with a lot of contracts (not counting opportunity cost of using up the contract slot).

Let's assume that I can take a contract which rewards 10k funds and complete it in a single day.  That means a contract which will probably take 300 days (guessing) should pay out 3 million.  Or at least double that 400k that they are offering.

+1 to this.  Not worth the trouble, effort, and calendar time, IMHO; there are plenty of other contracts worth that much or more which are less hassle.  The only reason to take that contract would be if you just wanted the challenge of it.

The "put a station in a solar orbit" contracts seem to be the sweet spot-- worth boatloads of cash, very quick to design, launch, and complete.

[GoSlash27]

9 hours ago, WuphonsReach said:

Personally, I'm not sure it's worth it -- but that's a "balance" complaint with a lot of contracts (not counting opportunity cost of using up the contract slot).

Let's assume that I can take a contract which rewards 10k funds and complete it in a single day.  That means a contract which will probably take 300 days (guessing) should pay out 3 million.  Or at least double that 400k that they are offering.

Actually, it's even worse than that. The transfer would take at least 3 years, 295 days.

 So I agree with the others above. If you take this contract and you are very careful, you might turn $60,000 profit right off the bat. Then (assuming everything works out according to plan) you'd have to wait nearly 4 years for the remaining $300,000. And of course if it fails you'll lose a lot of money and take a rep hit.

 I wouldn't do it.

 

Best,

-Slashy

Edited December 31, 2015 by GoSlash27

[Champ]

10 hours ago, WuphonsReach said:

Personally, I'm not sure it's worth it -- but that's a "balance" complaint with a lot of contracts (not counting opportunity cost of using up the contract slot).

Let's assume that I can take a contract which rewards 10k funds and complete it in a single day.  That means a contract which will probably take 300 days (guessing) should pay out 3 million.  Or at least double that 400k that they are offering.

Sorry, i don't understand.

Why would the paycheck be proportional to the in-game time it takes ? Why would you care about in-game time ? We can time warp, so we don't care about time. You missed that transfer-to-moho window ? Then just wait for the next one !

What is important (at least to me) is the real-life time you need to complete the contract.

Back to the subject :

14 hours ago, GoSlash27 said:

I ran the numbers for you.

Transfer from LKO to your desired apoapsis is 1,740 m/sec.

Inclination change at apoapsis is 2,916 m/sec

Circularization to raise periapsis is 1,688 m/sec

This brings the DV budget up to 6,344 m/sec in orbit.

But of course you have to get it to orbit, so you're lookin' at another 3,500 m/sec

This brings the grand total to 9,844 m/sec DV from the pad with no margin.

Figure you pad that out to 10 km/sec DV from the pad, maybe 2km/sec per stage, so 5 stages. And .8t payload or so. That's going to be a big rocket. About a hundred ton launch vehicle with a Mainsail. You could probably get it up there for $50,000 or so.

Best,

-Slashy

Wouldn't it be cheaper to do a bi-elliptic transfer ? You can use Jool's gravity assist to lessen the dV needed to raise Ap. You could even use Eve's gravity assit to get you to jool, but this would be too bad in terms of 'real-life time'

Edited December 31, 2015 by Champ
orthograph

[peteletroll]

A gravity assist can help a lot with the plane change too

[GoSlash27]

1 minute ago, Champ said:

Wouldn't it be cheaper to do a bi-elliptic transfer ? You can use Jool's gravity assist to lessen the dV needed to raise Ap. You could even use Eve's gravity assist to get you to Jool, but this would be too bad in terms of 'real-life time'

Champ,

 Yes, a bi- elliptic transfer would save some DV, but not very much. There's not a lot of eccentricity in this transfer.

 And you could play gravity assist pinball and save lots of DV, but that approach is harder to plan for in the building phase (it's hard to predict exactly how much DV is required) and (as you point out) extends the mission time.

Unless I'm mistaken and it has been fixed, there actually is a time constraint to KSP. The clock used to overrun after around 60 years. If that's still the case, then in- game time actually does matter.

Best,

-Slashy

[Snark]

1 hour ago, Champ said:

Why would the paycheck be proportional to the in-game time it takes ? Why would you care about in-game time ? We can time warp, so we don't care about time. You missed that transfer-to-moho window ? Then just wait for the next one !

Well, I wouldn't go so far as to say that the paycheck should be proportional to the in-game time it takes, but in-game time is definitely not "free", depending on one's play style.

If you don't care about calendar date, and if you tend to play linearly (run a mission, finish; run next mission, finish), then yeah, I could see someone not caring much about in-game time.  However, for a lot of players, this is not the case:

• Some folks care about in-game time for "challenge" or role-playing reasons.  I know I do; there's something psychological about it.  It bugs me if I'm in year 20 of my space program.  I want to get stuff done soon on the calendar, and I can't stop myself from viewing the calendar as part of my "score", and I want to get things done efficiently.  "Land a kerbal on every body in the solar system and return them to Kerbin" is one definition of "done" for a career that I've used, and I tend to view the "score" as how much calendar time has passed to do that.
• It's quite common for folks to run lots of missions in parallel.  You can't just timewarp to the next window because you have a lot of other stuff going on in the meantime.
• It's also fairly common to run mods such as life support which can make you really care about in-game time.

Back to the subject of contracts:  my own opinion is that the reward should be proportional (in some fashion, not necessarily linear) with the amount of player time it takes.  Arguments about in-game time notwithstanding, it's the player's own time that's the most valuable.  A contract that takes you a minute of your time is much more attractive than one that takes half an hour of your time, assuming the rewards are equal (and assuming that you're just doing it for the sake of the reward, rather than the fun challenge of the mission itself).

On that metric, this satellite contract isn't great.  Getting to that orbit will take a fair amount of your time, both in designing the rocket and in futzing around with the orbits.  A simple put-a-station-on-a-solar-orbit contract, for example, takes much less of your time and is worth more cash.  And has the added bonus of needing much less in-game time, for folks who care.

The best contracts, in terms of "funds per minute of player time", are the "Science data for space around X" ones.  Assuming you've been prudent enough to leave a satellite (with a science instrument) parked in orbit around each body you've visited, then completing one of these contracts literally takes only about thirty seconds of your time:  switch to the satellite, take a measurement, transmit, done.  The payoff is low, but the time required is so much lower that they're really worth it.

[WuphonsReach]

Yeah, way off-topic, but when thinking about contracts:

• Contract slots are limited in career mode, which means each contract competes against the others for the player's attention.  Even when you unlock the top-tier building for unlimited contracts, there's still a configuration file limit on the maximum number of contracts.  So having 10 multi-year contracts running, means fewer contracts to pick from in the meantime.
• If you're using KCT (Kerbal Construction Time) and/or a life-support and/or maintenance cost mod, then time becomes a large part of the player's calculation in "is it worth it".  Contracts with minimal payouts will not pay the bills.
• As Snark said, there's also the issue of the effort required by the player to fulfill the contract.  That involves designing a vessel, launching it, possibly a transfer orbit burn, mid-course correction, insertion burn, performing the task requested, and then possibly returning it to Kerbin (3 more burns, plus monitoring the return flight).  Even if you use tools like MJ to plot all those burns and maneuvers, the player still has to sit through that time watching the burn take place.  Or watching the launch / landing (or flying the launch landing).

So, hopefully contracts will get a rebalance in terms of rewards after 1.1 ships.

[StarManta]

That inclination change is overly pessimistic, IMO. It doesn't account for, for example, the possibility of a gravity assist, or starting the maneuver from a polar orbit. Starting from the right orbit could easily shave off 2 km/s from that. This would at least give you a comfortable margin.

It's still a difficult mission, don't get me wrong on that. I would say to build a huge many-staged launcher for a tiny satellite and go for it, but the requirement to have a materials bay makes it that much harder, so maybe not.

[GoSlash27]

6 minutes ago, StarManta said:

That inclination change is overly pessimistic, IMO. It doesn't account for, for example, the possibility of a gravity assist, ...

StarManta,

 This is true. This is just a Hohmann transfer. You could save a lot of DV by using gravity assists... but it's hard to predict exactly how much, exactly when to launch, and it will take a whole lot longer. It'd be cheaper in terms of player time, game time, and difficulty to simply build in the extra DV.

8 minutes ago, StarManta said:

or starting the maneuver from a polar orbit. Starting from the right orbit could easily shave off 2 km/s from that.

This, however, is incorrect. When you leave Kerbin's SoI, you will still be in the ecliptic plane.

Don't worry; I've made the same mistake myself.

Best,

-Slashy

[Snark]

2 hours ago, StarManta said:

Starting from the right orbit could easily shave off 2 km/s from that. This would at least give you a comfortable margin.

 

2 hours ago, GoSlash27 said:

This, however, is incorrect. When you leave Kerbin's SoI, you will still be in the ecliptic plane.

How so?  Admittedly, it depends on what your phase angle is when you do the maneuver.  If you're in a polar orbit, and you eject in the celestial north or south direction, you won't be in the ecliptic plane-- you'll be tilted at some inclination.  It would certainly be a dV savings over launching to equatorial orbit first, as long as you time it right.

Still a metric crapload of dV needed, and still (IMHO) not worth the contract, but if someone decided they were going to do it, a polar orbit can help a bit.

[GoSlash27]

25 minutes ago, Snark said:

 

How so?  Admittedly, it depends on what your phase angle is when you do the maneuver.  If you're in a polar orbit, and you eject in the celestial north or south direction, you won't be in the ecliptic plane-- you'll be tilted at some inclination.  It would certainly be a dV savings over launching to equatorial orbit first, as long as you time it right.

I'll have a hard time explaining this without pics...

 When you leave Kerbin's SoI, you're carrying it's orbital velocity of (roughly) 9300 m/sec. We have used this to our advantage in the initial burn (Oberth effect) so that we only need 1,700 m/sec to hit apoapsis.

If we instead expend that 1,700 m/sec perpendicular to Kerbin's orbital plane, that will only buy us 10° of inclination change (1.7k up and 9.3k east), and we haven't made any headway on raising apoapsis.

 We would now have to make an even bigger prograde burn to compensate for the fact that we wasted our Oberth effect; about 2,300 m/sec to get back to where we were instead of the original 1,700, so this approach just wasted the DV we had earmarked to get us 54° inclination change on 10° of inclination change. We have to spend an additional 1,800 m/sec to get us the rest of the way.

 You never want to make inclination changes at high speed. That's why I planned it where I did.

 The way to reduce the budget for the inclination change would be what Champ said; do a bi- elliptic transfer, but really even that doesn't buy you much.

Best,

-Slashy

[Snark]

6 minutes ago, GoSlash27 said:

 You never want to make inclination changes at high speed. That's why I planned it where I did.

Absolutely, yes.  I've got no quarrel with the proposed trajectory.  I was just addressing this particular statement,

4 hours ago, GoSlash27 said:

When you leave Kerbin's SoI, you will still be in the ecliptic plane.

which isn't strictly accurate, though now I understand what you were trying to get at.

And FWIW, I wasn't suggesting a polar orbit in this particular case (since the Api is significantly higher than Kerbin's orbit), just that this is a possible technique in the general case.  It would make sense, for example, if the target orbit were at a lower Pe than Kerbin's orbit.

[GoSlash27]

34 minutes ago, Snark said:

Absolutely, yes.  I've got no quarrel with the proposed trajectory.  I was just addressing this particular statement,

which isn't strictly accurate, though now I understand what you were trying to get at.

And FWIW, I wasn't suggesting a polar orbit in this particular case (since the Api is significantly higher than Kerbin's orbit), just that this is a possible technique in the general case.  It would make sense, for example, if the target orbit were at a lower Pe than Kerbin's orbit.

Yeah, I get what you're saying, but even in the case of a burn for lower Pe it's still not helpful to eject out- of- plane. You lose the benefit of the Oberth effect to help make your Pe burn cheaper and it works against you, making your inclination change more expensive.

It would be cheaper to eject retrograde and do the plane change after escaping Kerbin's gravity well when your velocity is at it's lowest.

Best,

-Slashy

[Snark]

1 hour ago, GoSlash27 said:

Yeah, I get what you're saying, but even in the case of a burn for lower Pe it's still not helpful to eject out- of- plane. You lose the benefit of the Oberth effect to help make your Pe burn cheaper and it works against you, making your inclination change more expensive.

It would be cheaper to eject retrograde and do the plane change after escaping Kerbin's gravity well when your velocity is at it's lowest.

Really?  Seems to me that to go from "Circular orbit at Kerbin's orbital radius" to "inclined orbit with Ap at Kerbin and Pe lower than that", there's a certain sun-relative dV needed for that, which lies in between retrograde and normal (or between retrograde and antinormal, depending on whether you're at the ascending or descending node).  Wouldn't it be the most efficient to eject in that direction directly?  i.e. combine both the "drop your periapsis" and "change your inclination" into a single burn?

I find it hard to believe that it would be more efficient to do two burns that aren't in the same direction.

[GoSlash27]

3 minutes ago, Snark said:

Really?  Seems to me that to go from "Circular orbit at Kerbin's orbital radius" to "inclined orbit with Ap at Kerbin and Pe lower than that", there's a certain sun-relative dV needed for that, which lies in between retrograde and normal (or between retrograde and antinormal, depending on whether you're at the ascending or descending node).  Wouldn't it be the most efficient to eject in that direction directly?  i.e. combine both the "drop your periapsis" and "change your inclination" into a single burn?

I find it hard to believe that it would be more efficient to do two burns that aren't in the same direction.

Snark,

 Intuitively it would seem so, but the Oberth Effect mucks it up.

 Your orbital velocity is vector added to Kerbin's velocity about the sun. Changing your periapsis under those circumstances is good, but changing your inclination is bad.

 By choosing a different orbital inclination about Kerbin, you are not only losing the free DV from launching East, but also losing the benefit of Kerbin's orbital velocity in proportion to the cosine of your inclination. Your inclination change faces the opposite problem; you're doing it at a moment when your velocity relative the sun is 2,300 m/sec higher than it would otherwise be.

Waiting until you have "climbed the gravity well" (so to speak) will soak up some of your speed, making the inclination change cheaper.

Firing retrograde in- plane and then doing the plane change gets you the best of both; full Oberth assistance in lowering your Pe and a lower velocity for the inclination change.

Best,

-Slashy

[Snark]

1 hour ago, GoSlash27 said:

Snark,

 Intuitively it would seem so, but the Oberth Effect mucks it up.

 Your orbital velocity is vector added to Kerbin's velocity about the sun. Changing your periapsis under those circumstances is good, but changing your inclination is bad.

 By choosing a different orbital inclination about Kerbin, you are not only losing the free DV from launching East, but also losing the benefit of Kerbin's orbital velocity in proportion to the cosine of your inclination. Your inclination change faces the opposite problem; you're doing it at a moment when your velocity relative the sun is 2,300 m/sec higher than it would otherwise be.

Waiting until you have "climbed the gravity well" (so to speak) will soak up some of your speed, making the inclination change cheaper.

Firing retrograde in- plane and then doing the plane change gets you the best of both; full Oberth assistance in lowering your Pe and a lower velocity for the inclination change.

Still not seeing it.  Let's work through an example.  Let's say I want to end up in an orbit with the same Ap as Kerbin, half the Pe, and a 45 degree inclination.

Kerbin's orbital velocity is 9284 m/s, directed in what I'm going to call the X direction (with Y being the normal axis of Kerbin's orbit).

Scenario 1:  Single burn

To be in an elliptical orbit as described, the orbital velocity would need to be 7581 m/s.  At a 45 degree angle, that would have an X component of 5360 m/s and Y component the same.  The X-axis dV would be 9284 - 5360 = 3924 m/s, and the Y-component dV would be the full 5360 m/s.  Put those together and we get a total dV relative to Kerbin (i.e. the speed with which we exit Kerbin's SoI) of 6643 m/s if we do a single burn.

Assuming an initial parking orbit at 90 km altitude, I make that a dV of 5105 m/s from LKO needed, assuming that the orbit is in the appropriate orientation.

Scenario 2:  Two burns

Now let's say we do a retrograde burn, then a normal one.  First we exit Kerbin's SoI with 5360 m/s, retrograde.  I make that a needed LKO burn of 3973 m/s.  Then, after leaving Kerbin's SoI, we do another 5360 m/s burn in the normal direction.  So the total dV needed is 3973 + 5360 = 9333 m/s of dV needed (again, assuming initial 90 km parking orbit).

So in scenario #1, that's 5105 m/s of dV, and for the two burns in scenario #2, that's 9333 m/s.  Therefore, unless I've goofed my math or incorrectly interpreted your suggestion, it seems to me that the single burn is a clear win over the two-burn solution.

(Yes, it's true you have to get to LKO first, and the equatorial orbit of the two-burn solution would be slightly cheaper than the inclined one of the one-burn solution.  But that would make only a fairly small difference, like a couple of hundred m/s at most, and wouldn't make up for the over 4000 m/s difference in dV between the two solutions.)

So... what am I missing, here?

Edited January 1, 2016 by Snark

[GoSlash27]

48 minutes ago, Snark said:

Still not seeing it.  Let's work through an example.  Let's say I want to end up in an orbit with the same Ap as Kerbin, half the Pe, and a 45 degree inclination.

Kerbin's orbital velocity is 9284 m/s, directed in what I'm going to call the X direction (with Y being the normal axis of Kerbin's orbit).

Scenario 1:  Single burn

To be in an elliptical orbit as described, the orbital velocity would need to be 7581 m/s.  At a 45 degree angle, that would have an X component of 5360 m/s and Y component the same.  The X-axis dV would be 9284 - 5360 = 3924 m/s, and the Y-component dV would be the full 5360 m/s.  Put those together and we get a total dV relative to Kerbin (i.e. the speed with which we exit Kerbin's SoI) of 6643 m/s if we do a single burn.

Assuming an initial parking orbit at 90 km altitude, I make that a dV of 5105 m/s from LKO needed, assuming that the orbit is in the appropriate orientation.

Scenario 2:  Two burns

Now let's say we do a retrograde burn, then a normal one.  First we exit Kerbin's SoI with 5360 m/s, retrograde.  I make that a needed LKO burn of 3973 m/s.  Then, after leaving Kerbin's SoI, we do another 5360 m/s burn in the normal direction.  So the total dV needed is 3973 + 5360 = 9333 m/s of dV needed (again, assuming initial 90 km parking orbit).

So in scenario #1, that's 5105 m/s of dV, and for the two burns in scenario #2, that's 9333 m/s.  Therefore, unless I've goofed my math or incorrectly interpreted your suggestion, it seems to me that the single burn is a clear win over the two-burn solution.

(Yes, it's true you have to get to LKO first, and the equatorial orbit of the two-burn solution would be slightly cheaper than the inclined one of the one-burn solution.  But that would make only a fairly small difference, like a couple of hundred m/s at most, and wouldn't make up for the over 4000 m/s difference in dV between the two solutions.)

So... what am I missing, here?

Snark,

 Your math is not including the Oberth effect.

 In your first example,  "The X-axis dV would be 9284 - 5360 = 3924 m/s" these values are not algebraically added, but rather vector added. sqrt(9284²-5360²)= 7580 m/sec, not 3294.

 The math is actually a hair more complicated than that. You take the excess velocity (that is, the difference between Kerbin's velocity and your desired Ap velocity) and square it. Then you take your velocity to escape Kerbin's SoI (which is the sqrt(2) times whatever your orbital velocity happens to be at the start of the maneuver) and square that. Then add or subtract as appropriate and take the square root Pythagorean style. Finally, you subtract your orbital velocity.

That's how transfers that occur in- plane are calculated when they cross a SoI. They are not added algebraically due to the Oberth effect.

 Inclination changes are simply twice the sine of  half the desired change times your instantaneous velocity at the moment. For example, a 60° inclination change costs whatever your velocity happens to be at the moment.

 So as you can see, doing a Pe change pays off at high instantaneous velocity, but an inclination change pays off at low instantaneous velocity. That's why it saves you DV to split up the burn; you have high velocity in LKO, and low velocity when you've exited Kerbin's SoI.

HTHs,

-Slashy

 

[Snark]

12 hours ago, GoSlash27 said:

 Your math is not including the Oberth effect.

 In your first example,  "The X-axis dV would be 9284 - 5360 = 3924 m/s" these values are not algebraically added, but rather vector added. sqrt(9284²-5360²)= 7580 m/sec, not 3294.

 The math is actually a hair more complicated than that. You take the excess velocity (that is, the difference between Kerbin's velocity and your desired Ap velocity) and square it. Then you take your velocity to escape Kerbin's SoI (which is the sqrt(2) times whatever your orbital velocity happens to be at the start of the maneuver) and square that. Then add or subtract as appropriate and take the square root Pythagorean style. Finally, you subtract your orbital velocity.

That's how transfers that occur in- plane are calculated when they cross a SoI. They are not added algebraically due to the Oberth effect.

 Inclination changes are simply twice the sine of  half the desired change times your instantaneous velocity at the moment. For example, a 60° inclination change costs whatever your velocity happens to be at the moment.

 So as you can see, doing a Pe change pays off at high instantaneous velocity, but an inclination change pays off at low instantaneous velocity. That's why it saves you DV to split up the burn; you have high velocity in LKO, and low velocity when you've exited Kerbin's SoI.

Hi Slashy,

Please read my post more carefully-- I do include Oberth effect, and I do add things in vector fashion.

Let me deconstruct it a bit more (I did some summarizing in my last post for the sake of brevity).  I'll start with the simple case, of being in a Kerbin-identical orbit but not actually at Kerbin.  Then I'll add in the extra bit that deals with Oberth effect and the fact that you're at Kerbin.

• You start with an X velocity of +9284 m/s and a Y velocity of 0.  (This is Kerbin's orbit around the Sun).
• To lower Pe down to half of Kerbin's orbit, and get the needed inclination, your velocity needs to be 7581 m/s, directed at a 45 degree angle above X.
• So the needed X velocity is 7581 * cos(45) = 5360 m/s.
• Your existing X velocity is 9284.  It needs to be 5360.  Therefore that's a change in your X velocity of 9284 - 5360 = 3924 m/s.
• Your existing Y velocity is zero.  It needs to be 5360.  Therefore that's a change in your Y velocity of 5360 - 0 = 5360 m/s.
• So we need an X component of dV of 3924 m/s, and a Y component of dV of 5360 m/s.  To add these in vector fashion, sqrt(3924*3924 + 5360*5360) = 6643 m/s.

Therefore, the velocity difference between Kerbin's orbit and the needed orbit is 6643 m/s.  If we were just orbiting the sun in an orbit like Kerbin's, and we wanted to change to the described inclined orbit, we'd need 6643 m/s of dV.

So what does this mean if we take into account the presence of Kerbin, and Oberth effect?

Well, what it means is that when we exit Kerbin's SoI, we need to be traveling at 6643 m/s relative to Kerbin.  So this gives us the question:  If we start in circular LKO at an altitude of 90 km, what dV do we need to exit the SoI with a velocity of 6643 m/s relative to Kerbin?

Well, doing the math, the orbital velocity at 90 km altitude is 2262 m/s.

Assume that we do a single impulse burn in LKO and then coast.  That means the ship's total energy (kinetic plus potential) will remain constant as it traverses Kerbin's SoI to the boundary.  Total energy is:  0.5*m*v² - GMm/R, where m is mass of ship, v is velocity of ship, GM is Kerbin's gravitational parameter, and R is radius from center of Kerbin.  Ship mass is constant, so this means that 0.5*v² - GM/R will remain constant.

At the boundary of the SoI, the velocity (calculated above) is 6643 m/s, and the SoI radius is 84159 km.  Plugging in those numbers, and then solving for the velocity at R = 690,000m (i.e. 90km altitude), that means the orbital velocity in LKO after the burn would need to be 7368 m/s.  Since we're already orbiting at 2262 m/s, that means that the dV needed at LKO would be 7368 - 2262 = 5105 m/s, as I said in my last post.  (Never mind the apparent off-by-one here, I'm rounding to the nearest m/s when I put the numbers here.)

The numbers I got for the two-burn scenario were calculated in similar fashion.  Won't go into them here unless you would like-- just want to see if you can point out if/where I go wrong in the above calculations.

Hope this helps!

Edited January 1, 2016 by Snark

[GoSlash27]

Snark,

 Apologies, I should've looked closer.

 I'll parallel your numbers this weekend (probably tomorrow) and get back to you.

Best,

-Slashy

[Snark]

2 hours ago, GoSlash27 said:

Snark,

 Apologies, I should've looked closer.

 I'll parallel your numbers this weekend (probably tomorrow) and get back to you.

Thanks!  And maybe I should actually try it in-game to see if it actually does what I think it does. 

 

[FancyMouse]

Did you guys forget argument of periapsis or I miss anything?