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Calculating Δv -- where am I going wrong?


Catbus
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I just installed the Outer Planets Mod and am trying to figure out what kinds of Δv's I'll need to reach those planets. I began setting up a spreadsheet to calculate the required orbital velocities, but something's not right. The standard equations are giving me a reasonably accurate orbital velocity of 2,295 m/s at 70 km above Kerbin, escape velocity of 950.7 m/s and orbital velocity around Kerbol of 9,282 m/s. However, when I plug in the average orbital distance of Duna (2.07 × 1010 m) and try to calculate the Δv necessary to raise myself from periapsis at Kerbin orbit to apoapsis at Duna orbit, no matter which variation of the equation I use (I've found three), I keep coming up with a Δv of about 918 m/s, rather than the 130 m/s listed on the KSP Δv map. I'm getting similarly inflated Δv's for every planet in the Kerbolar system. What on earth am I doing wrong?

The initial formula I used, provided by Scott Manley in one of his tutorial videos, is v2 = GM(2 / r - 1 / a). For M, I'm using the mass of Kerbol; for r, the orbital radius of Kerbin; and for a, the semimajor axis between the orbital radii of Kerbin and Duna. Variations found on Wikipedia provide the same result as the initial formula.

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The DeltaV map shows how much deltaV it takes to get to duna from an orbit that has a low periapsis and its apoapsis almost touches the end of Kerbin SOI, not from a low kerbin orbit. 130m/s from such an orbit is enough to get you to duna as long as you burn at periapsis.

Edited by TheDestroyer111
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I know that. That's not what I was asking. I'm trying to find the correct equation to calculate Δv to transfer from one planetary orbit to another, and I'm testing it with the Duna data, which is known information. The starting orbital altitude I'm using isn't my altitude above Kerbin, it's Kerbin's altitude above the sun -- operating from the assumption that I've already escaped Kerbin's gravity. And yet, using Kerbin's orbital altitude as the peri and Duna's as the apo, I'm getting a number that I know is wildly wrong. So the question is, if the seemingly correct equation isn't the correct one, what is?

Edited by Catbus
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Catbus,

What @TheDestroyer111 said. You've got the correct values, but you're putting them together incorrectly.  To model a transfer burn from low orbit requires excess hyperbolic velocity. Sounds daunting, but it's actually pretty easy.

DV=sqrt(Vxs^2+Vesc^2)-Vorb  where

DV= the delta-v of the burn

Vxs= excess hyperbolic velocity; the DV that would've been required had you started your transfer in Kerbin's orbit about the sun (vis-viva as you outlined above)

Vesc= escape velocity from your current orbit around Kerbin

Vorb= your current orbital velocity.

 

So in your case, your 950.7 isn't the escape velocity, but rather DV required to reach escape velocity. Your actual escape velocity is 2,295*sqrt(2)= 3,246.

Vxs= 918
Vesc= 3,246
Vorb=2,295

DV=sqrt(918^2+3,246^2)-2,295= 1,078 m/sec to Duna.

If you subtract out the DV required to escape Kerbin (950.7), that leaves 128m/sec, which shows all is in agreement between your models and the map (the map rounds up to the nearest 10).

 

As a side note, I think it's awesome that you are doing this.

Best,
-Slashy

 

 

 

Edited by GoSlash27
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13 minutes ago, GoSlash27 said:

@TheDestroyer111

So in your case, your 950.7 isn't the escape velocity, but rather DV to escape. Your actual escape velocity is 2,295*sqrt(2)= 3,246.

You're right, and I knew this. That was incorrect wording on my part.

 

Quote

Vxs= 918
Vesc= 3,246
Vorb=2,295

DV=sqrt(918^2+3,246^2)-2,295= 1,078 m/sec to Duna.

If you subtract out the DV required to escape Kerbin (950.7), that leaves 128m/sec, which shows all is in agreement between your models and the map (the map rounds up to the nearest 10).

Thanks. This must be the equation that I was missing. Does it have a name I could have been looking for?

Also, do I need to execute an inversion of that same formula to find my braking Δv at the destination planet?

Edited by Catbus
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8 minutes ago, Catbus said:

You're right, and I knew this. That was incorrect wording on my part.

 

Thanks. This must be the equation that I was missing. Does it have a name I could have been looking for?

Well... my version is a bit modified, but it's called "hyperbolic excess velocity"

@OhioBob has an excellent write-up here: http://www.braeunig.us/space/orbmech.htm#hyperbolic

Be sure to throw a quarter in his tip jar.

*edit* Stressing the importance of what TheDestroyer said upstream, if you attempt to do the burn in two stages (1 to escape Kerbin and 1 to transfer to Duna), then it really *will* work out to what you calculated initially.

Best,
-Slashy

Edited by GoSlash27
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Oh, wow, I just noticed something weird: It looks like the escape velocity at any given altitude is exactly √2 times the circular orbital velocity of that altitude. That's so simple, I'm surprised none of the pages I was reading mentioned it.

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1 minute ago, Catbus said:

Oh, wow, I just noticed something weird: It looks like the escape velocity at any given altitude is exactly √2 times the circular orbital velocity of that altitude. That's so simple, I'm surprised none of the pages I was reading mentioned it.

 Truth. I noticed that myself a while back, which is why I defined it as such above.

Point in fact, it is actually less than that in KSP due to it's patched conic model. It induces a slight error in the calculation, but the error works in your favor.

Best,
-Slashy

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35 minutes ago, Catbus said:

Also, do I need to execute an inversion of that same formula to find my braking Δv at the destination planet?

Yes... unless you plan to aerobrake.

1 minute ago, Catbus said:

So how do I use this snazzy new hyperbolic excess velocity concept to calculate braking Δv for capture?

Same exact way. Pretend you're doing a transfer from that planet home. The capture burn is the same DV as a transfer would be.

Best,

-Slashy

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48 minutes ago, Catbus said:

Also, do I need to execute an inversion of that same formula to find my braking Δv at the destination planet?

Yes, the formula works the same regardless of whether you are departing a planet or arriving at one.

Be advised, however, that Vxs is equal to the relative velocity between the spacecraft and the planet.  This is the difference between the velocity vectors, not simply the different between the scalar magnitudes.  This is important because it is rarely ever possible to achieve a pure Hohmann transfer, the velocity vectors at arrival will almost always be crossing at an angle.  It is necessary to take this into account when computing the relative velocity.

(edit)  You can find more information here:  http://www.braeunig.us/space/interpl.htm#hyperbolic

Edited by OhioBob
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10 minutes ago, Catbus said:

Aerobraking at Jool produced less than optimal results, so for the OPM, let's work from a non-aerobraking assumption. :-D

While we're on the subject...

 You picked up right away that escape velocity is a function of orbital velocity, and both are therefore a function of orbit radius.

If you rewrite the whole thing in terms of radius, you will notice that there's a happy spot where the DV hits a minimum, and that's not actually low Kerbin orbit. Taking the first derivative will yield the "gate orbit".

That opens up all sorts of new possibilities that don't fit on a DV map :wink:

Best,
-Slashy

 

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3 minutes ago, Catbus said:

Is there a way to determine what one's incoming velocity will be after executing one of these hyperbolic transfers?

Yes but it's not easy.  The entire web page that I referenced you to explains it all, but it is a very lengthy and involved process.  And each transfer is unique, so there's an unlimited number of variations.  When I first started playing KSP I wrote my own  launch window calculator, but it was way more involved than I can explain here.  Now I just use is this one: http://alexmoon.github.io/ksp/.  Note that that site acknowledges my web site as providing most of the math.  The online version doesn't include the OPM planets, but you can add a planet if you know its orbital elements.  Easier yet is to use the mod version, which imports all the OPM planets as well as the stock planets. 

 

 

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28 minutes ago, OhioBob said:

Yes but it's not easy.  The entire web page that I referenced you to explains it all, but it is a very lengthy and involved process.  And each transfer is unique, so there's an unlimited number of variations.

True, but you can "ballpark" it well enough for planning purposes. It's simply the orbital velocity + the DV of a theoretical transfer.

Doing a hypothetical transfer to Jool...

70 km above Kerbin to 100 km above Jool would hit Jool's atmosphere going 7,107 m/sec. There's variations, but that's a good ballpark.

 Assuming Catbus wants to avoid aerobraking, the cheap transfer to the Jool system is actually Pol. The gate orbit at Jool's end is 177Mm; conveniently close to Pol's 180Mm SMA. Vxs at Pol is only 1,242 m/sec, which reduces the DV for a Pol retroburn (and subsequent shot home) to 1,122 m/sec.

Best,
-Slashy

Edited by GoSlash27
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22 minutes ago, GoSlash27 said:

True, but you can "ballpark" it well enough for planning purposes. It's simply the orbital velocity + the DV of a theoretical transfer.

That's true.  And if we want to get a little more complex (but not go completely overboard), the best method is probably to treat the transfer as a one-tangent burn, as described here:

http://www.braeunig.us/space/orbmech.htm#maneuver

The examples provided describe transfers between Earth orbits, but the method can also be used for heliocentric interplanetary transfer orbits.  In the latter case, the values of ΔVA and ΔVB would be the values if Vxs at departure and arrival.  We would then use Slashy's equation to compute the ejection and insertion Δv.  Even this method is somewhat limited because it assumes the initial and final orbits (i.e. those of the planets) are circular and coplanar.

@Catbus, for what you are doing I think the one-tangent burn technique might be your best bet.  It provides a pretty good ballpark estimate for planning purposes and it's still relatively simple as far as the math is concerned.  Trying to be any more precise than this and you'll be getting into some really messy and complex math.

 

 

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20 hours ago, GoSlash27 said:

Yes... unless you plan to aerobrake.

Same exact way. Pretend you're doing a transfer from that planet home. The capture burn is the same DV as a transfer would be.

Best,

-Slashy

OK . . . so now I'm computing a capture Δv of 246.7 for Duna and a circularizing Δv of 368.8, both of which are pretty close to the figures on the Δv map (250 and 360, respectively). I think I have most of what everything I need now to calculate Δv's for the outer planets. Thanks for the help.

ETA: My max inclination change Δv formula worked right the first time. :-)

Edited by Catbus
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Aw, shucks. :-)

BTW, while under other circumstances I might be up for the challenge, I'm not sure I still have the math chops to determine the gate orbit equation. I had a hard enough time with calculus the second time I took it, and that was 30 years ago.

(Don't ask me about the first.)

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10 hours ago, Catbus said:

OK . . . so now I'm computing a capture Δv of 246.7 for Duna and a circularizing Δv of 368.8, both of which are pretty close to the figures on the Δv map (250 and 360, respectively). I think I have most of what everything I need now to calculate Δv's for the outer planets. Thanks for the help.

ETA: My max inclination change Δv formula worked right the first time. :-)

Sounds like a winner. If you get around to posting your findings as an extension of the DV subway map, I'm sure the community would be eternally grateful!

Best,
-Slashy

 

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Well, that would require computing Δv's for all the OP satellites as well, and I'm not there yet. But here's what I've calculated for the planets themselves (rounded up to the nearest 10 m/s):

  Kerbin Escape Δv (m/s) Transfer Δv (m/s) Max Inclination Change Δv (m/s) Capture Δv (m/s) Circularization Δv (m/s)
Sarnus 960 1,300 440 270 1,550
Urlum 960 1,530 150 340 910
Neidon 960 1,620 290 230 1,010
Plock 960 1,660 1,400 650 210
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