How to calculate delta V with direct travel from Mun to Duna?

[Sippitous]

Like the title says. Is there a way to calculate the delta V required for a transfer from a moon to a distant planet? For example, Mun to Duna, Without first stopping at the moons planet (Kerbin in this instance). It seems to save considerable delta V in most cases and for the life of me I can't find any examples of the math that would be used.

[Human Person]

how do you do the transfer? The most efficient way is to drop from the Mun to a low Kerbin periapsis and burn prograde there for your transfer. In this case, your Delta v needed is the delta v to exit the Mun towards Kerbin (310 m/s according to the map) and the delta v you'd normally need from LKO minus the velocity of a lunar transfer (860 m/s according to the map). Adds up to a net benefit of 550 m/s over a transfer from LKO.

Edited August 31, 2017 by Physics Student

[HebaruSan]

Most of the manual orbit calculators have an input field for the radius of your orbit. You can approximate the calculation you're requesting by inputting the moon's semimajor axis in that field. That gives you the "velocity at infinity (from the parent planet)" your ship would require to get on that trajectory. You can use that velocity to get the velocity at the edge of the moon's SOI (one of the few places where KSP's patched conics become relevant), then calculate an orbit that leaves the SOI with that velocity and is tangent to your current orbit with the vis viva equation. The speed at periapsis of that new orbit minus your orbital speed at that position is the delta V needed.

Regarding mods, Astrogator does this calculation if your ship is orbiting such a moon (see my forum signature). I don't think any of the other transfer calculators do. (In fact, I believe my forum signature's screenshot was captured in Mun orbit, so 391 m/s is a decent best guess for Mun->Duna.)

Edited August 31, 2017 by HebaruSan

[Gaarst]

(The following is always assuming perfect ejection angles and planetary alignement)

 

To go from a Kerbin escape trajectory to Duna, you need to get yourself in a transfer orbit between Kerbin and Duna, so with a periapsis of 13,599,840 km and an apoapsis of 20,726,155 km (Duna's SMA). This vis-viva equation tells you that for this orbit the velocity at periapsis is about 10,203 m/s.

Kerbin's orbital velocity being about 9,285 m/s, you need to add 918 m/s to that velocity to get into your transfer orbit correctly.

Now, let's suppose that you are orbiting at the Mun's altitude but are not in its SOI. Then, your orbital height is 12,000 km and your escape velocity is about 767 m/s. Using the definition of escape velocity and hyperbolic excess velocity, to escape Kerbin with 918 m/s relative velocity, you need to be going 1,197 m/s at the Mun's altitude. (V² = v_esc² + v_excess² = 767² + 918² = 1197²)

The orbital velocity of the Mun being 542 m/s, you need to add 654 m/s to that velocity at the Mun's SOI exit to get into the right orbit.

Supposing you are orbiting the Mun at 25 km over sea level (so with a SMA of 225 km), your escape velocity out of the Mun's SOI is 771 m/s. Using once again hyperbolic escape velocity, to escape Mun with 654 m/s relative velocity, you need to be going 1,003 m/s at 25 km over the Mun.

In a 25 km circular orbit around the Mun, your orbital velocity will be 538 m/s, you then need to add 465 m/s to that velocity to escape the Mun correctly, ie: you need 465 m/s of dV from a low munar orbit to get to Duna.

 

In the end, what has happened?

You are orbiting the Mun 25 km over the ground, going 538 m/s. You light up your engines in the right direction and burn 465 m/s of dV; your orbital velocity is now 1,003 m/s. You are now on an escape trajectory relative to the Mun.

When you exit the Mun's SOI, you have a velocity relative to the Mun of 654 m/s. This means that you are now going 1,197 m/s relative to Kerbin at 11,400 km over the ground. You are now on an escape trajectory relative to Kerbin.

When you exit Kerbin's SOI, you have a velocity relative to Kerbin of 918 m/s. This means that you are now going 10,203 m/s relative to the Sun at 13,599,840 km over the Sun's centre. You are now on an elliptical orbit with an apoapsis equal to Duna's SMA, if you've timed everything right, you now have an encounter with Duna, all for 465 m/s of dV!

That's 615 m/s cheaper than with a LKO transfer, and 65 m/s cheaper than the method Physics Student described above.

 

Is it worth it? No!

This method is more of a theoretical minimum than a practical thing. It assumes no relative inclination between Kerbin and Duna, Duna being at an altitude exactly equal to its SMA (happens only twice in an orbit), perfect alignment and infinite SOI (though taking finite SOI would actually reduce the total dV).
Setting such a manoeuvre would be a pain and launch windows would be never (you need Duna at the right place in its orbit, Duna and Kerbin at the right phase angle, and the Mun at the right place in Kerbin orbit, all at the same time), plus the dV saved over a more conventional escape is negligible (you'd lose that 65 m/s in inaccuracies and correction burns anyway).

[Sippitous]

Gaarst, thank you for the help. That was well answered. I think I can manage an equation out of all that info. (I am making my own dV map in excel.) I knew it was cheaper and I also knew it was not practical to hope that the moon would be in the right place at the right time to make the transfer at an optimal time. But I prefer to spend more dV to make an imperfect transfer than to try and time when the high eliptical orbit around kerbin (from the muner escape) is aligned with Kerbins orbital path so that burning at PE results in a clean Hohmann transfer (if that made sense). Also, Mun to Duna was just an example. If you are returning from one of the moons of Jool to kerbin, then it takes even less dV to fudge your transfer burn to meet up with Kerbin since Jools orbital speed Is so much slower than Kerbin.  ***Typed from my tiny phone. Sorry for spelling or any other errors.

[YNM]

I guess you need a specific case though. Unless Mun is always at the back of Kerbin every single Duna transfer, it'll vary depends on where Mun is taking you around Kerbin.

[Hesp]

I did some calculations and experiments on another old thread, but it was meant for a fuel station in Mun orbit: