�v for return calculation

[Floule]

Hi everybody!

Currently I'm learning the basics of calculating how much ðÂݴv I need for different things. So let's say i want to go to mun, land there and return. To get to the mun I need a total ðÂݴv of 6260 m/s, right? (according to this) So with this map, how do I calculate the ðÂݴv needed for return?

Regards

Floule

[Febin]

I believe you need about 650 to return and enter atmosphere. As long as you have a medium amount of fuel you should be fine, if worse comes to worse, leave the craft in circular lunar orbit and send a larger craft to rescue him.

[Umlüx]

you need roughly the same amount of dV for landing and takeoff of a moon or planet without atmosphere.

as for the return to kerbin.. i think you need roughly 300-350 m/s if i remember correctly? could be less..

[Boonestorm]

Look at the map, you can also read it the other way round. Going back needs roughly 640 for low orbit and another 210 for escape. The rest of the path can be neglected as long as your periapsis is low enough in kerbins atmosphere to aerobrake sufficiently (e.g. you can drop the 860 and 4550 you spend on the return).

Hope that helps.

[Floule]

you need roughly the same amount of dV for landing and takeoff of a moon or planet without atmosphere.

as for the return to kerbin.. i think you need roughly 300-350 m/s if i remember correctly? could be less..

Yes, about 310 m/s for return from mun. But how can I calculate it?

[numerobis]

A guide!

Calculate the Hohmann transfer between LKO (let's say at 100km) and Mun's orbit. That gives you two values: 841 m/s and 362 m/s. The first burn is what you need to do from your 100km orbit to intersect Mun's orbit. The second burn is what you need to do to circularize at Mun's orbit. Reverse the burns to go the opposite way, from Mun's orbit down to LKO.

You can improve on the 362 burn because you aren't randomly circularizing in the middle of space, you're diving into a gravity well. When taking off, you need to hit the limit of Mun's SOI at 362 m/s relative to Mun, going retrograde (conversely, if you're going to Mun, and you did the Hohmann transfer to get there, you'll hit the SOI at 362 m/s). If you're at a 10km circular orbit around Mun, you've already got a bunch of speed from your orbit, so you only need to thrust for 278 m/s. That explains why 310 m/s is a pretty safe assumption.

When you're returning from Kerbin, you can use the atmosphere to do the bulk of the 841 m/s "burn"; if you just want to land, you can have the atmosphere do all of it.

A theoretically optimum liftoff from a body with no atmosphere costs you as follows: first go into orbit at your landing altitude (about 570 m/s on Mun, +/- 9 m/s from the rotation of the Mun), then do a Hohmann to your parking altitude, a total of 575.4 m/s to get to 10km. That's assuming you have infinite thrust and no mountains to avoid; 600 m/s is a pretty good manual ascent.

There are various calculators around; I use my KSP-scripts. My python session for this post looks like:

>>> import planet
>>> planet.kerbin.hohmann(100000, 11.4e6)
(841.5988234288104, 362.37616808920507)
>>> planet.mun.soiBurn(10000, 362)
278.39651256887646
>>> planet.mun.orbitalVelocity(0)
570.6943031124456
>>> planet.mun.siderealRotationSpeed
9.041595157974422
>>> planet.mun.hohmann(0, 10000)
(6.917759278702847, 6.833885951939877)
>>> planet.mun.orbitalVelocity(0) - planet.mun.siderealRotationSpeed + sum(planet.mun.hohmann(0, 10000))
575.4043531851139

[KvickFlygarn87]

If you are merely doing a flyby, you could put yourself on a perfect retrograde Mun orbit and you can return to kerbin without burning any fuel! This is called a free return trajectory and it was used in the Apollo missions to the moon, in case an engine misfired.

[Floule]

A guide!

Calculate the Hohmann transfer between LKO (let's say at 100km) and Mun's orbit. That gives you two values: 841 m/s and 362 m/s. The first burn is what you need to do from your 100km orbit to intersect Mun's orbit. The second burn is what you need to do to circularize at Mun's orbit. Reverse the burns to go the opposite way, from Mun's orbit down to LKO.

You can improve on the 362 burn because you aren't randomly circularizing in the middle of space, you're diving into a gravity well. When taking off, you need to hit the limit of Mun's SOI at 362 m/s relative to Mun, going retrograde (conversely, if you're going to Mun, and you did the Hohmann transfer to get there, you'll hit the SOI at 362 m/s). If you're at a 10km circular orbit around Mun, you've already got a bunch of speed from your orbit, so you only need to thrust for 278 m/s. That explains why 310 m/s is a pretty safe assumption.

When you're returning from Kerbin, you can use the atmosphere to do the bulk of the 841 m/s "burn"; if you just want to land, you can have the atmosphere do all of it.

A theoretically optimum liftoff from a body with no atmosphere costs you as follows: first go into orbit at your landing altitude (about 570 m/s on Mun, +/- 9 m/s from the rotation of the Mun), then do a Hohmann to your parking altitude, a total of 575.4 m/s to get to 10km. That's assuming you have infinite thrust and no mountains to avoid; 600 m/s is a pretty good manual ascent.

There are various calculators around; I use my KSP-scripts. My python session for this post looks like:

>>> import planet
>>> planet.kerbin.hohmann(100000, 11.4e6)
(841.5988234288104, 362.37616808920507)
>>> planet.mun.soiBurn(10000, 362)
278.39651256887646
>>> planet.mun.orbitalVelocity(0)
570.6943031124456
>>> planet.mun.siderealRotationSpeed
9.041595157974422
>>> planet.mun.hohmann(0, 10000)
(6.917759278702847, 6.833885951939877)
>>> planet.mun.orbitalVelocity(0) - planet.mun.siderealRotationSpeed + sum(planet.mun.hohmann(0, 10000))
575.4043531851139

Okay, thank you very much! I'm a bit confused with those calculations. Whats the calculation for the return, according to the ðÂݴv Map I posted?

[Boonestorm]

From Floules numbers, 310 m/s to get from Munar Orbit to Kerbin intercept + 600 m/s to get into Munar Orbit. Totals 910 m/s.

You can also get that from the map; the way back is the same as the way to the mun:

640+210=850 m/s

(the numbers differ a bit because it also depends on your flying skills and Floules added some safety margins)

Usually you would need to add the 860 and 4550 m/s from Kerbin as well; however, the atmosphere will do that braking for you.

(Sorry for my posts taking a while to show up, but I still need to be verified by a mod )

Cheers,

Boonestorm.

[capi3101]

The delta-V map works both forwards and backwards - so to go to the Mün, it works like this:

Launch to 80k LKO: 4,550 m/s.

Munar Transfer from 80k LKO: 860 m/s.

14k LMO from Munar Transfer: 210 m/s.

Land from 14k LMO: 640 m/s.

Launch to 14k LMO: 640 m/s.

Transfer to Kerbin from 14k LMO: 860 m/s.

The total delta-V you need for a full round-trip mission is the sum of these values, in this case 7,760 m/s. 6,260 m/s does indeed get you there; another 1,500 m/s gets you back. When you transfer back, put your Kerbin periapsis around 30k and you'll land for free.

Bear in mind that the delta-V map assumes "best case scenario"; depending on how well you pilot, the values can vary significantly (and always higher), so pack extra fuel if you have low confidence in your piloting skills. Not too much though - 5% is generally sufficient within Kerbin's SOI; 25% is generous (read: wasteful) no matter where you're going.

[numerobis]

Okay, thank you very much! I'm a bit confused with those calculations. Whats the calculation for the return, according to the dv Map I posted?

My post was largely about how to calculate the return. I have never been able to really understand the deltaV maps.

Edited October 23, 2013 by numerobis

[numerobis]

Transfer to Kerbin from 14k LMO: 860 m/s.

No, that's the part that you can do in the atmosphere. From 14k LMO to Kerbin is symmetric to the 14k LMO from transfer, so 210 using your numbers (but that's a lot lower than what I calculate, which is more like 280 m/s).

[capi3101]

No, that's the part that you can do in the atmosphere. From 14k LMO to Kerbin is symmetric to the 14k LMO from transfer, so 210 using your numbers (but that's a lot lower than what I calculate, which is more like 280 m/s).

Hmm...you've got a point. Reading it backwards it would be 210 to kick from the 14k LKO back to a transfer orbit with a 80k Kerbin periapsis. The discrepancy (you reading 280) might be what it takes to get that down to the ground instead.

So hack 580 off of what I said earlier - you get 7,180 m/s for a round-trip Mün journey. I think 7,280 is the figure that everybody always throws, and now I know why my Mün rockets always come back to Kerbin with some fuel left in the tank......

[Floule]

Hmm...you've got a point. Reading it backwards it would be 210 to kick from the 14k LKO back to a transfer orbit with a 80k Kerbin periapsis. The discrepancy (you reading 280) might be what it takes to get that down to the ground instead.

So hack 580 off of what I said earlier - you get 7,180 m/s for a round-trip Mün journey. I think 7,280 is the figure that everybody always throws, and now I know why my Mün rockets always come back to Kerbin with some fuel left in the tank......

Those 210 m/s for the return sounds good. But the discrepancy can't be for getting down on the ground. Because you need less ðÂݴv for getting down on the ground than into orbit.

[numerobis]

It's a couple meters per second either way if you're leaving from any low lunar orbit to any low kerbin orbit. My personal experience suggests that 300 m/s is pretty good piloting. I suspect the map has a bug.