Jump to content

Transfer delta v


Recommended Posts

Hi.

why do all delta v tables around state the delta v required to go from just outside kerbin SOI to intercept, for instance, eve, is around 80m/s when one can easily see that, if I am on kerbol orbit at approx. the same distance than kerbin ( that is, just left from kerbin SOI) I need around 700 m/s of deceleration to intercept eve orbit (not counting with plane changes)?

Edited by Jaeleth
Link to comment
Share on other sites

http://forum.kerbalspaceprogram.com/threads/96985-1-0-2-WAC-s-Delta-V-Map-(24-05-2015)

this one says 430, it will definitely depends on the quality of the transfer window.

According to http://alexmoon.github.io/ksp/, if you start from 85000Km of Kerbin and target the same orbit on Eve, it will cost you about 1250 total (including insertion, 500+)

Link to comment
Share on other sites

That low delta-v number is not the Kerbin SOI to Eve transfer number.

It is the extra delta-v needed from LKO to get an Eve transfer, relative to the delta-v to exit Kerbin SOI.

From just outside Kerbin's SOI, on Kerbin's orbit, it costs several hundred m/s to intercept Eve,

but if you had initiated your Eve transfer from LKO, it would have cost less than 100 m/s more than Kerbin escape velocity to intercept Eve.

You get huge benefits from doing your transfer directly from low Kerbin orbit.

Edited by Yasmy
Link to comment
Share on other sites

http://forum.kerbalspaceprogram.com/threads/96985-1-0-2-WAC-s-Delta-V-Map-(24-05-2015)

this one says 430, it will definitely depends on the quality of the transfer window.

According to http://alexmoon.github.io/ksp/, if you start from 85000Km of Kerbin and target the same orbit on Eve, it will cost you about 1250 total (including insertion, 500+)

1250, more likely, 700 for transfer, 500 for insertion without aerobrake, of course,.430 I can't see how, whatever transfer window

Link to comment
Share on other sites

Most delta-v maps do not list the delta-v for the kind of transfer you are doing. Your method is more expensive than the standard insertion from low orbit around the original body.

Link to comment
Share on other sites

Looking at your first example, it says 430 from just outside Kerbin's SOI to Eve.

It says 90 if you combine your Kerbin escape burn with your transfer burn.

The best escape from kerbin burn should than be in a way that I am accelerating to escape kerbin while pointing BACKWARDS towards kerbin movement aroun the sun, right? If we consider kerbin heading while orbiting the sun to be 360, then, I should perform the escape burn at around radial 270, while on a normal, anti-clockwise orbit (W to E)?

Link to comment
Share on other sites

The best escape from kerbin burn should than be in a way that I am accelerating to escape kerbin while pointing BACKWARDS towards kerbin movement aroun the sun, right? If we consider kerbin heading while orbiting the sun to be 360, then, I should perform the escape burn at around radial 270, while on a normal, anti-clockwise orbit (W to E)?

Ignore that post. I was wrong, which is why I deleted it. The 430 m/s was 80 at kerbin (not in interplanetary space near kerbin, but LKO) + 350 plane change at the line of nodes in interplanetary space.

But yes, you should exit Kerbin retrograde to go to the inner planets.

- - - Updated - - -

If you want to follow most delta-v maps, do the Kerbin escape burn and the transfer burn in one go, while still in low orbit around Kerbin.

Then it will only take approximately 90m/s more than Kerbin escape velocity to drop your orbit down to Eve.

If you escape Kerbin, then perform the transfer burn, it will cost a lot more delta-v.

Link to comment
Share on other sites

Most delta-v maps do not list the delta-v for the kind of transfer you are doing. Your method is more expensive than the standard insertion from low orbit around the original body.

Can you point me to a description of standard insertion from LO around original body? Escaping kerbin SOI first and then decelerating to Eve was the only way I thought possible to transfer.

- - - Updated - - -

Can you point me to a description of standard insertion from LO around original body? Escaping kerbin SOI first and then decelerating to Eve was the only way I thought possible to transfer.

never mind, yasmy, you already answered above, thanks :)

Link to comment
Share on other sites

Glad to help. I'm sure someone will point you to a nice video tutorial.

I use one of the transfer calculators to find a transfer window, time warp to the window,

plop a maneuver node in LKO and fiddle around until the trajectory points along Kerbin pro/retrograde and it intercepts my target.

Sometimes you have to drop a second node in interplanetary space to perform a plane change maneuver before you can generate an encounter.

Link to comment
Share on other sites

You want to exit Kerbins SOI toward Kerbin's retrograde. However, you dont want to actually burn exactly on the day side of kerbin. Gravity will sling you around while you move towards the SoI edge. You need to burn at the right ejection angle.

Take a look here: http://ksp.olex.biz

This will tell you the ejection angle for your specific transferburn. Alternatively, you can always just drag the maeuvernode around on the orbit to align your exit vector with kerbin's orbit.

You actually want to do the escape and the transfer in a single burn in LKO. This way you take advantage of the Oberth effect and it will only take a hand full of m/s more than ascape velocity to get an encounter. That's why there are these low numbers on the delta v map. Those maps always assume you do things in the most efficient way.

Link to comment
Share on other sites

Just to toss in the word: Oberth Effect. This effect is the reason why it's more efficient to burn directly from low orbit into the transfer.

Shameless self-quote:

On the Oberth Effect.

I just found that the Wikipedia article on Oberth effect is not very descriptive, and I think that adding an example will help to illustrate.

For this, let's start from a circular low orbit around Kerbin. Our orbit has a certain radius r, and our ship has a certain velocity v0. We already used the formulas on the Wikipedia article on Hohmann transfers to calculate the dV needed to enter the Hohmann orbit to Duna (let's call it dVh), but that number applies only if we are already free of Kerbin's gravity. Naively one might now think, that one could simply add the dV needed to escape Kerbin to the dVh needed to go into the Hohmann transfer, but the situation isn't that easy. Thing is, there's the principle of conservation of energy, and energy is the quantity we can add here. Let's assume we burn and add to our current velocity v0 an additional velocity dV0. The kinetic energy of our ship after the burn will then be m*(v0+dV0)^2/2. In addition, we sit in Kerbin's gravity well, adding a (negative) energy of -GM*m/r, where GM is the standard gravitational parameter of Kerbin. After we leave Kerbin's gravity well, we would still like to have enough kinetic energy to enter the Hohmann transfer, so we'll need a kinetic energy of m*dVh^2/2. This gives us the energy balance:

m*(v0+dV0)^2/2-GM*m/r=m*dVh^2/2

The required dV for a burn using Oberth effect can now be found by solving this equation for dV0.

(I'm not sure if this is really correct. Please notify me if there's a mistake, I'll fix it asap.)

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...