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Ejection angle


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In tools like http://alexmoon.github.io/ksp/, it shows an "ejection angle" of for example 107º to Retrograde. How would one go about measuring that? Is that the angle I start my departure burn from LKO, or the angle I should be heading after I'm done burning?

I've been using MechJeb for this, but it's been doing a very bad job of getting me to Moho with something under a million dV for the capture burn.

Edited by GoatRider
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Enter the map view and move the camera so it's looking at the system from above.

Zoom out to see Sun and move it to comfortable position - I usually move it to the left, horizontally from my position:

OlYjZEx.png

Without moving the camera, zoom in the map view ("+" key on numpad or mouse wheel; I don't have anything in orbit at the moment, sorry)

You get 0° prograde straight up, 90° prograde straight to the right, 0° retrograde straight down, and 90° retrograde straight to the left.

e4YOqxK.png

With that, it is possible to guess the angle and place the maneuver quite well. I don't usually need to put it exactly to the right position as the necessary tuning to get the intercept is rarely more than a few m/s.

Next thing I usually do is I read and set the normal value of the maneuver (opens if you click on the "i" button next to dv value) using the normal/antinormal icon, and then I top it up to full dv using prograde maneuver icon. Last but not least, I fix the maneuver so it shows me actual intercept by experimenting with all maneuver icons.

Edited by Kasuha
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What about this part of the question that I also am interested in:

Is that the angle I start my departure burn from LKO, or the angle I should be heading after I'm done burning?
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I just compared these numbers with what MechJeb is using. It came out with numbers pretty much the same as the "Mid point correction" method from the alexmoon site. And the ejection node was a bit to the right of the line between Kerbin and the Sun. alexmoon said that angle should be 107º from Retrograde, so I was thinking of it wrong. I was thinking of the tangent that was headed that way, so I was 90º off. That's what was confusing me.

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What about this part of the question that I also am interested in:

Is that the angle I start my departure burn from LKO, or the angle I should be heading after I'm done burning?

Both.

The system assumes you're doing the burn instantly, i.e. you point your ship in the requested direction, and apply the impulse within one frame. If you don't and your burn is particularly long, you need to apply corrections for inaccuracies caused by long burn. Unfortunately the maneuver indicator on navball is extremely unhelpful in that because if you follow it, you'll end up ejecting in a very different direction than where you want to go. So if you don't have enough experience in making these corrections (and don't want to spend a long time learning them), it's recommended to pack sufficient thrust to execute the burn in under one minute. Even in that case, you should just point your ship in the requested direction and apply all the thrust in that direction.

Edited by Kasuha
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I tend to add the burn first then change the position, so you create the maneuver, set the DV to the required amount, then move the maneuver node around the planet until you will be travelling parallel with kerbins vector on kerbin SOI departure, assuming you have this all correct then as soon as you leave SOI the kerbol periapsis (or apoapsis depending on which planet you're shooting for) should be located on the place you performed the burn also while you've just finished the burn, the orbit around kerbol should be visible as a huge great yellow line passing through the kerbin SOI (but not showing an encounter) in the rare event of a "perfect" maneuver this line should pass right through kerbin. This would mean your departure from LKO is EXACTLY on the kerbol periapsis or apoapsis (depending on direction of travel) although even with a perfect departure, you might still need to tweak your orbit since the timing could be up to about 17 minutes out depending on the departure window.

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then move the maneuver node around the planet until you will be travelling parallel with kerbins vector on kerbin SOI departure

When I started to use the calculator, I spent some very frustrating moments trying to get the encounter this way. Then I realized the calculator sometimes sends you deliberately in a different direction because the optimum trajectory has some radial component.

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When I started to use the calculator, I spent some very frustrating moments trying to get the encounter this way. Then I realized the calculator sometimes sends you deliberately in a different direction because the optimum trajectory has some radial component.

I cant understand why there would be an additional radial component, surely in the event of a standard hohmann transfer the target body is always encountered at the opposite side of the orbit from the initial burn, why would this change for an interplanetary transfer?

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I cant understand why there would be an additional radial component, surely in the event of a standard hohmann transfer the target body is always encountered at the opposite side of the orbit from the initial burn, why would this change for an interplanetary transfer?

I can't comment on why exactly the calculator does that, the matter is it does. It happened to me the first time with Eeloo and I can remember quite well that the trajectory had periapsis below Kerbin orbit and apoapsis way behind Eeloo orbit but if I tried to change it to Hohmann transfer I couldn't get an intercept no matter what I tried. My guess is it's just how the math works out if your target has some significant inclination and/or eccentricity.

So my experience is that the most reliable approach is to eyeball the angle in map view and then, if needed, move the trajectory by adding some radial component rather than trying to move the maneuver. Pretty often I manage to set up the intercept a few m/s below what the calculator is telling me I'll spend.

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Perhaps its due to the discrepancy caused by the variable travel time from LKO to the edge of SOI, since kerbin would be moving around its orbit the actual kerbin escape vector is thrown off with different escape speeds the angle would vary.... or, something.

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