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Protractor - Rendezvous Plugin - Under New Management!


mrenigma03

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Hmm, it sounds like you may have. Also, make sure to put only one calculator on a given ship.

Just to be safe, reinstall the mod and let me know if that fixes it. The mod doesn't save anything (the plugin data is just where i put the icons for the button), so you won't lose any data or anything by reinstalling.

I bet i used symitry on the darn thing! Ill re-install it too, just to be safe.

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Between the ability to burn for mun orbits in the various planetary systems and the new minimize button, this is right up there with Mechjeb in terms of "Absolute, must have swiss army knives of the KSP universe". Don't leave home without the Protractor - Planetary/Lunar Rendezvous Mod. For everything else, there's eyeballin' it.

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If someone comes up with a reason why a 270-degree heading confers an advantage over a 90-degree heading' date='

[/quote']

Read mine fly history - http://kerbalspaceprogram.com/forum/showthread.php/21505-Race-to-the-Jool-%28and-back%29

You dont always start from planet :P.

I would prefer going 270-degree, cause gravity slingshot that way was lower.

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Read mine fly history - http://kerbalspaceprogram.com/forum/showthread.php/21505-Race-to-the-Jool-%28and-back%29

You dont always start from planet :P.

I would prefer going 270-degree, cause gravity slingshot that way was lower.

Thank you, that is a very good reason. I will now see what I can do about supporting it. In the meantime, I think using the angle's explementary works (subtract the ejection angle from 360).

Edited by mrenigma03
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There is a genuine need to do retrograde orbits to reach lower planets.

No, there isn't.

dHRDt.png

If you are in a retrograde orbit because you captured into one when you arrived from another starting body, that's one thing. If you are launching from KSP on a heading of 270 to put yourself in a retrograde orbit, you're wasting a serious amount of fuel.

When launching to the east from KSC, the only difference in heading "inward" or heading "outward" is which side of the planet you are on when you start your injection burn.

Edited by RoboRay
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No, there isn't.

dHRDt.png

If you are in a retrograde orbit because you captured into one when you arrived from another starting body, that's one thing. If you are launching from KSP on a heading of 270 to put yourself in a retrograde orbit, you're wasting a serious amount of fuel.

When launching to the east from KSC, the only difference in heading "inward" or heading "outward" is which side of the planet you are on when you start your injection burn.

Technically both images are a "retrograde" burn because you are heading in the opposite direction of the planet's prograde. What there is no need for (and what your illustration describes quite well) is a western orbit (except in the case of when you get into one at the end of an interplanetary transfer). The confusion is in part due to the terminology. Either way, the next update will adjust based on whether you're in an east or west orbit automatically, so if people want to waste the delta-v going west at launch, at least the plugin still works.

Thank you very much for the image, I think it will help clear up a lot!

Ok, I'm a bit puzzled. When the Ren angle is negative, at which point is the current angle considered matching? Absolute value? Ren angle + 360?

Yes, you're correct. For negative values, add 360. This is a bug I accidentally introduced while cleaning up some code, and it is already fixed. I'll be posting an update when I finish a couple more minor things.

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Technically both images are a "retrograde" burn because you are heading in the opposite direction of the planet's prograde. What there is no need for (and what your illustration describes quite well) is a western orbit (except in the case of when you get into one at the end of an interplanetary transfer). The confusion is in part due to the terminology.

Yes, people are mixing up the retrograde orientation of their craft, the retrograde direction of an orbit around Kerbin, and the retrograde direction of Kerbin's orbit around Kerbol. They are not the same thing, and people don't specify which one they mean, so people get confused.

However, what I was replying to was not about a retrograde burn, it was specifically saying you must be in a retrograde orbit to go inward, and this is simply not true. If it was true, you could never return a craft in orbit around Kerbin to Kerbin without burning to come to a complete halt relative to Kerbin's surface then burning back the way you came from.

(And I didn't make that image, by the way... found it on Reddit.)

Either way, the next update will adjust based on whether you're in an east or west orbit automatically, so if people want to waste the delta-v going west at launch, at least the plugin still works.

Hmm... I haven't actually used this (or any of the mods) yet for transfer injections, so I don't know exactly how they work (I've just been eyeballing things), but how would polar orbits affect things? I'm doing a lot of polar orbits for MapSats, and have made a couple of planet-to-moon polar orbit transitions, but it's a little tricky. I've managed it by lining up the moon and the craft (while passing over the equator) along the edge of the planet in Map view, just like burning at moon-rise. You end up arcing high above or below the moon's inclination plane and dropping back in to it just as the moon passes by.

It would be great to have some help for polar-orbit transfers.

Edited by RoboRay
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New version up.

Changes in 2.1.4

- detects 270° orbit and adjusts ejection angles accordingly

- fixed bug where ejection angles sometimes showed negative values

- code cleanup

It would be great to have some help for polar-orbit transfers.

I think if you're going interplanetary from a polar orbit, you're better off making it equatorial first. Otherwise, as long as you're at the equator when all the angles line up, the plugin will work. The delta-v will be wrong, but all the angles will work. The problem is you may never be in a position where you're at the equator when the angles match.

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In terms of Duna's Phase Angle, it seems it's calculations are a bit off by 2 degrees, which is pretty disastrous for an accurate transition. Is the eccentricity of Duna being calculated???

No, it isn't. If someone suggests a way to account for eccentricity, I'll try to implement it. The math gets really, really complicated when you start dealing with eccentric orbits.

Edited by mrenigma03
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I think if you're going interplanetary from a polar orbit, you're better off making it equatorial first. Otherwise, as long as you're at the equator when all the angles line up, the plugin will work. The delta-v will be wrong, but all the angles will work. The problem is you may never be in a position where you're at the equator when the angles match.

Yeah, you're probably right. It would be easier to transfer out to another local body and use it's gravity to swing you around into the plane of the ecliptic, for interplanetary missions. Otherwise, it might be years or never before the right launch window opens.

It's just getting to the other local body while in a polar orbit that's the challenge. :)

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In terms of Duna's Phase Angle, it seems it's calculations are a bit off by 2 degrees, which is pretty disastrous for an accurate transition. Is the eccentricity of Duna being calculated???

Yeah the problem with Duna is that not only does it have the SMALLEST sphere or influence of all the planets (smaller than some moons, even) but it's eccentricity is something like 0.3 AU. Good luck landing on it.

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While Duna a pretty small SOI (48 Mm radius) , Moho's is smaller, both in radius (11.2 Mm) and as a fraction of its semimajor axis(Duna 0.23%, Moho 0.21%). Moho has the added issue that it's moving so much faster that it can cross a distance equal to the diameter of its SOI in as little as 20 minutes. Largest moon SOI is Tylo's at 10.8 Mm radius.

Also, Duna's eccentricity is fairly low, at 0.051. Moho, by comparison, has an eccenricity of 0.2.

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In terms of Duna's Phase Angle, it seems it's calculations are a bit off by 2 degrees, which is pretty disastrous for an accurate transition. Is the eccentricity of Duna being calculated???

Just out of curiosity, when you say they are off by 2 degrees, where are you getting that information?

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I'm lost ... see my screenshot:

http://i132.photobucket.com/albums/q18/mikeallenbrown/Ashampoo_Snap_20120926_13h29m51s_003_.png

Directions say: Wait until current phase angle (first column) matches Ren angle (second column) for the body you want to visit. Then make the Eject angle (third column) match your current eject angle on the bottom. Then burn the delta-v amount shown in the last column.

So you can see that my angels for Moho are the same (.02 off by the time I took the pic) ....my question is how do I make the eject angle match my current eject angle?? I turned the ship all over the place and all the angle does is increase.

-Mike

Edited by mikeallenbrown
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I'm lost ... see my screenshot:

http://i132.photobucket.com/albums/q18/mikeallenbrown/Ashampoo_Snap_20120926_13h29m51s_003_.png

Directions say: Wait until current phase angle (first column) matches Ren angle (second column) for the body you want to visit. Then make the Eject angle (third column) match your current eject angle on the bottom. Then burn the delta-v amount shown in the last column.

So you can see that me angels for Moho are the same (.02 off by the time I took the pic) ....my question is how do I make the eject angle match my current eject angle?? I turned the ship all over the place and all the angle does is increase.

-Mike

The eject angle is the angle of your position relative to the planet you are orbiting. You just wait until you are in the right position to start burn.

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