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Intermediate Question: What is the ideal orbital altitude for Mun & Minmus transfer?


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And the slower you're going, the easier it is to deflect your direction of motion by a particular amount.

This is interesting, but I'm struggling to understand it. Can't the dimensions be handled as two separate vectors that add up to the final trip? There's the 47,000km horizontal trip and the 4900km vertical trip, and if you're going to have to cover the vertical leg one way or another, why does it matter at which point in the journey you make the burn? It's still going to have to be enough fuel to cover the 4900km at whatever speed is required to complete that leg before the horizontal leg comes to its end, isn't it?

(I hope I don't sound argumentative. I've learned that when I disagree with Maltesh, it's because *I'm* wrong. I just want to learn why I'm wrong. :) )

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Thanks for the response, sir. If I understand you, instead of merely adding 6° to my 90° (newb mistake) I should launch at 6°? That's great!

Not exactly. Minmus orbits on a plane inclination difference of 6 degrees from Kerbin. So you either add or subtract the 6 degrees to the heading depending on when you launch. I did a write up explanation on this for someone in another thread. Here it is with some photo's as well.

Minmus Inclination

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