How to plan a tylo gravity assist?

[Space Nerd]

I planned a jool transfer node and can't find a encounter with tylo that slowed me into orbit without spending ridiculous amount of dv , does anyone know how to do it?(I'm going to laythe)

Edited July 23, 2019 by Space Nerd

[bewing]

You are trying to do this before you've even done your ejection burn from Kerbin?

I don't think it really works that way. Just do the bare minimum transfer to Jool for now. You set up your Tylo gravity braking encounter when you are doing your mid-course correction. That's the only time you will have sufficient precision to actually make an encounter work.

[Zhetaan]

38 minutes ago, Space Nerd said:

[D]oes anyone know how to do it?

As @bewing said, you're not going to get the very best encounter until you're already on your way there--it is possible with the right timing, but even multiple-planet flyby assists (of the same kind as the Voyager Grand Tour) will require correction burns to get good encounters.  However, when the time comes, the encounter you choose will be critical.

The most important things to know about gravity assists are:

1. The key to slowing relative to Jool is the direction by which you exit from Tylo's sphere of influence.  No matter how you approach, you need to exit Tylo's sphere as close to Tylo's retrograde as you can get it.
2. The best gravity assists warp your path by 180 degrees, meaning that you approach Tylo from behind (so you and Tylo are both going in Tylo's prograde direction), you whip around Tylo as close to its surface as you can manage, and then you exit to Tylo's retrograde as I mentioned before.  This gives the maximum Jool-relative velocity reduction.  The reason to go so close to Tylo's surface is to take advantage of the stronger gravity to warp your trajectory to something close to 180 degrees.
3. You may find that a 180-degree turn is not possible with your approach (this is what usuallu happens).  In that case, you need to modify your approach, but in any case, the retrograde exit from Tylo's sphere is non-negotiable.  You can enter Tylo's sphere from a normal or radial direction, if you like, but that comes at the cost of limiting the velocity reduction.
4. You can also light the engines at your Tylo closest approach.  It stops being a gravity assist at that point, but you get the full benefit of the Oberth Effect.

Look at this for more information (and animations of the various ways assists can work).

Edited July 23, 2019 by Zhetaan

[Superfluous J]

As said above, do this at your mid course correction, and aim for the correct area.

HOW do you do that? Very lightly Use the new maneuver editor (in particular the slider that allows you to to make larger or smaller changes) or a mod and dial down your steps/precision as you get closer and closer.

Set Tylo as a target, and then fiddle with your encounter until your orbit is passing by just by it, to the right of Jool (in default map mode, make sure your ship is passing Tylo so you're going in the same direction it is - albeit much faster - as you pass by). You've got maybe a 1/10 chance of actually encountering Tylo here. If you to great, if not, use tiny (like, the smallest) prograde/retrograde and radial/antiradial steps to move Tylo around its orbit as you near. Each way prograde or retrograde or the radials will move both your orbit a bit and change the time you get to Tylo, causing it to be closer to you when you arrive. All you need to do is fiddle with each until you are showing up as Tylo is there.

Then you need to further tweak it as above, so your resultant orbit is nice. I'd try to shoot for one that goes lower than Vall at the top and not much lower than Laythe at the bottom, but don't sweat it too much, just cutting out that initial slowdown saves a ton of fuel. Sometimes many, many tons.

[Snark]

24 minutes ago, Space Nerd said:

I planned a jool transfer node and can't find a encounter with tylo that slowed me into orbit without spending ridiculous amount of dv , does anyone know how to do it?(I'm going to laythe)

Here's what you need:

1. Have a good transfer trajectory from Kerbin in the first place.
2. Understand the basic "how it works", i.e. what it is you're trying to do.
3. Know where Tylo is supposed to be upon your arrival, and what path your craft should be taking.
4. Adjust your path, midway on your trip from Kerbin, so that Tylo (and your craft) will be in the right place upon arrival.

Details on each of these, below.

 

Step 1:  Have a good transfer from Kerbin

You need to have an efficient Hohmann transfer from Kerbin.  This is because you want to arrive at Jool with the lowest possible Jool-relative speed.

I mean, obviously you want that anyway to try to lighten your dV requirements.  But also, it's especially critical for using a gravity assist.  Because the faster you go whizzing past a planet or moon, the less dV you can squeeze out of it in a gravity assist, because there's less time available for its gravity to bend your path.

I've done Tylo gravity assists, and they work great if set up properly with a decent Hohmann transfer:  i.e. can shave a few hundred m/s off the craft's speed, and actually capture to Jool without spending a single gram of propellant.  So it's totally doable. 

Just, make sure you have a pretty good transfer window, or a Tylo assist won't help you all that much, is all.

 

Step 2:  Understand the basic "how it works" (e.g. "why am I finding this so hard")

Long-winded details in spoiler, but  the key concept to bear in mind here is that timing matters.  Tylo has to be in the correct place.

Spoiler

Specifically:  If you have a craft that's arriving at Jool from interplanetary space, and you want to use a Tylo assist to slow down, that's only going to work if Tylo is in the right part of its orbit relative to the direction you're approaching Jool from.  Otherwise it's simply not going to work.

And that's why you're having trouble finding an intercept that works for you.  When you're setting up your ejection from Kerbin, you're probably trying to pick the right transfer window to get from Kerbin to Jool with minimum dV, amirite?  So you're already focusing on timing... for a different purpose.  You're focusing on the right timing to launch from Kerbin to Jool.  But if you've done that... then "where will Tylo be when I arrive" is essentially completely random.  That's a completely different (and unrelated) timing cycle.

It's not impossible... but it's tricky.

Fortunately, you don't have to care, because there's a straightforward workaround, as @bewing describes above. 

Here's the deal:

Tylo takes just shy of ten Kerbin days to make one orbit around Jool.  So, suppose you didn't do any planning and just arrived at Jool without trying to coordinate with Tylo.  That means that Tylo would be at some random position that could be within plus-or-minus five days (in its orbit) of the "correct" position for a gravity assist.

So, if you can adjust the timing of your arrival at Jool by up to plus-or-minus five days, you can set things up so that it will be in the right spot.  As bewing suggests, do a mid-course correction en route to Jool.  If you do this when you're still a few hundred days away from Jool, that means you're altering your trip time by only a few percent, so you won't need to spend large amounts of dV.

 

Step 3: Know where Tylo is supposed to be upon your arrival, and what path your craft should be taking.

(In the following discussion, I'll use "left" and "right" in the sense of "north is up" that KSP uses in space.)

When your craft is approaching Jool, you'll see the face of Jool moving from left to right as it rotates.  The moons are orbiting in the same direction:  from left to right as they pass in front of Jool, and from right to left as they pass behind it.

• You're going to want your craft to pass to Jool's right (i.e. you want to be orbiting in the direction relative to Jool's rotation).
• You want Tylo to be in front of Jool (not behind), i.e. between you and Jool:  you want to meet Tylo before your Jool Pe, not after.
• You want Tylo to be slightly to the right of Jool as you approach from interplanetary space.
• You want your craft to whizz past Tylo as close as possible to Tylo's surface, in order to maximize the dV transfer from the assist.
• You want your craft to pass to the right of Tylo (i.e. you want to go past its leading edge in its orbit), so that it will slow you down.

Put another way:  If Jool is the center of a clock dial, and Tylo is at the rim of the clock and orbiting counterclockwise, and your craft is approaching from the 6 o'clock direction, you want to pass just to Tylo's right when Tylo is in about the 5 o'clock position.

 

Step 4:  Adjust your arrival timing to make that happen.

Wait until you're around halfway from Kerbin to Jool, when you're still many hundred days away from Jool arrival.  Presumably you already have an intercept with Jool's SoI (since you set that up when you left Kerbin).  So now you just need to adjust things for your gravity maneuver.

You'll be doing this by plopping down a maneuver node, there at the midway point, and adjusting the handles.  Here's what you do:

1. Focus the map view on Jool, so you can see your path through the system.
2. You need to know where Tylo's going to be, given your current timing.  To do this, just use the controls on your maneuver node to scan your arrival trajectory back and forth from left to right through the Jool system, from the left edge of Tylo's orbit to the right edge of Tylo's orbit.  Somewhere in there, you'll see your trajectory show a Tylo intercept.
3. Okay, now you know where Tylo's going to be at the time of your arrival, if you didn't do any tinkering with the timing.  This tells you what you have to do about the timing.  Maybe you got super lucky and Tylo's already in the right place, in which case yay, you're done.    But it probably isn't.  So, unless you got really lucky, it will be one of the following two cases:
   • Tylo is behind where it needs to be (in which case you need to delay your Jool arrival to give Tylo time to move farther along in its orbit).
   • Tylo is ahead of where it needs to be (in which case you need to make your Jool arrival earlier, so that you'll catch Tylo before it moves along too far).
4. Okay.  So now you know whether you need to delay or advance your arrival, you need to make that happen.
5. Take a look at your Jool Pe marker to see when it is.  Note the time.
6. Start playing with the controls.  The goal is to adjust the time of your Jool Pe, while keeping its location where you want it.  Note that for each of these pairs of controls ( on the one hand, on the other), one of them will make your arrival path sweep to the left, and the other will make it sweep to the right.  The way you adjust your time without adjusting your path is to use those controls in turn.  For example, if you use to sweep your path a bit to the left, and then use to sweep it back to the right again to the original location... the time of your Pe won't be the same as it was.  It will be ahead or behind where it was, by a bit.  So just keep repeating that operation to shift your time until Tylo gets to where you want it to be.

[Snark]

37 minutes ago, Zhetaan said:

The best gravity assists warp your path by 180 degrees

I believe that this is incorrect, and is actually one of the worst gravity assists.  I think the optimal case is when you're deflected by something like 90 degrees, not 180.

Yes, this may seem counterintuitive.  Details in spoiler.

Spoiler

The angle of deflection is a function of how fast you travel past the body.  The slower you go relative to that body, the more it deflects your path.  For example, if you arrived at Tylo's SoI going at some insanely high speed relative to Tylo, you'd just go whizzing past it like a bullet in a nearly straight line, and it would hardly deflect your path at all-- a really crappy gravity assist, I think we'll both agree.

So, a slower pass, and a greater angle of deflection, would make a better gravity assist, right?  More bendy is better, right?

Well, yes... but only up to a point.  Beyond that, it gets worse.  In the extreme case-- a 180-degree bend-- you get nearly zero benefit, same as if you'd done a high-speed-with-almost-zero-degree-deflection.

Here's why a 180-degree gravity assist is ineffective:

Unless you do a burn while in the midst of a gravity maneuver (more on that below)-- i.e., if you're simply doing a passive gravity assist-- then you will leave a body's SoI at the exact same speed as you entered it, relative to that body.  The body's gravity will change your direction, but due to conservation of energy it won't affect your speed (relative to that body).

And the only way to get a 180-degree deflection is if you enter the body's SoI at very, very low relative speed.  That is, for example, you're only going to make a 180-degree U-turn around, say, Tylo, if you entered its SoI at nearly zero speed relative to Tylo.

And if you enter its SoI at nearly zero relative speed... you'll leave that way, too, at nearly zero relative speed.

Which means that the velocity change will be very low, too.  You went from "low speed in one direction" to "low speed in the opposite direction", which is a tiny dV change.

In other words:  To get the largest possible velocity change from a gravity maneuver, you need two things:

• The greatest possible deflection angle
• The greatest possible speed relative to the body as you're traversing its SoI.

Unfortunately, those two goals are in direct opposition.  The faster you go, the less you deflect.  It turns out that you get minimum benefit for "extreme high speed, extreme low deflection", and also minimum benefit for "extreme low speed, extreme high deflection".  The best result is a happy medium somewhere in the middle.

I've never tried to do the math to work out exactly what the optimum deflection angle is, but my guess is that it's somewhere in the neighborhood of 90 degrees, and anecdotally from playing KSP and doing gravity assists, my impression is that that's not too far off the mark.

 

Note that all of the above discussion assumes that you're doing a passive gravity assist.  If you do a burn when you're at your low Pe over Tylo (or whatever body you're using for the assist), then "exit speed equals entry speed" doesn't apply, and things shift around.  I haven't made any attempt to address that in the above-- I'm simply talking about passive assists.

 

[Empiro]

You are correct that from a practical standpoint, a 180 degree turn will usually only happen if you're moving very slowly with respect to the body you're getting the assist from. However, for any given velocity, the closer to 180 degrees, the better the potential benefit.

At 180 degrees, you pick up 2x the difference in velocity with respect to the parent. For example, if you were stationary with respect to Jool, Tylo would approach you at 2000 m/s (its orbital velocity). If you managed a 180 degree deflection from Tylo, you'd exit Tylo moving 2000 in the opposite direction, and end up moving 4000 m/s with respect to Jool.

The problem is that you can't get 180 degree deflection from Tylo while moving that fast. You get a greater deflection if you pass close to Tylo, but at some point, you'll crash into it. However, if you replaced Tylo with a black hole with the same mass, you would be able to approach much closer, and it would likely be possible to get a 180 degree deflection, even when moving very quickly.

In general, the best places to do gravity assists are around objects that are both massive and moving fairly quickly. That's why moons like Tylo are a good choice, and why both Eve and Jool are used as well. 

This article also does a really good job explaining how gravity assists work with pictures.

Edited July 24, 2019 by Empiro

[Zhetaan]

21 hours ago, Snark said:

I believe that this is incorrect, and is actually one of the worst gravity assists.  I think the optimal case is when you're deflected by something like 90 degrees, not 180.

You make an interesting point.  I agree that there is a difference between the theoretical ideal and the realistically achievable, and to that end, a perfect 180-degree assist simply isn't going to happen in most cases.  However, the impossibility of instantaneous velocity changes doesn't stop us from assuming them when describing Hohmann transfers (even when we include ejection angles for interplanetary transfers), and there are other approximations that we make for the sake of computational ease at the expense of absolute purity of physics, so while I will concede that 90-degree assists where you enter the sphere from a radial direction and leave in a retrograde direction are probably the best practical case for gravitational capture, that does not mean that 180-degree assists, where possible, are the worst possible.  There is one detail that you missed:

21 hours ago, Snark said:

Here's why a 180-degree gravity assist is ineffective:

Unless you do a burn while in the midst of a gravity maneuver (more on that below)-- i.e., if you're simply doing a passive gravity assist-- then you will leave a body's SoI at the exact same speed as you entered it, relative to that body.  The body's gravity will change your direction, but due to conservation of energy it won't affect your speed (relative to that body).

And the only way to get a 180-degree deflection is if you enter the body's SoI at very, very low relative speed.  That is, for example, you're only going to make a 180-degree U-turn around, say, Tylo, if you entered its SoI at nearly zero speed relative to Tylo.

And if you enter its SoI at nearly zero relative speed... you'll leave that way, too, at nearly zero relative speed.

Which means that the velocity change will be very low, too.  You went from "low speed in one direction" to "low speed in the opposite direction", which is a tiny dV change.

The amount of velocity change has nothing to do with our velocity relative to Tylo.  As you said, we enter and exit the sphere of influence at the same speed regardless of what that speed is.  The key velocity is not our velocity relative to Tylo, but Tylo's velocity relative to Jool.  Remember that Jool orbit is the destination; Tylo is, at best, a waypoint that we pass on the way.

Tylo orbits Jool at 2030.8 m/s.  Let's say that I approach Tylo from its retrograde at a Jool-relative 2031 m/s, for a Tylo-relative difference of .2 m/s.  Let's also assume that I can get a perfect 180-degree assist from this.  I will leave Tylo's sphere of influence at a Tylo-relative .2 m/s, but in the opposite direction from that by which I entered.  That amounts to 4061.6 m/s of total change, and if I subtract the 2031 I had to begin, then that means I leave Tylo at 2030.6 m/s Jool-relative, in the retrograde direction.  Try using the engines to switch from a prograde to a retrograde orbit while at Tylo's orbital altitude about Jool and I promise you that you will see how tiny a delta-V change it is not.

Edited July 24, 2019 by Zhetaan

[Snark]

3 hours ago, Zhetaan said:

The amount of velocity change has nothing to do with our velocity relative to Tylo... The key velocity is not our velocity relative to Tylo, but Tylo's velocity relative to Jool.

These are the same thing.  If your velocity relative to Tylo changes by <amount>, then your velocity relative to Jool (and everything else in the universe) also changes by <amount>.  Velocity change is velocity change.  Sure, it changes from J+X1 to J+X2 in the Jool frame, and it changes from T+X1 to T+X2 in the Tylo frame... but in both cases, the velocity change is X2-X1.  It's the same change.

3 hours ago, Zhetaan said:

Let's say that I approach Tylo from its retrograde at a Jool-relative 2031 m/s, for a Tylo-relative difference of .2 m/s.  Let's also assume that I can get a perfect 180-degree assist from this.  I will leave Tylo's sphere of influence at a Tylo-relative .2 m/s, but in the opposite direction from that by which I entered.

That's... um... not how it works.  Might want to check your geometry, there.

I'll take your example and use your numbers.

• Initially, just before entering Tylo's SoI:
  • Tylo is traveling 2030.8 m/s relative to Jool.
  • To simplify the math and avoid having to do a bunch of sines and cosines, let's say that at the time of the encounter, Tylo's direction is in the +X direction relative to Jool, with no Y or Z component.
  • Your spacecraft is very, very slowly approaching Tylo from behind.
  • Its velocity relative to Jool is 2031 m/s in the +X direction.  Its velocity relative to Tylo is 0.2 m/s in the +X direction.
• Just after entering Tylo's SoI:
  • Its velocity inside Tylo's SoI is 0.2 m/s in the +X direction.
  • Its velocity relative to Jool is 2031 m/s in the +X direction.
• After making a 180-degree U-turn around Tylo, and just about to leave its SoI:
  • Its velocity inside Tylo's SoI is 0.2 m/s in the -X direction.
  • Its velocity relative to Jool is 2030.6 m/s in the -X direction.
• Just after leaving Tylo's SoI:
  • Your spacecraft is very, very slowly fleeing Tylo in the -X direction.
  • Its velocity relative to Tylo is 0.2 m/s in the -X direction.
  • Tylo's velocity relative to Jool is still 2030.8 m/s in the +X direction.
  • Your ship's velocity relative to Jool will equal its velocity relative to Tylo, plus Tylo's velocity relative to Jool.  It's basic arithmetic.
  • In this case, the former is -0.2 m/s and the latter is +2030.8 m/s.  So we add -0.2 to 2030.8.
  • So we find that the ship's velocity relative to Jool is 2030.6 m/s in the +X direction.

The total velocity change you get, from your entire Tylo gravity assist, is... 0.4 m/s.  Not 4061.6 m/s.  You've overstated the benefit by literally more than 8,000 times.  Unless you're asserting that Tylo itself somehow made a U-turn during this maneuver and started orbiting Jool in the opposite direction?

 

Here's another way to think about it.  During your ship's sojourn inside Tylo's SoI, it started at 0.2 m/s in the +X direction (relative to Tylo), and ended at 0.2 m/s in the -X direction (relative to Tylo), for a total of 0.4 m/s velocity change (relative to Tylo).  We both agree on that, right?

3 hours ago, Zhetaan said:

Let's say that I approach Tylo from its retrograde at a Jool-relative 2031 m/s, for a Tylo-relative difference of .2 m/s.  Let's also assume that I can get a perfect 180-degree assist from this.  I will leave Tylo's sphere of influence at a Tylo-relative .2 m/s, but in the opposite direction from that by which I entered.

...Well, if you think that that's all that's needed to get a velocity change of >4000 m/s relative to Jool, then you wouldn't actually have to go through the tedious business of swinging all the way around Tylo, now would you?  You could just gently enter Tylo's SoI at 0.2 m/s relative, and then immediately do an engine burn of 0.4 m/s to reverse your Tylo-relative direction, and then exit the way you came.  Right?  It's exactly the same thing; you're just burning fuel to accomplish the 0.4 m/s change, instead of letting Tylo's gravity do it for you over a period of hours.  So... are you really saying that you think that just by dipping your toe in Tylo's SoI for a few seconds and doing a 0.4 m/s burn that you can achieve a 4000 m/s velocity change relative to Jool?

 

Look, if you still don't believe me, just try it.  Do the maneuver you describe.  Go on, try it.  Give it a shot and let us know what happens.  We're not talking about something subtle, here-- the difference between your predicted result and mine is starkly dramatic.  Try it and see if, afterwards, you do in fact end up completely reversing your direction around Jool and orbiting it retrograde, as you claim.  If I'm wrong, I'd be delighted to find out.

Go on, I'll wait. 

[Aelipse]

I didn't do the math, but just from an intuitive point of view; as much as 180° gravity turns would be effective, they're impossible at high SOI insertion speeds, and the direct transfer from Kerbin to Jool produces some rather insane insertion speeds. 

From a slightly different perspective; the Sun's (Kerbol's) perspective to be precise, it works like this. You're reaching Jool's SOI along an ellipse, which ideally has its apoapsis right at Jool's orbit. The reason why you have such a huge speed relative to Jool is because you're actually moving very slow relative the the Sun (you're literally about to fall back to Kerbin's orbit), while Jool is whizzing past you on its almost circular trajectory. 

What you wanna do is hang out in Jool's gravitational well, or the gravitational well of any object that travels with Jool, like its moons, for as long as possible. That's because the longer you spend in Jool's gravity (and the stronger the gravity the better), the more it pulls you behind, granting you extra velocity and effectively circularising your orbit relative to the Sun. If you can encounter more than one of Jool's moons, all the better.

The mechanics of encountering Tylo has been described in great lengths above. What you wanna do with the assist then is have Tylo redirect you back towards Jool. Remember, you want to spend as much time as possible in the strongest possible gravitational field. If you have some fuel left, you can perform a deceleration burn at your periapsis above Tylo to further increase the deflection angle towards Jool (or theoretically ideally retrograde to Tylo's orbital velocity, as suggested above, but I cannot really image how that could be possible). It is also more effective to reach Tylo "from behind" than encountering it in the opposite direction to its direction of motion. Remember, you want to spend as much time in their gravity as possible, so whizzing by it at 5000 m/s won't do much for you.

In the meantime, I'll be waiting for corrections from people who actually understand the mechanics of gravity assists.

Edited July 24, 2019 by Aelipse

[Snark]

1 hour ago, Aelipse said:

I didn't do the math, but just from an intuitive point of view; as much as 180° gravity turns would be effective, they're impossible at high SOI insertion speeds

Agreed.

1 hour ago, Aelipse said:

the direct transfer from Kerbin to Jool produces some rather insane insertion speeds. 

It does.  However, if you hit Tylo's SoI with the right geometry (so that you're approaching it from "behind" in its orbit around Jool), then Tylo's own orbital velocity helps you, here.  It's fleeing you at over 2000 m/s (its orbital velocity around Jool), so your velocity relative to Tylo is 2000 m/s less than your velocity relative to Jool, upon your arrival at Jool's SoI.

That's helpful, since it increases your dwell time in Tylo's gravity well as you use it for the gravity assist.

1 hour ago, Aelipse said:

What you wanna do is hang out in Jool's gravitational well, or the gravitational well of any object that travels with Jool, like its moons, for as long as possible.

Yes and no.

Simply dwelling in Jool's SoI doesn't help you capture to Jool.  It is of absolutely zero use in a passive-trajectory situation.  (You could use it to your advantage if you go for an Oberth maneuver, i.e. if your trajectory takes you to a low Pe over Jool and then you do a burn there.  That's a valid navigational approach, though in my experience a Tylo assist works to do the capture and is absolutely free in terms of dV, no burn required.)

Dwelling in Jool's SoI would help you if you were using Jool for a passive gravity assist to go somewhere else (e.g. if you're just swinging past Jool for an assist to elsewhere in the solar system).  But that's not what we're talking about here.

However, dwelling in Tylo's SoI does indeed help for a passive capture to Jool, as you describe.  It gives Tylo's gravity more time to act, in order to bend your trajectory and thereby affect your Jool-relative speed.

1 hour ago, Aelipse said:

That's because the longer you spend in Jool's gravity (and the stronger the gravity the better), the more it pulls you behind, granting you extra velocity and effectively circularising your orbit relative to the Sun.

Arriving at Jool with lower relative speed does help, but not for this reason.

Hanging out passively in Jool's SoI will affect your Sun orbit, if you swing past Jool without burning or encountering its moons.  But it won't help with capturing to Jool itself.  If you enter Jool's SoI with some speed, and don't burn, and don't encounter any moons, then you will leave Jool's SoI with the same relative speed.  Conservation of energy and momentum.

However, it does help, not because you're increasing your dwell time in Jool's SoI, but rather because it means you can increase your dwell time (and get more path deflection) in Tylo's.  (Or Laythe, which also can work well in the same way as Tylo.  It has different advantages and disadvantages, depending on your intended mission profile.)

1 hour ago, Aelipse said:

It is also more effective to reach Tylo "from behind" than encountering it in the opposite direction to its direction of motion.

True, but it's a much stronger statement than that-- "more effective" isn't quite the phrase I'd use.

• If you approach Tylo from behind and pass in front of it, then its gravity will retard your speed relative to Jool and help you capture thereto, which is what you want.
• If you approach Tylo from in in front of it and pass behind it, then its gravity will accelerate your speed relative to Jool, and fling you out of the Jool system even faster than you arrived.

In other words:  doing the latter is "less effective" in the strict technical sense, I suppose, but a better way of putting it would be that it does the exact opposite of what you want.

Here's how to think of it:

• Picture your path going past Tylo.
• Draw an arrow.  The start of the arrow is right at your Pe marker on your path past Tylo.  The head of the arrow (i.e. the direction the arrow is pointing) is at Tylo's center.
• That arrow is the overall velocity change, relative to Jool, that Tylo is giving you for free.
• Whether Tylo helps you or hurts you, and by what amount, depends on the direction of that arrow relative to Tylo's motion around Jool.
• The ideal best case-- where the Tylo assist helps you the most-- is when that arrow is pointing directly opposite Tylo's motion around Jool.
• The worst case-- where it hurts you the most-- is when that arrow is pointing directly in the same direction as Tylo's motion around Jool.
• In between those two cases-- where the arrow is perpendicular to Tylo's orbital path-- is the "neutral" case that neither increases nor decreases your speed relative to Jool.  It does, however, affect your direction, i.e. increasing or decreasing the eccentricity of your orbit around Jool, which may or may not be helpful depending on what kind of navigation you're trying to do.

 

Everything I've said up to this point (including my previous messages in this thread) is based on actual real-world physics, so it applies to the real world just as much as it applies to the simulation inside KSP.  In addition, however, there's another effect that you can leverage, which is an artifact of KSP's imperfect simulation (basically, an exploit of the patched conics model that KSP uses).  Details in spoiler.

Spoiler

In the real world, there's no such thing as a literal "sphere of influence" with a sharp boundary.  No matter how close you get to the Moon, both the Moon's gravity and Earth's are still affecting you (and, for that matter, the gravity of every other object in the universe).  Earth's gravity doesn't magically and totally disappear when you cross some "boundary line" near the Moon.

However, that's not how it works in KSP.  For various good reasons that would be extremely time-consuming to go into here, KSP (like a lot of orbital-simulation software) uses a patched conics model to reduce the complexity to something that's feasible for the software (and your CPU) to handle.

That's why "spheres of influence" are a thing in KSP.  It means that Tylo has a spherical radius around it-- its SoI-- such that the moment you cross into that sphere, Jool's gravity completely ceases to exist and you're affected only by Tylo's.

This allows an exploit in KSP that wouldn't be possible in the real world.  Specifically, you can exploit the difference in Jool altitude between the point where you enter Tylo's SoI and the point where you exit.

Let's consider two cases:

1. You arrive in Jool's SoI, and meet Tylo before your Jool Pe, so that you're still falling inwards towards Jool as you transit Tylo's SoI.
2. You arrive in Jool's SoI, swing past your Jool Pe, and after that you meet Tylo, so that you're already rising outwards away from Jool as you transit Tylo's SoI.

Think about what the above two cases look like from Jool's point of view:

• In case #1, you enter Tylo's SoI at a high Jool-relative altitude, and exit it at a much lower Jool-relative altitude.
• In case #2, it's the other way around:  you enter Tylo's SoI at low altitude, and exit at high altitude.

If KSP actually modeled realistic N-body physics-- i.e. if it behaved perfectly like the real world-- then both of these two cases would be, in principle, equally effective at slowing you down with a gravity assist, assuming that the geometry is set up correctly.

However, in KSP... they're not.  There's actually a pretty big difference.  Here's why:

Let's consider case #1, and compare it.  Not to case #2, but to our null hypothesis of "what would have happened if Tylo weren't there at all, and you just fell farther into Jool's gravity well."

Let's say you  enter Tylo's SoI at some altitude R_high, and leave it at some altitude R_low, as you fall inwards towards Jool.

If you hadn't met Tylo, and simply fell into Jool's SoI from R_high to R_low... then you would be going faster at R_low than R_high, simply because you're "going downhill".  Jool's gravity is pulling you in.

On the other hand... the moment you enter Tylo's SoI, Jool's gravity completely ceases to exist, unlike the real world.  Even though you exit Tylo's SoI at a much lower Jool-relative altitude than you entered it... you don't gain any speed thereby.

To take an analogy:  it's the difference between jumping off the top of a building, versus getting a free elevator ride to the ground floor.  You arrive at the bottom a lot faster in the former case than the latter.     Tylo is giving you a free elevator ride, because of the patched conics.

Similarly, case #2 works the other way around.  In that case, Tylo's giving you a free elevator ride up instead of down.  Which is not what you want.  Because you're trying to reduce your energy in the Jool reference frame, not increase it.  Downward elevator rides are good, upward ones are bad.

That's why, when you're settting up a Tylo assist to capture to Jool, it's better to do that on the inward part of your path past Jool than the outward part.  In the real world, it wouldn't matter; in KSP,  it does.

To be clear, this effect is completely unrelated to the overall "slingshot effect" that we've been discussing everywhere in this thread up to now, other than inside this spoiler.   It's a separate thing that simply increases (or decreases, if you do it "wrong") the effectiveness of the slingshot.

 

1 hour ago, Aelipse said:

so whizzing by it at 5000 m/s won't do much for you.

Absolutely true on that one. 

 

[Zhetaan]

2 hours ago, Snark said:

The total velocity change you get, from your entire Tylo gravity assist, is... 0.4 m/s.  Not 4061.6 m/s.  You've overstated the benefit by literally more than 8,000 times.  Unless you're asserting that Tylo itself somehow made a U-turn during this maneuver and started orbiting Jool in the opposite direction?

You have my apologies; I transposed two figures.  I meant to write .2 m/s Jool-relative.

I stand by the rest, however.  If I enter Tylo's sphere at .2 m/s Jool-relative, then I enter it at 2030.6 m/s Tylo-relative (Tylo's 2030.8 minus my .2).  I leave that sphere at 2030.6 m/s Tylo-relative in the opposite direction, and re-enter Jool's sphere at 4061.4 Jool-relative.

I don't deny that this is impractical, if it's even possible.  At the least, approaching a body from retrograde normally necessitates travelling faster than that body, so I think we can safely relegate my example to the realm of frictionless surfaces, instantaneous burns, and perfectly spherical cows.  Nevertheless, I see people use Jool-based assists to get solar-retrograde orbits (usually for rescue contracts) often enough that there is clearly enough practical merit to the idea that it can be done even if it's done in a less-effective actual trajectory.

[Aelipse]

Thanks for clarification. I'll have to read through your answer a couple more times though to hopefully fully understand.

[Snark]

18 minutes ago, Zhetaan said:

If I enter Tylo's sphere at .2 m/s Jool-relative, then I enter it at 2030.6 m/s Tylo-relative (Tylo's 2030.8 minus my .2).

Okay, with you so far.

18 minutes ago, Zhetaan said:

I leave that sphere at 2030.6 m/s Tylo-relative in the opposite direction, and re-enter Jool's sphere at 4061.4 Jool-relative.

No.  You do not.

Because if you enter Tylo's SoI at 2030.6 m/s Tylo relative, then you do not make a 180 degree turn.  You go streaking past Tylo like a bullet, and your path is only bent by a slight angle.

That's my point.  180-degree gravity turns are ineffective because the only way to make a 180-degree turn is to enter the SoI at near-zero relative velocity to that body.  Which results in the scenario you originally stated above, and which I explained doesn't actually help you by any appreciable amount.

You've described a scenario (enter at high speed and do a 180 degree turn) that's physically impossible.

18 minutes ago, Zhetaan said:

I don't deny that this is impractical, if it's even possible.

It's not merely impractical-- it's mathematically impossible.

18 minutes ago, Zhetaan said:

At the least, approaching a body from retrograde normally necessitates travelling faster than that body, so I think we can safely relegate my example to the realm of frictionless surfaces, instantaneous burns, and perfectly spherical cows.

We are in that realm.  We're in a simulator with a simple patched-conics model.  There's no friction, there's no burn at all, everything is perfectly spherical.  Heck, it's not even N-body; KSP's patched-conic approximation means that at all times it's just a simple two-body problem that is precisely solvable with basic algebra.  This is, indeed, the ideal theoretical case.  It doesn't get any more ideal than this.

And it still doesn't work. Because even in the ideal case, the math says it won't.

You can't have it both ways.  It's physically impossible to encounter a body at high speed (i.e. substantially in excess of its escape velocity) and make a 180 degree turn, if you're not doing a burn.  You can do one or the other.  Pick one.

You make a 180 degree turn if you encounter it right precisely at escape velocity (which, in KSP-land, translates to "enter its SoI with near-zero relative velocity").  If you're going higher than escape velocity (i.e. if you enter its SoI with a relative velocity substantially higher than zero), then your orbit past it will be a hyperbola rather than a parabola, and the deflection angle will be less than 180 degrees.  The higher the excess over escape velocity, the less the deflection angle.

Thus spake Sir Isaac Newton, over three centuries ago, and he's still right. 

[Aelipse]

22 minutes ago, Snark said:

Agreed.

...

Absolutely true on that one. 

 

OK, I actually think I grasped everything you said and it makes complete sense! I don't know why I haven't thought about the total delta-v acquired in a gravity assist as the gravity-induced centripetal acceleration integrated over time. Your explanation with arrows totally nailed it for me.

The "exploit" due to SoI is also really clever and it had never even occurred to me. Thanks for going to the lengths of explaining it all. You're not a physics teacher by any chance?

Edited July 24, 2019 by Aelipse

[Zhetaan]

4 hours ago, Snark said:

if you enter Tylo's SoI at 2030.6 m/s Tylo relative, then you do not make a 180 degree turn.

That is true, but I was using Tylo as an illustration to make a point about a theoretical limit.  In the limit, a 180-degree turn is still the ideal.  Given that a gravitational trajectory is, in essence, identical to a perfectly elastic collision, a vessel designed of the right kind of unobtainium could simply bounce off of the assistor and achieve the same result.  The fact that that is not possible for any known material (including lithospheric material--a collider of this sort would simply sink into the crust and be swallowed) is not the point and never was.

4 hours ago, Snark said:

We are in that realm.  We're in a simulator with a simple patched-conics model.  There's no friction, there's no burn at all, everything is perfectly spherical.  Heck, it's not even N-body; KSP's patched-conic approximation means that at all times it's just a simple two-body problem that is precisely solvable with basic algebra.  This is, indeed, the ideal theoretical case.  It doesn't get any more ideal than this.

No, we are not in that realm.  The ideal realm also has no heat, no drag, and no other bodies but the spacecraft and its primary.  KSP is indeed a simulator, and when comparing it to reality, I don't know exactly how it rates, but I think it's safe to say that it is not a good simulator.  But it is still an environment based on rules and a lot of those rules are derived from reality.  For example, the theoretical limit of a gravity assist depends on the gravity being a point source.  KSP actually does model gravity as a point source--there are no heterogenous mass distributions to perturb orbits--but it also adds a massless surface a particular distance from that point.  Thus the reality of assists being approach-limited by the presence of a planetary surface is, if you'll pardon the pun, realised, even though the underlying factors that give rise to that limit are completely outside of what is physically possible.

But there are correct solutions to Newton's laws that allow for a 180-degree turn, and I will not give on that point until ...

4 hours ago, Snark said:

the math says it won't.

... you show me the maths.  Please accept my earnest encouragement to do so, because although I cannot say that I enjoy being proven wrong, I would prefer to be helpful, and I cannot be that with bad information.  Until then, I'm very much afraid that we are at an impasse, and while I think our discussion so far to be on-topic, I fear that I am no longer being helpful to the conversation or supportive of its original end of guiding Space Nerd to a Tylo-assisted capture, so I shall stop here.

Edited July 24, 2019 by Zhetaan

[Snark]

1 hour ago, Zhetaan said:

In the limit, a 180-degree turn is still the ideal.

Sure, in a hypothetical universe where the laws of physics are different and orbital mechanics don't obey Newton's laws.

That does not pertain, however, either in real life or in KSP, so it doesn't strike me as an especially useful generalization to make.

In a Newtonian model with 1/R² gravity, such as KSP.... no, it's not.  Because a 180-degree turn around a planetary body happens only when the encounter is at extremely low speeds and therefore very little net benefit.

Remember the topic of this thread.  It's about getting a Tylo assist.  "What's the best way to do that."  And trying to get a 180-degree turn out of Tylo is absolutely not a good strategy-- it's a terrible one, it'll give nearly zero benefit.  In your initial example-- which you say you misstated, but which is in fact basically the only way to get a 180-degree U-turn around Tylo-- the benefit is under 1 m/s, which doesn't seem to me like helpful advice to be giving someone who's looking to save dV with a gravity assist.

1 hour ago, Zhetaan said:

Given that a gravitational trajectory is, in essence, identical to a perfectly elastic collision

No, it's not, quite.

It's pretty similar, yes.  A gravitational trajectory in this sense does behave like an elastic collision... with one important difference.  The angle of deflection of an elastic collision is independent of the speed of the encounter.

You can have an elastic collision that causes a deflection from 180 degrees down to approaches-zero degrees.  And that full range of results is possible at any speed.

Whereas with a gravitational encounter, the angle of deflection is a function of the encounter speed.  And there's no way around that.

1 hour ago, Zhetaan said:

The ideal realm also has no heat

Correct, and KSP does not, either (at least not in any way that affects orbital mechanics).

1 hour ago, Zhetaan said:

no drag

Correct, and KSP does not, either, if a ship's not in an atmosphere.

1 hour ago, Zhetaan said:

no other bodies but the spacecraft and its primary

Correct, and that also describes KSP, too.  It uses a 2-body model in which the only gravity a spacecraft experiences is that of its primary.

So yes.  This is the ideal. You haven't given a single factor in your "ideal" situation that doesn't apply to KSP, nor have you shown any reason why KSP doesn't meet that criterion.

1 hour ago, Zhetaan said:

KSP is indeed a simulator, and when comparing it to reality, I don't know exactly how it rates

It matches up pretty well.  And where it diverges, it does so because KSP is actually the ideal situation you speak of, and reality isn't.  Reality has lots of factors that aren't modeled at all in KSP-- those "non-ideal" factors you've mentioned, plus N-body interactions, plus general relativity that can alter trajectories by measurable amounts (though after quite a few decimal places).

KSP has none of that.  A KSP ship that's not in atmosphere or using any thrusters moves on a perfect, mathematical-ideal conic-section trajectory.  They've correctly modeled that.  It'll be either an ellipse, a parabola, or a hyperbola, depending on whether the ship is traveling under, at, or above escape velocity, respectively.

1 hour ago, Zhetaan said:

For example, the theoretical limit of a gravity assist depends on the gravity being a point source.

The closer you can get to a body, the sharper the turn you can make.  So yes, if Tylo were a point source of gravity and not in fact a sphere of radius 600 km, then yes, you could dive to a super-duper-low altitude and make a sharper turn.  I have no argument there.

But we're not talking about a black hole, or theoretically what you could do with one.  We're talking about Tylo.  Because, lest we forget, the whole point of this thread is to help someone who's playing KSP and wants to know how to do a gravity assist to accomplish what he wants to do.

And Tylo, like all KSP bodies-- and like all real-world bodies that aren't black holes-- has a physical extent that limits the minimum possible radius that an encounter can reach.

1 hour ago, Zhetaan said:

Thus the reality of assists being approach-limited by the presence of a planetary surface is, if you'll pardon the pun, realised

Exactly.  So, when the OP was asking,

Quote

can't find a encounter with tylo that slowed me into orbit without spending ridiculous amount of dv , does anyone know how to do it?

...and you said,

On 7/23/2019 at 8:14 AM, Zhetaan said:

The best gravity assists warp your path by 180 degrees, meaning that you approach Tylo from behind (so you and Tylo are both going in Tylo's prograde direction), you whip around Tylo as close to its surface as you can manage, and then you exit to Tylo's retrograde as I mentioned before.  This gives the maximum Jool-relative velocity reduction.  The reason to go so close to Tylo's surface is to take advantage of the stronger gravity to warp your trajectory to something close to 180 degrees.

...then you were, in fact, giving him incorrect advice, i.e. describing a situation that is physically impossible, and which, if he had attempted to follow it, would have resulted in frustration and confusion.

If you had amended your comment to say "Well, if you're planning on visiting a black hole instead of Tylo, here's what you would do", then your advice would have been technically correct, though not applicable to the OP's actual problem.

Everything that I (and, as far as I can tell, everyone else in this thread) said up to this point in the thread has been assuming we're talking about a body with an actual surface, i.e. Tylo.  Sorry, I assumed that was obvious.  You didn't mention until just now that you were talking about theoretical limits of infinitely dense point sources and not Tylo or other planetary body.

1 hour ago, Zhetaan said:

But there are correct solutions to Newton's laws that allow for a 180-degree turn

Sure, but they're completely irrelevant here, of course.

Because such solutions assume you can get much closer to the gravity source than a planetary surface permits, which naturally excludes them from consideration here, because we're talking about Tylo and not a black hole.

1 hour ago, Zhetaan said:

... you show me the maths.

They're laid out fairly nicely here:

https://en.wikipedia.org/wiki/Hyperbolic_trajectory#Eccentricity_and_angle_between_approach_and_departure

With a bit of algebraic rearranging of the equations on that page, you'll find that the equation for the sharpest possible turn you can make is given by:

θ_∞ = cos^-1[-1 / (1 + μr_p / v_∞²)]

where:

• μ is Tylo's standard gravitational parameter (2.82528e12 m³/s²)
• r_p is Tylo's radius (6e5 m)
• v_∞ is the hyperbolic excess velocity of your Tylo encounter.

Derivation of the above equation in spoiler:

Spoiler

The angle is given by the following, in terms of eccentricity (where e > 1 for hyperbolic orbits):

θ_∞ = cos^-1(-1/e)

Periapsis distance (i.e. Tylo's radius, for the sharpest possible turn) is given by

r_p = a(1 - e)

...where a is the semi-major axis, which in turn is given by

v_∞² = -μ/a

...where μ is the standard gravitational parameter.

Rewriting and combining a bit, we get

a = -v_∞²/μ

..and therefore

r_p = -v_∞²(1 - e)/μ

...which becomes, when we solve for e,

e = 1 + μr_p / v_∞²

...and plugging that back into the original equation up top,

θ_∞ = cos^-1[-1 / (1 + μr_p / v_∞²)]

...so, there you go.  Plug in your excess hyperbolic velocity of your Tylo encounter, and that'll give you the sharpest possible turn you can manage.

 

So just plug in the velocity, and that'll tell you how sharp a turn you can make in the ideal case where you're scraping your toes on Tylo's surface.

You'll find that if you want a very high angle (i.e. close to 180 degrees), you need a very low velocity.

(The only way to get a very high angle without a very low velocity, would be to choose a very tiny r_p that's actually smaller than the planetary radius.  But in that case, of course, lithobraking is involved, which I suspect is not in line with the OP's plans.) 

[Space Nerd]

Well, I tried once and is successful , but then I have to redo it because I didn't have enough fuel.

Now no matter how I changed my encounter, I can't get into jool orbit (my interception is sort of on the "sideways" of tylo),can anyone help?

Also what's the most efficient way to enter laythe orbit from an elliptical jool orbit?

[Snark]

7 hours ago, Space Nerd said:

Now no matter how I changed my encounter, I can't get into jool orbit (my interception is sort of on the "sideways" of tylo),can anyone help?

Could you post a screenshot of the map view of Jool system, showing your incoming trajectory and where / how you're intercepting Tylo?  Would make it easier to give specific "you need to do X, Y, and Z" type instructions. 

The technique is going to end up being essentially what I described at length here, but I realize that's awfully long and written in completely general terms.  Having your specific example will make it easier to give concrete, specific advice.

8 hours ago, Space Nerd said:

Also what's the most efficient way to enter laythe orbit from an elliptical jool orbit?

Aerobraking.  Laythe has an atmosphere-- use it, if you can!

What sort of vessel are you flying?  Does it have a heat shield, i.e. can it survive going through an atmosphere at well over 3000 m/s?  Or does it have to worry about overheating and blowing up easily?

Anyway, here's what you want to do (I'm assuming that your Jool Ap is something way up high, i.e. much higher than Laythe's orbit, to start with), if you're trying to minimize your dV expenditure and are fine with it taking a lot of time on the KSP clock:

 

Step 1:  Set up the geometry of your Laythe encounter.

TL;DR:  Coast up to Ap, then adjust your Pe to match the radius of Laythe's orbit, and make sure you're zero inclination relative to Laythe.  Details in spoiler.

Spoiler

You want to encounter Laythe tangentially, i.e. so that the Pe of your big elliptical Jool orbit just "kisses" Laythe's circular orbit.  You don't want to cross it.

In other words, you want to set up your Pe so that it's the same radius as Laythe's orbit.  So, how do you do this in a minimal-dV way?

Well, the important principle to remember for large, eccentric elliptical orbits is that Pe changes are cheapest when you do a burn at Ap, and Ap changes are cheapest when you do a burn at Pe.

So, in this case, you want to adjust your Pe.  So, wait until you're way out at Ap, then do a or burn to adjust your Pe so that it matches the radius of Laythe's orbit.

And when you do that burn out at Ap... do some burn as well, if needed, to match your inclination with Laythe's.  Because Ap is the cheapest place to make inclination changes.

I expect that the above burn (to adjust Pe and change your inclination) will likely be a pretty small one.  The higher your Ap is, the smaller the needed burn will be.

 

Step 2:  Set up the timing of your Laythe encounter.

TL;DR:  Coast down to Pe, then do a small burn to adjust the timing of your orbit so that you'll encounter Laythe at next Pe.  Details in spoiler.

Spoiler

After you've adjusted your Pe as described above,

1. Make sure you've set Laythe as your target (if you haven't already).
2. Since you're already coplanar and have your Pe matching Laythe's (from the previous step), the map view should show you the little pale-blue indicators that show "here's where Laythe is at your closest approach, and here's where you'll be".
3. Drop a maneuver node at your Pe, or just a smidgeon past it.
4. Give the node some .  Note that when you do that, the "where will Laythe be at your closest approach" marker moves.  If your Ap is something very high, then even a slight amount of burn will move Laythe's position by a lot.
5. So just add enough that Laythe slides around to the point where you'll get an encounter at your next Pe.
   • Extra bonus points:  If you're really pressed to save dV, it's possible to make an even smaller burn in order to get a Laythe encounter in 2, 3, or more orbits into the future.  The more orbits into the future, the smaller the burn you'll need.  But this is more complicated to set up so I won't go into it unless you want to learn that more "advanced" technique.
6. Fine-tune it so that your Laythe encounter will have a low Pe.
   • Just how low is appropriate depends on your ship.
   • If you're heat-resistant and can aerobrake, you'll want Pe to be down in Laythe's atmosphere.  Put it something fairly high-up in the atmosphere, you can fine-tune it later.
   • If you're not heat-resistant and can't aerobrake, aim for a Laythe Pe that's as low as you can get without hitting atmosphere.
7. You're all set!  Coast down to your Pe, execute the maneuver node, and you're ready to meet Laythe on your next orbit.

 

Step 3:  Encounter Laythe

TL;DR:  Either aerobrake to Laythe orbit, or else do a big burn at your Laythe Pe to capture.  Details in spoiler.

Spoiler

Here's where you hit a bit of a "fork in the road", where the most efficient thing to do depends on whether your craft is capable of aerobraking (without burning up) or not.

If you can aerobrake, that's best (free dV!), but if you can't, that's okay too.

 

If you can aerobrake:

1. Your previous step above set your Pe to something high in Laythe's atmosphere.
2. As soon as you enter Laythe's SoI, stop and hit F5, because judging "what's the right altitude" is really hard and you'll probably need to re-try it several times to get right.
3. Do the aerobraking thing.  The "sweet spot" you're aiming for is, after the aerobrake, you want your Ap to be something that's low, but above the atmosphere.
   • If you aerobrake and it turns out you braked too much and end up heading down to the surface... F9 to load your quicksave, do a tiny burn to raise your Pe a bit, and try again.
   • If you aerobrake and it turns out you didn't brake enough and end up flying away and escaping Laythe...F9 to load your quicksave, do a tiny burn to lower your Pe a bit, and try again.
4. Assuming that you get it just right, then coast up to Ap and do a small burn to lift your Pe out of the atmosphere and circularize.
5. Yay, you're done! 

 

If you can't aerobrake because your ship would blow up:

1. Your previous step above set up your Pe to something really low over Laythe, that's just above the atmosphere.
2. Coast down to your Pe, and then do a burn to capture to Laythe and lower your Ap down to what you like (e.g. to circularize).
   • This will be the biggest burn of the entire process (that's why aerobraking is so good, if you can manage it-- so that you can avoid this big burn).  But you're doing it in the most efficient place, so this is as good as you can do.
3. Yay, you're done! 

 

[Space Nerd]

Ok,I will give photo of my ship(is quite big but the lander has a 10m heat shield facing outwards) and orbit later.

I want to use laythe to drop me into an lower orbit to reduce relative speed and capture, will it work?(Brought 1800m/s for that, but I want to save some fuel)

Edited August 2, 2019 by Space Nerd

[Space Nerd]

Never mind, I did it!

[Snark]

11 hours ago, Space Nerd said:

I want to use laythe to drop me into an lower orbit to reduce relative speed and capture, will it work?(Brought 1800m/s for that, but I want to save some fuel)

Glad to hear you sorted out your problem!

For anyone else who may be reading this thread and wondering the answer to Space Nerd's question above:  yes, Laythe can definitely do that.  Laythe reverse assists can also be very helpful, just like Tylo's.  The instructions for a Laythe assist are pretty much the same as for Tylo, as described above.  I've never tried to quantify exactly how much dV one can squeeze out each one, but it's "hundreds of m/s" in both cases, and both of them are good enough to capture to Jool from an interplanetary trajectory.

The main difference is that Laythe has an atmosphere, which means that aerobraking's another potential tool on the table, depending on ship design-- which obviously isn't a thing on vacuum-clad Tylo.

Which one's "better", in general, depends on circumstances and intended mission profile.  Laythe has aerobraking, which is great if you're equipped to take advantage of it.  But Tylo has a higher orbit and is easier at capturing to a less-eccentric Jool orbit, which can save you some dV if your destination is one of the upper Jool moons.

[Space Nerd]

I mean I'm doing the laythe assist after I'm in jool orbit to reduce laythe capture dv.

Now my laythe lander broke its jet engines when splashed down at 5m/s, what did I do wrong?(ps. Any advice for how to land on land?)

edit: @Snark maybe split the discussion about laythe into another thread?

Edited August 4, 2019 by Space Nerd