Difference between revisions of "Talk:Tutorial: Gravity Assist"

Latest revision as of 07:11, 21 September 2023

This guide does not adequately explain how gravity assists actually work. For just KSP as it is now, this will suffice. But it is off the mark in explaining technically how they operate, which would be fine, if the authors did not attempt to get technical in their explanation of the physics, even including equations in the diagrams. But since the contributors went there, some corrections or at least clarifications are necessary.

It is not the body's momentum, or trajectory as described here, that generates the assist. It is the body's rotation that does. The trajectory is only important in plotting the correct course to get the assist you want (from rotation) and end up going in the direction you want as you exit the boost.

Why is this important? Well for KSP in current form, it is not important because all of the planets and moons rotate in the same direction. For real life, or if changes are made in KSP or the upcoming KSP2, then these distinctions become very important, even crucial in getting these maneuvers right for people who don't understand the mechanics of it.

If a body has a reverse spin as Kerbin, then your description of plotted maneuvers here is also reversed. So everything mentioned here is opposite of the desired outcome. Take a real life planet in our own solar system, for example. Venus spins in the opposite direction of Earth. In the opposite direction of every other planet in our system, for that matter. If you were to perform these maneuvers, as explained here, against Venus, you would be getting a very undesired result.

It is also worth noting that the rotation "pull" against the vessel is strongest at the equator and weakest at the poles. Therefore, maximizing the effect of this maneuver would need to be done as close and as parrelel to the equator as possible.

A gravity assist maneuver, as described here, would have almost no impact at all, if performed in a polar, or near-polar plane, even if going parallel to the body's trajectory. The only thing this maneuver would do, if conducted on a polar plane is throw your vessel wildly off course, gaining very little or no momentum if you want it, perhaps even losing some with sloppy piloting. Or even gaining slight momentum, when you don't want it and are using the maneuver to try and deccelerate.

Think of the gravity well around a body as a whirlpool. If the vessel is flying with the flow of the vortex, it grabs some of that energy and gains momentum. If the vessel is flying against the flow of the vortex, it is losing momentum, like swimming against a tide.

I find it odd that the very most important aspect of this fundamental concept was completely ignored in this article. While the word "trajectory" was used 13 times, "spin" or "rotation" was never even mentioned once. This article depicts that it is the body's forward motion, its momentum, or as you describe it, simply its trajectory that creates the boost. The planet's trajectory has virtually nothing to do with generating the energy behind the gravity boost.

Again, think of the manipulation of space around a celestial body as a whirlpool. The deeper you go into that whirlpool, the faster the current flow and the more energy you steal with this maneuver. Out near the edges of a whirlpool, water (space, in this case) moves relatively slowly. That is why you get your periapsis as close as safely possible to the body as you can. By doing so, you are going deeper into the whirlpool, where the waters are flowing very fast.

Finally, your diagrams clearly demonstrate this lack of understanding. If drawn properly, they would portray a planet from a polar angle/view (from its axis, to be more precise), looking down at the planet where the equator surrounds the edges of planet. They would include a circular arrow reflecting the direction of the planet's rotation or spin. Your images show us looking at a planet from the equatorial plane. That is like looking at a whirlpool from precisly water level. You wouldn't see it. You can only see a whirlpool from above or below. The best view of the whirlpool is from directionly above, at its axis, 90 degrees up from the water plane.

EDIT: As a follow-up, it is fine to retain the current diagrams shown at the equatorial plane. But only with some changes that reflect how the force of pull from rotation is strongest at the equator, gradually getting weaker as the approach moves toward either of the poles. That is how equatorial diagrams can be useful. But since that was not done here, these diagrams are insufficient and rather pointless, bordering on misleading.

UPDATE: Upon closer inspection, I see the diagrams presented are from the axis. However, without knowing the spin direction, you still don't know whether you should be performing this meneuver in front of or behind the planet to either accelerate or decelerate. With Venus, or if a Venus-like planet is introduced in KSP2, this will be absolutely critical to know.

Again, to demonstratate how it is not momentum, but rotation, that generates the boost, I will use the Sun or Kerbol as an example. They are relatively stationary within our solar system. Everything else revolves around them. Yet you get the greatest gravity assists from the solar center because it has the most mass and is generating the biggest, strongest whirlpool in the system by its rotation. The most can be gained from that maneuver, yet it is not moving in any direction.

It is also worth noting, that is why most experienced players, and why NASA start angling their ascents in an easterly, 90 degree direction. This technique not only sets yourself up for an easier orbital burn. It is also taking advantage of a small gravity assist. Because your vessel is starting to go with the flow of that gravitational whirlpool of space generated by spin as it angles in that direction, instead of against it, if it were to angle in the opposite direction. That is also why NASA has all of its launch sites in the southern-most parts of the its territory. You want those liftoff sites to be as close to the equator as possible to get the most out of that assist because as mentioned before, that is where the pull on your vessel by rotation is the strongest.

This is also why when you intend to circularize an orbit and land on any body in the Kerbol system, you set up your encounter to approach against the spin. As this tuturial suggests, make your intercept periapsis appear anywhere ahead of the planet or moon. That way your retrograde burn for both circulurization and deorbit/landing is benefiting from your vessel going against the flow of that whirlpool, gaining a natural decceleration. You will require less delta v from your first encounter to your final descent. If fuel usage is your greatest concern, shift your plane to be as parralel to the equator as your fuel budget will allow. When you liftoff the body, do so just like on Kerbin and gradually get your vessel pointing in the 90 degree direction to benefit from the gravitational spin, just as you do during liftoff from Kerbin. Of course, even the advice in this paragraph is reversed if the planet spins in the opposite direction. You would approach from behind the planet in that case and liftoff towards the 270 degree heading.

And I just realized my own advice fell victim to the same flawed thinking of this tutorial. I assumed the player is moving outward in the solar system. If the player is doing an interplanetary transfer inwards, my own advice on approaching the encounter was backwards. If you want to deccelerate for orbital circularization and landing on a planet closer to the sun than your position, you would approach from anywhere behind the planet to gain natural, gravity assisted decceleration. But that highlights a simple summary of gravity assists in general that even the most inexperienced player can understand. All of this can be summarized like this: "If you want acceleration, go with the spin of the planet or moon. If you want decceleration, go against it. The closer you get to the body, the greater the assist. Affects are maximized nearest the equator."

That is why a spin arrow on the diagrams is so important. I does not matter what direction a planet is moving in. It only matters which direction it is spinning in. KSP developers get it. That is why encounter icons on the map conveniently include a very handy spin arrow of the body, as you enter and leave that body's sphere of influence.

Evaluation of the above by Hotel26:

Readers be warned that the above opinion by Worban is likely to be incorrect and misleading, particularly in its conjecture about "rotation" and "spin".

To quote the article: "using the orbital velocity and gravity of a satellite".

Wikipedia says the same: "makes use of the relative movement (e.g. orbit around the Sun) and gravity of a planet or other astronomical object to alter the path and speed of a spacecraft...

...The 'assist' is provided by the motion of the gravitating body as it pulls on the spacecraft."

Note that the central body of the frame of reference (be it Kerbol or Sun) cannot provide any change in orbit by its gravity alone. Gravity provides the influential connection but it is the relative motion of the assisting body within the frame of reference (Kerbol) that imparts a change in total energy to the vehicle receiving the assist. --Hotel26 (talk) 00:43, 21 September 2023 (UTC)

How To Do It: #3: "the moment it intersects the orbit of Kerbin"

Like any Hohmann transfer, raising AP is a prograde maneuver and lowering PE is a retrograde maneuver.

I would think that "the moment it intersects the orbit of Kerbin" either likely incorrectly signifies an orthogonal intersection with Kerbin's prograde (incorrect) or less likely ambiguously indicates a "tangent" to Kerbin's orbital direction.

Either way, not very helpful.