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Using vectors to estimate orbital plane change by OTV Booster on 19 Feb, 2022 21:50
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Over in: https://forum.nasaspaceflight.com/index.php?topic=49622.msg2343157#msg2343157
Somewhere around message 3750 we got into an off topic discussion on orbital plane changes, so I'm moving it here.
In the discussion I proposed a quick and dirty vector approach for estimating the dV needed for a plane change from lunar equatorial to lunar polar orbit. Cut to its bare essentials, I suggested that the initial vector had to be completely negated as the vector for a polar orbit has no equatorial component, and then to establish the polar orbit a burn of the same magnitude would be needed. Total dV would be twice the initial orbital velocity.
Adding back a complicating factor, the orbit was assumed to be highly elliptical. I further asserted that the most economical location for the plane change would be at apolune*.
Other complicating factors are the impact of a theoretical instantaneous change and combining the two separate burns into one burn.
Now, no one is suggesting that calculus be abandoned or mission planning be done by vector. What I am suggesting is that for two body problems and an extremely limited number of three body problems, vectors offer a quick and dirty BoE for a sanity check. And it appeals to the math challenged nerd inner me.
Argument was made that the total burn should only require 1.414 times the orbital velocity as this is the magnitude of the resultant vector. This is a recap of another's argument. If this is mangled, I apologize. I think I see why this would not be adequate but don't want to bother if there's no interest in the topic.
OTOH, if there's interest, let's explore the limits of vectors in working orbital mechanics.
*More properly termed aposelene, but I speak little Greek or Latin and apolune is easier to punch in on a phone. -
#1 by cdebuhr on 19 Feb, 2022 22:27
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Over in: https://forum.nasaspaceflight.com/index.php?topic=49622.msg2343157#msg2343157
I suppose for BOTE level sanity checks you're proposal to fully negate the orbital velocity vector before imparting the new one has some utility insofar as it will impose an absolute hard upper limit on the required deltaV for the proposed maneuver. That said, it is a really pessimistic estimate in general, and is only strictly true in the "turn that satellite around and get back where you came from!" sense (i.e., a 180-degree plane change). E.g., in the example discussed on the other thread, executing a 90 degree turn (as in a 90 degree plane change in a circular orbit) requires devta-V of 1.414 x your current velocity (for a 90 degree turn, you can just use Pythagoras).
Somewhere around message 3750 we got into an off topic discussion on orbital plane changes, so I'm moving it here.
In the discussion I proposed a quick and dirty vector approach for estimating the dV needed for a plane change from lunar equatorial to lunar polar orbit. Cut to its bare essentials, I suggested that the initial vector had to be completely negated as the vector for a polar orbit has no equatorial component, and then to establish the polar orbit a burn of the same magnitude would be needed. Total dV would be twice the initial orbital velocity.
Adding back a complicating factor, the orbit was assumed to be highly elliptical. I further asserted that the most economical location for the plane change would be at apolune*.
Other complicating factors are the impact of a theoretical instantaneous change and combining the two separate burns into one burn.
Now, no one is suggesting that calculus be abandoned or mission planning be done by vector. What I am suggesting is that for two body problems and an extremely limited number of three body problems, vectors offer a quick and dirty BoE for a sanity check. And it appeals to the math challenged nerd inner me.
Argument was made that the total burn should only require 1.414 times the orbital velocity as this is the magnitude of the resultant vector. This is a recap of another's argument. If this is mangled, I apologize. I think I see why this would not be adequate but don't want to bother if there's no interest in the topic.
OTOH, if there's interest, let's explore the limits of vectors in working orbital mechanics.
*More properly termed aposelene, but I speak little Greek or Latin and apolune is easier to punch in on a phone.
As long as your burns don't take "too long" you don't need calculus here, just vector algebra, at least for a pretty good approximation. In the limit of an instantaneous impulse, its no longer an approximation - it's exactly true, as it should be for non-instantaneous burns in flat space. I can't say with confidence off the top of my head how much calculus gets involved in orbit, but it'll depend on how long your burn takes relative to your orbital angular velocity. Take your final velocity vector and subtract your initial velocity. The difference vector represents the required deltaV for the maneuver.
If your orbit modification is done through a series of short burns, the overall analysis is more complicated, but at the level of the individual burn, the same rules as above apply. Now, if you're doing a long, low-thrust burn that takes a good bit of an orbit (or multiple orbits ... think orbit raising with ion drive), it'll get more complicated.
Plane changes are cheapest at apoapsis for the simple reason that the initial and final velocity vectors are shorter. -
#2 by OTV Booster on 21 Feb, 2022 17:58
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I haven't been ignoring you. Been killing trees doing vector diagrams and kept getting tripped up with frame references on non-instantaneous burns. So I just kept it simple and looked at an instantaneous 45 deg plane change which would be at the mid point of a 90 deg burn. See diagram.Over in: https://forum.nasaspaceflight.com/index.php?topic=49622.msg2343157#msg2343157
I suppose for BOTE level sanity checks you're proposal to fully negate the orbital velocity vector before imparting the new one has some utility insofar as it will impose an absolute hard upper limit on the required deltaV for the proposed maneuver. That said, it is a really pessimistic estimate in general, and is only strictly true in the "turn that satellite around and get back where you came from!" sense (i.e., a 180-degree plane change). E.g., in the example discussed on the other thread, executing a 90 degree turn (as in a 90 degree plane change in a circular orbit) requires devta-V of 1.414 x your current velocity (for a 90 degree turn, you can just use Pythagoras).
Somewhere around message 3750 we got into an off topic discussion on orbital plane changes, so I'm moving it here.
In the discussion I proposed a quick and dirty vector approach for estimating the dV needed for a plane change from lunar equatorial to lunar polar orbit. Cut to its bare essentials, I suggested that the initial vector had to be completely negated as the vector for a polar orbit has no equatorial component, and then to establish the polar orbit a burn of the same magnitude would be needed. Total dV would be twice the initial orbital velocity.
Adding back a complicating factor, the orbit was assumed to be highly elliptical. I further asserted that the most economical location for the plane change would be at apolune*.
Other complicating factors are the impact of a theoretical instantaneous change and combining the two separate burns into one burn.
Now, no one is suggesting that calculus be abandoned or mission planning be done by vector. What I am suggesting is that for two body problems and an extremely limited number of three body problems, vectors offer a quick and dirty BoE for a sanity check. And it appeals to the math challenged nerd inner me.
Argument was made that the total burn should only require 1.414 times the orbital velocity as this is the magnitude of the resultant vector. This is a recap of another's argument. If this is mangled, I apologize. I think I see why this would not be adequate but don't want to bother if there's no interest in the topic.
OTOH, if there's interest, let's explore the limits of vectors in working orbital mechanics.
*More properly termed aposelene, but I speak little Greek or Latin and apolune is easier to punch in on a phone.
As long as your burns don't take "too long" you don't need calculus here, just vector algebra, at least for a pretty good approximation. In the limit of an instantaneous impulse, its no longer an approximation - it's exactly true, as it should be for non-instantaneous burns in flat space. I can't say with confidence off the top of my head how much calculus gets involved in orbit, but it'll depend on how long your burn takes relative to your orbital angular velocity. Take your final velocity vector and subtract your initial velocity. The difference vector represents the required deltaV for the maneuver.
If your orbit modification is done through a series of short burns, the overall analysis is more complicated, but at the level of the individual burn, the same rules as above apply. Now, if you're doing a long, low-thrust burn that takes a good bit of an orbit (or multiple orbits ... think orbit raising with ion drive), it'll get more complicated.
Plane changes are cheapest at apoapsis for the simple reason that the initial and final velocity vectors are shorter.
This is a representation of a one body model and makes perfect sense. A burn of .707 V at 135 deg, results in a 45 deg vector, V=.707. As you pointed out, as long as the burn is short (instantaneous in this case) vectors are fine for a two body model.
Here is where I see the problem. The resultant is .707 and orbital velocity is 1. This burn lowers perigee (assuming earth orbit). The total angular momentum is lowered. As R. Heinlein pointed out, There Ain't No Such Thing As A Free Lunch. For any orbital plane change, 90 deg or higher, the initial vector has to be completely zeroed and the new vector has to gain the same magnitude.
At less than 90 deg some part of the initial vector is kept. In the diagram it's equal to 1/2 of the equatorial V. The polar component is also 1/2. The hypotenuse is .707 which means the orbit is decaying.
Extending the instantaneous burn to 1.414V does give a polar orbit with a V of 1 but these instantaneous plane changes are really rough on the crew. So I guess the upshot is that dV of 1.414 is a minimum and dV of 2 is the maximum. In the real gravity well it will be somewhere in between. This loops us right back into the SS/SH engineering thread and a discussion of how many engines and the throttle setting that are practical for the LSS in maneuvering.
I guess this is what you and others have been saying.
Now, back to our regularly scheduled programming and (maybe) the end of this thread.
Edit: I'm still scratching my head over loosing V in one direction, then adding the same V in another direction and it potentially being a dV as low as 1.414V. It smacks of a free lunch.
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#3 by Jim on 21 Feb, 2022 18:47
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Here is where I see the problem. The resultant is .707 and orbital velocity is 1. This burn lowers perigee (assuming earth orbit).
No, you got your math wrong. True plane changes don't affect perigee and apogee.
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#4 by TheRadicalModerate on 21 Feb, 2022 19:14
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The magnitude difference between two vectors is always just the Law of Cosines. If you're trying to figure out delta-v:
Δv = sqrt(v1ē + v2ē - 2v1v2cos(δ))
where δ is the angle between the two vectors. If you're trying to change the orbital periapse (or apoapse) and do a plane change at the same time, and you're at apoapse (or periapse), then δ = Δi, and you're done. (If you're somewhere else in the orbit then δ also has a component between the initial and final flight path angles in addition to the inclination change and things are more complicated. Let's assume we aren't doing this.)
If you use the identity:
cos(Δi) = 1 - 2sinē(Δi/2)
this becomes:
Δv = sqrt((v1-v2)ē + 4v1v2sinē(Δi/2))
If you are doing a pure plane change while keeping the same periapse and apoapse (i.e., with no velocity change) then v1=v2=v, and you get:
Δv = sqrt(4vēsinē(Δi/2)) = 2vsin(Δi/2)
Source: Howard Curtis, Orbital Mechanics for Engineering Students, Chapter 6.9. -
#5 by Jim on 21 Feb, 2022 19:22
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90 degree plane change requires 11.5 squares (1.414) of velocity.
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#6 by cdebuhr on 21 Feb, 2022 19:34
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@ OTV Booster - I think you may be overthinking this. Note that in the following, I'm assuming we're either dealing with a circular orbit or you're doing your plane change at either apogee or perigee (i.e., tangential velocity only). I'm not 100% on the implications of doing this at any other point in an elliptical orbit ... my intuition tells me that the math will be the same, but you only consider the tangential part of your velocity (resulting in a purely tangential burn), leaving the radial velocity to take care of itself. Not sure if this is correct or not. I'm further assuming instantaneous impulse. This puts us into spherical cow country, but suspect its not a terrible approximation for most realistic plane changes. I just checked the IXPE launch video, and they manages a 28 degree plane change in about a minute, and while the direction of the gravity vector will have changed during that minute, it didn't change that much. So don't use instantaneous approximations for detailed mission planning, but for BOTE, I think they're fine, as long as your burn doesn't take an appreciable fraction of an orbit (i.e., don't do this with an ion thruster). I've attached a diagram putting it as simply as I can.
Enough for the preamble ... lets say you want to rotate your orbital plane 90 degrees (extreme yes, but it makes the math and graphical depiction really easy, so I'm going with it) from circular equatorial to circular polar. The blue vectors on the attached diagram are the initial and final velocity you want just before and after your burn., respectively. The red vector is the shortest path from one to the other. Just as in position space, the shortest path between two points in velocity space is a straight line (er ... vector). While you could fully negate your velocity in one direction and the head off on your new orbit (at a total cost of 2x initial velocity), you should be able to see that that is not as efficient as just directly imparting deltaV equal the the vector difference in the initial and final state, costing 1.414 x initial velocity.
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#7 by 1 on 21 Feb, 2022 21:27
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Here is where I see the problem....the initial vector has to be completely zeroed and the new vector has to gain the same magnitude.
Hi OTV, glad you see you continuing this in a place where it's less likely to be trimmed.
I'm highlighting this particular sentence because I believe this is the root of your misunderstanding. This phrasing, which has appeared a few times now, suggests to me that in your mind, you're assigning a "this-then-that" order of operations in your mind. The initial velocity needs to be zero-ed out, then the new velocity must be increased by the same amount. This will work, and it will cost 2x your delta-v as you say. But burning at a 45 degree angle (135, really), you can change both at the same time, and this is simply more efficient.
Let me try the 'distance' analogy again, but in a different way.
Imagine you're on the northwest corner of an intersection, and you want to go to the southeast corner. You can first cross south if you wish, and then cross east. That will work. But it's better to cross diagonally. Rather than crossing the equivalent of two streets (1 south then 1 more east), you've achieved your goal by taking a path that effectively only requires crossing the equivalent of 1.41 streets; heading south AND east simultaneously. No free lunch needed.
I suspect that you have little difficulty 'seeing' in your mind's eye why crossing both at the same time is faster: the distance is physically shorter. Velocity works the same way. -
#8 by TheRadicalModerate on 21 Feb, 2022 22:01
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Imagine you're on the northwest corner of an intersection, and you want to go to the southeast corner. You can first cross south if you wish, and then cross east. That will work. But it's better to cross diagonally. Rather than crossing the equivalent of two streets (1 south then 1 more east), you've achieved your goal by taking a path that effectively only requires crossing the equivalent of 1.41 streets; heading south AND east simultaneously. No free lunch needed.
In space, no one can ticket you for jaywalking. -
#9 by Eer on 21 Feb, 2022 22:34
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Imagine you're on the northwest corner of an intersection, and you want to go to the southeast corner. You can first cross south if you wish, and then cross east. That will work. But it's better to cross diagonally. Rather than crossing the equivalent of two streets (1 south then 1 more east), you've achieved your goal by taking a path that effectively only requires crossing the equivalent of 1.41 streets; heading south AND east simultaneously. No free lunch needed.
In space, no one can ticket you for jaywalking.
Unless your path crosses a solid boundary on your way to the new plane. -
#10 by OTV Booster on 24 Feb, 2022 15:49
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Yeah, I get it. It still smacks of a free lunch. Don't get me wrong. There's nothing bad about a free lunch.
Exit stage left, mumbling.
