Author Topic: Electron : NASA TROPICS cubesats : Wallops LC-2 : NET 1 May 2023  (Read 3414 times)

Online trimeta

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"Launch Complex 2" seems to be just a Rocket Lab in-house identifier.  MARS has "Launch Pad 0A" (Antares) and "Launch Pad 0B" (Minotaur) and on its web site identifies the new Electron pad as "Launch Pad 0C". 
https://www.vaspace.org/our-facilities

Wallops Flight Facility already has/had a "Launch Area 1" and a "Launch Area 2" which were used for sounding rockets.  "Launch Area 3" handled Scout back in the day.   And so on.

 - Ed Kyle
When SpaceX took over CCAFS Launch Complex 13 to build their landing zone, they renamed it LZ-1 (and later added LZ-2). I view Rocket Lab Launch Complex 2 the same way: they built it, they get to name it, even if that's inconsistent with the name scheme used by other nearby pads.

If it helps, they're consistent about saying "Launch Complex," so it's distinct from any nearby Launch Areas.

Offline Steven Pietrobon

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From a 185 km insertion orbit, the required delta-V to go to a 550 km circular orbit with an 8º plane change (38 to 30 degrees) is 1088.1 m/s.

Enter initial perigee height (km): 185
Enter initial apogee height (km): 185
Enter required inclination change (deg): 8
Enter required perigee height (km): 550
Enter required apogee height (km): 550

Burn at   185.0 km: theta1 =  2.33 deg, dv1 =  335.8 m/s
Burn at   550.0 km: theta2 =  5.67 deg, dv2 =  752.3 m/s
dv = 1088.1 m/s

Using the attached program, a delta-V of 1088.1 m/s is equivalent to reaching a height of 2220.3 km without a plane change. Extrapolating from the Electron users guide, that gives a payload of 140 kg for an equivalent 40º circular orbit. Two TROPICS cubesats are 12 kg total, so Electron should have performance in reserve.

ha = 2220.3 km
dv = 1088.1 m/s

I used the last two points to generate the following linear extrapolation formula. For 400 km, the formula gives 272 kg, pretty close to the 270 kg value in  the graph, so it seems to be good over a wide range.

mc = -0.07222*h + 300.6
h = height in km
mc = payload in kg
« Last Edit: 11/24/2022 05:42 am by Steven Pietrobon »
Akin's Laws of Spacecraft Design #1:  Engineering is done with numbers.  Analysis without numbers is only an opinion.

Online ZachS09

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Why can't they launch all four cubesats in one go?
« Last Edit: 11/24/2022 05:17 am by ZachS09 »
Liftoff for St. Jude's! Go Dragon, Go Falcon, Godspeed Inspiration4!

Offline Steven Pietrobon

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Why can't they launch all four cubesats in one go?

They are in different orbital planes.
Akin's Laws of Spacecraft Design #1:  Engineering is done with numbers.  Analysis without numbers is only an opinion.

Offline edzieba

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #24 on: 11/24/2022 11:07 am »
Edit: It occurred to me after thinking for a while that there is one potential difference: how long between launch and performing the trajectory-correction maneuver. If you're just trying to dodge populated areas, you can change as soon as you're past them, but if you're trying to hit a lower inclination than your launch latitude, you need to wait at least until you're under the appropriate latitude, if not until you hit the equator (I don't understand orbital mechanics enough to know whether making the change at the equator is extra efficient). The longer you wait to make the change, the faster you're going and the harder it is to change direction, so I could see "hitting extra-low inclinations" costing more than simple population avoidance maneuvers in general.
Trying to plane change at time of launch is inefficient (you waste energy gaining velocity you then need to waste more energy cancelling out again), plane changes are most efficient at apoapsis. You launch to a transfer orbit at the minimum inclination achievable (38°) then perform the plane change from that transfer orbit. The higher you can get your apoapsis (and therefor the lower the orbital velocity at apoapsis) the less energy needed for the plane change - but the more energy needed for the eccentricity change, so there is a tradeoff. This is why many GSO launches launch to supersynchronous GTO transfer orbits in order to perform the plane change with minimum propellant usage.

A dogleg is not a plane change, it is a temporary shifting of ground track (or more accurately, steering of the IIP).

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #25 on: 11/24/2022 12:05 pm »
Edit: It occurred to me after thinking for a while that there is one potential difference: how long between launch and performing the trajectory-correction maneuver. If you're just trying to dodge populated areas, you can change as soon as you're past them, but if you're trying to hit a lower inclination than your launch latitude, you need to wait at least until you're under the appropriate latitude, if not until you hit the equator (I don't understand orbital mechanics enough to know whether making the change at the equator is extra efficient). The longer you wait to make the change, the faster you're going and the harder it is to change direction, so I could see "hitting extra-low inclinations" costing more than simple population avoidance maneuvers in general.
Trying to plane change at time of launch is inefficient (you waste energy gaining velocity you then need to waste more energy cancelling out again), plane changes are most efficient at apoapsis. You launch to a transfer orbit at the minimum inclination achievable (38°) then perform the plane change from that transfer orbit. The higher you can get your apoapsis (and therefor the lower the orbital velocity at apoapsis) the less energy needed for the plane change - but the more energy needed for the eccentricity change, so there is a tradeoff. This is why many GSO launches launch to supersynchronous GTO transfer orbits in order to perform the plane change with minimum propellant usage.

A dogleg is not a plane change, it is a temporary shifting of ground track (or more accurately, steering of the IIP).
It still seems to me like any maneuver which changes the angle the rocket is traveling at will change the orbital inclination, and thus plane, relative to where the rocket would have gone absent said maneuver. However, I'll grant that performing this maneuver early into the first-stage flight, while the rocket is still gaining speed, is very different from doing so with a relight of the second-stage engine at apoapsis. Perhaps to the point where I have to abandon calling the latter a "dogleg," since the ground track doesn't change until half an orbit away from the launch site, thus looking nothing like the quick bend associated with a dogleg.

Offline edzieba

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #26 on: 11/24/2022 12:19 pm »
Edit: It occurred to me after thinking for a while that there is one potential difference: how long between launch and performing the trajectory-correction maneuver. If you're just trying to dodge populated areas, you can change as soon as you're past them, but if you're trying to hit a lower inclination than your launch latitude, you need to wait at least until you're under the appropriate latitude, if not until you hit the equator (I don't understand orbital mechanics enough to know whether making the change at the equator is extra efficient). The longer you wait to make the change, the faster you're going and the harder it is to change direction, so I could see "hitting extra-low inclinations" costing more than simple population avoidance maneuvers in general.
Trying to plane change at time of launch is inefficient (you waste energy gaining velocity you then need to waste more energy cancelling out again), plane changes are most efficient at apoapsis. You launch to a transfer orbit at the minimum inclination achievable (38°) then perform the plane change from that transfer orbit. The higher you can get your apoapsis (and therefor the lower the orbital velocity at apoapsis) the less energy needed for the plane change - but the more energy needed for the eccentricity change, so there is a tradeoff. This is why many GSO launches launch to supersynchronous GTO transfer orbits in order to perform the plane change with minimum propellant usage.

A dogleg is not a plane change, it is a temporary shifting of ground track (or more accurately, steering of the IIP).
It still seems to me like any maneuver which changes the angle the rocket is traveling at will change the orbital inclination, and thus plane, relative to where the rocket would have gone absent said maneuver.
No.
A dogleg does not change the inclination you can reach from a launch site by any appreciable value. It minimises the risk of dropping debris over population areas, but if those keep-out areas were not present then you could launch directly to that inclination form the same launch site. The velocity you add as part of the dogleg manoeuvre almost always ends up as part of the velocity component of the final orbit. 
A plane change is fundamentally different. You cannot launch from a 38° latitude launch site to a 30° inclined orbit no matter what direction you point the rocket. If you try and 'fly towards the equator' then turn 90° to point towards your desired inclination, you need to cancel out the 'northwards' velocity you gained in doing so. That means you waste energy gaining velocity and then cancelling that velocity again.

Online trimeta

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #27 on: 11/24/2022 01:31 pm »
No.
A dogleg does not change the inclination you can reach from a launch site by any appreciable value. It minimises the risk of dropping debris over population areas, but if those keep-out areas were not present then you could launch directly to that inclination form the same launch site. The velocity you add as part of the dogleg manoeuvre almost always ends up as part of the velocity component of the final orbit. 
A plane change is fundamentally different. You cannot launch from a 38° latitude launch site to a 30° inclined orbit no matter what direction you point the rocket. If you try and 'fly towards the equator' then turn 90° to point towards your desired inclination, you need to cancel out the 'northwards' velocity you gained in doing so. That means you waste energy gaining velocity and then cancelling that velocity again.
I didn't say it changes the inclination you can reach. I said it changes the inclination you would be headed to if you didn't use the dogleg. If the rocket initially launches at 40°, and then 20 miles downrange changes to be going to 50°, it will end up in a different orbital plane vs. if it continued at 40°. Yes, it could also have reached the 50° inclination by going straight and not worrying about the ground track, but it didn't, to avoid overflying populated areas. So in this case, the dogleg changed where the rocket ended up.

Again, I'm not saying that the dogleg "plane changed" because it enabled accessing an inclination otherwise inaccessible due to latitude. That inclination could have been reached without the dogleg, by just overflying land. But it wasn't, because the rocket didn't start out aiming for the correct inclination. The dogleg changed where it ended up, relative to its initial trajectory. What do you call it if the initial trajectory went to one plane, then something happened to change it to another one?

Offline edzieba

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #28 on: 11/24/2022 03:28 pm »
No.
A dogleg does not change the inclination you can reach from a launch site by any appreciable value. It minimises the risk of dropping debris over population areas, but if those keep-out areas were not present then you could launch directly to that inclination form the same launch site. The velocity you add as part of the dogleg manoeuvre almost always ends up as part of the velocity component of the final orbit. 
A plane change is fundamentally different. You cannot launch from a 38° latitude launch site to a 30° inclined orbit no matter what direction you point the rocket. If you try and 'fly towards the equator' then turn 90° to point towards your desired inclination, you need to cancel out the 'northwards' velocity you gained in doing so. That means you waste energy gaining velocity and then cancelling that velocity again.
I didn't say it changes the inclination you can reach. I said it changes the inclination you would be headed to if you didn't use the dogleg. If the rocket initially launches at 40°, and then 20 miles downrange changes to be going to 50°, it will end up in a different orbital plane vs. if it continued at 40°. Yes, it could also have reached the 50° inclination by going straight and not worrying about the ground track, but it didn't, to avoid overflying populated areas. So in this case, the dogleg changed where the rocket ended up.

Again, I'm not saying that the dogleg "plane changed" because it enabled accessing an inclination otherwise inaccessible due to latitude. That inclination could have been reached without the dogleg, by just overflying land. But it wasn't, because the rocket didn't start out aiming for the correct inclination. The dogleg changed where it ended up, relative to its initial trajectory. What do you call it if the initial trajectory went to one plane, then something happened to change it to another one?
You've missed the entire second half of the post, which explains the critical difference between a dogleg and a plane change.

Online TrevorMonty



Edit: It occurred to me after thinking for a while that there is one potential difference: how long between launch and performing the trajectory-correction maneuver. If you're just trying to dodge populated areas, you can change as soon as you're past them, but if you're trying to hit a lower inclination than your launch latitude, you need to wait at least until you're under the appropriate latitude, if not until you hit the equator (I don't understand orbital mechanics enough to know whether making the change at the equator is extra efficient). The longer you wait to make the change, the faster you're going and the harder it is to change direction, so I could see "hitting extra-low inclinations" costing more than simple population avoidance maneuvers in general.
Trying to plane change at time of launch is inefficient (you waste energy gaining velocity you then need to waste more energy cancelling out again), plane changes are most efficient at apoapsis. You launch to a transfer orbit at the minimum inclination achievable (38°) then perform the plane change from that transfer orbit. The higher you can get your apoapsis (and therefor the lower the orbital velocity at apoapsis) the less energy needed for the plane change - but the more energy needed for the eccentricity change, so there is a tradeoff. This is why many GSO launches launch to supersynchronous GTO transfer orbits in order to perform the plane change with minimum propellant usage.


These are type maneuvers Electrons's high endurance kick stage is ideal for.


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From a 185 km insertion orbit, the required delta-V to go to a 550 km circular orbit with an 8º plane change (38 to 30 degrees) is 1088.1 m/s.

Enter initial perigee height (km): 185
Enter initial apogee height (km): 185
Enter required inclination change (deg): 8
Enter required perigee height (km): 550
Enter required apogee height (km): 550

Burn at   185.0 km: theta1 =  2.33 deg, dv1 =  335.8 m/s
Burn at   550.0 km: theta2 =  5.67 deg, dv2 =  752.3 m/s
dv = 1088.1 m/s

This  is great!

When doing calculations like this in the past I've always assumed the burns would happen at equator crossings, because otherwise you can't get to a 0 deg inclination orbit. But if the target orbit is 30 deg then in theory you could do the plane change from 38 deg inclination as soon as you reach 30 deg latitude, all in one burn. If the initial apogee height were 185 km and the initial perigee height were -1000 km (i.e. sub-orbital), the single burn could conceivably require lower total deltav than the two burn approach. (Of course I should do the math before posting like this; oh, weakness!)
« Last Edit: 11/24/2022 05:18 pm by sdsds »
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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #31 on: 11/24/2022 09:51 pm »
No.
A dogleg does not change the inclination you can reach from a launch site by any appreciable value. It minimises the risk of dropping debris over population areas, but if those keep-out areas were not present then you could launch directly to that inclination form the same launch site. The velocity you add as part of the dogleg manoeuvre almost always ends up as part of the velocity component of the final orbit. 
A plane change is fundamentally different. You cannot launch from a 38° latitude launch site to a 30° inclined orbit no matter what direction you point the rocket. If you try and 'fly towards the equator' then turn 90° to point towards your desired inclination, you need to cancel out the 'northwards' velocity you gained in doing so. That means you waste energy gaining velocity and then cancelling that velocity again.
I didn't say it changes the inclination you can reach. I said it changes the inclination you would be headed to if you didn't use the dogleg. If the rocket initially launches at 40°, and then 20 miles downrange changes to be going to 50°, it will end up in a different orbital plane vs. if it continued at 40°. Yes, it could also have reached the 50° inclination by going straight and not worrying about the ground track, but it didn't, to avoid overflying populated areas. So in this case, the dogleg changed where the rocket ended up.

Again, I'm not saying that the dogleg "plane changed" because it enabled accessing an inclination otherwise inaccessible due to latitude. That inclination could have been reached without the dogleg, by just overflying land. But it wasn't, because the rocket didn't start out aiming for the correct inclination. The dogleg changed where it ended up, relative to its initial trajectory. What do you call it if the initial trajectory went to one plane, then something happened to change it to another one?
You've missed the entire second half of the post, which explains the critical difference between a dogleg and a plane change.
I agree there are two different types of maneuvers. I'm saying that both of them change the plane. Maybe formally, only one is called a "plane change." But just because you could reach a given inclination from a particular launch site without any sort of maneuver (if you were willing to overfly populated area), if for one specific launch the initial heading does not go to the inclination you actually want (because you're performing a dogleg maneuver), then changing your heading later changes the plane. And no amount of telling me "it's only a plane change if you use it to hit an inclination that would otherwise have been physically impossible from that latitude" will make me think that changing the ground track of the rocket has no impact on the plane. If you launched due east from a 38° north latitude launch site, then 50 miles later turned the rocket due north, are you going to end up in a 38° inclination orbit? The fact that you could have aimed the rocket due north when you launched it doesn't change that for this launch, you didn't.

(And yes, that particular example is especially silly. The example isn't supposed to be efficient or directly represent any sort of realistic trajectory that one would ever use. It's supposed to be an extreme example to illustrate what I'm getting at.)

Offline sdsds

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Has Rocket Lab or NASA indicated anything about whether the TROPICS launch windows will be instantaneous? ULA is rightly proud of their launch systems' ability to do "RAAN steering" into a target orbital plane. See e.g.:
https://blog.ulalaunch.com/blog/lucy-trajectory-technique-gives-atlas-v-time-to-launch

I suppose it might be worth using a term like "inclination steering" rather than "dogleg" to describe the (simpler) process of targeting an inclination that would be otherwise unavailable from a launch site.
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Offline edzieba

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Re: Electron : NASA TROPICS cubesats : NET 1 May 2023
« Reply #33 on: 11/25/2022 10:48 am »
No.
A dogleg does not change the inclination you can reach from a launch site by any appreciable value. It minimises the risk of dropping debris over population areas, but if those keep-out areas were not present then you could launch directly to that inclination form the same launch site. The velocity you add as part of the dogleg manoeuvre almost always ends up as part of the velocity component of the final orbit. 
A plane change is fundamentally different. You cannot launch from a 38° latitude launch site to a 30° inclined orbit no matter what direction you point the rocket. If you try and 'fly towards the equator' then turn 90° to point towards your desired inclination, you need to cancel out the 'northwards' velocity you gained in doing so. That means you waste energy gaining velocity and then cancelling that velocity again.
I didn't say it changes the inclination you can reach. I said it changes the inclination you would be headed to if you didn't use the dogleg. If the rocket initially launches at 40°, and then 20 miles downrange changes to be going to 50°, it will end up in a different orbital plane vs. if it continued at 40°. Yes, it could also have reached the 50° inclination by going straight and not worrying about the ground track, but it didn't, to avoid overflying populated areas. So in this case, the dogleg changed where the rocket ended up.

Again, I'm not saying that the dogleg "plane changed" because it enabled accessing an inclination otherwise inaccessible due to latitude. That inclination could have been reached without the dogleg, by just overflying land. But it wasn't, because the rocket didn't start out aiming for the correct inclination. The dogleg changed where it ended up, relative to its initial trajectory. What do you call it if the initial trajectory went to one plane, then something happened to change it to another one?
You've missed the entire second half of the post, which explains the critical difference between a dogleg and a plane change.
I agree there are two different types of maneuvers. I'm saying that both of them change the plane. Maybe formally, only one is called a "plane change." But just because you could reach a given inclination from a particular launch site without any sort of maneuver (if you were willing to overfly populated area), if for one specific launch the initial heading does not go to the inclination you actually want (because you're performing a dogleg maneuver), then changing your heading later changes the plane. And no amount of telling me "it's only a plane change if you use it to hit an inclination that would otherwise have been physically impossible from that latitude" will make me think that changing the ground track of the rocket has no impact on the plane. If you launched due east from a 38° north latitude launch site, then 50 miles later turned the rocket due north, are you going to end up in a 38° inclination orbit? The fact that you could have aimed the rocket due north when you launched it doesn't change that for this launch, you didn't.

(And yes, that particular example is especially silly. The example isn't supposed to be efficient or directly represent any sort of realistic trajectory that one would ever use. It's supposed to be an extreme example to illustrate what I'm getting at.)
If you count every change in vehicle direction as a plane change, then every single Shuttle mission, every Apollo mission, every Falcon mission, etc (basically every vehicle other than Soyuz with its rotating launch table) conducted a 'plane change' immediately after liftoff due to the roll program and gravity turn manoeuvres being conducted at the same time but not taking the same amount of time.
It is a nonsensical definition of 'plane change' to apply it to every vehicle attitude change, and makes about as much sense as trying to call any vehicle pitch change an eccentricity variation.

Online edkyle99

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"Launch Complex 2" seems to be just a Rocket Lab in-house identifier.  MARS has "Launch Pad 0A" (Antares) and "Launch Pad 0B" (Minotaur) and on its web site identifies the new Electron pad as "Launch Pad 0C". 
https://www.vaspace.org/our-facilities

Wallops Flight Facility already has/had a "Launch Area 1" and a "Launch Area 2" which were used for sounding rockets.  "Launch Area 3" handled Scout back in the day.   And so on.

 - Ed Kyle
When SpaceX took over CCAFS Launch Complex 13 to build their landing zone, they renamed it LZ-1 (and later added LZ-2). I view Rocket Lab Launch Complex 2 the same way: they built it, they get to name it, even if that's inconsistent with the name scheme used by other nearby pads.

If it helps, they're consistent about saying "Launch Complex," so it's distinct from any nearby Launch Areas.
SpaceX originally called it "Landing Complex 1", which conflicted slightly with the original LC (Launch Complex) 1 at the Cape.  They even put up a sign.  Then the name changed to "Zone", possibly to de-conflict with Air Force naming conventions.  At Wallops, Electron will be flying from Pad 0C because that's the MARS identification.   

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Online Comga

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Edit: It occurred to me after thinking for a while that there is one potential difference: how long between launch and performing the trajectory-correction maneuver. If you're just trying to dodge populated areas, you can change as soon as you're past them, but if you're trying to hit a lower inclination than your launch latitude, you need to wait at least until you're under the appropriate latitude, if not until you hit the equator (I don't understand orbital mechanics enough to know whether making the change at the equator is extra efficient). The longer you wait to make the change, the faster you're going and the harder it is to change direction, so I could see "hitting extra-low inclinations" costing more than simple population avoidance maneuvers in general.
Trying to plane change at time of launch is inefficient (you waste energy gaining velocity you then need to waste more energy cancelling out again), plane changes are most efficient at apoapsis. You launch to a transfer orbit at the minimum inclination achievable (38°) then perform the plane change from that transfer orbit. The higher you can get your apoapsis (and therefor the lower the orbital velocity at apoapsis) the less energy needed for the plane change - but the more energy needed for the eccentricity change, so there is a tradeoff. This is why many GSO launches launch to supersynchronous GTO transfer orbits in order to perform the plane change with minimum propellant usage.

A dogleg is not a plane change, it is a temporary shifting of ground track (or more accurately, steering of the IIP).

Is any of this correct?
A dogleg IS a plane change.
Period
Plane changes are more efficient when done at lower velocity.  That means at apogee (apoapsis in general) not perigee. That’s why supra-synchronous transfer orbits work well.
Or it means soon after launch like the gravity turn from getting out of the atmosphere to gaining orbital velocity.
All energy is not equal between stages.
Photon may have capacity to change altitude and self-deorbit, but it’s not enough for much of  a plane change.
The specific optimum requires math, obviously, calculus of variations and such, which I can no longer do.  But shy of calculating (“just an opinion”) Electron likely will launch to the southeast, and yaw eastward when it gets near 30 deg N, making much of total plane change as early as possible. 
We have members on NSF who can do that calculation, and I hope they do.

The “roll programs” are completely separate, distinct from any change in direction. Those are to simplify internal guidance calculations.
« Last Edit: 11/25/2022 04:06 pm by Comga »
What kind of wastrels would dump a perfectly good booster in the ocean after just one use?

Offline edzieba

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A dogleg IS a plane change.
Period
No. A dogleg is a manoeuvre to shift the ground track for a launch azimuth to an inclination that is achievable at the launch site latitude but required ground avoidance. A plane change is when you change the plane of an orbit to one you could not achieve from that latitude (otherwise you would have launched to that inclination).
Quote
Plane changes are more efficient when done at lower velocity.  That means at apogee (apoapsis in general) not perigee. That’s why supra-synchronous transfer orbits work well.
Or it means soon after launch like the gravity turn from getting out of the atmosphere to gaining orbital velocity.
All energy is not equal between stages.
No, plane changes are not efficient during ascent. That is why nobody performs plane changes during ascent, and do it once in a parking or transfer orbit instead.
Quote
Photon may have capacity to change altitude and self-deorbit, but it’s not enough for much of  a plane change.
Photon (and the kick-stage, which is likely what will be used as the full Photon capability is not required) has sufficient delta-V for the required plane change, as calculated earlier in the thread.
Quote
The “roll programs” are completely separate, distinct from any change in direction. Those are to simplify internal guidance calculations.
Vehicles do not leap off the pad directly into the final launch azimuth. They first need to clear the physical structure around the pad, then the pad complex, then the launch complex (for US sites, that means getting out over the ocean as all sites are coastal), then you can start being concerned with final launch azimuth.

For example, Shuttle first had to clear the launch mounts, then pitch over 'backwards' (which because the MLP could not rotate, meant the vehicle was now pitched due south), then clear the launch complex, and only then could it start the roll program to bring it to the final launch azimuth.
By the "any vehicle direction change is a plane change" idea, that would mean that every single Shuttle launch was a 0° polar orbit launch that conducted a plane change to the final 35°-120° azimuth.

::EDIT:: Attached is another Shuttle-era example: a dogleg that performs no inclination change at all. This is also what Atlas V's 'RAAN steering' also does.
« Last Edit: 11/25/2022 06:02 pm by edzieba »

Offline LouScheffer

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It still seems to me like any maneuver which changes the angle the rocket is traveling at will change the orbital inclination, and thus plane, relative to where the rocket would have gone absent said maneuver.
The mathematician in me forces me to say this is not true that a change in flight angle always results in a change in inclination.  For example, suppose you launch from the cape.  When you reach the equator your trajectory is 28.5 degrees to the equator, going south.  Then suppose you do an enormous plane change, until you are pointing 28.5 degrees north of the equator. Both orbits have exactly the same inclination.   So a huge change in angle, 57 degrees, results in no change in inclination.  Of course this is an enormously expensive and completely pointless maneuver, since the same resulting orbit can be obtained simply be launching earlier or later.  But it's mathematically possible.

More practically, any burn that is within the plane of the existing orbit (but not straight ahead or straight back) will change the direction of flight without changing the inclination or orbital plane.  It can change the apogee, perigee, eccentricity, and so on, and is often used for this purpose.


Online trimeta

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It still seems to me like any maneuver which changes the angle the rocket is traveling at will change the orbital inclination, and thus plane, relative to where the rocket would have gone absent said maneuver.
The mathematician in me forces me to say this is not true that a change in flight angle always results in a change in inclination.  For example, suppose you launch from the cape.  When you reach the equator your trajectory is 28.5 degrees to the equator, going south.  Then suppose you do an enormous plane change, until you are pointing 28.5 degrees north of the equator. Both orbits have exactly the same inclination.   So a huge change in angle, 57 degrees, results in no change in inclination.  Of course this is an enormously expensive and completely pointless maneuver, since the same resulting orbit can be obtained simply be launching earlier or later.  But it's mathematically possible.

More practically, any burn that is within the plane of the existing orbit (but not straight ahead or straight back) will change the direction of flight without changing the inclination or orbital plane.  It can change the apogee, perigee, eccentricity, and so on, and is often used for this purpose.
Sure, I wasn't talking about things in the same "direction" as gravity turns or whatnot. Although I hadn't considered edzieba's point about launches from non-rotating pads where infrastructure (and possibly making a beeline to the coast) means you're not headed to the target inclination from T-0. I still think there's a little difference between doing this turn moments after you clear the tower and tens of miles downrange, but (and I say this as a non-rocket scientist) I could see the calculations being very similar, since you're making the adjustment during the initial boost phase in either case.

Although I'd say that ULA's "RAAN steering," if it means turning twice as in the picture, doesn't count as "changing the inclination" in the same way a dogleg does, simply because it changes it once but then changes it back to what it started at. Of course, the scale in that picture could be wrong, perhaps the initial trajectory (before the first turn) is different from the final heading (after the second turn), in which case it's doing the same thing as a dogleg.

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