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Launch Trajectories Q&A
by
fatjohn1408
on 10 Jun, 2011 14:43
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Hello, I have some questions regarding how to model a launch trajectory.
For now, let me just start with the following one:
Why would a reference atmospheric model be chosen over a standard atmospheric model and vica versa.
Is it just that reference models are more accurate but more complicated? And is there a way to quantify the error when using for instance US1976?
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#1
by
fatjohn1408
on 06 Jul, 2011 13:18
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Another question then, this one about the losses during flight.
Gravity loss is defined as the integral of g*sin(gamma)
Where gamma is the flightpath angle.
This is only the case during thrusting. First thing i do not get, what does that have anything to do with it.
How about you only push your rocket with 0.02 g's when you are almost flying at orbital velocity... does gravity loss then accumulate?
My gut feeling is that the flightpathangle should be defined wrt the instantaneous orbit. instead of wrt earth.
I also have trouble to comprehend the whole thrust loss explanation.
Anyone that can guide me to good info about these losses please do. Thanks.
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#2
by
baldusi
on 06 Jul, 2011 17:26
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That can't be the gravity loss, because it doesn't considers the conservation of momentum of an orbit and you don't consider the decrease in gravity pull from distance.
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#3
by
Proponent
on 08 Jul, 2011 09:59
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The flight-path angle (FPA) is defined wrt to the local horizontal.
Acceleration at 0.02 G is likely to be done when flying nearly horizontally, hence the FPA and its sine will be very small, resulting in little gravity loss.
The gravitational acceleration, g, appearing in the formula is the *local* acceleration of gravity, hence the decrease in gravity with altitude *is* taken into account.
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#4
by
baldusi
on 08 Jul, 2011 14:59
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The flight-path angle (FPA) is defined wrt to the local horizontal.
Acceleration at 0.02 G is likely to be done when flying nearly horizontally, hence the FPA and its sine will be very small, resulting in little gravity loss.
The gravitational acceleration, g, appearing in the formula is the *local* acceleration of gravity, hence the decrease in gravity with altitude *is* taken into account.
No, what's taken into account is the incidence of gravity to the angle wrt the local horizontal. It would need to have a distance from Earths center of gravity to take into account the loss of gravity pull with the square of the distance. In other words is an approximation during the ascent phase, where the d^2 to the CG is "small". Once you get to high orbits, you have to use orbital mechanics.
Now that I've understood better the gamma, it does takes into account the momentum.
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#5
by
Proponent
on 08 Jul, 2011 15:16
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The quantity g is a function of distance from the center of the earth.
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#6
by
fatjohn1408
on 22 Jul, 2011 22:10
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The flight-path angle (FPA) is defined wrt to the local horizontal.
Acceleration at 0.02 G is likely to be done when flying nearly horizontally, hence the FPA and its sine will be very small, resulting in little gravity loss.
The gravitational acceleration, g, appearing in the formula is the *local* acceleration of gravity, hence the decrease in gravity with altitude *is* taken into account.
This local horizontal, is that the defined here?
http://www.ehartwell.com/afj/LVLH_(Local_Vertical/Local_Horizontal)
Ah, I think I start to get it. So gravity loss will also account for the difference in speed between a low and a high orbit. If you just do an impulse from Leo to Geo gravity loss between those orbits will amount to the difference between 7800 m/s and 3500 m/s (what I thought to be Geo speed, could be wrong). Okay, I think i understand.
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#7
by
Antares
on 22 Jul, 2011 22:35
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No, gravity loss is the component of F/m against the gravity vector carrying mass that is not payload (dry stage mass or propellant to be spent). Or at least that's how I describe it at the end of a long week before going to cocktails.
Kinetic energy traded for potential energy is not gravity loss.
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#8
by
fatjohn1408
on 28 Jul, 2011 11:12
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Kinetic energy traded for potential energy is not gravity loss.
I think that is a part of what it is.
if you define the equation: dVlauncher = dVgloss + dVdragloss + dVthrustloss + flightpath speed
Then where else would that transfer of energy be accounted in if it is not in gravity loss?
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#9
by
Antares
on 28 Jul, 2011 16:55
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Kinetic energy and potential energy trading back and forth is an orbit. It doesn't get lost.
There would be zero gravity loss if all propellant could be instantaneously used for thrust to a higher orbit. Gravity loss comes from impulse spent on mass carried along that will be spent later.
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#10
by
fatjohn1408
on 30 Jul, 2011 23:20
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Kinetic energy and potential energy trading back and forth is an orbit. It doesn't get lost.
There would be zero gravity loss if all propellant could be instantaneously used for thrust to a higher orbit. Gravity loss comes from impulse spent on mass carried along that will be spent later.
That is because the equation of gravity loss would involve the integral over zero time. The equation INT g sin(gamma) would amount to the same change from kinetic to potential if you set the boundaries wider than the burn time (impulse) itself. Gravity loss comes from impulse not directed perpendicular to the gravity vector. Your definition is fine too, though.
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#11
by
sdsds
on 21 Aug, 2011 01:35
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One aspect of launch trajectory planning that's easy to overlook is the location of drop zones for expended stages. I'm particularly interested in ascending node EELV and Falcon launches towards ISS. Is it clear that any spent solid boosters (e.g. for Atlas V 412) would come down relatively close to shore? Where would the drop zones for the first stages be located? Presumably north of FL, but still close to shore? Or is the vehicle far out over the North Atlantic by then?
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#12
by
Proponent
on 14 Feb, 2015 15:38
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No, gravity loss is the component of F/m against the gravity vector carrying mass that is not payload (dry stage mass or propellant to be spent). Or at least that's how I describe it at the end of a long week before going to cocktails.
Kinetic energy traded for potential energy is not gravity loss.
Until a few days ago, I was in complete agreement with this. But then something occurred to me.
Imagine we're launching to a 200-km circular orbit from a planet that's just like Earth except that it has no atmosphere and doesn't rotate. A perfect trajectory (I would once have described it as "lossless") entails two instantaneous impulses. Firstly we take off with a purely horizontal delta-V of 7965 m/s to enter a 0 x 200-km transfer orbit. Then we coast up to apogee, slowing to 7723 m/s. Circular speed at 200 km is 7783 m/s, so we complete the ascent to orbit with a second instantaneous horizontal impulse of 60 m/s.
The total ideal delta-V, then, is 7965 m/s + 60 m/s = 8025 m/s. Our actual delta-V, though, is just the final speed, namely 7783 m/s. Somehow we've lost 242 m/s. That's despite the fact that the powered maneuvers were lossless, which is seen as follows. Obviously, there are no drag losses, since there's no atmosphere. Nor do the manuevers generate any gravity losses: not only are they instantaneous, but the flight path is horizontal in both cases. There are no steering losses, because the thrust is applied solely in the direction of motion. So where does the 242-m/s loss arise?
Of course, what's happening is that during the coast to apogee, we trade kinetic for potential energy. Another perfectly accurate way of looking at it is that the component of gravity along the flight path slows us down. We could invent another category of loss (is this what people mean by "shaping loss"?) to describe the missing 242 m/s, but it's really just gravity.
To put it another way, imagine a very slightly different ascent to orbit. It's just like the previous one, except that between the initial and final impulses, we turn on a 1-piconewton thruster and apply continuous thrust in the direction of motion. Such a tiny thrust will have practically no effect on the trajectory. Because the entire ascent to orbit is now under power, however, we can, by conventional definitions, describe the disappearing 242 m/s as a gravity loss.
My point is that it's somewhat artificial to say that gravity losses accrue only when engines are on. I think it can make sense to accumulate gravity losses over an entire trajectory, even if some parts are unpowered.
EDIT: Corrected numerical error; made tenses consistent.
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#13
by
MP99
on 14 Feb, 2015 19:16
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It's not a gravity loss, it's a conversion of kinetic energy to potential energy, which you even mentioned.
http://en.wikipedia.org/wiki/Vis-viva_equationIf it was a true gravity loss, the energy would be lost. But the combined kinetic & potential energy of the elliptical orbit is no different at apogee than it had been at perigee. You can get the potential energy back by reducing the perigee.
cheers, Martin
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#14
by
Proponent
on 16 Feb, 2015 16:47
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It seems to me that the whole point of tallying up velocity losses is to be able to add a correction to the rocket equation to obtain the actual burn-out velocity. I doesn't seem to me that the fact that energy is conserved is relevant. Keeping track of the energy is very tricky, anyway, because you'd really need to track the energy of the exhaust as well as of the rocket.
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#15
by
pericynthion
on 17 Feb, 2015 00:57
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MP99 is correct.
Proponent: Consider an extension of your hypothetical scenario. Launch from your nonrotating spherical airless Earth to a parabolic escape trajectory (C3 = 0). This requires a single impulsive delta-V of 11.2 km/s. An infinite amount of time later, you arrive at an infinite distance, with zero relative velocity. Are you going to claim that the "real" delta V was zero, and the 11.2 km/s was all gravity loss? That's simply not the case, by the commonly understood meaning of the term.
To me, the "essence" of gravity loss is best understood by considering a rocket hovering in place above the ground, with zero velocity. For every second that it stands there, 9.8 m/s of delta V is being thrown away to gravity loss. This is truly loss as it's not adding to the total energy of the vehicle. Any nonimpulsive, non-horizontal launch trajectory has a component that is akin to that hovering rocket.
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#16
by
deltaV
on 18 Feb, 2015 05:08
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It seems to me that "gravity loss" is a fuzzy concept like "simple", not a precise one like "specific impulse", "delta vee" or "energy". It's still a useful concept and can usefully be quantified but definitions that are useful in some contexts don't work in other contexts. For example one popular definition of gravity loss is the integral of 1 - cosine(angle of thrust above horizontal) d(delta vee), which makes sense when launching to LEO but not for suborbital launches. For vertical suborbital flight a useful definition of gravity loss is little-gee times time spent thrusting, which again doesn't work in other contexts.
One reasonably universal definition may be the delta vee actually used minus aerodynamic drag minus the delta vee that the optimal impulsive trajectory from initial to final conditions would require, but this definition seems too complicated to be very useful. There are also a lot of details to fill in; for example when launching to LEO are the final conditions the position and velocity at engine cutoff or the apogee and perigee of the orbit? This matters because the optimum impulsive trajectory to the final orbit is presumably a Hoffman-type transfer which doesn't reach orbital height until half an orbit later and hence is a lot different from a real launch.
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#17
by
Proponent
on 20 Feb, 2015 21:10
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It seems to me that gravity loss should be defined in a way that's useful. Velocity losses are useful if they can be subtracted from ideal delta-Vs to get actual delta-Vs. For example, if I'm back-of-the-enveloping LEO launch vehicles, I'll want to know the aggregate velocity loss on the way from the launch pad to orbit so that I can factor it into my estimate of a the performance of a given design. In this case, I
do want to regard the loss of speed that occurs during non-thrusting periods, such as staging, as a loss.
On the other hand, in pericynthion's example of a payload that's going from Earth to a C3 of zero, it's likely that any maneuvers taking place after escaping Earth will be performed by a different propulsion system than the one that put the payload on an escape trajectory in the first place. In this case, I would agree that thinking of the slowing of the craft as it coasts away from Earth as a velocity loss is not useful.
Attached is one-page description of my take on useful definitions of losses -- gravity, steering, drag and pressure. It's not the only possible framework, but it is at least self-consistent.
Another twist on the whole thing is that the losses depend on the reference frame used. You might think that just after lift-off, when a rocket is climbing vertically, there are no steering losses (which arise whenever there is a component of thrust normal to the velocity vector). But if you do the calculation in an inertial reference frame, then the steering loss is non-zero during vertical climb, because the rocket has a few hundred meters per second of horizontal velocity just sitting on the pad.
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#18
by
MP99
on 21 Feb, 2015 12:14
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It seems to me that gravity loss should be defined in a way that's useful. Velocity losses are useful if they can be subtracted from ideal delta-Vs to get actual delta-Vs. For example, if I'm back-of-the-enveloping LEO launch vehicles, I'll want to know the aggregate velocity loss on the way from the launch pad to orbit so that I can factor it into my estimate of a the performance of a given design. In this case, I do want to regard the loss of speed that occurs during non-thrusting periods, such as staging, as a loss.
On the other hand, in pericynthion's example of a payload that's going from Earth to a C3 of zero, it's likely that any maneuvers taking place after escaping Earth will be performed by a different propulsion system than the one that put the payload on an escape trajectory in the first place. In this case, I would agree that thinking of the slowing of the craft as it coasts away from Earth as a velocity loss is not useful.
Attached is one-page description of my take on useful definitions of losses -- gravity, steering, drag and pressure. It's not the only possible framework, but it is at least self-consistent.
Another twist on the whole thing is that the losses depend on the reference frame used. You might think that just after lift-off, when a rocket is climbing vertically, there are no steering losses (which arise whenever there is a component of thrust normal to the velocity vector). But if you do the calculation in an inertial reference frame, then the steering loss is non-zero during vertical climb, because the rocket has a few hundred meters per second of horizontal velocity just sitting on the pad.
Launch to a circular LEO staging orbit is a one-way thing. You're not going to recover the potential energy by lowering perigee. Nevertheless, some of the dV imparted does convert into potential energy, and that is a straight 1:1 during non-thrusting periods - I don't believe it changes the dV that the vehicle must impart to reach orbit.
ISTM manoeuvres from the initial LEO (to GTO, to escape) should be accounted differently because they offer the opportunity to recover potential energy is kinetic. Also, they burns are close to impulsive, and you can't say the drop in orbital velocity from GTO impulse to GTO apogee is a gravity loss. You only imparted the dV because that was the only way to pump potential energy into the payload.
cheers, Martin
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#19
by
sugmullun
on 21 Feb, 2015 13:50
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I suck at analytical? math so I love graphs, charts, vector diagrams...
I've assumed that gravity loss is the "tax" for avoiding drag loss for a launch. Wrong?
If Earth had no atmosphere then a launch could be horizontal with no grav loss.
So, if that's a valid assumption ISTM that one could possibly make a graph showing
the relationship, grav loss vs drag loss, to an orbit with varying launch profiles?
?
I would be fascinated in seeing such.
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#20
by
sugmullun
on 21 Feb, 2015 14:27
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On the same subject- am I correct in assuming that a much larger vehicle would have a better
ballistic coefficient and so with less atmospheric drag wold be able to start it's gravity turn
sooner..or does the math say that the present trajectories in use still give the best performance?
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#21
by
Proponent
on 21 Feb, 2015 19:27
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Launch to a circular LEO staging orbit is a one-way thing. You're not going to recover the potential energy by lowering perigee. Nevertheless, some of the dV imparted does convert into potential energy, and that is a straight 1:1 during non-thrusting periods - I don't believe it changes the dV that the vehicle must impart to reach orbit.
A coasting period does change the delta-V needed. If it didn't, there would be no Oberth effect.
It seems to me that you are quite focused on the conservation of energy. I don't think conservation of energy is a very useful concept when analyzing rockets, because it's very difficult to keep track of all of the energy: each bit of exhaust is ejected at a different speed (kinetic energy) and altitude (potential energy). Just tracking velocity is much easier and produces straight-forward corrections to the rocket equation.
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#22
by
Proponent
on 21 Feb, 2015 19:47
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On the same subject- am I correct in assuming that a much larger vehicle would have a better
ballistic coefficient and so with less atmospheric drag wold be able to start it's gravity turn
sooner..or does the math say that the present trajectories in use still give the best performance?
Generally, yes, a larger vehicle will have a higher ballistic coefficient and will suffer less drag loss. It's not immediately obvious to me that that correlates strongly with how soon the gravity turn begins; it may correlate more with how large the pitch kick at the beginning of the turn is. Both are definitely correlated with the thrust-to-weight ratio, the pitch kick very strongly so. If you like charts and graphs, you could have a look at figures 50 and 51 of the
Flight Performance Handbook for Powered Flight Operations, which show those quantities, or something related, as a function of thrust-to-weight ratio for LEO missions. Variations in ballistic coefficient doesn't seem to be important enought to be worth treating, though I'm sure that if you went to extreme values there would be some effect.
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#23
by
Proponent
on 21 Feb, 2015 19:55
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I suck at analytical? math so I love graphs, charts, vector diagrams...
I've assumed that gravity loss is the "tax" for avoiding drag loss for a launch. Wrong?
If Earth had no atmosphere then a launch could be horizontal with no grav loss.
So, if that's a valid assumption ISTM that one could possibly make a graph showing
the relationship, grav loss vs drag loss, to an orbit with varying launch profiles? ?
I would be fascinated in seeing such.
Well, that's sort of the topic of this whole minidebate we've been having over the last few posts. The situation you describe (no air, horizontal launch) is
precisely the one I analyzed. The the total delta-V needed to reach a 200-km circular orbit exceeds the final speed (i.e., the orbital speed). In that sense, there is a velocity loss, and it seems to me it's fair to call it a gravity loss. Others disagree.
There is a trade-off between gravity loss and drag loss in that if you ascend very slowly, you'll keep drag losses low while racking up huge gravity losses (I think we'd all agree on that). But there are (at least in my definition) other kinds of losses, and it's not clear to me that there is a simple relationship between them.
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#24
by
deltaV
on 22 Feb, 2015 02:15
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According to
http://en.wikipedia.org/wiki/Delta-v_budget launch to LEO typically takes 9.3-10.0 km/s of delta vee, including 1.5-2.0 km/s of gravity and aerodynamic losses. Other sources give similar values, e.g.
http://forum.nasaspaceflight.com/index.php?topic=9959.msg189860#msg189860 . A rocket sitting on the pad at Cape Canaveral has a speed of about 0.409 km/s from rotation of the Earth and needs to be going at 7.784 km/s for a 200 km LEO so for a due-east launch delta vee required should be roughly 7.4 km/s plus losses. Adding that to the 1.5-2.0 km/s loss figure gives 8.9-9.4 km/s of delta vee, which is a lot lower than the 9.3-10.0 km/s figure. What's going on? Does the 9.3-10.0 km/s figure assume launch to polar orbit (where the initial velocity from the rotation of the Earth is useless)?
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#25
by
denis
on 24 Feb, 2015 19:48
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Well, that's sort of the topic of this whole minidebate we've been having over the last few posts. The situation you describe (no air, horizontal launch) is precisely the one I analyzed. The the total delta-V needed to reach a 200-km circular orbit exceeds the final speed (i.e., the orbital speed). In that sense, there is a velocity loss, and it seems to me it's fair to call it a gravity loss. Others disagree.
The problem is that you are applying the rocket equation and its simple corrections (gravity/drag losses) to a context to which it is not applicable.
What you really want to track is neither velocity nor energy. What you want is integrate over time the total force acting on the vehicle.
The rocket equation is a simple solution to that, in the trivial context of no external force acting.
As such, as gravity is always therefore, it's only really useful for "instantaneous" changes in delta-V.
For launchers, some simple corrections have been introduced (gravity and drag losses) to be able to do some simple assessment of launcher performances. These corrections are only a crude way of taking those into account and they are only valid for the short propulsive phases during which these effects are important.
However you cannot generalize these concepts to a significant part of an orbit (like from end of propulsive phase to the start of the next propulsive phase at apogee). In that case, you need to consider each "short" propulsive phase individually and link them using orbital mechanics considerations. Lumping the latter into the gravity losses is just meaningless.
...
To put it another way, imagine a very slightly different ascent to orbit. It's just like the previous one, except that between the initial and final impulses, we turn on a 1-piconewton thruster and apply continuous thrust in the direction of motion. Such a tiny thrust will have practically no effect on the trajectory. Because the entire ascent to orbit is now under power, however, we can, by conventional definitions, describe the disappearing 242 m/s as a gravity loss.
No, this would not qualify as a gravity loss, because the concept is not applicable to this situation.
What you describe is like a satellite with a Hall/Plasma thruster. You are certainly not going to use the rocket equation in this context!
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#26
by
jebbo
on 24 Feb, 2015 19:56
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On continuous thrust trajectories, are there any GMAT plugins available for such things?
--- Tony
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#27
by
Proponent
on 25 Feb, 2015 18:15
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Well, that's sort of the topic of this whole minidebate we've been having over the last few posts. The situation you describe (no air, horizontal launch) is precisely the one I analyzed. The the total delta-V needed to reach a 200-km circular orbit exceeds the final speed (i.e., the orbital speed). In that sense, there is a velocity loss, and it seems to me it's fair to call it a gravity loss. Others disagree.
The problem is that you are applying the rocket equation and its simple corrections (gravity/drag losses) to a context to which it is not applicable.
What you really want to track is neither velocity nor energy. What you want is integrate over time the total force acting on the vehicle.
The rocket equation is a simple solution to that, in the trivial context of no external force acting.
As such, as gravity is always therefore, it's only really useful for "instantaneous" changes in delta-V.
For launchers, some simple corrections have been introduced (gravity and drag losses) to be able to do some simple assessment of launcher performances. These corrections are only a crude way of taking those into account and they are only valid for the short propulsive phases during which these effects are important.
To integrate force (or, more precisely, acceleration) over time is to track velocity. The breakdown of velocity losses that I
attached to a previous post in this thread is based on precisely such an integration.
Applying velocity-loss corrections to the rocket equation can produce performance estimates accurate to about 10% for many vehicles. Given the complexity of performing full numerical performance calculations, I'd say being able to estimate velocity losses is pretty handy.
However you cannot generalize these concepts to a significant part of an orbit (like from end of propulsive phase to the start of the next propulsive phase at apogee). In that case, you need to consider each "short" propulsive phase individually and link them using orbital mechanics considerations. Lumping the latter into the gravity losses is just meaningless.
I'm inclined to agree: once in orbit, the concept of gravity loss during unpowered flight is unlikely to be useful. I'm just arguing that sometimes its useful to accrue gravity loss even during unpowered flight. An example would be during the few seconds' unpowered coast that might occur during staging.
...
To put it another way, imagine a very slightly different ascent to orbit. It's just like the previous one, except that between the initial and final impulses, we turn on a 1-piconewton thruster and apply continuous thrust in the direction of motion. Such a tiny thrust will have practically no effect on the trajectory. Because the entire ascent to orbit is now under power, however, we can, by conventional definitions, describe the disappearing 242 m/s as a gravity loss.
No, this would not qualify as a gravity loss, because the concept is not applicable to this situation.
What you describe is like a satellite with a Hall/Plasma thruster. You are certainly not going to use the rocket equation in this context!
Obviously, nobody's going to use very-low-thrust engine during ascent to LEO. I was simply using this example to to illustrate that choosing to accrue or not to accrue gravity losses depending on whether thrust is being applied, which is what I had previously thought made sense, isn't obviously always sensible.
I might disagree, though, that nobody's going to use the rocket equation in the context of a low-thrust transfer. It's a well-known result, for example, that the actual delta-V needed for a low-thrust transfer between two circular orbits is simply the difference in orbital speeds. So, if I wanted to transfer from a low orbit with a speed of 7800 m/s to geosynch at 470 m/s, the ideal delta-V needed would be 7330 m/s.
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#28
by
allamarein
on 27 Feb, 2015 23:21
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It seems to me that "gravity loss" is a fuzzy concept like "simple", not a precise one like "specific impulse", "delta vee" or "energy". It's still a useful concept and can usefully be quantified but definitions that are useful in some contexts don't work in other contexts. For example one popular definition of gravity loss is the integral of 1 - cosine(angle of thrust above horizontal) d(delta vee), which makes sense when launching to LEO but not for suborbital launches. For vertical suborbital flight a useful definition of gravity loss is little-gee times time spent thrusting, which again doesn't work in other contexts.
One reasonably universal definition may be the delta vee actually used minus aerodynamic drag minus the delta vee that the optimal impulsive trajectory from initial to final conditions would require, but this definition seems too complicated to be very useful. There are also a lot of details to fill in; for example when launching to LEO are the final conditions the position and velocity at engine cutoff or the apogee and perigee of the orbit? This matters because the optimum impulsive trajectory to the final orbit is presumably a Hoffman-type transfer which doesn't reach orbital height until half an orbit later and hence is a lot different from a real launch.
This is the point. Gravity loss definition (1-cos(pitch) can lead to a unintuitive solution. Indeed, if I could choose, I should fire my rocket nosedive, in order to have a positive contribution? Let's suppose I have already reached a sufficient altitude to not crash on land, perhaps just after crossing the apogee, in order to raise the perigee altitude.
On the contrary, it is intuitive a solution derived from Hohmann transfer, with a boost across the apogee.
Which solution would you prefer?
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#29
by
denis
on 01 Mar, 2015 16:31
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To integrate force (or, more precisely, acceleration) over time is to track velocity. The breakdown of velocity losses that I attached to a previous post in this thread is based on precisely such an integration. Applying velocity-loss corrections to the rocket equation can produce performance estimates accurate to about 10% for many vehicles. Given the complexity of performing full numerical performance calculations, I'd say being able to estimate velocity losses is pretty handy.
I'll agree with that.
I'm inclined to agree: once in orbit, the concept of gravity loss during unpowered flight is unlikely to be useful. I'm just arguing that sometimes its useful to accrue gravity loss even during unpowered flight. An example would be during the few seconds' unpowered coast that might occur during staging.
Ok, yes I agree it is useful in this case (staging). What I was more arguing against is the idea to maintain the concept of gravity losses for long unpowered phases, like between boots from the upper stage at different point in the - temporary - orbit.
Obviously, nobody's going to use very-low-thrust engine during ascent to LEO. I was simply using this example to to illustrate that choosing to accrue or not to accrue gravity losses depending on whether thrust is being applied, which is what I had previously thought made sense, isn't obviously always sensible.
Ok, then I agree that you cannot accrue just based on the fact that you are thrusting or not.
I might disagree, though, that nobody's going to use the rocket equation in the context of a low-thrust transfer. It's a well-known result, for example, that the actual delta-V needed for a low-thrust transfer between two circular orbits is simply the difference in orbital speeds. So, if I wanted to transfer from a low orbit with a speed of 7800 m/s to geosynch at 470 m/s, the ideal delta-V needed would be 7330 m/s.
But how is that using the rocket equation ?
Indeed you can use the rocket equation to compute the required propellant for a given delta-V, whatever the manoeuvre type (short manoeuvre or long, low thrust thrust). However, I don't think it makes that much sense to introduce the concepts of velocity losses (obviously not drag loss, but even gravity or steering losses) in this context. [How would you formulate it then ? The 'ideal' rocket equation produces 7330 m/s of positive delta-V (because thrusting parallel to velocity vector) and you have 14660 m/s of gravity loss such that the final velocity is 7330 m/s lower than the initial one ? It just seems very confusing]
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#30
by
allamarein
on 01 Mar, 2015 16:46
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You should do your delta-v budget.
The inertial velocity in LEO is about 7 km/s.
At this value, you must add gravity, steering and aerodynamic losses. These values are typical for the family of the family of rockets. The total delta-v is the value you have to put in Tsiolkowsky equation to find the propellant you need.
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#31
by
sdsds
on 02 Mar, 2015 05:51
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If Earth had no atmosphere then a launch could be horizontal with no grav loss.
Unless your horizontal launcher could instantaneously reach orbital velocity at ground level, it would suffer some pretty severe "litho-losses."
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#32
by
Proponent
on 02 Mar, 2015 23:41
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I might disagree, though, that nobody's going to use the rocket equation in the context of a low-thrust transfer. It's a well-known result, for example, that the actual delta-V needed for a low-thrust transfer between two circular orbits is simply the difference in orbital speeds. So, if I wanted to transfer from a low orbit with a speed of 7800 m/s to geosynch at 470 m/s, the ideal delta-V needed would be 7330 m/s.
But how is that using the rocket equation ?
Indeed you can use the rocket equation to compute the required propellant for a given delta-V, whatever the manoeuvre type (short manoeuvre or long, low thrust thrust). However, I don't think it makes that much sense to introduce the concepts of velocity losses (obviously not drag loss, but even gravity or steering losses) in this context. [How would you formulate it then ? The 'ideal' rocket equation produces 7330 m/s of positive delta-V (because thrusting parallel to velocity vector) and you have 14660 m/s of gravity loss such that the final velocity is 7330 m/s lower than the initial one ? It just seems very confusing]
That's exactly right: the gravity loss is turns out to be twice the ideal delta-V, a fact that can be derived by integrating the first equation in the attachment a few posts upthread in the limit of a vanishingly small thrust applied tangentially. I confess that in this case the result (that the necessary ideal delta-V is the difference in orbital speeds) is probably more useful than the characterization of gravity loss, but at least the framework remains consistent.
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#33
by
denis
on 04 Mar, 2015 23:17
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That's exactly right: the gravity loss is turns out to be twice the ideal delta-V, a fact that can be derived by integrating the first equation in the attachment a few posts upthread in the limit of a vanishingly small thrust applied tangentially. I confess that in this case the result (that the necessary ideal delta-V is the difference in orbital speeds) is probably more useful than the characterization of gravity loss, but at least the framework remains consistent.
Ok, I fully agree the framework remains consistent but I'll keep my prerogative of not using the term "gravity loss" in this context
?