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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?
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I would be fascinated in seeing such.
?