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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? -
#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. -
#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. -
#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?
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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. -
#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 -
#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.QuoteHowever 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.Quote...
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. -
#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? -
#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.QuoteI'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.QuoteObviously, 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.QuoteI 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] -
#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." -
#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. -
#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
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