Are ships heat shielded?
- Thread starter whartung
- Start date
esampson
SOC-13
Yes, that's assuming a vertical path. Gliding would add to the time but there wouldn't be a whole lot of purpose to gliding, beyond the fact that you might have to make some adjustments because you aren't arriving directly over your destination.
I would imagine that one of the jobs of the navigator is to try and calculate an exit point from J-space and a vector that would line the ship up for its arrival point. However, since there's some variation to exactly when and where the ship emerges there would need to be some course corrections.
I would imagine that one of the jobs of the navigator is to try and calculate an exit point from J-space and a vector that would line the ship up for its arrival point. However, since there's some variation to exactly when and where the ship emerges there would need to be some course corrections.
GypsyComet
SOC-14 1K
Only *really* time sensitive jumps need to do that, and the variability of jump transit time makes it all a craps shoot anyway. The reality is that even the slowest of starships is capable of 1G continuous maneuver, which is immense by the time you are within 100 diameters of a mainworld. 800,000 miles falls pretty quickly even from a standing emergence, so those last 8000 miles (or 12,000 since you have to go around) are trivial. The stellar jump horizon is also frequently a factor, so popping out above the daylight side of a habitable world is not easy,
kilemall
SOC-14 5K
Another major variable would be streamlined vs. the 'partially streamlined', who thanks to their shapes could generate more heat friction and/or not be fully controllable on a 'hot' reentry.
The NASP was a testbed for flowing hydrogen over the hull for cooling/energy recovery purposes, so that could be a technique for reducing wear and tear on hulls.
On hot vs. not, I would think the more pop and higher quality the starport, the more hot reentry would be discouraged, for the NIMBY environmental noise/light issues, and also safety.
But you couldn't eliminate them entirely, so I would think starports would tend to be located next to oceans, deserts, ice caps etc. that allow a hot approach.
One some of my IMTU planets reaction drive takeoff and landing is forbidden, must be grav.
I think repulsor landing would be not a good idea due to the potential for repulsor failure at critical instances, and the fact you would have to manage a LOT of repulsor 'lanes' that would have to be active for several minutes at a time.
But repulsor 'catapults' that help a ship launch fast and reduce time and space usage to allow more ships to egress the same atmo windows would be highly desirable.
The NASP was a testbed for flowing hydrogen over the hull for cooling/energy recovery purposes, so that could be a technique for reducing wear and tear on hulls.
On hot vs. not, I would think the more pop and higher quality the starport, the more hot reentry would be discouraged, for the NIMBY environmental noise/light issues, and also safety.
But you couldn't eliminate them entirely, so I would think starports would tend to be located next to oceans, deserts, ice caps etc. that allow a hot approach.
One some of my IMTU planets reaction drive takeoff and landing is forbidden, must be grav.
I think repulsor landing would be not a good idea due to the potential for repulsor failure at critical instances, and the fact you would have to manage a LOT of repulsor 'lanes' that would have to be active for several minutes at a time.
But repulsor 'catapults' that help a ship launch fast and reduce time and space usage to allow more ships to egress the same atmo windows would be highly desirable.
esampson
SOC-13
Actually, it probably isn't just time sensitive jumps that do that. If you pop out 100D from Earth on an optimum vector with a 1G ship it would take 5 hours, 39 minutes, 24 seconds to make land fall. On the other hand if you are at a dead stop relative to your arrival destination it takes an additional 2 hours, 20 minutes, and 35 seconds to reach your destination.
On the surface it would seem you are right, the difference between best and worst is only 2 and a half hours. However, being stopped relative to your destination is far from the worst case scenario. Assume that you exit jump space with the same velocity but a vector 180 degrees off and you add another 7 hours, 59 minutes, and 59 seconds to the trip. That's an entire shift that people could be enjoying on the planet, and that's not even truly worst case.
If the navigator pays no attention to the exit vector the ship could easily be moving even faster than that in the wrong direction, especially if you had to travel a lot further out to jump because your point of departure was close to a primary or was the moon of a gas giant.
So since the navigator has to pay attention to their arrival vector they might as well choose one that is fairly efficient. Now, yes, they aren't actually going to be trying to hit a point that requires them to drop out at exactly 100D from the planet after 168 hours of jump with a velocity that bleeds exactly to 0 as they touch down. They will put in safety margins and may have other considerations such as making sure the exit point is outside of 100D of a star or gas giant if the arrival point is within one of those wells, but the idea remains; a navigator should be targeting an exit point and vector for their jump that gets the ship to its destination in a fairly efficient manner.
GypsyComet
SOC-14 1K
The idea that emergence velocity is preserved or predictable through jump has generated a great deal of discussion over the years and editions, so the assumption that you can do these things is frequently campaign specific. For others, the assumption has no relevance.
Also, you can aim into a jump horizon and initiate (riskier) jump from within one, but you *always* emerge at the edge (assuming you hit it at all). Jump shadows are a thing, so if the target world is behind its sun's 100 diameter limit relative to your jump initiation point, you will be stopped at the sun's jump horizon and need to use M-Drive from there.
BlackBat242
SOC-14 1K
It also helps when the vessel never gets anywhere near orbital velocity - which is the case in all your examples (as well as the X-15). They all reached less than 20% of orbital velocity.
Therefore, unless you can use M-drive to drop your speed from orbital to M4 or less those examples are useless in describing a Traveller re-entry.
If you can do so, then those examples are valid.
Thinking of other reasons for low-velocity atmospheric insertion, there's the planetary defence angle (sic): Objects coming in at high velocity are identifiable by expert systems in the defence network as potential HiVel Impactors (rocks, comets, debris, etc), and are therefore targeted on hi-pop/hi-ind worlds for automatic destruction - local traffic control rules would tend to emphasise coming in at sub- (whatever they feel comfortable with) velocities to avoid potentially terminal misidentification 
esampson
SOC-13
You would be stopped at the star's jump horizon, yes, but that's all the more reason to try and plan a clean exit vector. If the ship needs to cover 20 million km then you are looking at a difference of around 7 hours and 20 minutes between a standing start and a ship with an 'optimum' vector (and around 25 hours if you exit jump going the wrong way).
However, yes, you are correct that there is debate as to whether velocity is preserved and how predictable the jump variances are, which makes that a bit campaign specific. I believe in T5 Marc has pretty well spelled out that velocity is preserved (although it is also possible to alter your velocity during your time in Jump space) and there is a strong implication that the variances are not predictable, but that only covers T5 and that only covers cases where the Referee isn't making their own house rules.
Always thought this was canon. It's explicit in JTAS #24 that the vector is maintained. Whatever vector you entered jump space with, you exit it with, and that vector is universal, not relative to the arriving system.
Simply that means that whatever vector differential there is between the two systems (notably their vector orbiting the galaxy) is added to your vector when you arrive (since you entry vector is very likely relative to the entering system).
esampson
SOC-13
It is spelled out in various locations. However, it isn't put anywhere nice and neat and I believe it is contradicted in other, similarly messy, locations. Also JTAS would only be a ruling for CT and wouldn't affect MT, TNE, T5, etc., although it could be taken as corroborating evidence.
That, incidentally, is how I assume J-Space works so I'm not trying to discount your position with a nebulous 'there's contradictory stuff' without providing support because I don't wish to relinquish my position. It's just that I also realize there are some other interpretations out there and that I do believe they come from a position of some support.
All of that aside, I went ahead and found some formulas for estimating atmospheric pressure (and density) at various altitudes. With that in hand I was able to put together a Google docs spreadsheet that illustrates relative forces a craft would experience landing on Earth if it had a constant deceleration. Force is expressed as V2 * Air Density. 318 m/s is the amount of force experienced by an object travelling at sea level at the speed of sound or at twice the speed of sound with an atmospheric density of 1/4 normal. Times expressed are the length of time to touch down from 200 km. For vertical landing it is assumed that the vertical speed will be 0 at touch down. For glide slopes it is assumed that the craft touches down with an airspeed of 70 m/s (landing speed of a 747). I choose 3 degrees as that is the typical glide slope for most airports according to Wikipedia and 5 degrees because that is the glide slope for London (and there are airports with much steeper glide slopes, so 5 degrees seems very reasonable for a space craft).
Of course depending on how maneuver drives work in your interpretation of the Traveller universe you might not be able to make the stronger deceleration while actually gliding since it may require your engine to be contrary to your direction of travel but I included them for people who interpret maneuver drives as being able to push in directions other than the aft of the ship.
It is interesting to note that the lower the angle the longer it takes to land and the greater the heating. There is no real upside to such an approach. Simply put, unless there is a requirement for the ship to land like an aircraft (most likely because the drive is not capable of directly overcoming gravity) you actually want the steepest approach possible.
You can in all but TNE.
mike wightman
SOC-14 10K
You can in TNE too...
mike wightman
SOC-14 10K
What about the galactic vector?
Once you start trying to pin jump drives down to a frame of reference you find several concepts do not work as described.
For example, jump is accurate to within 1000km across parsecs.
Wrong - if the temporal uncertainty is in game rather than meta game then you can arrive at where you want to be but not when - your actual target could be +/- a few hours and therefore several thousands of km distant from where you intended.
No, you usually can in TNE. The need for reaction mass makes aerodynamic re-entry far more important in TNE. But then, TNE also has the ship suffering only 2% of local G's, thanks to contragrav, so that can be surprisingly effective.
In CT, MT, T20, T5, GT, MGT and HT, almost all spacecraft have gravitic thrust, so as long as the PP is running, they can do so.
In TNE, by contrast, it's often quite common for ships (Ok, at least for PC ships) to be out of fuel by reentry, so descent can be controlled/impelled by turning off the contragrav for a short, gaining a few m/s downward, then putting it back on as you hit the desired speed, so that your slow descent is balanced at a desired speed. And in TNE, descents can be hours long, if you don't have enough fuel for the powered stop at the end.
T4 includes both modes... gravitic thrust and contragrav, so ships with both can descend at any speed they can brake from...
Unless you're jumping "intergalactic", then your previous vector was relative to the galactic vector already, and would remain so when you popped out. So, only the vector of the system you're arriving in relative to the galactic vector is particularly important.
Also, since the systems are so close together, the system vector isn't some huge insurmountable offset, based on what I've see for different star trajectories calculated from here on Earth. If you consider it similar to leaping from one particle in a tornado to another, while the particles are going very fast compared to the ground, compared to each other, their differentials are not as great.
JTAS #24 addresses that tool. It says it's accurate to within 3000km per parsec jumped with the primary factors affecting the accuracy being a combination of computer accuracy and drive tuning.
Clearly this accuracy is relative to some absolute system of measurement, not "relative" to some object hurtling in space.
esampson
SOC-13
More to the point, while in TNE fuel (or more specifically reaction mass) is a consideration it is nothing like it is in the real world. Fuel is measured in 30 minute turns. Since most ships have in excess of 50 turns of fuel they can do a continuous burn for 25 hours. Our modern craft are only able to produce full thrust for about 30 minutes (excepting things like the Dawn Space Probe which is capable of producing full thrust for a much greater period of time but which only produces a small fraction of a G).
So yes, it is possible to need to do a fiery aerobrake re-entry, but that is probably the TNE equivalent of someone driving at a parking space at 60 mph and then jamming on the brakes; possible to do but really only done either because you are being extremely flashy or else because the navigator has really screwed up their job.
