Starship maintenance costs... what are they?

[aramis]

Not sure where MT get's its time from but I'm guessing there's a different limiting factor or something on jumping (been a long time since I looked at those rules).

Let me know if I'm getting something wrong here:

Size A planet has a diameter of 16,000 km.
100D out from 16,000 km is 1,600,000 km.
Half that distance is your 'turnaround' where you have to start decelerating, so 800,000 km or 800,000,000 m.
800,000,000m / 1/2 of 10m (for convenience) = 160,000,000
160,000,000^.5 = 12,649
12,650 * 2 (since we will spend as long after turn around as we did before it) = 25,300
25,300 / 3600 = 7 hours, 1 minute, 40 seconds.

So by my math that is way under 19 hours. What's more, that's assuming you are coming in from a dead stop, which to me is crazy because you retain your initial velocity. Let's assume you jumped in from a fairly normal size 5 world. You would have had to fly 800,000 km away from the planet which at 1 G would have taken roughly 3.5 hours. When you jumped you would have had a velocity of about 63 km/s. You would continue to accelerate towards your destination for another 48 minutes before turn around and then decelerate for 4 hours and 18 minutes.

Trip time from precipitation to planet would be 5 hours and 36 minutes.

All of this assumes that the relative velocities of the planets are identical. If there is any difference at all then the time will be decreased because the ship can jump in on a vector either ahead of or behind the target planet, depending on what is best.

Of course there's also some safety margins since you can't be sure exactly where or when you'll drop out and possibly you wouldn't be able to take the most optimal course because of masses in the way during the jump but you are looking at somewhere between a bit over 5 1/2 hours for a suicidally confident astrogator and 7 hours for an overly cautious one, not 19.

That is unless I've missed something, which is entirely possible.

Double for the round trip. Note that the tables in MTIE are for 100 to 10, and 10 to "orbit" (presumably a low orbit of under 0.1 diameters)

Step 10 of the flow chart is travel from 100 diameters
6.7 hours at 1G to a Size A. Roll for encounter.
No instruction to skip Step 11.

Step 11 is from 10 diameters, shows 2.1 hours to a Size A
Roll for encounter again.
Again, no instruction to skip step 12...

Step 12 is from orbit to surface. Shows 42 min (0.7 hours).
roll for encounter AGAIN.

Now, times outbound (steps 4-6) match these for both times and sequential nature...

So, double the time for the trip to account. Size A to Size A is thus (6.7+2.1+0.7)+(6.7+2.1+0.7)
=2*(6.7+2.1+0.7)
=2*(8.8+0.7)
=2*(9.5)
=19

Because you HAVE to count both in and out.
If you want a consistent schedule block, you have to allow the longest routed time.

Why it doesn't match the acceleration formula? Probably because they're adding rotational velocity so that you ARE in orbit. (ISS orbit is 7.6km/s, using the "Traveller G" of 10m/s/s, that's a 760G-seconds... 12:40 minutes:seconds... and that would be higher for larger worlds... but it really should not be added... in short, because you can, most of the time, simply not brake... tho' at other points, this may result in undesired vectors.

I'll agree the tables are not a good match to the formulae unless they're each to/from Low Orbit, but that's not how they're presented.

And it's slower still in TNE...
Where a 1 G-burn to 100 diameters for a size A world would be 12.6 hours by itself... plus the 42min surface to orbit. (Aside from the fuel used column, itself broken, the surface/orbit table is the same as MT's.) So, for TNE, 210.6hrs{=168+16.8+2*(12.6+0.7)=184+26.6}... plus load/unload. Call it 214 hours. 8.9 days, rather than 8.6 or so.

 

[Murdoc]

If you are at 100D and travel for 6.7 hours ~ 1G you have covered 100D's of distance & are at the planet. NOT still 10D's out. o:

IE p93 Step 10: "10 Emerge From Jumpspace ..
The ship emerges from jumpspace. The ship emerges at the limit of the gravity well (at about 100 diameters out) of the destination world that the navigator designated in jump preparation. The ship must be taken from 100 diameters in to ten diameters."

Step 11: "11 Travel From Ten Diameters ..
Take the ship from ten diameters in to orbit."

 

[rancke]

The MT numbers are based on the assumption that a ship will maneuver so as to be at rest when it reaches 10 diameters. Afterwards it has to accelerate again until midpoint and decelerate to arrive in orbit around the world.

Whereas the CT number (420 minutes) is based on the assumption that ships do the entire 100 diameters in one go.

You can, of course, postulate that traffic control requires ships to come to rest at 10 diameters, but if you do, you should at least point out that that's why ships waste time like that and that the tables don't apply to systems without a traffic control (systems with class E starports and probably also systems with class D starports).

As far as I know no Traveller travel rules about in-system travel have taken the movement of the world itself into account. Apparently ships arrive at rest relative to the destination world.


Hans

 

[esampson]

. . .
Why it doesn't match the acceleration formula? Probably because they're adding rotational velocity so that you ARE in orbit. . . .

Actually. It looks an awful lot like the formula they used to compute time is (d/9.8)^.5x2.

This is quite understandable as it is the time it would take a 1G ship to fly to a specific destination


. . .if you decelerated for the second half of the trip so you could slow to a stop.
:rofl:

 

[esampson]

. . .
As far as I know no Traveller travel rules about in-system travel have taken the movement of the world itself into account. Apparently ships arrive at rest relative to the destination world

I think that it is generally disregarded because the calculations required to work out the relative velocities are horrible. You need to figure the relative velocity of the two stars to one another then the velocity of each planet relative to that vector. The relative velocity of the stars probably would not change significantly out near the edges of the spiral arm but the velocity of the planets relative to that vector would be constantly changing as they orbit, so to accurately model it you would need to accurately model the system and keep track of exactly what day it is.


And in the end you would end up with a number that would almost certainly be cancelled out through a combination of proper astrogation (when you depart the first planet you don't fly towards your jump destination but along your goal vector so you can begin to match relative vectors) and using the maneuver drive in jump space (even at only 1% efficiency a 1 G ship can build up nearly 60 km/s in a week).

 

[Licheking]

T5 has 3 types of re-entry/launch movements based on the amount of friction and time required to land/launch.

If you do it fast then time down to and up from a planet is the Size in minutes.
If you do it normally then the time is Size in hours
and if you do it safely it is the Size + Atmosphere in hours.

I can't remember the page number off hand (At work) but the times don't seem to be modified by different G ratings which i would have thought would need to be done. It doesn't mention what the difficulty for the Pilot roll is for these re-entry/launch maneuvers but the operations section of the VehicleMaker say that re-entry/launch is 3d difficulty so i have ruled that Fast is 4d and slow is 2d taking the middle one as the standard type of re-entry/launch maneuver. It is entirely possible that your re-entry and launch will take longer than the flight from 100D.

The other thing to remember in T5 is that not all ships can come in at 100D it depends on the type of Jump Field they employ and the skill of the pilot and the tech level stage of the J-Drive. A jump bubble for example (the default) is 120D as standard add in an average Astrogator with Professional level skill Edu=7, Astrogation-3 and it becomes 117.

 

[HG_B]

T5 has 3 types of re-entry/launch movements based on the amount of friction and time required to land/launch.

and if you do it safely it is the Size + Atmosphere in hours.

T5 must has crap for space craft or, hamsters for pilots.

 

[andycowley]

Licheking,

The other thing to remember in T5 is that not all ships can come in at 100D it depends on the type of Jump Field they employ and the skill of the pilot and the tech level stage of the J-Drive. A jump bubble for example (the default) is 120D as standard add in an average Astrogator with Professional level skill Edu=7, Astrogation-3 and it becomes 117.

I was personally assuming the jump field type only affects safe jump distance on the OUTBOUND leg .... and that the INBOUND leg would always occur when the jump field (regardless of type) intercepted the 100D of the target object (star/planet/object etc.) ....

 

[esampson]

. . .The other thing to remember in T5 is that not all ships can come in at 100D it depends on the type of Jump Field they employ and the skill of the pilot and the tech level stage of the J-Drive. A jump bubble for example (the default) is 120D as standard add in an average Astrogator with Professional level skill Edu=7, Astrogation-3 and it becomes 117.

True, but because of how acceleration works those additional distances don't become significant. The length of time is increased by the square root of the distance, so to go 117% of the distance only takes 108% of the time. On a 10 hour trip (pretty much the longest possible) that would add 48 minutes.

As for the launching and landing times, those are deceptive. What they call a 'fast reentry' heats the hull to 2000/friction modifier degrees, although there's no discussion on what the effect of that temperature is to a ship. However, an unstreamlined ship has a friction modifier of /2 so it would only be experiencing 1000 degrees for a fast reentry (there's nothing specifying whether this is Kelvin or Celsius, but that doesn't matter. We will assume the worst case condition which is Celsius).

While that might sound like a lot, it really isn't. Tiles on the space shuttle are suppose to handle up to 1250 degrees Celsius (2300 degrees Fahrenheit), so pretty much the majority of Traveller ships should be able to handle fast reentry with no problem. The only ones that can't are things like Cluster, Braced Cluster, and Planetoid, which you probably wouldn't normally be landing anyway. Streamlined ships (really the worst case for the majority of ships people would be landing) would have it even easier with a temperature of 667 degrees, or roughly half the space shuttle's limits, airframes would be only 500 degrees and you could practically stick your head out the window of an lifting body at 400 degrees (ok, an exaggeration, but you get the point).

So really, the problem isn't so much the time that the launch and reentry charts give you, it's their names. Fast reentry should be called 'normal' with the other two called 'slow' and 'glacial' (although truthfully, given the temperatures fast reentry is probably too fast. I would probably house rule it as World Size * 5 minutes or something close and then make a new 'really fast reentry' which would have a much higher friction temperature).

 

[Licheking]

2000/friction modifier degrees, although there's no discussion on what the effect of that temperature is to a ship.

Apparently this is supposed to be the amount of hits inflicted on the ships hull, and assumes that the Hull AR is multiplied by 100 against friction and cold which is stated in the rules.

I would probably house rule it as World Size * 5 minutes or something close and then make a new 'really fast reentry' which would have a much higher friction temperature).

I personally ruled that fast entry/boost time was done regardless of G rating, whereas the other 2 types were divided by the G-Rating to get you final time.

 

[HG_B]

. Fast reentry should be called 'normal' with the other two called 'slow' and 'glacial' (although truthfully, given the temperatures fast reentry is probably too fast. I would probably house rule it as World Size * 5 minutes or something close and then make a new 'really fast reentry' which would have a much higher friction temperature).

You cold do it faster without any heat problem.

http://www.cnn.com/2012/10/14/us/skydiver-record-attempt

 

[Murdoc]

The current rules for re-entry are ridiculous, so I've been working on my own house rules for a while incorporating things like more realistic hull configuration stats, Friction rules, etc. (Short version: any ship should be able to reach the surface of a planet with absolutely no friction in about 30 minutes.) I've just finished and posted them so check them out if you want.

 

[esampson]

You cold do it faster with any heat problem.

http://www.cnn.com/2012/10/14/us/skydiver-record-attempt

There was quite a bit of hyperbole in Baumgartner's jump when they said he was jumping from space. He was only around 40 km high, which is about twice as high as the Armstrong limit, so he did need a space suit. However America uses a definition of 50 miles (80 km) for someone to be classified as an astronaut. FAI uses the Karman line (100 km) for deciding when someone is entering space and NASA considers re-entry to occur at about 120 km.

The last is probably the most significant factor because that's where a ship will begin hitting the atmosphere. Obviously this is for a size 8 world with an atmosphere of 7 so that height will change, but assuming a ship is coming in with engines firing at a full 1G it would take them 156 seconds to reach the ground if they entered in at a velocity of about 1.5 km/s, which is roughly 4.75 times the speed of sound.

I don't know what the velocity of the space shuttle use to be when it hit 120km but I can reasonable extrapolate that it was around 5 times as fast as that (based on the velocity you would need to maintain an orbit of 120km).

 

[HG_B]

However America uses a definition of 50 miles (80 km) for someone to be classified as an astronaut. FAI uses the Karman line (100 km) for deciding when someone is entering space and NASA considers re-entry to occur at about 120 km.

Doesn't matter. Even at those higher alt. figures, you won't burn up in a space suit. Once again, the writer didn't consider that a pilot would NOT enter the atmosphere at orbital velocity when piloting a Trav ship. If whoever wrote these rules would sit and consider before writing we wouldn't be dealing with stupid rules like this.

 

[esampson]

Doesn't matter. Even at those higher alt. figures, you won't burn up in a space suit. Once again, the writer didn't consider that a pilot would NOT enter the atmosphere at orbital velocity when piloting a Trav ship. If whoever wrote these rules would sit and consider before writing we wouldn't be dealing with stupid rules like this.

Not if they had any sense at least. I was trying to think of a case where a TU ship might do a powered landing but still have to do a de-orbit. The closest I could come up with was a ship without lifters that was orbiting at around 120 km before landing (probably because they were in a holding pattern). Even that didn't work, however, because what you would do is use a portion of your thrust to offset planetary gravity while the rest of your thrust is used to slow you to a safe speed before you descend. The only time that wouldn't work is when the planet's gravity is higher than the drive rating, but then it is very unlikely you would be trying to land on said planet anyway (and in fact it would be impossible to do so unless you had an airframe or lifting body configuration).

 

[Original_Carl]

The divergent discussion on de-orbiting and burn out is interesting, but I keep hoping that someone will chime in with a working model for ship maintenance. "Annual Maintenance" is all over the book, but there's nothing concrete.

What was the conclusion on ship maintenance costs? 0.1% of ship price per year?

What's the effect of ignoring maintenance? QREBS reduction? More frequent checks to see if components fail?

I could just make some $#!} up, but where's the fun in that?

 

[Licheking]

I do 0.1% annually or when the QREBS rating would indicate. If you fail to do the maintenance when required then the QREBS rating does decrease by one. And i have decided that Staterooms cost 10% of the ticket price in Life Support and Luxuries where applicable.