Up by the bootstraps or, Interstellar Empire sans gravitics

[Rikki Tikki Traveller]

Several things related to propulsion are tied up in the idea of Specific Impulse (Isp). Essentially it is a measure of how powerful your rocket is.

While your rules seem a workable compromise, ISP is closer to a measure of 'Fuel Economy'. Thrust is the measure of how powerful your rocket is. The two tend to be inversely related (low thrust = high ISP, high thrust = low ISP) but that is of no use in Traveller which requires high thrust and high ISP.

True, but I was trying to simplify the details. Specific impulse is measured as Exhaust Velocity divided by gravity.

High Isp requires lots of power (which is our problem now). With Fusion power, higher Isp's and higher Vexh would be possible very quickly (material sciences need to improve too don't get me wrong).

By listing the numbers the way I did, someone could have a little pseudo-science to justify using the PP fuel like Jame suggested. The Vexh of his system is probably a good fraction of the speed of light, but not impossible like HePlaR.

 

[DCAnsell]

So, the fusion drive has an equivalent weapon factor of 1 per drive letter. Range is limited to CLOSE since it disperses faster than a standard weapon.

Attacks are made via the Pilot's skill (taking one of his actions for the combat round) and done at a -2 DM to hit on all other weapons being used on that ship, since the pilot will be aiming the drive, not helping the gunners, in fact, he will be hurting the gunners.

That makes the drive a weapon of last resort, so it doesn't really impact the existing designs.

That's not bad, the only thing I'd do differently is model the damage based on 1d6 per 25 tons of M-drive. Of course that means that a really huge, fast ship would do tremendous damage with its drives. A 100,000 ton 6-G ship would do 3250/25= 130 dice of damage!

 

[Rikki Tikki Traveller]

Well, that is why HG has barrages!

If you throw in some DMs to make it pretty hard to hit with that drive, then it is POSSIBLE to use as a weapon, but not really practical.

Your Universe, Your Call.

 

[Ishmael]

I dunno... the fuel ( reaction mass ) numbers don't look right to me....

Using the standard 100dton ship and 1g using L-hyd reaction_mass and an exhaust velocity of 30,000km/s, here are the numbers I got... ( I'll ignore mass ratios and handwave that the 10tonne*dton ship's mass assumes a half-full fuel tank and a certain cargo density ))

100dton ship masses 1000tonnes. This is seems to be a common convention used by rule-sets that don't figure ship mass directly....... 1,000,000Kg

To accelerate at 1g, this ship requires 10,000,000 newtons of force ( assuming 1 g =10m/s^2 )
force = mass_flow_rate*exhaust_velocity
In this example the Exhaust_velocity = 30,000,000m/s
therefore the mass_flow_rate must be .333333Kg/s

As 1 dton of fuel masses 1000kg, it would last 3000secs or 50 minutes of constant 1g accel.

A type "S has 40 dtons of fuel. If all of it were available as reaction mass, the ship could accelerate at 1 g for only 33.3hours

to attain that exhaust velocity would require 1.5e14 watts of power
( .33333Kg/s * .5 * 30,000,000^2 )
I think worrying about heat loading would make things even worse, so don't bother too much about it; handwave it away

Isp is proportional to Ve ... high Isp requires less mass_flow_rate for a given thrust, but lots of power
on the flip side.... low Isp requires more reaction_mass but much less power for a given thrust
This example engine has an Isp of 3,000,000sec


Hopefully, I didn't make any mistakes

 

[DCAnsell]

Well, that is why HG has barrages!

If you throw in some DMs to make it pretty hard to hit with that drive, then it is POSSIBLE to use as a weapon, but not really practical.

Your Universe, Your Call.

Yeah, I think it would be darn near impossible to hit a moving target with your drive exhaust, but boy if a big ship did, ouch! In my little universe, I think I'd reserve exhaust damage for relatively immobile targets, giving it a -4 DM to hit, in addition to the close range limitation. and another DM (to be named later) based on the relative sizes of the ships, or maybe an opposed task of Pilot skills, or both!

 

[Rikki Tikki Traveller]

Ishmael, we were using the same info.

The only difference is I assumed the entire ship had the density of liquid hydrogen, so the force needed was 1MN not 10MN as in your example. That is where we were different.

My 100 dton ship massed 100,000 kg.

As you can see, even changing some basic assumptions can make a big difference in the final numbers.

 

[atpollard]

The fuel (by definition) is 1 metric ton per displacement ton.
The fuselage of a passenger aircraft is also close to 1 metric ton per displacement ton (making this value reasonable for the bridge and accommodations).

Hulls are variable (based on Striker and MegaTraveller values) with higher TLs reducing the weight - but heavily armored ships are HEAVY.

Engineering (all drives combined) roughly averages a much higher 30 metric ton per displacement ton (based on MegaTraveller values).

Cargo runs the range from 1 to 30 tons per dton but averages around 3 to 4 tons per dton.

All this balances out to yield passenger ships with low armor and performance near the 1 ton per dton 'minimum' and warships above 10 tons per dton.

I am not trying to make any point, just offering statisics to anyone who might care. Since the heavy items are fictional (drives, armor, interstellar trade goods) this is an area subject to a great many IMTU type of assumptions.

IMHO, both the Rikki Tikki Traveller and Ishmael analysis are valid for MgT, they just represent opposite ends of the same spectrum.

 

[Rikki Tikki Traveller]

Interesting, I don't have MT, so these numbers weren't available to me.

So, my base ship was a civilian/scout and Ishmaels is a Military ship. Good to know if anyone wants to use reaction drives.

Thanks Arthur!

 

[Ishmael]

The only difference is I assumed the entire ship had the density of liquid hydrogen, so the force needed was 1MN not 10MN as in your example. That is where we were different.

My 100 dton ship massed 100,000 kg.

I figured it was something like that.
I don't quite agree with that one assumption though.
That would make the ship ( as a whole ) have a specific gravity of .071, which is roughly half that of balsa wood. A 100% scale model of the ship carved out of balsa wood would mass twice as much. ( balsa wood specific gravity = .16 ). The ship would have an overall density of light extruded polystyrene ( styrofoam ).

I use a specific gravity of .71 for ships in rough guesstimates of mass.
This seems to fit many MT ships and is easy to remember and figure out.
MT does state in the design evaluation section that cargo masses 1 tonne per cubic meter ( same as water ).

After looking around on the 'net for more info, I found this:
http://www.freelancetraveller.com/features/shipyard/tonnage.html

Based on this, ship's mass would probably be between 3-7 tonnes per dton.
However, a quick perusal of MT ships, which have mass as part of their design sequence, tend to run between 5-12 tonnes per dton, with the main differences being the amount of cargo/empty space on the ship. A type 'S' has its equipment 'packed in' tighter than a 400dton fat trader.
This is likely because MT requires an AV of at least 40 for ships ( FFS1 requires 10*g rating ). I don't use those limits myself, but I do base hull capacities on structure ( my own ruleset ).
Most armored warships would actually sink in water.

I'm guessing that the best method would be to make ship's mass = dtons * armor value*scaling constant
( I don't use Mgt ship rules, so I can't guess what the best way to make the ship's mass be proportional to armor rating )

 

[atpollard]

MT does state in the design evaluation section that cargo masses 1 tonne per cubic meter ( same as water ).

While MegaTraveller does say that, 13.5 tonnes per dTon is too high for a realistic cargo.

Containerized cargo on a ship is about 22 tonnes maximum per 4 dTon container - about 5.5 tonnes per dTon max and 2-3 tonnes per dTon average.

The Freight version of a 747 is designed for cargo at 10 pounds per cubic foot (156 kg per cubic meter or about 2 tonnes per dTon).

Frankly, most things we encounter are less dense than water if we include the enclosed air space.

 

[Ishmael]

That's funny
The only thing you questioned in my last post was the only thing I could quote out of a rule book!

According to Maersk container dimensions,

max capacity for 40ft containers runs around 7 tonnes per dton
max capacity for 20ft containers runs up to and a little past 14 tonnes per dton

I can imagine thats mostly limited by container construction strength and lift/trucking capacity. If so, the future containers may be able to carry heavier things
This is probably the sort of cargo a player's ship might carry.
If we abandon containers and use commodity freight trains, a lot of the cargo is denser than water for sure. (coal, taconite, lime, et al )

How about this then....
assume the cargo masses the same number of tonnes/dton as the rest of the ship (loaded )
assume there's nothing there for empty

as an example, a 100 ship, and 30 dtons of cargo volume; At 5 tonnes per dton, it masses 500 tonnes loaded (100*5 )
but only 350 tonnes empty (70*5 )
use the average of the 2 values to figure performance
(425 tonnes in this example )

I don't think the aircraft analogy works as modern aircraft are as flimsy as soap bubbles compared to the way Traveller envisions starships ( at least in MT anyways )
Running around on a plane with battledress would probably destroy the flooring and pretty much any kind of bullet would penetrate the hull... not exactly how spaceships are portrayed in the game.....
CT just punted the whole issue by pretending that mass didn't exist.

 

[Rikki Tikki Traveller]

Well, no matter what you assume your ship density to be (and I don't disagree with anything you have said), the principle is the same as what I presented.

If you believe that starships will be about 10 tons per Dton or more or less, then use that number. The only thing that changes it the required thrust. For a 100 ton ship, 1g of acceleration requires 1MegaNewtons of thrust. If your ship masses 1000 tons, then you need 10 MN of thrust.

Plug the factor of 10 increase in thrust into the equations (like you did) and you will get higher fuel consumption.

If you keep your Vexh below about 50% speed of light, you are probably not bending things too badly from a material science point of view.

The hardest thing about using reaction thrust is going to be figuring out how much fuel you will NEED for the ship's mission. Assigning 1 day of thrust seems reasonable, but for a warship, in combat, I personally don't know if that is too much or too little. Someone would have to do some serious studying work.

 

[atpollard]

Does anybody have handy formulas for the 'golfball' approach (brief acceleration, long coast, brief deceleration)?

I know A WAY to calculate it, but was wondering if there might be other (simpler) ways. With increasing reaction fuel requirements, it seems likely that lower energy concepts would become more acceptable ("OK, so it takes 3 days instead of 1 day, I can live with that.")

The minimum energy (without getting into transfer orbits or gravity slings) would probably be accelerate to escape velocity, cruise, decelerate to a stop. Some quick idea of the tradeoff between reaction mass and travel time for longer accelerations would be useful.

 

[Ishmael]

here's what I came up with....

Time_travel = Time_acc + (Dist_total-(g*(time_acc/2)^2))/(g*(time_acc/2))

Time_travel is the total travel time is seconds
Time_acc is the total burn time, in seconds, where half is acceleration and half is deceleration back to a stop after turnaround..
g is the acceleration in m/sec^2
Dist_total is the distance to be travelled in meters

From the total distance, I subtracted twice the distance travelled by the ship during half the total burn time because half of the burn time represents the acceleration and half the burn time represents the deceleration; doubled because the acceleration distance and the deceleration distance should be equal . This is the distance travelled during the coasting phase of the trip.
The coasting distance is divided by the velocity of the ship after the initial acceleration; the velocity after half the total burn ( initial acceleration ). This gives the coasting time.
This time is added to the length of the total burn to give the total time.

I just whipped this up, but the units of measure work out to give seconds if using meters and seconds. You'll have to convert to whatever units you like on your own.
Ya'll can give me grief if I screwed up.

 

[Madarin Dude]

I was thinking on an alternative method of getting small ships shuttles off planet sans gravitics. Have giant rail guns that shoot standard sized craft into orbit. Thus to get to escape velocity you have no need for on board propellant. Of course this would only be on more civilized planets that can support such a thing and it would require some engineering like having standard diameters and limits to mass for the ships in question but would save a bit on the burning sensation of liftoff and most landings would be similar to the shuttle dead stick type or at least a reentry with limited thruster use.

 

[Rikki Tikki Traveller]

Does anybody have handy formulas for the 'golfball' approach (brief acceleration, long coast, brief deceleration)?

I know A WAY to calculate it, but was wondering if there might be other (simpler) ways. With increasing reaction fuel requirements, it seems likely that lower energy concepts would become more acceptable ("OK, so it takes 3 days instead of 1 day, I can live with that.")

The minimum energy (without getting into transfer orbits or gravity slings) would probably be accelerate to escape velocity, cruise, decelerate to a stop. Some quick idea of the tradeoff between reaction mass and travel time for longer accelerations would be useful.

Ishmael gave you the math, but I think I have a better way to represent it in game terms.

CONCEPTS:
The idea of a G-hr as a velocity. This is the velocity your ship will build up by accelerating at 1g for 1 hour.

1 G-Hr = 35.316 km/sec

I have converted this into more useful units:

1 G-Hr = 127138 km/hr = 0.610 AU/month

If you accelerate for 2 hours, you get twice that speed, if you accelerate at 2gs, for 1 hour, you get twice that speed, so it is completely linear in calculating this velocity for a variety of thrust and time intervals.

However, very rarely will you be able to actually travel in a straight line. Almost every transfer will be in an ellipse, even if it is a shallow one, after all your ship is orbiting a planet or the sun and that must be accounted for.

Therefore, I suggest using the following velocities to account for this non-linear travel:

1 G-Hr = 120,000 kph = 0.5 AU/Mo

This seems backwards, but rather than calculating a longer distance, we just assume a slightly lower velocity and the end result (trip time) will be about the same.

Now, we have to consider two different types of maneuvering.

1. Orbital Maneuvers
This is travel from one orbit to another around a planet. For example, travelling from low planetary orbit to a moon or from one moon to another moon about a planet.

2. Interplanetary Maneuvers
This is travel from one planet to another

Each situation uses the same basic math, but because the time scales are so different, certain simplifying assumptions can be made about condition #2.

In each case, I have made an assumption that the acceleration time is the same as the deceleration time. This is not exact, real orbital mechanics must account for the differences in orbital velocity, but that makes things WAY to complicated. As a base assumption, I suggest you add 1 hour and 1 G-Hr of thrust to account for any variations in orbital velocities; this will be more than enough expect for very special circumstances.

So:

CONDITION 1: Orbital Travel

Step 1: Determine Straight Line Distance in Kilometers
You must know the straightline distance between your starting point and your endpoint. For my example, I will use a trip from Low Earth Orbit (LEO) to Earth's moon Luna. Distance is from 360,000 to 401,000 km.

I will select a distance of 380,000 km.

Step 2: Determine G-Hr expenditure for the trip.
This is selected by the captain of the ship. You must specify both the Thrust (Gs) and the Time (Hours)

I will select 1 G-hr of thrust (1 G for 1 Hr)

Step 3: Calculate Cruise Speed
Given my 1G-Hr, I know that my cruise speed is 120,000 kph

Had I used 2 G-Hr, the Cruise Speed would have been 240,000 kph; had I used 0.5 G-Hr Cruise Speed would be 60,000 kph.

Step 4: Determine Base Travel Time
Simply divide your distance by the cruise speed.

Base Time = 380,000/120,000 = 3.17 Hours

Step 5: Determine Acceleration Time and Distance.
It takes time to go from zero velocity to Cruise Speed, you also travel during that acceleration time. Conveniently, by assuming symmetrical thrust at beginning and end, it is possible to figure this out very quickly and easily. The distance travelled during acceleration and deceleration is the same distance that you travel during 1 hour at Cruise Speed. So, to account for this extra time to go the first unit of distance, just add the acceleration time (from Step 2) to the Base Travel Time.

So, in my example, I accelerated for 1 hour

STEP 6: Calculate Total Trip Time
To calculate total trip time, I add 1 hour to the Base Travel Time for the Acceleration Time.

In my example:

Trip Time = 3.17 hours + 1 = 4.17 hours

So, a trip from Earth orbit to Luna using a 1 G-Hr thrust pattern would take 4 hours and 10 minutes. Two of those hours (one at the beginning and one at the end) would be under 1G thrust and the remaining 2 hours 10 minutes would be under Zero-G.

CONDITION 2: Orbital Travel
This is actually a lot simpler, since we are dealing with travel times of WEEKS or MONTHS, we can ignore the acceleration/deceleration time and just assume instantaneous change in velocity to Cruise Speed and only worry about the Base Time.

So, you follow Steps 1 through 4 above and you are done.

For example, assume a trip from Earth to Mars. Earth and Mars can very in distance from 0.52 AU to 2.52 AU.

Step 1: Determine Distance
I will assume 1.5 AU distance

Step 2: Determine Thrust Pattern
I will use the same 1 G-Hr as above, this time though, I don't worry about how I got that, it could be 2G for 0.5 hours or whatever, it doesn't matter.

Step 3: Calculate Cruise Speed
Cruise speed is 0.50 AU/Mo

Step 4: Detemine Travel Time
Travel time is distance divided by Cruise Speed.

Travel Time = 1.5/0.50 = 3.0 Months

If I wanted to reduce that time by accelerating longer, I could.

Lets assume I have 24 G-Hrs of fuel to burn for this trip. I can use 22 G-Hrs for the trip, leaving me 1 G-Hr to match orbital speeds and 1 G-Hr of fuel for emergencies.

22 G-Hrs total means that I can use 11 G-Hrs to accelerate and 11 G-Hrs to decelerate

Step 1: Same Distance: 1.5 AU

Step 2: Thrust Pattern: 11 G-Hrs (11 hrs at 1G)

Step 3: Cruise Speed:
Cruise Speed = 0.50 * 11 = 5.5 AU/Mo

Step 4: Travel Time:
Travel Time = 1.5/5.5 = 0.273 Months = 8.2 Days

 

[Rikki Tikki Traveller]

Another example of Orbital Travel:

Travel from Io to Ganymede (Jupiter's moons)

Step 1: Straight Line Distance:
Io orbits Jupiter at an orbital radius of 421,700 km
Ganymede orbits Jupiter at an orbital radius of 1,070,400 km

Distances can be from 648,700 km to 1,492,100 km

For this example, I will use 1,000,000 km (I like round numbers)

Step 2: Thrust Pattern:
I will assume 0.6 G-Hrs of thrust (1 Gs for 0.3 hours)

Step 3: Cruise Speed:
Cruise Speed = 120,000 * 0.6 = 72,000 kph

Step 4: Base Travel Time:
Base Travel Time = 1,000,000/72,000 = 13.89 hrs

Step 5: Acceleration Time:
0.3 hours from Step 2.

Step 6: Total Travel Time:
Total Travel Time = 13.89 + 0.6 = 14.19 Hours

 

[FallingPhoenix]

Thanks everyone for the input, this is exactly the sort of stuff I was looking for!

Problem 2 that came to mind while reading through the responses:

Surface to orbit: these advanced fusion drives sound like they might be a bad idea for this part of the trip. Thoughts, anyone?

Also, if anyone would like to share some numbers, like those already given, for calculating surface to orbit times and fuel?

FP, who would like to visit Mars.

 

[Rikki Tikki Traveller]

I would require some other means to get to and from the surface of a planet. Fusion drives are VERY bad for atmospheres.

Avenger recently published a book about archaic ships and stations. There is a nice little table in there if you want to use chemical rockets. It basically tells you what percentage of your ship you have to dedicate to fuel for a given world size. It is for low TL, but you could easily throw a TL factor to reduce the fuel for more efficient high-tech engines. The pdf is called "Golden Age Starships #5: Archaic Ships and Stations".

One thing that will come out of this type of drive system is that the ships built for interplanetary and interstellar travel will NOT be the same ship that lands on the planet. There just won't be enough room for both engines and all that fuel. So interface craft (shuttles) will be VERY important. The modular cutter/shuttle would be very important in this kind of setting. Your cargo would be carried in these modules and then the modules would be ferried down to the planet's surface.

 

[Rikki Tikki Traveller]

BUMP

I am bumping this back up so that Mithras can use it as reference for his One Star System setting if he likes.

It might give you some ideas on drives.