I noticed that the M-drive in Traveller⁵ has a 1000D limit, in that it is only operational within 1000 diameters of a gravity source. For the solar system, the drive would cease functioning (or function at 1% efficiency) at ~9.3AU. How is it envisioned that outer or remote system objects are visited in T5? Are ships operating in that region expected to have a NAFAL drive supplementing their maneuver drive?
T5 M-Drive distance limit
- Thread starter sudnadja
- Start date
- Option #1: Install an H-Drive (HEPlaR) or N-Drive (NAFAL) or O-Drive (Orion-Drive) alongside the M-Drive if the ship is likely to do much out-system maneuver. Alternatively, install a T-Drive (Reactionless Thruster) which is not limited by G-fields.
- Option #2: Accelerate at full maneuver the entire way to the 1000-D limit, then coast (or accelerate at 1.0% drive efficiency) to a predefined point, then roll over and decelerate at full maneuver when you come within the 1000-D limit of your destination. (Adjustments will have to be made, of course, as to the exact roll-over point to accommodate correct deceleration into destination orbit).
- Option #3: Use your Jump Drive instead, from 100-D limit of departure point to 100-D limit of destination.
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That is why I said you will have to adjust the roll-over: maybe you do not accelerate to the full 1000-D limit, but only to part of that limit (however the mathematics works out). It is a much more complex maneuver solution using this method.
Just off the top of my head, if there is enough gravity to keep a planet in orbit, wouldn't there be enough to react with a gravitic drive? Just a wild hair idea.
mike wightman
SOC-14 10K
So you flip and decelerate at 1% of drive efficiency until you enter 1000D of your target.
You may have to limit your velocity while within 1000D of your original gravity source so that you can decelerate enough to arrive at your target with the velocity you desire.
It would also be possible to use any planets/asteroids/large rocks on you course to adjust your velocity while within their maneuver zones.
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Here's the situation in particular:
A hypothetical ship, nominally capable of 2g in M-Drive, needs to transit to a remote system sednoid currently in orbit of a sunlike star, from the habitable garden world. Present distance is 94 AU, and the Sednoid has a radius of 127349 meters (~127 km). The incoming ship can, at maximum, decelerate over 4000x the radius of the world, which means the maximum velocity change it can do at full efficiency is 141352 m/s, which is therefore also the velocity it will need to transit the system at - which makes a transit time of 3 years, 2 months.
I had it in my head that M-Drives worked out to about Oort Cloud distances, but that was mistaken, and it seems the more appropriate transit method would be a micro jump (7 days certainly beats 3 years). Is this sort of "expected" in the T5 universe, that there are a lot of in-system micro jumps or that ships frequently have other propulsion systems than M-drives installed?
mike wightman
SOC-14 10K
You can accelerate/decelerate at 1% efficiency for the entire journey beyond the garden planet's 1000D limit...
The first thing you do is actually accelerate in the opposite direction to the sednoid. You then accelerate towards the sednoid using the full 2000D to build velocity.
You can then accelerate at 1% until you need to flip in order to decelerate at 1% until you enter the sednoid's maneuver space.
So calculations required so far:
1 - time to the 1000D limit in the direction away from the sednoid.
2 - velocity achieved traversing 2000D of garden world towards the sednoid and time elapsed.
3 - velocity required to slow down upon arrival within the sednoid 1000D limit
4 - acceleration/deceleration required in the 1% zone between the garden world and the sednoid and time elapsed.
5 - time elapsed while maneuvering in the sednoid 1000D zone.
I'm not sure of the author's intent in requiring quite this amount of maths for a simple journey that in CT could be achieved easily with a quick calculation...
The first thing you do is actually accelerate in the opposite direction to the sednoid. You then accelerate towards the sednoid using the full 2000D to build velocity.
You can then accelerate at 1% until you need to flip in order to decelerate at 1% until you enter the sednoid's maneuver space.
So calculations required so far:
1 - time to the 1000D limit in the direction away from the sednoid.
2 - velocity achieved traversing 2000D of garden world towards the sednoid and time elapsed.
3 - velocity required to slow down upon arrival within the sednoid 1000D limit
4 - acceleration/deceleration required in the 1% zone between the garden world and the sednoid and time elapsed.
5 - time elapsed while maneuvering in the sednoid 1000D zone.
I'm not sure of the author's intent in requiring quite this amount of maths for a simple journey that in CT could be achieved easily with a quick calculation...
Simple... 1000D is enough to get a pretty decent speed up.
Remember:
So, let's take earth... roughly 12,742,000 m So 1000× that and, say, 1G=10m/s²
D=12,742,000,000=0.5AT²
12,742,000,000=0.5 × 10 × T²
12742000000×2=2×0.5 × 10 ×T²
25484000000=10 × T²
25484000000/10=10/10 ×T²
2548400000 = T²
Sqrt(2548400000) = T
50481 sec = T
And thus V=AT
V = 10 × 50481.7
V = 504817 m/s
1.496e+8 km/AU
296345 s per au
82:19:05 H:M:S per A
3d 10:10:05 per AU
Close enough to 3.5 days
Mars at closest approach is just over 0.52 AU, at at opposition, just about 2.5 AU but needing a 0.1 AU deviation.
However, since sol's radius is 1/215 AU, and the diameter is thus 2/215, and 1000D is 2000/215 or 9.3 AU, and saturn's semimajor is 9.58 AU....
You can coast to saturn pretty easily as if just a 9.3 AU course but with 0.28 AU's of coasting during the deceleration leg...
A pure burn course to saturn would be 210:19:4 (rounded up to nearest second) - but that's past the solar cutoff. (and is not actually closest approach, nor fully right angle, either. most of the year is longer than this, and about a 5th is notably shorter...)
An interrupted burn... saturn's 60268000 m radius, thus the acceleration limit is 120536000000m; at 1G is 451784 sec, for 4,517.840 km/s coast able to be soaked easily. (More can by flying past and using both sides of the sphere plus a little more by bending the course around saturn)
we need to coast 0.28 AU or 41888000000 m, dividing that by speed 92717 s
25:45:17 coast, plus the 9.3 AU burnable... (9.02 before the coast, 0.28 after)
which is 207:13:18, for 232:58:35
which is only 22:38:59 hours longer than a pure burn.
Not to scale, but conceptual illustration
So instead of
AAAAAAAAAAFDDDDDDDDD
you get
AAAAAAAAAFDDDDDDDCCD.
Remember:
- distance crossed D is D=0.5AT²
- Speed at distance is V=AT
So, let's take earth... roughly 12,742,000 m So 1000× that and, say, 1G=10m/s²
D=12,742,000,000=0.5AT²
12,742,000,000=0.5 × 10 × T²
12742000000×2=2×0.5 × 10 ×T²
25484000000=10 × T²
25484000000/10=10/10 ×T²
2548400000 = T²
Sqrt(2548400000) = T
50481 sec = T
And thus V=AT
V = 10 × 50481.7
V = 504817 m/s
1.496e+8 km/AU
296345 s per au
82:19:05 H:M:S per A
3d 10:10:05 per AU
Close enough to 3.5 days
Mars at closest approach is just over 0.52 AU, at at opposition, just about 2.5 AU but needing a 0.1 AU deviation.
However, since sol's radius is 1/215 AU, and the diameter is thus 2/215, and 1000D is 2000/215 or 9.3 AU, and saturn's semimajor is 9.58 AU....
You can coast to saturn pretty easily as if just a 9.3 AU course but with 0.28 AU's of coasting during the deceleration leg...
A pure burn course to saturn would be 210:19:4 (rounded up to nearest second) - but that's past the solar cutoff. (and is not actually closest approach, nor fully right angle, either. most of the year is longer than this, and about a 5th is notably shorter...)
An interrupted burn... saturn's 60268000 m radius, thus the acceleration limit is 120536000000m; at 1G is 451784 sec, for 4,517.840 km/s coast able to be soaked easily. (More can by flying past and using both sides of the sphere plus a little more by bending the course around saturn)
we need to coast 0.28 AU or 41888000000 m, dividing that by speed 92717 s
25:45:17 coast, plus the 9.3 AU burnable... (9.02 before the coast, 0.28 after)
which is 207:13:18, for 232:58:35
which is only 22:38:59 hours longer than a pure burn.
Not to scale, but conceptual illustration
So instead of
AAAAAAAAAAFDDDDDDDDD
you get
AAAAAAAAAFDDDDDDDCCD.
Yes, within the solar cutoff is a very easy problem to solve, outside of the solar cutoff it becomes a bit more difficult. In this case, the destination is well outside the solar cutoff (~90 AU distant). If you accelerate continuously at 2g to the solar cutoff you will have too much velocity to bleed off at 0.02g for the rest of the trip even including the 1000D 2.0g deceleration available at the sednoid (thanks to how small the sednoid is), so you have to turnover to start the deceleration within the solar cutoff somewhere - within the 9.3AU or whatever it is.
The first steps aren't that hard, and if the turnover point is outside the solar cutoff it's very staightforward and can be treated as a simple algebra problem ( {(90 AU - r2) == 1/2 9.8 2 t^2, v == 9.8 2 t ,
1/2 v^2 == 9.8 2 r1 + 9.8 0.02 (r2 - r1)} - basically an acceleration, kinetic energy and acceleration potential energy system of equations (which could be simplified) - I'm solving for r2 in this case, which is the turnover point inbound to the sednoid. In this example, v -> 2.2881*10^6, t -> 116740., r2 -> 1.33304*10^13 (SI base units) but it doesn't work, as the turnover is within 1000D of the sun.
Compounding the problem is determining where the exit point is from the solar 1000D sphere. My target is inclined vs the garden world's ecliptic and the direction to the target isn't a straight line from the star through the planet to the target. Additional compounding the problem is I do account for gravity of the star and planet, so there is some deceleration vs the 2g then 0.02g of the ship, and not in the line of flight of the ship. [I can ignore that for now, but eventually would like to build it in]
(inclination can be easier seen from an angle)
The first steps aren't that hard, and if the turnover point is outside the solar cutoff it's very staightforward and can be treated as a simple algebra problem ( {(90 AU - r2) == 1/2 9.8 2 t^2, v == 9.8 2 t ,
1/2 v^2 == 9.8 2 r1 + 9.8 0.02 (r2 - r1)} - basically an acceleration, kinetic energy and acceleration potential energy system of equations (which could be simplified) - I'm solving for r2 in this case, which is the turnover point inbound to the sednoid. In this example, v -> 2.2881*10^6, t -> 116740., r2 -> 1.33304*10^13 (SI base units) but it doesn't work, as the turnover is within 1000D of the sun.
Compounding the problem is determining where the exit point is from the solar 1000D sphere. My target is inclined vs the garden world's ecliptic and the direction to the target isn't a straight line from the star through the planet to the target. Additional compounding the problem is I do account for gravity of the star and planet, so there is some deceleration vs the 2g then 0.02g of the ship, and not in the line of flight of the ship. [I can ignore that for now, but eventually would like to build it in]
(inclination can be easier seen from an angle)
I'm not quite sure what you did there, amount of velocity change from given distance and acceleration is v² = 2 a x, where x is distance traveled. 4000x saturns radius and 10 m/s² would give a velocity change of about 2195 km/sec - that's how much deceleration saturn could absorb - presuming an overshoot and return, getting 2x the sphere diameter.
120536000000 at 1 g (9.8 m/s²) would be 120536000000 = 1/2 a t², and solving for t gives 156841 seconds, not 451784 seconds, right?
There is always the classic Transfer Orbit.
Accelerate to the required difference between source and destination orbital velocities plus the available deceleration from the target world 1000 diameters, intercept the target world on an elliptical orbit and decelerate as you approach the destination, then one last push to circularize your orbit around the destination as you match location and velocity at the point of intercept.
Accelerate to the required difference between source and destination orbital velocities plus the available deceleration from the target world 1000 diameters, intercept the target world on an elliptical orbit and decelerate as you approach the destination, then one last push to circularize your orbit around the destination as you match location and velocity at the point of intercept.
Timerover51
SOC-14 5K
I would tend to agree with you and simply ignore the 1000D limit with respect to the system's star gravity. Then you also have multi-star systems where you could have a range of varying gravity inputs.
I had inadvertently done that, but Traveller⁵ does quite specifically call out the 1000D limit and mentions almost this exact problem, just because you can accelerate within 1000D of the Sun does not mean you will be able to bleed off that velocity within 1000D of Pluto was specifically cited.
That's why I sort of wondered what the Traveller⁵ universe looked like. Was it an attempt to shift focus away from M-drives and toward Heplar or NAFAL, or was it an attempt to focus spaceflight as a star-to-star activity.
Timerover51
SOC-14 5K
Ask Marc.
Condottiere
SOC-14 5K
I don't know how viable slingshots are, though I'd be cautious about coming to sudden stops.
I guess future would be black hole drives, that both create the conditions for gravitational push and pull, and as a byproduct produce energy.
I guess future would be black hole drives, that both create the conditions for gravitational push and pull, and as a byproduct produce energy.
Wrong. you've mistaken distance for speed. V (velocity, vector) is distance per time, with time always defaulting to seconds if not noted otherwise.
Distance defaults to meters.
For traveller purposes, A is shifted to a flat 10 per G,
Anyway, if you know the acceleration, total distance (Dt), and two of short leg distance (Ds), long leg distance (Dl), and coast distance (Dc), you can figure the total Time.
Burn and flip is generally T=Sqrt(4D/A)
But we need to not count the distance coasting, because that is time we aren't accelerating. (While Mike is absolutely correct about the reduced thrust past limit, it makes the math much uglier. So, ignoring it.)
We need distance under thrust, let's call it Da. the time taken is Ta, and the velocity is constantly changing. Peak may be relevant if the ref's a realism nut, and Vap (Peak Velocity under thrust) = ATa/2 (because at half the time, you flip and start reducing).
So
Da = Dt - Dc
Ta = Sqrt(4Da/A)
We cannot figure coasting speed until we know Ts, because we need to know Vs. But note that this is a one way burn, so
Ds=ATs²/2
2Ds=2ATs²/2
2Ds=ATs²
2Ds/A = ATs²/A
2Ds/A = Ts²
Sqrt(2D/A) = Sqrt(Ts²)
Sqrt(2D/A) = Ts
Ts = Sqrt(2D/A)
Now, for V... ATs=Vsp (peak V on short leg) which also is Vc
Now, plugging that in to get coasting time... wich is distance over velocity
Tc = Dc/Vc
Sum the two times
Ttotal=Ta+Tc
It's not hard, but you have to remember the various portions.
Why doesn't Ts get added? Because it's part of the acceleration time already. We ONLY need it to know how fast the short leg maximum speed is.
It doesn't matter which end the coasting is at, either.
Where did I mistake distance for speed? I'm using x for distance, v for velocity, and a for acceleration. Position given initial position, initial velocity and acceleration is x = x0 + v0 t + 1/2 a t², if you're only worried about change in position (initial velocity and position are 0) you get x = 1/2 a t².
Velocity change given acceleration and position (you don't have time) is v² = 2 a x or v = Sqrt[2 a x].
I think we're trying to solve different problems. You're looking for a coast distance, I'm looking to be under acceleration the entire time, with the turnover at the appropriate time such that the remaining available deceleration will bring the ship to zero velocity at its destination.
That is specifically why I posted. As you note, the problems are simple to solve given constant thrust and not that difficult given acceleration continuously varying on a function. It is more difficult to solve given particular zones of available variable thrust. But to continue with this case -
We can already solve for how much velocity can be absorbed in the short distance, as we know that v² = 2 a x, or v=Sqrt[2 a x]. Checking that in the case of Saturn, which has a radius (that you're using) of 60268000, thus 2000 diameters is 4000x that, so 241072000000. The question is "how much velocity can be absorbed in that distance given acceleration of 10 m/s². The approach I would take:
Total velocity change available given that distance is 2.19548*10⁶ m/s, or about 2195 km/sec. Verifying that a different way, I'll set up an ODE with the second derivative of position/time (that is, acceleration) to be -10, initial condition of the first derivative of position/time (velocity) to be what is implied above (2195 km/sec), and initial position to be zero, and solve for when velocity is zero:
it takes 219578 seconds after which velocity is zero, and at that time position is 2.41072x10¹¹ meters, or, as expected, 4000 saturn radii. I did not do an order reduction specifically so there's no confusion that y[t] is position and y'[t] is velocity.
So, entry velocity to the destination 1000D sphere is, as originally implied by the v² = 2 a x equation, v = Sqrt[2 a x] given a is acceleration and x is how much distance you have to work with (2000 diameters) you will get the entry velocity v that can be reduced to 0.
You definitely took a different approach than I did and ended up with a different number for that component of it. I'll break up the responses a bit to deal with each component in turn.
(and, position vs velocity plot for the above test case of saturn)
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mike wightman
SOC-14 10K
Think backwards.
Calculate the maximum velocity change within the 2000D maneuver range of the sednoid.
Treat the AU distance between the sednoid 1000D limit and the 1000D limit of the garden world as your typical burn and flip but at 0.2m/s^2.
Your maximum velocity leaving the garden world maneuver space is whatever you calculated for the sednoid maximum velocity change, it is also the velocity you will arrive at the sednoid with.
Now solve for time elapsed.
Simples.
The game is meant to be fun, not work...
Calculate the maximum velocity change within the 2000D maneuver range of the sednoid.
Treat the AU distance between the sednoid 1000D limit and the 1000D limit of the garden world as your typical burn and flip but at 0.2m/s^2.
Your maximum velocity leaving the garden world maneuver space is whatever you calculated for the sednoid maximum velocity change, it is also the velocity you will arrive at the sednoid with.
Now solve for time elapsed.
Simples.
The game is meant to be fun, not work...
