Acceleration times

[Leitz]

Okay, a simple question for those of you who do math(s). For some of us, it takes both brain cells a lot of time to figure this stuff out. Please check my assumptions.

Given Time in turns, Distance in units, Acceleration as M-Drive rating, and Momentum as cumulative Acceleration over time. Assume a ship with M-1.

Turn               Distance            Acceleration         Momentum
                     Moved

0                      0                     1                    0   
1                      1                     1                    1
2                      2                     1                    2
3                      5                     1                    3
4                      9                     1                    4

At the beginning of Turn 4, the ship would have moved 9 Distance Units and have a Momentum of 3. Thus, at the beginning of Turn 5, assuming no Maneuver activity, the ship would move 3 more Distance Units in the direction it was going at the beginning of Turn 4, correct?

The Travel Times listed in CT Book 2, page 10, assume mid-point turn around and deceleration. Is there a correlation between Momentum at the start of jump and Momentum when re-entering realspace at the end of the jump?

 

[AnotherDilbert]

The Travel Times listed in CT Book 2, page 10, assume mid-point turn around and deceleration. Is there a correlation between Momentum at the start of jump and Momentum when re-entering realspace at the end of the jump?

Yes, momentum is maintained in Jump. You leave jump with the same velocity in the same direction as when you entered jump. See JTAS#24.

I believe you mean Velocity, not Momentum. Velocity (speed and direction) is the direct result of acceleration, and leads to distance travelled. Momentum is velocity times mass, more related to kinetic (motion) energy than just velocity.

Given Time in turns, Distance in units, Acceleration as M-Drive rating, and Momentum as cumulative Acceleration over time. Assume a ship with M-1.

Turn               Distance            Acceleration         Momentum
                     Moved

0                      0                     1                    0   
1                      1                     1                    1
2                      2                     1                    2
3                      5                     1                    3
4                      9                     1                    4

At the beginning of Turn 4, the ship would have moved 9 Distance Units and have a Momentum of 3. Thus, at the beginning of Turn 5, assuming no Maneuver activity, the ship would move 3 more Distance Units in the direction it was going at the beginning of Turn 4, correct?

Basically yes, using the truncated turns of LBB2. (At the end of turn 2 the ship should have moved 3 "distance units"; 2 more than at the end of turn 1.)

But I think you confuse turn 4 and 5 here.

At the beginning of turn 4 the ship has moved 6 "distance units". During turn 4 it moves another 4 "distance units" (assuming constant acceleration), so ends the turn 10 "distance units" away from the start position.

At the end of turn 4 (=the start of turn 5) (accelerating all the way) the ship has moved 10 "distance units" and has a velocity vector of 4 "velocity units" (not momentum).

With no further acceleration, it will move an additional 4 "distance units" during turn 5, since that is its velocity, ending turn 5 a total of 14 "distance units" from the start.

Adding to you table:

Turn               Distance            Acceleration           Velocity
(end of)             Moved

0                      0                     1                    0   
1                      1                     1                    1
2                      3                     1                    2
3                      6                     1                    3
4                     10                     1                    4
5                     14                     0                    4

It is perhaps easier to see it as:

Turn             Acceleration       Velocity   =   Distance           Distance            
(end of)                                           this turn           total

0                     1                0                0                  0
1                     1                1                1                  1
2                     1                2                2                  3
3                     1                3                3                  6
4                     1                4                4                 10
5                     0                4                4                 14

I hope that made some sense...




For straight acceleration not approximated into turns the basic formulæ are:

acceleration = A      (assumed constant)
velocity     = A×t    (note velocity, not momentum)
distance     = ½A×t²

where t (time) is measured in seconds, A (acceleration) in m/s² (≈G×10), and distance in meters.

 

[whartung]

It doesn't quite work like that.

Simple example.

1 turn is 30m, or 1800s. 1G is 10m/s².

A ship, starting at 0, accelerated at 1G for 3 turns.

v = v0 + a * t

So,

v = 0 (initial v) + 10 (1G) * 1800 (1 turn) * 3 (number of turns)

v = 0 + 10 * 1800 * 3

v = 54,000 m/s

So, during 1 turn, this ship will travel

d = v * t

d = 54,000m/s * 1800s

d = 97,200,000m or 97,200km

Now, let's accelerate for another turn.

v = v0 + a * t

v = 54,000 + 10 * 1800

v = 72,000

BUT, how far did the ship travel during that period?

Under acceleration:

d = ½a * t²

But they're already moving at 54,000, so we have to account for that as well.

d = v * t + ½a * t²

d = 54,000 * 1800s + ½ 10 * 1800²

d = 97,200,000m + 16,200,000m

d = 113,400,000m

Now if you coast at 72,000m/s, you will move 129,600,000.

But during the acceleration phase, you only move 113,400,000. Essentially, you actually move "half" your velocity change during the acceleration phase.

Pretty much every game for the sake of ease of play gets this wrong.

The simplest mechanic is to take your current velocity, add "Gs" of acceleration to it, and move the "future" marker by that many hexes.

So, if your velocity is 5 hexes per turn, and you do a 2G burn, you make your future position 7 hexes ahead, and the vector length is your new velocity of 7 (5 +2).

In fact, for the first turn, you should only move the ship a single hex (to account for the acceleration), but then keep the 7 for velocity.

But nobody does this. It's too much math for a hex board and cardboard chits, and it doesn't much matter in game play terms anyway.

 

[Condottiere]

I think originally you had to deal with twenty minute blocks, and being constrained to reactionary rockets with limited fuel, it was probably easier to calculate in distance units.

 

[mike wightman]

Original CT LBB2 77 edition did in fact have the use of real world physics as an option on page 37.
There is of course one problem with this, namely the mapping of vectors rather than displacement...

 

[Leitz]

I think originally you had to deal with twenty minute blocks, and being constrained to reactionary rockets with limited fuel, it was probably easier to calculate in distance units.

I'm using abstracts like "units" and "turns" to let myself focus on the math. Seems like I need to focus more.

This question is mostly because I don't really have an ingrained perspective on "how long would it take to get to..." when I game and write. Trying to fix that.

 

[kilemall]

I was under the firm impression CT turns were 1000s, which correlated with 1-G accel giving 10 m/s and at the end of the 1000s that is 10 km/s velocity.



If the ship does not otherwise accelerate or run into a major gravity field, it keeps going at 10 km/s or 10000 km per turn.

 

[sudnadja]

Page 10 of the LBB book 2 says that turns are 1000 seconds long.

all of the discussion above boils down to d²r/dt² = a, and set ic's and integrate appropriately, but that probably won't match rules as written.

 

[whartung]

This question is mostly because I don't really have an ingrained perspective on "how long would it take to get to..." when I game and write. Trying to fix that.

Well, that's pretty easy for most cases.

For point to point, its t = 2 * sqrt(d / a). This is "burn to the middle, turn, decel to the destination". You start and stop at velocity "0".

And, just use 10 for A. So, 6G is 60.

Example: Earth -> Jupiter (closest approach) is ~558Mkm at 1G

t = 2 * sqrt(558,000,000,000 / 10)

t = 484974s/3600 = 134.7hrs/24 = 5.6 days

With a 2G drive, its ~4 days.

So, the inner system is a week or less of travel. The outer system, you jump. It's faster.

100D from earth.

12,000km for earth diameter. * 100 is 1,200,000km

t = 2 * sqrt(1,200,000,000m / 10) = about 6hrs.

So, once you work the numbers, with Traveller drives, space shrinks quite a bit, truthfully. The moon is less than 4hrs at 1G. Heck, I can't get to Vegas in 4hrs.

 

[Leitz]

Well, that's pretty easy for most cases.

For point to point, its t = 2 * sqrt(d / a). This is "burn to the middle, turn, decel to the destination". You start and stop at velocity "0".

That's what I don't really want. I'm looking at "I need to be there as soon as I can, how long is that?" Turn and burn assumes you end up with zero velocity.

 

[AnotherDilbert]

That's what I don't really want. I'm looking at "I need to be there as soon as I can, how long is that?" Turn and burn assumes you end up with zero velocity.

That is generally what you want. If you accelerate all the way you won't "be there", you will fly by (or crash into) at very high speed, without any real chance of interacting with the target.


But if you just want to accelerate all the way it is D=½A×t² or time t=SQRT(2D/A).

So, one million km (=one billion m) at 3 G (≈30 m/s²) would take t = SQRT(2×1000000000/30) ≈ 8164 s ≈ 136 minutes.

At that point you would travel at 30 m/s² × 8164 s ≈ 244949 m/s ≈ 245 km/s ≈ 881816 km/h or approx. 550000 mph.

 

[Grav_Moped]

That's what I don't really want. I'm looking at "I need to be there as soon as I can, how long is that?" Turn and burn assumes you end up with zero velocity.

That is generally what you want. If you accelerate all the way you won't "be there", you will fly by (or crash into) at very high speed, without any real chance of interacting with the target.


But if you just want to accelerate all the way it is D=½A×t² or time t=SQRT(2D/A).

So, one million km (=one billion m) at 3 G (≈30 m/s²) would take t = SQRT(2×1000000000/30) ≈ 8164 s ≈ 136 minutes.

At that point you would travel at 30 m/s² × 8164 s ≈ 244949 m/s ≈ 245 km/s ≈ 881816 km/h or approx. 550000 mph.

If Leitz is asking something like "how long does it take to get to the 100D limit if you're jumping 'hot' (a 'running jump')" it makes some kind of sense.

Though if you assume away relative motion between origin and destination (and simplicity says you might want to), the "turn and decelerate" point could be on either side of the Jump. Exactly where, depends on the relative world sizes (that is, the 100D limits for each world may be different, and the trip distance is 100D_origin + 100D_destination). If the origin world is larger than the destination world, you flip and retro-burn before Jumping; if it's smaller, you flip after Jump exit.

 

[Werner]

If Leitz is asking something like "how long does it take to get to the 100D limit if you're jumping 'hot' (a 'running jump')" it makes some kind of sense.

Though if you assume away relative motion between origin and destination (and simplicity says you might want to), the "turn and decelerate" point could be on either side of the Jump. Exactly where, depends on the relative world sizes (that is, the 100D limits for each world may be different, and the trip distance is 100D_origin + 100D_destination). If the origin world is larger than the destination world, you flip and retro-burn before Jumping; if it's smaller, you flip after Jump exit.

If someone is chasing you, if you flip and start decelerating, he will catch you, but since you are accelerating to a jump point, there is nothing physical to crash into, you just reach the required distance and you press the jump button. Of course when you drop out of jump space at the 100 diameter limit you may have a problem, because you are still going at that speed, which you need to slow down from if you do not wish to crash into your destination. You know where this is a real problem, when you are trying to accelerate to the 100 diameter limit of a gas giant and your destination is a terrestrial planet. Do you have two choices, you can accelerate so you miss the planet and then slow down, accelerate towards the planet and then slow down again or you decelerate much faster that you accelerated away from that gas giant, and I'm assuming since it was a chase, your ship was accelerating away at its maximum capacity, so this second option is not usually available.

 

[whartung]

So, for the 100D Earth scenario at 1G:

t = sqrt(2d/a)

t = sqrt(2 * 1,200,000,000 / 10)

t = sqrt(240,000,000)

t = 15492s / 3600 = 4.3 hours

Velocity is v = at

So,

v = 10m/s² * 15492s

v = 154,920 m/s

If you happen to arrive at the destination system with it's Earth like body.

t = d/v

t = 1,200,000,000 / 154,920

t = 7,745s / 3600 = 2.1hrs before you "arrive" at the planet, going far to fast to do anything.

This is not a problem, however. When you plot your jump, you simply orient the ship to arrive not vectored toward anything dangerous as you work on slowing yourself down, ideally out of danger from whatever is pursuing you.

With Book 2, a turn is 1000 seconds.

So, in the first scenario, there's potentially 15 turns of combat that can take place on your way to Jump.

In Mayday, a turn is 100 minutes, 6000 seconds. That means there's 3-4 turns of combat.

I would ask your pirates to use the Mayday rules instead of the Book 2 rules .

Better chance of surviving 3-4 rounds of continual laser fire than 15.

In Brilliant Lances, it's 30m per turn, so 9 turns.

 

[Grav_Moped]

If someone is chasing you, if you flip and start decelerating, he will catch you, but since you are accelerating to a jump point, there is nothing physical to crash into, you just reach the required distance and you press the jump button. Of course when you drop out of jump space at the 100 diameter limit you may have a problem, because you are still going at that speed, which you need to slow down from if you do not wish to crash into your destination. You know where this is a real problem, when you are trying to accelerate to the 100 diameter limit of a gas giant and your destination is a terrestrial planet. Do you have two choices, you can accelerate so you miss the planet and then slow down, accelerate towards the planet and then slow down again or you decelerate much faster that you accelerated away from that gas giant, and I'm assuming since it was a chase, your ship was accelerating away at its maximum capacity, so this second option is not usually available.

If you're running, you burn the whole way to Jump Limit and expect to overshoot the destination. The extra fuel burn to finish slowing down as you pass the destination world, then travel back to it from your stopping point, is just the cost of doing business.

The other thing here is that if "running" jumps are standard, the flight path ought to give you a chance to predict the ship's destination based on velocity at Jump. The vector (handwaving the relative stellar motions issue) tells you something about the relative size of the destination world compared to that of the origin world.

 

[Condottiere]

Running depends on how fast the pursuer is closing, and most commercial spaceships don't appear to have a larger engine than factor two.

 

[Werner]

Running depends on how fast the pursuer is closing, and most commercial spaceships don't appear to have a larger engine than factor two.

Of course the players might be running from Imperial Patrol Cruisers because they are pirates and don't want to be caught.

 

[Condottiere]

Time to either change the transponder, or go into stealth mode.

 

[sudnadja]

So, for the 100D Earth scenario at 1G:

t = sqrt(2 * 1,200,000,000 / 10)

I don't know that it is important to your analysis, but it looks like you are using about 94 diameters rather than 100, and you don't seem to be accounting for force applied by Earth's gravity. Velocity at 100D from the surface (201 radii from the center) would be more like:

1/2 v² = ∫ -G earthmass/r² + 10 {r,6378135,201 * 6378135} and that ends up being about v=159337 m/s. Time to 100 diameters from the surface is more like 18555 seconds, unless planetary gravity is handwaved - but page 37 of book 2 of the LBB implies that it should not be.

How is this typically done in practice in most games?

 

[Condottiere]

Most inhabited worlds in Traveller are less than Terran norm.

So default one gee drives have some room to manoeuvre.