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February 22nd, 2021, 08:37 AM


Baron


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Acceleration times
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 MDrive rating, and Momentum as cumulative Acceleration over time. Assume a ship with M1.
Code:
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 midpoint turn around and deceleration. Is there a correlation between Momentum at the start of jump and Momentum when reentering realspace at the end of the jump?

February 22nd, 2021, 10:20 AM

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Quote:
Originally Posted by Leitz
The Travel Times listed in CT Book 2, page 10, assume midpoint turn around and deceleration. Is there a correlation between Momentum at the start of jump and Momentum when reentering 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.
Quote:
Originally Posted by Leitz
Given Time in turns, Distance in units, Acceleration as MDrive rating, and Momentum as cumulative Acceleration over time. Assume a ship with M1.
Code:
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:
Code:
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:
Code:
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:
Code:
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.
Last edited by AnotherDilbert; February 22nd, 2021 at 10:52 AM..

February 22nd, 2021, 11:57 AM

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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.

February 22nd, 2021, 01:08 PM

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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.

February 22nd, 2021, 01:48 PM


Noble


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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...
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February 22nd, 2021, 02:12 PM


Baron


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Quote:
Originally Posted by 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.

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.

February 22nd, 2021, 02:20 PM

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I was under the firm impression CT turns were 1000s, which correlated with 1G 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.
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February 22nd, 2021, 04:26 PM

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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.
Last edited by sudnadja; February 22nd, 2021 at 05:12 PM..

February 22nd, 2021, 05:28 PM

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Quote:
Originally Posted by Leitz
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.

February 22nd, 2021, 07:48 PM


Baron


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Quote:
Originally Posted by whartung
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.

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