StuffOnline (Jump Drive)

Notes

Introduction

Below is an article first published on the TML on 31 Jan 1995. Following the subsequent debate on the TML a revision was in order. That revision is appended below the main article. (Since "G" is used in the formula below, the unit of "G" is expressed in "gees" to avoid confusion.)

Original Article (aka "Simple Gravity" model)

Here's a thought or two ...

1) The jump drive's "100 diameter" safe jumping distance is supposed to be due to the gravity from large masses affecting the ability of the jump drive to form a jump field correctly. But gravity will vary with planetary density. According to "From Port to Jump-point" (JTAS 22) this number is a thumbnail rule anyway so what's the real rule?

2) In MT the jump task is 'Routine' outside this limit, 'Difficult' between 10 and 100 diameters, and 'Formidable' within 10 diameters. Each of these are a gap of 4 on the dice roll, yet the effect is described as gradual ("Starship Operator's Manual Vol 1") ... so does that mean its 'Difficult+2' at 50 diameters?

I propose that the phrase "100 diameter limit" is merely a rule of thumb provided by ship's pursers to their passengers and that seasoned space travellers have a more accurate value for the MSD (minimum safe distance) to form a jump point.

First some boring math so you can follow my reasoning if interested:

Using the formula

L = 64 (M / G)^0.5

and

R = 8 D

from the original LBB rule books (where L is distance in mapping units, M is mass in earth masses, G is gravity in gees, R is world radius in mapping units, D is world UPP size code) you can deduce that 1 mapping unit = 62.5 miles (or approx 100 km as stated in the original book). So to express the formula for L in miles ...

L (in miles) = 4000 (M / G)^0.5

or put another way

G = M / (L / 4000)²

Not wanting to be too terra-centric lets pick Vland as our reference example. (As an aside the specified mass for Vland given in ""Aliens Vol 1: Vilani & Vargr"" is incorrect - given a diameter of 14,850 km and a density of 1.02 the mass should be 1.5661792.)

So G at 100 diameters is 2.94185 x 10^-5 gees. Remember from your high school physics that 1 gee is 9.80665 m/s/s and you get G = 2.88497 x 10^-4 m/s/s.

Now because the the chance of a world being exact like that is small lets use some pseudo-science to pull a little slight-of-hand. The universally constant speed of light c = 2.99792458 x 10⁸ m/s and so G/c is 9.62322 x 10^-13 ... thats so close to 1 x 10^-12 that maybe it should be! Now say that space stress is measured in units of S or in this case pS (p = 'pico' = 10^-12) where 1 pS = 3.05703 x 10^-5 gees.

So the 1 pS ('one pico ess') limit is the so-called '100 diameter' limit for Vland. Recalculating back gives 98.1 diameters.

So what was the point of all that (apart from wearing out my calculator)? Calculate the 1 pS boundary for Terra and you get 91.3 diameters ... not 100.


Definition:	Minimum safe distance to initiate Jump is any point in space where the gravitational stress factor is less than 1 pS.


Definition:	1 pS is the stress on space by a gravitational force of 3.05703 x 10^-5 gees.

Okay, so now we've dealt with the first point and can add some colour to individual systems.

Sample Minimum Safe Distances
World	Icy Body	Rocky Body	Molten Core	Heavy Core
Size	(k=0.32)	(k=0.64)	(k=0.96)	(k=1.45)
S	6.5	9.1	11.2	13.8
1	18.3	25.9	31.8	39.0
2	25.9	36.7	44.9	55.2
3	31.8	44.9	55.0	67.6
4	36.7	51.9	63.5	78.1
5	41.0	58.0	71.0	87.3
6	44.9	63.5	77.8	95.6
7	48.5	68.6	84.0	103.3
8	51.9	73.3	89.9	110.4
9	55.0	77.8	95.3	117.1
A	58.0	82.0	100.4	123.4

MSD in planetary diameters at 1 pS boundary
(incorrectly referred to as "100 diameter limit")

Other Minimum Safe Distances
Other Bodies	Size	MSD
Small Gas Giant (k=0.20)	60	112.3
Large Gas Giant (k=0.20)	170	189.0
Star (Sol) (k=0.26)	865	482.6

Note that Sol's MSD of 482.6 is equavalent to 4.49 AU
(almost as far as Jupiter's orbit)!

Repeating this process for the "10 diameter" limit and we find its actually the 100 pS boundary. Hey, the numbers go up with the difficulty! A geometric progression gives a step of 3.1622777 or ...

Jump Difficulty
Stress Factorxxx		MSD	Difficulty
< 1 pSxxxxx		100%	Routine
1 pS -	3 pS	56%	Routine+1
3 pS -	10 pS	32%	Routine+2
10 pS -	32 pS	18%	Routine+3
32 pS -	100 pS	10%	Difficult
100 pS -	316 pS	6%	Difficult+1
316 pS -	1,000 pS	3%	Difficult+2
1,000 pS -	3,162 pS	2%	Difficult+3
3,162 pS -	10,000 pS	1%	Formidable
10,000 pS -	31,623 pS		Formidable+1
31,623 pS -	100,000 pS		Formidable+2
100,000 pS -	316,228 pS		Formidable+3
316,228 pS <xxxxx			Impossible

The MSD column assumes only one distorting object.

So if your players are running for the "100 diameter" MSD jump point while being persued but for feel the need to go early, just compare their distance with the table above to determine the actual risk at that point.

Another issue: Are Lagrange points (points of gravitational stability) safe jump points within the 1 pS boundary? No. In none of the published Traveller works is there any mention of this. The term ""stress"" is important, at these points the gravitational forces may be in balance but the fabric of space is still stressed (like the rope in a tug-o-war between two evenly matched teams). Therefore, even when two or more gravitational sources are balanced their effects are additive. Compute the pS value of all likely objects and add them together before consulting the Jump Difficulty Table.

ATTENTION

In general, the Lagrange points don't have gravitational forces balanced. That can only happen at L1, and even there only if the 2 masses are equal. All other cases balance the 2 gravitational forces and the centrifugal force of the orbit.

Range Stress
MSD	Stress Factor
5%	400.000
10%	100.000
15%	44.444
20%	25.000
25%	16.000
30%	11.111
35%	8.163
40%	6.250
45%	4.938
50%	4.000
55%	3.306
60%	2.778
65%	2.367
70%	2.041
75%	1.778
80%	1.563
85%	1.384
90%	1.235
95%	1.108
100%	1.000

Also note that this space stress does not simply end at 1 pS, but continues out indefinately getting smaller and smaller. It is theoretically possible to be beyond MSD for all objects yet their combined gravitational pulls to still exert more than 1 pS!

Range Stress (cont)
MSD	Stress Factor
110%	0.826
120%	0.694
130%	0.592
140%	0.510
150%	0.444
160%	0.391
170%	0.346
180%	0.309
190%	0.277
200%	0.250

For example, at 140% MSD the stress factor is 0.510. So if a starship were at 140% MSD for one planet and 140% MSD for another planet, the total stress factor would be 1.100. This makes the jump task 'Routine +1'.

One final issue: When player characters are feeling particulaly suicidal/desperate/destructive they sometimes wonder what would happen if they tried to jump while still landed in the starport. On a 1 gee planet the stress factor at ground level is 32,711.463 pS which makes such a task 'Formidable +2'. Now you know!

Revision (aka "Tidal Force" model)

In the debate that followed this article being posted to the TML it was concluded that a simple gravity model for jump difficulty deviated from canon too much. The problem was that most Earth-like worlds were embedded very deeply within their star's MSD: calculating a mainworld's MSD was irrelevant. The reason is that simple gravity decreases with the square of the distance, making the MSD of large objects too big.

An alternative would be to use tidal force instead of simple gravity. Tidal force decreases with the cube of the distance and thus avoids this problem.

Boring math bit ...

For tidal force we'll use the formula:

F = N M / D³

And for no good reason we'll keep Vland's MSD at 98.1 diameters and the measure of stress at MSD as 1 pS (as in the original article). That means ...

D = 14850 x 98.1 = 1456785

1 pS = N x 1.5661792 / 1456785³

therefore N = 1973990

F = 1973990 M / D³

or D = (1973990 M / F)^(1/3)

(where D is in kilometers, M is in earths)

Since M is calculated from size, and we want D in terms of size, this factor drops out leaving us with ...

D = (925565 K / F)^(1/3)

(where D is distance in diameters, K is density in earths, F is stress in pS)

Example: Vland K=1.02, F at MSD = 1 then ...

D = (925565 x 1.02 / 1)^(1/3) = 98.1


D = (925565 K / F)^(1/3)	F = 925565 K / D³
(where D is distance in diameters, K is density in earths, F is stress factor in pS)

So with the revised formula lets review the sample MSDs.

Sample Minimum Safe Distances
World	Ice Body	Rocky Body	Molten Core	Heavy Core
Size	(k=0.32)	(k=0.64)	(k=0.96)	(k=1.45)
any	66.7	84.0	96.1	110.3

MSD in planetary diameters at 1 pS boundary
(incorrectly referred to as "100 diameter limit")

Other Minimum Safe Distances
Other Bodies	Size	MSD
Small Gas Giant (k=0.20)	60	57.0
Large Gas Giant (k=0.20)	170	57.0
Star (Sol) (k=0.26)	865	61.8

Sol's MSD by this method comes out as 61.8 diameters ... equivalent to 0.57 AU, or just past the orbit of Mercury. A big improvement on the original figure. Also, although there is now no variation based on world size there is still variation based on density ... and for a density of 1.08 the MSD really is 100 diameters!

Reanalysing for the "10 diameter" limit we find the stress is 1000 pS (not 100 pS). The stress factor is still increasing but step is now 5.623414. Recomputing the jump difficulties ...

Jump Difficulty
Stress Factorxxxxxxx		MSD	Difficulty
< 1 pSxxxxxxxxx		100%	Routine
1 pS -	6 pS	56%	Routine+1
6 pS -	32 pS	32%	Routine+2
32 pS -	178 pS	18%	Routine+3
178 pS -	1,000 pS	10%	Difficult
1,000 pS -	5,623 pS	6%	Difficult+1
5,623 pS -	31,623 pS	3%	Difficult+2
31,623 pS -	177,828 pS	2%	Difficult+3
177,828 pS -	1,000,000 pS	1%	Formidable
1,000,000 pS -	5,623,415 pS		Formidable+1
5,623,415 pS -	31,622,791 pS		Formidable+2
31,622,791 pS -	177,828,027 pS		Formidable+3
177,828,027 pS <xxxxx			Impossible

The MSD column assumes only one distorting object.

Finally, here is the recalculated Range Stress table:

Range Stress
MSD	Stress Factor
5%	8000.001
10%	1000.000
15%	296.296
20%	125.000
25%	64.000
30%	37.037
35%	23.324
40%	15.625
45%	10.974
50%	8.000
55%	6.011
60%	4.630
65%	3.641
70%	2.951
75%	2.370
80%	1.953
85%	1.628
90%	1.372
95%	1.166
100%	1.000
110%	0.751
120%	0.579
130%	0.455
140%	0.364
150%	0.296
160%	0.244
170%	0.204
180%	0.171
190%	0.146
200%	0.125

Conclusion: Using a tidal force model rather than a simple gravity model gives us a set of figures which solve the original intent and conforms much more closely to canon. The only draw back is that without the G/c trick of the simple gravity model the MSD now seems to be arbitarily set. (But I guess we can't have everything.)