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Web mines; a study of snares
By Richard McAteer


Due to the unusual nature of web mines, travel through them can be more difficult than in regular mines, especially for groups of ships. The need to conserve fuel causes one to wish to spend the least amount of time in the field, but the loss due to collision is high. The following tables were computed in order to help me travel through and sweep web mines, and though the crystals are one of my favorite races, I figure it may be of help to those poor souls facing them.

The webmine has several important differences from regular mines; firstly, it drains fuel from any ship within the field. Secondly, a minehit will stop a ship of any size. Thirdly, minehits will drain 16th of the fuel on a vessel, or 50kT, whichever is greater. Fourthly, unless defaults have been changed you must sweep them with beams, as they are not affected by regular minefields. Fifthly, cloaking does not help you avoid them. The good news is that the mines do only a fraction of the damage that a regular mine inflicts.

The fuel drain while in a webminefield is 25kT of fuel per field. (i.e. a ship within 2 webminefields loses 50 kT of fuel per turn.) A collision with a webmine will drain 50kTs of fuel or 1/6th of your fuel, whichever is greater. Collisions should be avoided :)

As is always the case when facing a mine using opponent, heavy phasers are the way to go. I would recommend always producing a few minesweeping ships no matter what race you are playing. Ships with mass over 100kTs are useful against regular minefields so that they can survive a minehit, and large fuel tanks are best against webmines. Those large fuel tanks lose a lot of fuel when a webmine is struck, however, so having a smaller ship tow the minesweeper is a good idea.

Commonly, when ships are in a webminefield and must sweep their way out, we choose to have them occupy the same space, in order to protect themselves from the crystal warships as well as to conserve fuel. Since fuel drain occurs before movement and after minesweeping, a ship with only 1 kT of fuel can still sweep mines. This allows a large ship to hold the bulk of the fuel, transferring fuel 1 kT per turn to the small ship. In this manner the loss of fuel is not 25 kT of fuel per ship, but 1 kT from one ship and 25 from the other. This will allow fuel tanks to be conserved over a longer period of time. If the crystals are allied to a race with cloaking, or have their own cloakers, this will make your ships vulnerable to a cloaked tow; you will be fuel-less, so either will be tow-captured or helpless. you run out of fuel right away. Randomly varying whether you transfer 1 kT or a whole bunch will discourage cloaking ships from trying this tactic.

In order to save fuel while moving through the minefield, many players will tow the large capacity vessel, transferring just enough fuel to the tower each turn for its movement. In this manner the fuel drained by a collision is minimized, and better progress can be made through a minefield.

The tables below assume that the standard 5% chance of hitting a mine per LY travelled are in effect. The formulae however, will work for any value of p, where p is the webmine hit odds, as a probability (default is thus .05), and q = 1-p (.95 default)


Table 1:

Prob. of safely flying n LY; (no minehits)
(for overlapping webmines raise the probability
to the power of # of minefields)

P(n) = q^n

n= 1 2 3 4 5 6 7 8 9 10
P(n) 0.95 0.9025 0.85737 0.81451 0.77378 0.73509 0.69834 0.66342 0.63025 0.59874
n= 11 12 13 14 15 16 17 18 19 20
P(n) 0.5688 0.54036 0.51334 0.48767 0.46329 0.44013 0.41812 0.39721 0.37735 0.35849
n= 21 22 23 24 25 26 27 28 29 30
P(n) 0.34056 0.32353 0.30736 0.29199 0.27739 0.26352 0.25034 0.23783 0.22594 0.21464
n= 31 32 33 34 35 36 37 38 39 40
P(n) 0.20391 0.19371 0.18403 0.17482 0.16608 0.15778 0.14989 0.1424 0.13528 0.12851
n= 45 50 55 60 65 70 75 80 81 162
P(n) 0.09944 0.07694 0.05954 0.04607 0.03565 0.02758 0.02134 0.01652 0.01569 0.00025

Table 2:

Probability of hitting a mine in the nth LY (halting after n LY)

P(n) = [q^(n-1)](p)

n= 1 2 3 4 5 6 7 8 9 10
P(n) 0.05 0.0475 0.04513 0.04287 0.04073 0.03869 0.03675 0.03492 0.03317 0.03151
n= 11 12 13 14 15 16 17 18 19 20
P(n) 0.02994 0.02844 0.02702 0.02567 0.02438 0.02316 0.02201 0.02091 0.01986 0.01887
n= 21 22 23 24 25 26 27 28 29 30
P(n) 0.01792 0.01703 0.01618 0.01537 0.0146 0.01387 0.01318 0.01252 0.01189 0.0113
n= 31 32 33 34 35 36 37 38 39 40
P(n) 0.01073 0.0102 0.00969 0.0092 0.00874 0.0083 0.00789 0.00749 0.00712 0.00676
n= 45 50 55 60 65 70 75 80 81 162
P(n) 0.00523 0.00405 0.00313 0.00242 0.00188 0.00145 0.00112 0.00087 0.00083 1.3E-05

Table 3:

Probability of successfully traveling n LY (either by hitting a mine in the nth, or arriving safely) (this is also 1 -the sum of the chances of hitting a mine in each previous LY) This can be used to determine the odds of reaching a desired position.

P(n) = [q^(n-1)](p) + q^n

n= 1 2 3 4 5 6 7 8 9 10
P(n) 1 0.95 0.9025 0.85737 0.81451 0.77378 0.73509 0.69834 0.66342 0.63025
n= 11 12 13 14 15 16 17 18 19 20
P(n) 0.59874 0.5688 0.54036 0.51334 0.48767 0.46329 0.44013 0.41812 0.39721 0.37735
n= 21 22 23 24 25 26 27 28 29 30
P(n) 0.35849 0.34056 0.32353 0.30736 0.29199 0.27739 0.26352 0.25034 0.23783 0.22594
n= 31 32 33 34 35 36 37 38 39 40
P(n) 0.21464 0.20391 0.19371 0.18403 0.17482 0.16608 0.15778 0.14989 0.1424 0.13528
n= 45 50 55 60 65 70 75 80 81 162
P(n) 0.10467 0.08099 0.06267 0.04849 0.03752 0.02904 0.02247 0.01738 0.01652 0.00026

Table 4:

Odds of completing a flight of n LY with an interceptor also completing the flight (thus, both could hit a mine in the last LY, or either one, or neither). For more interceptors, replace the ^2 with ^x+1, where x is the number of interceptors. This table is very useful for determining the odds of a pair (or fleet) reaching a position together (for an attack, for example). This would also be the probability of flying through two overlapping webmine fields and reaching the destination, with the possibility of hitting one or even two mines in the last LY.

P(n) = {[q^(n-1)](p) + q^n}^2

n= 1 2 3 4 5 6 7 8 9 10
P(n) 1 0.9025 0.81451 0.73509 0.66342 0.59874 0.54036 0.48767 0.44013 0.39721
n= 11 12 13 14 15 16 17 18 19 20
P(n) 0.35849 0.32353 0.29199 0.26352 0.23783 0.21464 0.19371 0.17482 0.15778 0.1424
n= 21 22 23 24 25 26 27 28 29 30
P(n) 0.12851 0.11598 0.10467 0.09447 0.08526 0.07694 0.06944 0.06267 0.05656 0.05105
n= 31 32 33 34 35 36 37 38 39 40
P(n) 0.04607 0.04158 0.03752 0.03387 0.03056 0.02758 0.02489 0.02247 0.02028 0.0183
n= 45 50 55 60 65 70 75 80 81 162
P(n) 0.01096 0.00656 0.00393 0.00235 0.00141 0.00084 0.0005 0.0003 0.00027 6.7E-08

Table 5:

This last table is the most useful for simply getting out of minefields; it show the odds of keeping a pair (or fleet) of ships together. If one ship is intercepting another, they will arrive at the same position only if the intercepting ship can fly as far as the first ship successfully flew. As a result, the following table shows the odds of staying together for a plotted flight of n LY. The destination may not be reached, but the intercepting ship will arrive in the same place, allowing for fuel transfer and defense. The table shows values for 1 intercepting ship, the x in the formula is the # of interceptors.

A good example of a time to use this table is when you have two big ships being towed through a webminefield, and wish to keep them together for protection. This way both ships can minesweep, but are at no risk for collisions themselves.

P(n) = q[q^n+( q^(n-1))p] + sum i = 1 to n-1 of {[(q^(n-1))p]*[(q^(n-1))p + q^n]} or for n>2 P(n) = 1 - (sum from i =2 to n of .95^(2n-3).05); for P(1) = 1

n= 1 2 3 4 5 6 7 8 9 10
P(n) 1 0.9525 0.90963 0.87094 0.83603 0.80451 0.77607 0.75041 0.72724 0.70634
n= 11 12 13 14 15 16 17 18 19 20
P(n) 0.68747 0.67044 0.65507 0.6412 0.62868 0.61739 0.60719 0.59799 0.58969 0.58219
n= 21 22 23 24 25 26 27 28 29 30
P(n) 0.57543 0.56932 0.56382 0.55884 0.55436 0.55031 0.54665 0.54335 0.54038 0.53769
n= 31 32 33 34 35 36 37 38 39 40
P(n) 0.53526 0.53308 0.5311 0.52932 0.52771 0.52626 0.52495 0.52377 0.5227 0.52174
n= 45 50 55 60 65 70 75 80 81 162
P(n) 0.51816 0.51602 0.51473 0.51397 0.51351 0.51323 0.51307 0.51297 0.51295 ???

As we can see, the odds fall off, but seem to be approaching about .5; this is due to the increasing odds that the lead ship will hit a mine; It is quite safe to travel 8 LY at a time; over 75% of the time the two ships will occupy the same end square, and they will never end up more than 7 LY apart :) We know then that the odds of the trailing ship catching up is at least 73%, from table 3. In this manner, we can figure that if we tell two ships to fly 8 LY the odds are that 2 turns later they will be together at least 1-(.25)(.27)=93.25% of the time.


I hope these tables help the rest of you as much as they helped me :)
If you have comments about the math or text please email me at rmcateer@chat.carleton.ca;


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