Step 6 - Offensive Systems |
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Step 6 - Offensive Systems |
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Weapon Mounts
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Vehicle Weapon Systems
Section incomplete
Bombs
Use the standard 2300AD 200Kg HE bomb as a guideline for bombs. Calculate EP as 0.75 *
the total mass of the bomb. Thus a 100kg bomb is EP 75.
Assume a length of about 1.2 m and a diameter of 0.17 m ( or 1/7th of the length ) for the standard 200kg bomb. This gives a volume of 0.03 cubic metres.
One bomb can be fitted per wing hardpoint on an aircraft, or several bombs can be mounted in a rack. Bomb racks weigh add 15% to the total mass and volume of the bombs, and require a surface area equal to 5 times the sum of the squares of the bomb diameters - for example, the 200kg bomb if mounted in a rack of five such bombs would require an extra 150 kg and 0.045 cubic metres for the rack, plus 0.723 square metres for the "hatch".
Crew Served Weaponry
Crew Served Weapons are usually heavy calibre machine guns or similar "squad support"
weapons. The weapons in the Adventurers' Guide can be used without any modification. As a
rule of thumb, assume that the volume occupied by a Crew Served Weapon is approximately
equal to the weapon length in metres x 0.01. Ammunition has a volume in cubic metres equal
to its mass in tons. In addition to the weapon there must be provision made for a standard
weapon mount. Contrary to the indication given by the name, Crew Served
Weapons may be mounted in remote mounts.
Lasers may be powered from a vehicle power plant. They have a power requirement equal to their discharge energy in joules multiplied by 3 - thus the Rorttmann LK-1 requires 1.05 MW input to fire. This is because I rule that lasers have an energy conversion efficiency of about 33% and normally take a second or so to "charge" up. Since 1 joule for 1 second is one watt, 0.35 megajoules in one second of charge up is 0.35 MW. Assuming 33% efficiency, this is 1.05 MW input. For more about this, see the section on FD-LMS Cells.
A small anomaly exists between lasers and ship mounted lasers. The DPV of a small arm laser is three times the delivered pulse in Mj - so a 0.2 Mj pulse has a DPV of 0.6. The DPV of a ship mounted laser is equal to the delivered pulse in Mj ( so 20 Mj is a DPV of 20 ). This is not a major problem as the pulse length and frequencies are different between the two. Small arm lasers have a pulse length of a hundredth of a second and are probably set in the IR or UV frequency range. Starship lasers have a shorter pulse length ( typically one thousandth of a second or less ) due to the stutterwarp movement of their targets and are fired in the X-Ray frequency range. I believe that these differences go some way to explaining this discrepancy.
Suitable Crew Served Weapons include the Type-81 Storm Gun, the Rockwell 12-81 Magnum, all Autoguns, all lasers and all plasma guns ( man-portable and heavy ).
Depth Charges
Lasers
Mass Drivers
A mass driver is a large calibre electromagnetic slug thrower. The two main factors for
such a weapon are the calibre of the projectile in centimetres and the length of the barrel.
Missiles
Particle Accelerators
Plasma Cannon
Torpedoes
Starship Weapon Systems
There was an excellent article in Challenge magazine which was an extract from Fire, Fusion and Steel which pointed out the sheer unpleasant mathematics behind the use of lasers in space combat. The following is a reworking of the concepts presented in that article for the purposes of 2300AD.
First of all, let's consider the problem of the firing solution. Let's assume we are stationary and perpendicular to course of a target vessel travelling at warp one. The target vessel is dead ahead as we start this calculation, at a range of 600,000 km ( or one Star Cruiser hex ). We will fire with a shipboard laser ( not a remote detonation laser, they are covered further below ).
A warp one vessel inside a 0.0001g field travels at 1/10,000th of the speed of light in a day, or 6.324 AU per day. This is 946056930 km per day or about 656,984 km per minute. This is about 10,950 km per second.
Let's assume that we have a laser that has a pulse duration of 1/1,000th of a second. In this time our target will travel 10.95 km. If we assume further than warp one is 100,000 cycles per second of about 109.5 metres per "stutter" then this represents one hundred such cycles.
Now, a range of 600,000 km is 2 light seconds. It takes 4 seconds for our active sensor pulse to reach the target and echo back, and a further two seconds for the laser pulse to reach it's target. By this time, six seconds have passed in total and our target vessel has moved 65,700 km along its course. This increases the range to 603,586 km, a mere 0.6% increase.
Now let's consider the delivered energy. The laser must strike the target with sufficient energy per square centimetre to penetrate the armour. 1 megawatt per square centimetre of hard steel is enough to penetrate 1 cm of such armour. Based on this calculation we can determine that we need to have 2.7 megawatts per cm2 to penetrate the minimum armour value of 2.70 ( which we equated to SCAR 0, above ). This is a DPV of at least 2.7.
Using the formula presented in Challenge and elsewhere, we have -
IR2= P / ( L / D )2 or D = ( L2IR2 / P )0.5
where -
I = Delivered Power Density in watts per square centimetre
P = Discharge Power in watts
L = Wavelength of light in centimetres
D = Mirror focussing element diameter in centimetres
R = Range to target in centimetres
If we have I=2.7x106 W cm-2, P= 2x107 W, L=1.30979x10-8 cm ( X-rays ), and R = 60,358,600,000 cm then D is about 2.5 metres. This seems reasonable for a laser on this scale.
The laser has one minute to charge before firing, as lasers fire once per one minute turn in Star Cruiser. This means that if we have a standard 1 MW input shipboard laser, we have an input energy to the laser of 60 seconds x 1 megajoule per second or 60 MW. Assuming that we have the same 33% efficiency as small arm lasers, then the discharge is in the order of 20 MW. ( Consider that a small arm laser requires a power input equal to three times the discharge energy - see FDLMS cells for more information on laser efficiency ).
So we have a laser "dish" 2.5 metres across firing at our target ship. Now the next problem is that of keeping the laser tracked onto the target. Remember that each 1/100,000th of a second the target moves 109.5 metres instantaneously. Each such change represents about 1/100,000th of a degree. Such fine adjustments must be performed every "jump" and are beyond mechanical calibrations. Fortunately, we are dealing with X-ray "Free Electron Lasers" or "FELs", which can be deflected by altering magnetic fields. This is just about theoretically possible. In my universe I assume that stutterwarp ships cannot significantly change vector over the course of 1/1000th of a second. If you allow random direction hops, then laser combat would never work.
Our delivered energy decreases with the inverse square law as follows -
| Range in km | Delivered Energy (Mj) |
| 221,750 or less | 20 |
| 240,000 | 17.08 |
| 270,000 | 13.49 |
| 300,000 | 10.93 |
| 330,000 | 9.03 |
| 360,000 | 7.59 |
| 390,000 | 6.47 |
| 420.000 | 5.58 |
| 450,000 | 4.86 |
| 480,000 | 4.27 |
| 510,000 | 3.78 |
| 540,000 | 3.37 |
| 570,000 | 3.03 |
| 600,000 | 2.73 |
| 1,200,000 | 0.68 |
| 2,400,000 | 0.17 |
| 4,800,000 | 0.04 |
| 9,600,000 | 0.01 |
Summary
I see no need to change
anything concerning lasers in the Naval Architect's Manual. Although
the actual surface area of a laser is about 4.9 m2
with a 2.5 m diameter, the mount will be larger. The above calculations
do not alter any other statistics and allow the Director to use
starship weapons on a personal combat level if required.
Particle Accelerators
Starship mounted particle accelerators in 2300AD always caused me
a slight problem. They are sublight weapons, in that the beam does
not travel at the speed of light, but at a substantial percentage
of it. The time lag between detection and a hit is therefore longer
than for a laser, and the weapon did not seem worthwhile.
However, a closer analysis of the available particle beam weapons
in the Naval Architects Manual has convinced me that they are viable
weapons for starship combat.
I assume that the beam velocity is about 75% of light speed. This is
a tremendous velocity, but I think that given the very small mass we
are talking about it is probably achievable.
If we take our laser example above, we find that the total time taken
for the issue of an active sensor pulse, the return of the pulse
echo from the target, and the arrival of a 0.75c particle beam is
about (2 + 2 + (2/0.75)) = 6.67 seconds. In this time, our target ship
travels about 73540 km, and the range increases by about 4490 km, or
only 0.75 %. This is not wildly different from the starship laser range.
( Actually, the time taken is slightly more than this, but this will do
for the purposes of this discussion ).
A x3 particle beam weapon has an energy requirement of 3 megajoules. This is 180MW input power
for the one minute duration of a Star Cruiser turn. I assume that particle weapons
are also about 33% efficient like shipboard lasers, and so this leads to a discharge
power of 60 MW. The DPV of such a weapon at point blank is therefore 60.
Interestingly enough, the example x3 particle beam weapon in NAM is the
"Allen BMZ 150 MW PBWS" which makes me wonder if the 150 MW
in the name bears some connection to this value of 180MW input power.
I am not sure that the delivered energy intensity will be high enough to cause
damage, as I do not know the equations for the divergance of particle beams.
However, all the evidence suggests that these weapons would work as described
and are consistent with shipboard lasers. You can therefore use the statistics
from NAM for particle beams unaltered.
Detonation Lasers
We can make some approximations concerning the explosive power of the nuclear bombs
that power detonation lasers, given that we are correct in taking 20 Mj per laser damage multiplier
in Star Cruiser.
For example, a 14 x 2 Kafer X-ray detonation laser provides 14 x 2 x 20 = 560 Mj of discharge
energy. Now, this energy is only a fraction of the actual nuclear explosion used
to power it. Let's assume that it is 1/1,000,000th of the total output
of the nuke. This means that the total energy output of the nuclear explosion was
in the order of 1,000,000 times 560 Mj, or 560,000,000,000,000 joules. Since
4.1868 x 109 joules are produced by the equivalent of 1 kilogram of TNT,
the detonation laser was about 0.134 kilotons.
It's probably quite tricky to make larger detonation lasers ; certainly the Kafer X-ray
is near the top end of the scale, or there would be larger and more powerful devices
available. I would think that 0.5 kilotons is a good limit. Larger devices have truly
frightening damage values ( 1 MT is 4.1868 x 1018 Joules ; 1/1,000,000th
of this is 4.1868 x 1012 Joules, which is 209340 x 20 Mj, or a weapon factor of
about 69780x3 !!! ) and would unbalance the game considerably.
Of course, I am assuming that 1/1,000,000th of the explosive energy output pumps
the detonation laser. It is possible that the fraction is much smaller, which in turn would
increase the "tonnage" of the nuclear explosive. Personally, I dislike the idea
of giving players large tactical nuclear weapons already conveniently mounted in missiles !!
A detonation laser warhead is based on a normal nuclear warhead - multiply the volume by 1.5,
the mass by 1.2 and the price by 5. Note that detonation lasers output X-ray and
perhaps even gamma ray lasers, and have much smaller focal diameters than their shipboard
equivalents. As a result, there is no surface area requirement for the warhead - in fact, the
lasers probably just punch straight through the missile's hull ( perhaps at predefined
"weak points" ) when they fire.
Tamper Proof Warheads
I suggest that detonation lasers should be fitted with a number of anti-terrorist
safeguards. Otherwise, such weapons could be bought and used as nuclear bombs for
terrestrial mayhem. Thus, the average remote detonation laser should be tamperproof
and only detonate in a given set of circumstances.
Stutterwarp Missiles
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