**Section 2 Technology****2.3 Weaponary**

**2.3.3 Explosive Yield of a Photon Torpedo.**

By the 24^{th }century the explosive
yields of weapons are described in terms of isotons as compared
with today's convention of using kilotons. However, given that we
know the quantity of antimatter and matter used, we should be
able to make some assumptions about the comparative yields of the
weapons of the Federation.

The current convention is to measure yields in multiples of
the explosive yield of 2,4,6-trinitrotoluene CH_{3}C_{6}H_{2}(NO_{2})_{3}.
We know that TNT has an explosive yield of 700 calories per gram,
(*Encyclopaedia Britannica*) so we can begin by converting
into SI units; specifically the joule. To do this we will
multiply the given figure by 4.1868.

- 1 gram of TNT = 700 calories = 700*4.1868 joules.
- 1 gram of TNT = 2930.76 joules

Now we need to know how this works in tons. Given that 1 ton is equal to 2240 pounds or 1016.046909 kg we should be able to scale up the explosive yield simply.

**Note***It is convention to use ton compared to short ton, net ton American or tonne (1000 kilogrammes.)*

1 kg of TNT = 2930.76j*1000g = 2930760 joules.

1 ton = 2930760j*1016.046909kg = 2,977,789,639.02084 joules.

1 megatonne = 2,977,789,639,020,840 joules.

According to Einsteins’s theory of special relativity the enrgy E, and momentum, p, of a particle are related by the equation :

E^{2}=p^{2}c^{2}+m^{2}c^{4}

However, given that the antimatter is injected into the matter and held by fields around each packet of antimatter until the detonation signal it would seem that the matter is not in movement and therefore p tends towards zero.

E^{2}=0c^{2}+m^{2}c^{4}

E=mc^{2}

Where c is the velocity of light and m is the mass of the particle when at rest.

What we will do is calculate for a detonation of 100% efficiency for simplicity. The amount of matter is therefore 3kg
(1.5 antimatter mass combined with 1.5 mass of matter, energy
must be conserved in any reaction) and the constant c is the
speed of light which we will approximate to 299,792,500 ms^{-1
}

Using the mks unit system for SI measurements:

E_{(joules)} = 3 _{(kg)} * 299 792 500 ^{2}
_{(ms}-1_{)}

E_{(joules)} = 3 _{(kg)} *
89,875,543,056,250,000

E_{(joules)} = 269,626,629,168,750,000 joules.

E_{(joules)} = 2.6962662916875*10^{17} joules.

Now if 1 megaton = 2,977,789,639,020,840 joules; then 1 photon torpedo yields 90.5458953968987946834375526278729 megatons. (90.55 to 2 D.P.)

This has been upgraded recently to increase yield by 5%. (*Zimmerman
et al. 1998*)

This gives 95.73 (to 2 D.P.) megaton yield which rates at 18.5 isotons. This would mean that 1 isoton is approximately 5.14 megatons to 2D.P.

A typical quantum torpedo has increased the typical yield to 21.8 isotons approximately 112.03 (to 2 D.P.) megatons but by using the zero point motion the yield increases to an equivalent of at least fifty isotons. This would give a megaton yield of 257 megatons. In 1987 the Soviet and American stockpile of nuclear arsenals measured 15 gigatons. This is only 58 times the yield of a single quantum torpedo. In fact the detonation of a single torpedo would release more energy than 13 times the explosives ever used in war from the discovery of gunpowder to the 1990’s.

The yield rate for a photon torpedo is therefore 95.73 megatons per 3 kilograms. This is equal to 31.91 per kilogram or 14.51 megatons per pound. This compares with modern weapons as below:

Name | Date | Weight | Yield/weight | Y/W ratio kton per pound | Destructive Area |

Little Boy | 1945 | 9000 lb | 1 kton to 600 lb | 0.0017 | 3 sq. miles |

MX missile | 1986 | 800 lb | 1 kton to 2.4 lb | 0.417 | 234 sq. miles |

Photon torpedo | Circa 2374 | 410.74 lb | 1 kton to 0.000069 lb | 14510 | unknown |

Groombrdige 273-2A Test | 2355 | unknown | 1 kton to 0.000025 lb | 40730 | Confined to 900 metre sphere. |

**Comparative Destructive Size**

An asteroid of 2000 metres diameter would have an equivalent
yield of one million megatons and would be sufficient to cause a
global cataclysm. The crater at Chicxulub was caused by a
considerably larger asteroid of almost 9000 metres across and had
an explosive yield of 10^{8} megatons (*Wynn 1998*) (*further
calculations below*).
Even the impressive yield of Starfleet weapons and global
extinction events pale compared to mines used by the Borg
Collective. In the first episode of season 4 of Voyager, Scorpion
Part 2, (Braga & Menosky 1997), 7of9 wishes to use a Borg
multi-kinetic-neutronic mine with an explosive yield of 5 million
isotons. The equivalent of 25,700, gigatons.

While Starfleet was testing the Marx IX warhead it was determined that a theoretical limit of 25 isotons had been reached for matter/antimatter explosions. However the Type 6 torpedoes that came into use in 2371 and which were utilised onboard intrepid class ships had a considerably larger explosive yield. The Borg examination of Voyager (Braga & Menosky 1997) described the type 6 as having a maximum yield of 200 isotons. Later that year a torpedo (Diggs and Kay 1998) was modified with a gravimetric charge for a 54 isoton explosion though this charge was not standard issue. The charge was later upgraded to 80 isotons.

Another weapon of
mass destruction that is worth considering is the Long Range Tactical Unit from
Star Trek Voyager’s Warhead. The Unit is a self aware artificially intelligent
weapon found on a planet; the destruction caused by the units companion weapon
allow us to calculate its explosive yield.

To calculate the energy we could compare with an incident meteor impact, from
these we can deduce the size of crater and using simple kinetic energy
equations, calculate the energy required to excavate a crater of said size.

First let us consider the dimensions of the impactor at Chicxulub.

Given the uncertainties let us begin by considering the arbitary figure of

10,000 metres (usually stated diameter approx. 6 miles)

we will round down due to unknown dimensions with a 20% volume loss due to

deviation from perfect sphere and consider a radius of 4000 metres.

Volume of a sphere is 4/3 (pi r^{3})

4000 cubed = 64000000000

4/3 * pi * 64000000000 = 268082573106.329 (3dp)

volume = 268082573106.329 m^{3}

Mass is given by v*d

(density The densities of Ceres, Pallas, and Vesta are 2.3, 3.4, and 4.0 grams
per cubic centimetre)

Considering a density of 3 for a type S meteorite.

Converting into SI 3,000,000 grams per cubic metre which is 3,000 kg per cubic
metre.

268082573106.329 * 3000 = 804247719318987.069 (3 d.p.) kg

Given that normal velocity of impact would average 25 kms^{-1}

*
This impact velocity is probably too high, most impacts are slower than this and
are severely retarded in the atmosphere, the Chicxulub event would not have been
significantly retarded so we will keep this value for the larger body.**
(For a discussion on this see Wynn and Shoemaker 1998)*

Energy at impact is:

Ke = 1/2mv^{2}

25000^2 * mass/2 = 625000000 * 804,247,719,318,987.069/2 =
251,327,412,287,183,459,077,011.471 (3d.p.) joules

The kinetic energy at impact is therefore: 251 *10^{21} joules

Now given that 1 megatonne =

1 gram of TNT = 700 calories = 700*4.1868 joules.

1 gram of TNT = 2930.76 joules

1 kg of TNT = 2930.76j*1000g = 2930760 joules.

1 ton = 2930760j*1016.046909kg = 2,977,789,639.02084 joules.

1 megatonne = 2,977,789,639,020,840 joules.

2,977,789,639,020,840 joules

The impact is roughly 84,400,660.474 megatonnes.

A two hundred kilometre crater would require above 100,000,000 megatonnes.

Now the total annihilation of 1 kilogramme of antimatter with 1 kilo of matter
will release:

E = mc^{2}

(this assumes all useable energy, all matter destruction, 100% efficiency and no
inert particulate matter generation.

179,751,086,112,500,000 joules of energy or 60.364 megatonnes.(3 d.p.)

Therefore for the crater seen in Warhead we would need a payload of:
1,656,618.440 kilos or 1,656.6 tonnes of antimatter. However, care must be given
to the shape of the crater. Though we do not know about the photogramteric
accuracy of the Voyager sensors and the orthographic corrections, the crater
still looked very deep given the width; far deeper than a normal blast crater.
This could effect the over-all energy of the impact.

The interesting side note becomes, if the Tactical Unit is an antimatter device (seemingly indicated by Tuvok) then how are two crewmen able to carry the unit, it is possible that the units explode and whatever they use for explosive yield does not detonate or that they some how replicate their own supply of antimatter before detonation.