What is the yield of a Federation photon torpedo?

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Robert Walper
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What is the yield of a Federation photon torpedo?

Post by Robert Walper »

Mike Wong's calculation for Federation torpedo yield is based solely upon information gathered from the TM, which isn't considered even a official source. So, what visuals and events can one extrapolate photon torpedo yield?
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Re: What is the yield of a Federation photon torpedo?

Post by SirNitram »

Robert Walper wrote:Mike Wong's calculation for Federation torpedo yield is based solely upon information gathered from the TM, which isn't considered even a official source. So, what visuals and events can one extrapolate photon torpedo yield?
Pegasus. The relevent dialogue:

SCRIPT: The ship nears the Devolin system -- a solar system that never coalesced into discrete planetary bodies. It's a swirling disc of rock and other matter.
RIKER: I recommend we destroy the asteroid. It would take almost all our photon torpedoes, but it would preclude any possibility of the Pegasus falling into Romulan hands.

Scaling pictures:

Image

Image

At about 5 Km wide, it's a good sized chunk.

The Implications:

The script, copied above, suggests this asteroid was never a part of a planetary body(Unlike the Hoth Asteroid Field in ESB, which was made by two colliding), and seems to be rock. Michael Wong's asteroid calculator indicates it would take 125MT to blast it if it's Igneous, or 30MT to shatter it if it's granite.

Divide by 255(Stated torpedo capacity, the reference is Conundrum), and you get between 110Kilotons and 450Kilotons.

This does not take into account the asteroid was hollow.
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Re: What is the yield of a Federation photon torpedo?

Post by Robert Walper »

SirNitram wrote: At about 5 Km wide, it's a good sized chunk.

The Implications:

The script, copied above, suggests this asteroid was never a part of a planetary body(Unlike the Hoth Asteroid Field in ESB, which was made by two colliding), and seems to be rock. Michael Wong's asteroid calculator indicates it would take 125MT to blast it if it's Igneous, or 30MT to shatter it if it's granite.

Divide by 255(Stated torpedo capacity, the reference is Conundrum), and you get between 110Kilotons and 450Kilotons.

This does not take into account the asteroid was hollow.
A good example. Any others?
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Post by Howedar »

ST5, in which a torpedo hit is insufficient to mess up Kirks hair :D

I don't know of any episodes offhand that support low dozens of MT for torpedos, although I happen to harbor that belief myself.
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Post by Doomriser »

The episode "Night Terrors" is also used as an example of torpedo firepower:

[from the CDB]

"Season 4, Ep# 91: "Night Terrors"

DATA: When Tyken was trapped in a Rift, his analysis determined that a massive energy release might overload and dislocate the anomaly. Fortunately, his cargo included anicium and yurium, which he used to detonate a massive explosion. He then escaped through the ruptured center of the Rift.
GEORDI: But we aren't carrying anything that could produce that kind of explosion. Even our photon torpedoes wouldn't be enough."

Naval Weapons: the yield of the Enterprise's entire complement of weaponry is inferior to that of a large chemical explosion, such as Tyken was able to create with his volatile cargo.

"DATA: Sir, Commander La Forge and I have come up with a potential solution to our predicament. Perhaps the modifications we used to increase firepower against the Borg might be effective here.
PICARD: Channeling power into the main deflector dish...
DATA: I believe within six hours we can generate a concentrated burst of energy which might disrupt the Tyken's Rift.
...
DATA: We are ready to discharge the main deflector, Captain.
...
GEORDI: Deflector power coils charging...
...
WORF: Discharge in three seconds... two... one...
...
RIKER: Nothing... it just fizzled out."

Naval Weapons: they try the special weapon they used against the Borg, and it fails. Not only is the entire weapons complement of the USS Enterprise inferior in yield to a large chemical explosion, but their special high-yield deflector-dish beam is also inferior in yield to a large chemical explosion.

"Season 4, Ep# 91: "Night Terrors"

DATA: What we must discover is a means by which, working together with the aliens, we can produce an explosion more intense than either of us could achieve alone.
TROI: What is it you're looking for?
DATA: These are the elements we have available. Some of them could be used in the creation of an explosive reaction. If we can communicate this inventory to the other ship ...
...
TROI: "One... moon... circles... "
DATA: Yes, Counselor, one electron circles one proton... this is a hydrogen atom.
TROI: "One moon circles"... that's what they have been telling me... over and over ...
DATA: Perhaps the aliens are thinking as we are... to collaborate in producing an explosion. If hydrogen is combined with certain elements... calendenium, for example, it is extremely volatile."

Naval Weapons: the chemical reaction of hydrogen (from the Enterprise's Bussard collectors) and "calendinium" (from the aliens' ship) will exceed not only the firepower of the Enterprise's phasers and photon torpedoes, but also its specialized anti-Borg weapon.

No wonder the Borg weren't impressed with their defensive capabilities.
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Post by Ted C »

And if we look at "Deja Q" as an example of the maximum amount of power that the ship can channel into any single effort (they were using "warp 9" power in their effort to shift the planet's moon), we can assume that the combined power output of the phasers and photon torpedoes must be less (since we also assume that the anti-Borg deflector blast will be more powerful). This would limit a single photon torpedo to less than 2 megatons (assuming that the moon was a 2.5km chunk of solid iron ore).
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