Starfleet Jedi

I came across this quote from Apocalypse Rising:

DAMAR: Personally, I think we'd be better off launching an orbital assault on Gowron's command centre. A full spread of photon torpedoes would take care of him, the Klingon High Command and everyone else within a few hundred kilometres.
ODO: You should ask Dukat for some shore leave. I think you've been in space too long.
DAMAR: Why? Because I'm willing to spill a little Klingon blood to get the job done?
O'BRIEN: Shelling Ty'Gokor won't get the job done. You'd be lucky to launch one torpedo before they shot you down. Besides, even a dozen won't penetrate the shielding around the command centre.

Now this is a rather interesting quote. It's best to remember that the ship that Dukat has is a BoP and its only launcher is the forward one. He suggests a full spread of photon torpedoes. For the most part, we've seen that consist of 5 photon torpedoes from the Enterprise D.

Now, as far as we've seen, the BoP is only capable of firing one torpedo per volley, so this isn't anything like the GCS's version of a spread, nor do I believe that the complex is so large that it would require that the torpedo's target would vary over such a large area that it would mean such a massive shift in each torpedo's target.


To cause enough damage that everyone within a couple hundred kilometers (going with 200 for minimum yields), the torpedoes would each have to provide 1025 megatons, which would mean people up to two hundred kilometers would suffer 3rd degree burns (ie, this is the widest known effect that I can find for it and fits the description fairly well.

Personally, I think it all matches up and it works fine, but I want to do this the right way. The first thing is, is there any reason why the Bird of Prey would want to space out the torpedoes? Surely, one might argue that they would try and surround the target so they could hit it with shockwaves on all sides, but I don’t see why it would be largely significant since the farther away the shockwaves, the less power each of them has and I would think that the fireballs would effective in helping taking down the shields.

The second is that while I’m pretty confident that the Bird of Prey will probably fire a its torpedoes at the structure ever five or seven seconds, I’m not entirely sure that there’s nothing we’ve witnessed that might show the BoP’s launcher acting differently (large spread use of torpedoes) or something else that might significantly change the results.

Thoughts?

I'm going to try pulling this as low as possible, just to see what kind of low end figure we can rely on.

First and foremost, we must agree that this is a technically reliable figure.

Now, the bad news is, the lowest interpretation I can concoct doesn't require more than kilotons per torpedo.
This is going with the idea that the complex is composed of several packets of discrete buildings, each group locked on by a torpedo. As the packets are spread over dozens and dozens of kilometers, each aimed shot would devastate the blocks and effectively kill everyone, perhaps everyone of importance, within hundreds of kilometers.

The second option, which can be associated to the first, is that there are torpedo silos and/or an AM reactor located within the compound that could power up a theater shield. Their destruction would release their respective amount of AM and amplify each blast. That said, this option would likely require hundreds of kilotons if not low megatons to make sure the structures are properly damaged.

As we keep climing the ladder of firepower, we now can work with widespread destruction from Wong's nuke calculator.

From there, let's assume the target is a stretching series of military and administrative structures some 200 km long and only a dozen km wide, eye shaped.
With five torpedoes, it requires each one's destructive effect to cover at the very least a 20 km radius.
The calculator returns 22 megatons a piece.

From there, you can play with higher yields and obtain overlapping blast radii.
The numbers obviously increase in all cases if the area is more than 200 km wide.

This quote was actually brought up by me over a year ago another thread comparing TOS and TNG power.

Even low megaton torpedo detonations would not accurately describe the effect of devastating an area 200-900 km across. Hell, it would be difficult even assuming Damar ment square km. There is another factor here as Ty'Gokor is described and shown in dialog to be a planetoid, not a big Class-M planet like Earth in which the fireballs effects have a substantial atmosphere to interact with. On the other hand, a smaller body means less gravity, but still those effects are rather impressive. And even if we attribute only kiloton range firepower to the torpedoes, it speaks rather impressively of the fortress' shields that they could handily take that much firepower without effort.
-Mike

Mr. Oragahn wrote:I'm going to try pulling this as low as possible, just to see what kind of low end figure we can rely on.

First and foremost, we must agree that this is a technically reliable figure.

Now, the bad news is, the lowest interpretation I can concoct doesn't require more than kilotons per torpedo.
This is going with the idea that the complex is composed of several packets of discrete buildings, each group locked on by a torpedo. As the packets are spread over dozens and dozens of kilometers, each aimed shot would devastate the blocks and effectively kill everyone, perhaps everyone of importance, within hundreds of kilometers.

Although a good idea and it is aimed at making it the lowest possible calculation, I find this hard to believe that their base is so wide spread out. Unfortuantely there is no image of the base, but we do get a rather interesting indication of what perhaps, they were thinking:

SISKO: Starfleet Science thinks that exposure to polaron radiation will have a destabilising effect on changeling physiology.
ODO: In other words, if we use these on Gowron and he is one of my people, he won't be able to retain his humanoid shape.
SISKO: That's the theory.
DAX: The problem is, for it to work properly, you have to activate all four emitters at once.
SISKO: (holding the remote) With this.
O'BRIEN: I hate prototypes.
DAX: Plus, too much polaron radiation can be fatal. Which means we can only expose a person once. Any more than that, changeling or not, and they could die from radiation poisoning.
SISKO: The plan is to smuggle these into Ty'Gokor. Once they're set up, they can cover about twelve thousand cubic metres.
O'BRIEN: So let me get this straight. All we have to do is get past an enemy fleet, avoid a tachyon detection grid, beam into the middle of Klingon headquarters and avoid the Brotherhood of the Sword long enough to set these things up and activate them in front of Gowron.
WORF: If we succeed, there will be many songs sung in our honour.
O'BRIEN: Let's hope we're there to hear them.
KIRA: The first problem is how to get you safely to Ty'Gokor.
SISKO: That's one I think I can solve.

Now, twelve kilometers is rather excessive for such a small area. Incredibly excessive. The only reason they would bother using something like that would be to expose Gowron on any part of the base. It would also be of adequate size of a massive military base.

The second option, which can be associated to the first, is that there are torpedo silos and/or an AM reactor located within the compound that could power up a theater shield. Their destruction would release their respective amount of AM and amplify each blast. That said, this option would likely require hundreds of kilotons if not low megatons to make sure the structures are properly damaged.

Well, there's a bit of a problem with that, as I believe this implies more thinking than what was actually done. No doubt that Damar knew what their weapons could do, since he's the second officer and has been on the BoP for some time. However, his comment was in fact, deprived of consideration of his target's defensive capabilities, one that O'Brian pointed out as being incredibly flawed.

As we keep climing the ladder of firepower, we now can work with widespread destruction from Wong's nuke calculator.

From there, let's assume the target is a stretching series of military and administrative structures some 200 km long and only a dozen km wide, eye shaped.
With five torpedoes, it requires each one's destructive effect to cover at the very least a 20 km radius.
The calculator returns 22 megatons a piece.

From there, you can play with higher yields and obtain overlapping blast radii.
The numbers obviously increase in all cases if the area is more than 200 km wide.

[/quote]


Although nice for a lower limit, we know that the NX Enterprise had 50 MT photonic torpedoes.

Mike DiCenso wrote:This quote was actually brought up by me over a year ago another thread comparing TOS and TNG power.

Even low megaton torpedo detonations would not accurately describe the effect of devastating an area 200-900 km across. Hell, it would be difficult even assuming Damar ment square km. There is another factor here as Ty'Gokor is described and shown in dialog to be a planetoid, not a big Class-M planet like Earth in which the fireballs effects have a substantial atmosphere to interact with. On the other hand, a smaller body means less gravity, but still those effects are rather impressive. And even if we attribute only kiloton range firepower to the torpedoes, it speaks rather impressively of the fortress' shields that they could handily take that much firepower without effort.
-Mike

Um, where do you get this idea from? No such thing is stated in dialogue and the image we see is not one that depicts a planetoid (Memory Alpha is what you'd see as a planetoid):

SISKO: According to Starfleet intelligence, Chancellor Gowron has relocated Klingon military headquarters to Ty'Gokor.
WORF: That will make our job more difficult. Ty'Gokor is located in an asteroid field deep in Klingon space. It is probably the most heavily fortified installation in the Empire.
SISKO: There are at least thirty warships stationed there at any given time and the entire asteroid field is protected by a tachyon detection grid.
O'BRIEN: Which means there's no way we can get a cloaked ship within transporter range.

Sure we see there is an asteroid field, but there is no mention of a planetoid here or anywhere else.

My mistake, the place is described in the original shooting script as a planetoid. Either way, in both versions, it is not described as a planet, and it is located in an asteroid field, which may or may not allow for a full planetary body or terrestrial planet sizes to exist.
-Mike

Mith wrote:Although nice for a lower limit, we know that the NX Enterprise had 50 MT photonic torpedoes.

Allow me to differ. We do not. I went over this with l33telboi at length, and it was a mess and certainly not an easy issue to solve. The crucial element, though, being that the crater is produced is a completely negligible-gravity environment.
For example, the depth of surface explosion's transient crater, which is the temporary one, deeper than the final one, has a depth that is only a percentage of the surface-level radius.

When you look at Wong's ADC, input 3,000 meters for the width of the asteroid, and see what you get for cratering on the surface of an asteroid. The low end values are 1.3 MT for an ice asteroid, and 4.9 MT for a lump of typical rock.

There are mitigating factors. One being that the crater depth mentioned in the calculator could either be the transient or final. It is not clear enough. As pointed out above, as even the transient depth, the lowest albeit temporary one, would be a percentage of the final ground-level radius, we should input a smaller diameter for the asteroid. Also, if the crater depth in question is the final one, then the value that is to be entered has to be much smaller.

Now, it's a crater blasted in an asteroid that's actually as wide as the asteroid is large, so there's not so much matter to push through beyond the fracturing range of the blast itself, since it's gouging into a sphere.
If that was not enough, burying the warhead increases the coupling (distribution of energy into surrounding matter) which enhances the crater diameter.

And if this wasn't enough, if the asteroid were to be a loose pile of rubble, then fragmentation and compression could increase spectacularly. If there were vacuum, shock would be absorbed however, but the fragility of the material and the lack of gravity to maintain it in place could allow the sudden formation of cracks, with a destruction logically more random in its shape and asymmetric, while in the case of a similar asteroid, but with holes filled with frozen water, the shock wave would be better transmitted.
I tried to find a copy of the book Wong got his equation from but couldn't.

Then there was another equation, the one l33telboi used, and that's the one we argued over a lot. One reason, among others -- and bear with me, that's an old topic so I'm working from memories -- was that even the gravity-less equation was obtained from the observation of empirical data of explosions against dry soil terrains, which density would be greater than that of many of the asteroids I picked as examples, large or small. Also lacking from the version of the equation, as provided at SBC, was that the real equation from the early 70s nuclear book, which got updated in 77, and featured the a factor of 60, and above all formulated in feet. So much that the equation was R_feet = 60 * Y^0.3, with Y in kilotons, R_feet the radius of the crater at ground level in feet.
Then, with Y = (R_feet/60)^(1/0.3), and with 1500 meters being 4921.259843 feet, the yield would be 2,397,453 KT, which is frankly already ludicrous, and the number would get even more stupid if you'd remove 60 from the equation.

Gravity is also very important because not only it acts against the removal of the crater's mass from the volume it's supposed to exit, but gravity applied to the surrounding material is all the more force that maintains the ground in place and acts against the blast's strength over the edges of the crater to a degree that is not dismissible, even if gravity is a force that is more and more perpendicular the more you look at the edge of the bowl-shaped volume that is soon to be a crater.

Then I recall there was some equation about calculating a crater on the Moon, returning a number around +40 MT or something. But then, again, the Moon has a significant gravity and its soil is rather firm.

Mike DiCenso wrote:My mistake, the place is described in the original shooting script as a planetoid. Either way, in both versions, it is not described as a planet, and it is located in an asteroid field, which may or may not allow for a full planetary body or terrestrial planet sizes to exist.
-Mike

It was probably changed later in production, otherwise they would have adjusted the script to allow us to know that.

More likely the visual FX people botched it up along the way.
-Mike

Mr. Oragahn wrote:

Mith wrote:Although nice for a lower limit, we know that the NX Enterprise had 50 MT photonic torpedoes.

Allow me to differ. We do not. I went over this with l33telboi at length, and it was a mess and certainly not an easy issue to solve. The crucial element, though, being that the crater is produced is a completely negligible-gravity environment.
For example, the depth of surface explosion's transient crater, which is the temporary one, deeper than the final one, has a depth that is only a percentage of the surface-level radius.

When you look at Wong's ADC, input 3,000 meters for the width of the asteroid, and see what you get for cratering on the surface of an asteroid. The low end values are 1.3 MT for an ice asteroid, and 4.9 MT for a lump of typical rock.

There are mitigating factors. One being that the crater depth mentioned in the calculator could either be the transient or final. It is not clear enough. As pointed out above, as even the transient depth, the lowest albeit temporary one, would be a percentage of the final ground-level radius, we should input a smaller diameter for the asteroid. Also, if the crater depth in question is the final one, then the value that is to be entered has to be much smaller.

Now, it's a crater blasted in an asteroid that's actually as wide as the asteroid is large, so there's not so much matter to push through beyond the fracturing range of the blast itself, since it's gouging into a sphere.
If that was not enough, burying the warhead increases the coupling (distribution of energy into surrounding matter) which enhances the crater diameter.

And if this wasn't enough, if the asteroid were to be a loose pile of rubble, then fragmentation and compression could increase spectacularly. If there were vacuum, shock would be absorbed however, but the fragility of the material and the lack of gravity to maintain it in place could allow the sudden formation of cracks, with a destruction logically more random in its shape and asymmetric, while in the case of a similar asteroid, but with holes filled with frozen water, the shock wave would be better transmitted.
I tried to find a copy of the book Wong got his equation from but couldn't.

Well, if he had the option I would think that he would have used the transiet depth since it would be the one that would be used most for debates given that most calculations wold occur just after the detonation in sci-fis.

But that would be sketchy at best.


Then there was another equation, the one l33telboi used, and that's the one we argued over a lot. One reason, among others -- and bear with me, that's an old topic so I'm working from memories -- was that even the gravity-less equation was obtained from the observation of empirical data of explosions against dry soil terrains, which density would be greater than that of many of the asteroids I picked as examples, large or small. Also lacking from the version of the equation, as provided at SBC, was that the real equation from the early 70s nuclear book, which got updated in 77, and featured the a factor of 60, and above all formulated in feet. So much that the equation was R_feet = 60 * Y^0.3, with Y in kilotons, R_feet the radius of the crater at ground level in feet.
Then, with Y = (R_feet/60)^(1/0.3), and with 1500 meters being 4921.259843 feet, the yield would be 2,397,453 KT, which is frankly already ludicrous, and the number would get even more stupid if you'd remove 60 from the equation.

Gravity is also very important because not only it acts against the removal of the crater's mass from the volume it's supposed to exit, but gravity applied to the surrounding material is all the more force that maintains the ground in place and acts against the blast's strength over the edges of the crater to a degree that is not dismissible, even if gravity is a force that is more and more perpendicular the more you look at the edge of the bowl-shaped volume that is soon to be a crater.

Then I recall there was some equation about calculating a crater on the Moon, returning a number around +40 MT or something. But then, again, the Moon has a significant gravity and its soil is rather firm.

Do you know how this discussion ended or where I could find it?

Mike DiCenso wrote:More likely the visual FX people botched it up along the way.
-Mike

Given that they've botched up so many things for ST in the past, I'll be happy to accept a mistake in our favor for once.=p

It's not that much of a benefit since a mostly airless planetoid with hundreds of km wide fireballs exploding on it would potentially make for higher yields, though there is something to be said for gravity and firestorms on a much larger body.
-Mike

Mike DiCenso wrote:It's not that much of a benefit since a mostly airless planetoid with hundreds of km wide fireballs exploding on it would potentially make for higher yields, though there is something to be said for gravity and firestorms on a much larger body.
-Mike

I think it's best as it is. =)

Mith wrote:Do you know how this discussion ended or where I could find it?

It's here, somewhere. Perhaps a thread involving the NX, but I'm not sure. It's a long one as well, made of trials and errors. The end? Like a balloon farting out its last cubic millimeters of moist air.

Mr. Oragahn wrote:

Mith wrote:Do you know how this discussion ended or where I could find it?

It's here, somewhere. Perhaps a thread involving the NX, but I'm not sure. It's a long one as well, made of trials and errors. The end? Like a balloon farting out its last cubic millimeters of moist air.

Damn.

Well, that aside, I think the lowest possible figure we could get with this is your first option (the second one just requires more effort than what was actually put into it), but the devices sort of cast a bit of doubt. However, we do know Picard has orded them before. Maybe if I can get the specs on what happened there, we can get something a bit more accurate.

EDIT: Well, this is what I've found so far from The Nth Degree:

BARCLAY [OC]: Maximum yield, full spread.
RIKER: We're too close.
BARCLAY [OC]: I'm certain the shields will hold.
PICARD: Mister La Forge?



LAFORGE: I don't know how he did it, but shield strength has been increased by three hundred percent.



LAFORGE [OC]: It should be enough, Captain.
PICARD: Mister Worf. Photon torpedoes. Maximum yield, full spread.
WORF: Aye.
(three direct hits and a big light display. The ship rocks)
PICARD: Thank you, Mister Barclay.

And this is what we get:



I was hoping for something a bit larger, since the torpedoes just seemed to be aimed at this one spot. Granted, it seems here that the targeting is aroung the edges and if we take the Booby Trap as another full spread (which it was, save it only used 4 torpedoes) and I do recall one indication of the Enterprise firing at enemies using a full spread, which targeted each ship.

My best guess then is that maybe the torpedoes would be spred out by 12 kilometers or so from the center of the target.

But again, I need to find more examples of full spreads and see how large of an area they usally cover.