Minutes? I won't even ask for proof of that, since the fact that it needed minutes is all I need to tell you that there wasn't enough gees generated by the overall mass of that asteroid to prevent a shuttle from maneuvering, especially out.Mike DiCenso wrote:Except that the molten material does flow "downwards' into the asteroid within minutes of the warbird's firing on it.Mr. Oragahn wrote: Surely, that's another problem. You density that's required to generate gees that would overpower engines which can leave Earth within a few minutes tops would be incredible.
How the asteroid in question didn't reach hydrostatic equilibrium is odd.
The whole thing is perhaps even more contradicted by the time the Romulan ship melts the cavity's entrance with disruptors. Did we see any gout of molten material fall towards the core of the asteroid, and hit the E-D for example?
Surely, if the gravity is so strong that a shuttle couldn't control entry and even exit, damaging the asteroid's integrity would have surely resulted in something much more hazardous than a cleanly sealed entrance. I would have expected something more like a rain of debris and a newly formed giant stalactite made of asteroid material, pointing "corewards".
Then, perhaps, and I may say once again, we're dealing with more Trek physics, aka pure nonsense.
The only other explanation is that the warbird pushed the material in with focused weapons fire, or with a tractor beam.
We perfectly know that the writers did this because they wanted the E-D inside. They could have pulled out an excuse like no, we won't send people in a shuttle, the environment is too risky and they couldn't defend themselves from the Romulans if things went wrong.As for the nonsense of it, I agree, but nonetheless, it is a canon fact. So do we charge this up to FX incompetence for not making an asteroid that was larger and more spherical, or assume that Data is an idiot?
But no, Trek has to to do with physics nonsense. That's the problem.
The other problem is that nothing in that episode agrees with that excuse.
What I tend to do, but again that's not glorious, is ignore the one silly detail.
Again, the Stargate example I love to use is characters, including Carter and McKay, not raising an eyebrow when Hammond says that a low gigaton explosion could cause some mass extinction event on Earth.
The point is that there's a margin of error anyway, from the edges selected by viv to make the measure, from the blurriness of the picture, to the fact that taking the E-D's length while we see the portside nacelle sticking out behind the other, the fact the asteroid may be seen at an angle as well, etc.I did some work on it. Roughly speaking I get a distance of warbird to camera in the 126 image of 7 km and from the E-D to camera of 886 meters. So about 6,100 meters seperates the two ships. Now that seems a bit small, and my math may be a bit faulty, but then I realized that the two ships are at an angle of some 30-45 degrees off the asteroid's center and this may throw everything off even more, as this may make the thing far larger since the distance would cover only across that section, not the whole asteroid.
As for the E-D as reliable yardstick, I would point out that the height of the ship is skewed here and no one source agrees on it how tall it is: some placing it only at 137 meters, while others estimate it at 150 meters, and the DS9 TM is way out there with 193 meters (not possible within the proportions we see). However within a meter every source places the ship at 641 to 642.5 meters. Call it 642 meters.
Using your images:
Measuring the chasm opening of 2.51" to the E-D length 1.85" (measuring from the narrowest parts of the chasm) I get 873 meters. Again using the images provided here, dividing 362.4 pixels by 34 I get a ratio of 10.66 to 1. 10.66 x 873 = 9.306 km. Measuring the wider parts of the chasm I can get the asteroid length up to 10 km. So that's a third to a full km or more longer using the ship's length, not height. Of course using one of the larger estimate averages for the ships height I can put it in that 10 km range. 160 meters x 5.78 = 924 m x 10.66 = 9,858.36 m.
Starting to get my point now?
I'm fine with the idea that it's around 9 km, more or less 1 km, I won't be retentive on that.
It still doesn't make much difference since contrary to you, I don't believe Riker pictured a giant plasma ball.
You say it doesn't make sense and legitimate this claim by picking interpretation from another episode?Mike DiCenso wrote: Among other things people in that thread, most suprisingly l33telboi missed, was that Kirk's statement was that a 97.835 MT explosion results from overloading a single starship's impulse engine. Thus 98 MT is a very lower limit there. But I digress. The interesting thing is the photon torpedo explosion examples which show that torpedoes rarely leave large fragments behind when they destroy asteroids and starships. The "Rise" incident was an anomaly, and was even as part of the plot acknowledged as one that was throughly investigated by the Voyager crew, and they soon discovered the asteroid there was made of artifical alloys and one natural one that was prone to fragmentation.
Thus one has to conclude that when Riker means the asteroid in "The Pegasus" is to be destroyed, he means mostly vaporized or reduced to tiny pieces vastly smaller than 10 meter chunks.That makes no sense. Assuming 100 megaton torpedoes as per the rise calculations, then 250 torpedoes x 100 MT = 25,000 MT or 25 gigatons. That's more than enough to signficantly vaporize a spherical asteroid of those dimensions, never mind an irregularly shaped one.Mr. Oragahn wrote:Nope. Because small asteroids largely vanish after being hit, it doesn't automatically translate in Riker thought that the big ass one would be largely vaporized.
Let me tell you that in Pegasus, there's just no problem with the idea that Riker didn't necessarily envision turning the asteroid into a plasma ball. Torpedo figures are probably stretched over a certain range.
Not to say that several gigatons is good enough to vaporize such a large asteroid because it's done efficiently. Wong's calculator assumes a perfecly homogeneous and spherical asteroid with a charge in its middle. That's totally different than firing volley after volley at the surface.
Now, assuredly, the best part of 250 torpedoes, even rated at 10 megatons, would surely turn this piece of rock into mincemeat no matter what.
He has enough torpedoes, even if rated at 10 MT, to blast all pieces into bits so small than the fused Pegasus couldn't hide in any of them.On the other hand, if we look at it the other way, knowing that vaporization is possible such that only tiny bits remain, then 768 gigatons divided by 250 torps = 3 gigatons per torpedo. Riker's thought, as the dialog shows, is to utterly destroy the thing and the Pegasus to ensure not a single bit remained for Pressman or the Romulans. Under your scenario, Riker is an idiot who will just be happy with leaving behind large chunks and possibly large pieces of the ship.
Riker is no specialized gunner, explosive engineer nor math genius either. You're putting too much stock into a man who was emotionally and professionally involved in this event, and just threw a figure in haste in order to convince people around him that it could be done and was the way to do it.On top of that, as noted earlier, Riker has no idea exactly what it will take. He gave an off the cuff option, one that ensures total destruction. We also cannot quantify exactly what "most of" the 250 torpedo loadout means. Is it 150, 165, 200 , or 249 torpedoes to do the job? Brian Young assumes 275 were expended (the number per the TNG TM, not the canon 250 number given in "Conundrum").
So what this comes down to is at very minimum "The Pegasus" points to low single digit megatons, and on the upper range low single digit gigatons.
-Mike
Were it coming from a non-involved technician reading what the computers tell him/her, there would be much less ambiguity and much less issues, although we would still not know what "destroy" meant.