In the RotS novelization, we hear of the Invisible Hand using "hydrofoamed permacrete" and, later, a note about "a pile of rubble: shattered permacrete, hydrofoamed to reduce weight." Indeed, Kenobi, unconscious, was Force-launched by Dooku "into the wall so hard the hydrofoamed permacrete buckled and collapsed onto him", though as Anakin found later, "no bones were broken: this was a concussion, no more."
Needless to say, then, this hydrofoamed permacrete doesn't sound like some super-tough, super-dense material. It would certainly be more substantial than the normal styrofoam we're all used to, but it doesn't seem any more substantial than, say, a sheetrock wall, and given the complete collapse it seems as if it wasn't especially well framed. (Meaning, a sheetrock wall is anchored to framing studs that are spaced fairly closely, meaning even if you launch a guy through the sheetrock in-between the studs, you'll usually just have a guy-sized hole in the sheetrock rather than a collapsed wall of sheetrock or, worse, collapsed wall altogether.)
I am reminded of other similar techniques used today. I've seen foam sections used as building material for ornamental purposes, covered in stucco or similar to provide the finishing touches. And indeed, there is a material commonly referred to as 'eps concrete' where, instead of using sand and rock for the aggregate material of the concrete, simple polystyrene foam is used instead, reducing weight. "Hydrofoamed" is a word of not perfectly clear meaning, though it suggests something involving water. Perhaps ice was used as aggregate for a low-temperature concrete mix and then, once hardened, the water is melted, and the wall was dried? That would be an interesting technique for low-cost, low-weight concrete if your hydrofoamed concrete factory is located in arctic climates, anyway. Maybe I should patent that. The last few sentences are hereby copyrighted!
In any case, we know of permacrete from other prequel novelization references, too. For instance, on Naboo the palace at Theed featured permacrete on the interior that was blasted into shards along with the glass by Panaka's blaster, and the ray shield preventing Obi-Wan from reaching Qui-Gon was said to be as good as a three meter wall of permacrete insofar as keeping Obi-Wan out, suggesting that a wall of such impressive proportions is sufficiently impervious as to keep a Jedi out for an indeterminate length of time. In the later prequel novelizations, permacrete seems to be referred to as walkway material for Coruscant, the landing pads of Kamino, and the Separatist landing platform of Utapau. The buildings of Coruscant are said to be durasteel and permacrete canyon walls after the Battle of Coruscant. The stuff's even slippery when wet.
In short, we have no great reason to suspect permacrete is terribly much stronger than concrete. Stronger, yes, insofar as it allows for the multi-kilometer buildings of Coruscant, but relatively simple steel-reinforced high-density concrete was used for the base of the Burj Khalifa (formerly Burj Dubai) in this century, and that building is almost a kilometer in height. I say "relatively simple" because although it was an engineering challenge to get the mix and pours just right, it did not require a totally unprecedented material bearing no resemblance whatsoever to the concretes that had come before. We might not be able to extend that to a multi-kilometer structure today, but I also rather doubt that it would take a thousand years of development to arrive at such a solution.
There is also reference to "duracrete" in the novelizations, identified as the material of the plaza around the Senate building in the RotS novel, but with no other information provided. But given the presumable age of the plaza, I would presume it is a lesser material than permacrete.
2014-02-08
2014-01-26
Nerd Culture Podcast Interview
If you'll forgive a touch of self-promotion, I let a persuasive bloke
talk me into a full-on guest spot on his podcast.
http://nerdculturepodcast.com/2014/01/podcast-episode-76/
Also, I forgot almost every single thing I wanted to say, so it is thus my natural rambling. (Suffice it to say, I'm better in text.) But fortunately the editing process involved cutting out about half of it so I almost sound like I stayed on topic.
Enjoy!
http://nerdculturepodcast.com/2014/01/podcast-episode-76/
Bo and David are joined by special guest Robert from ST-vSW.net and discuss all things Star Trek vs Star Wars, including the difference in Fans, Canon rules and changes, the imposter Spock & who would win in a dogfight between the Millennium Falcon and the Defiant…and much much more!In it, you'll hear me imitating a popsicle . . . this due to letting a stray but talkative cat in from the cold, meaning I ended up having to go out into it for the purpose of the interview. So, pardon me for sounding shiver-y.
Also, I forgot almost every single thing I wanted to say, so it is thus my natural rambling. (Suffice it to say, I'm better in text.) But fortunately the editing process involved cutting out about half of it so I almost sound like I stayed on topic.
Enjoy!
2014-01-20
Death Star Explosions: The Foliage Factor
More calculatory bliss.
On a forum it was recently asked just how powerful the Death Star explosion over Yavin IV could've possibly been . . .that is, "what is the most energy released by the Death Star that Yavin IV could have survived?"
Well, that sounded like fun.
A beginning estimate would be to basically figure out a level of energy that would observably damage the moon on the surface. This is useful because of three factors.
1. As the Rebels take off to go fight the Death Star, the red gas giant is visible in the sky. This suggests that the Yavin base ought to have been within line of sight of the Death Star blast.
2. After the battle, we get two useful scenes.
A. We see a brief shot of the planet from a great distance as the surviving Rebels fly home.
B. We see a close shot of the jungle near the Rebel base with its impenetrably dense stone before the medal ceremony.
So, as a result, we know that the condition of the base and nearby jungle was basically the same before and after the Death Star was destroyed. With that fact in mind, we can estimate an energy intensity that would have produced observable effects, then rock the inverse square law back out to the range of the Death Star. Since the planet and jungle are not observably affected, this would serve as a back-of-the-envelope upper limit.
- Atmosphere
Now, in reality, there would be a number of other factors to consider, such as atmospheric attenuation (especially in the sense of what parts of the spectrum make it through), cloud cover more specifically, and so on. So while we could use surface effects and ignore the atmosphere, technically the actual energy would need to be greater. For sunlight on Earth, it is said that about 50 percent makes it to the surface to be absorbed, 30 percent gets reflected at the surface, and 20 percent is absorbed by the atmosphere. That 20 percent is what we need to zoom in on at the moment, because I presume the 30 percent figure is a general value including such things as oceans (what with their being 75 percent of the planet and all), which are not relevant to our inquiry.
Just in case, though, let's assume that a third of the radiative emission of an exploding Death Star would be absorbed by the atmosphere . . . that means we need to increase our relevant values below by 33 percent.
- Range
As for the range to the Death Star from Yavin, most estimates are in the 150,000km range, which I believe is true for both the SFJ pages and Wong's SDN. (Also, some nincompoop on SDN claimed I am the source for a very specific figure from Michael January's pages, but that's just a funny thing I've been wanting to share and is not relevant right now.)
I haven't personally studied the matter, but we'll roll with this figure for now as it is not really inconsistent with the ~77,000km range of the Alderaan shot.
- Damaging the Surface
For ease, we can use estimates from http://glasstone.blogspot.com/2006/04/ignition-of-fires-by-thermal-radiation.html . . . that for a nuclear explosion, the energy required to ignite assorted things like dry and dark rotted wood falls somewhere in the range of 15-25 calories per square centimeter (my own selected range from the wide assortment of figures he reports). Now, Yavin IV didn't exactly look dry, so we could crank this up further to 50 just to try to be sure, and to try to cover the 33% atmosphere absorption mentioned above.
I don't see where a precise time factor is provided, but we can presume for the purpose of this that the nuclear weapon and the Death Star blast would radiate for about the same amount of time.
Translating to joules per square centimeter, then, with a calorie being 4.184 joules, that's 209 J, which in square meters would be 2,090,000 . . . so, two megajoules per square meter. (The total energy incident on Yavin would be interesting but not relevant here.)
- The Inverse Square Law
The inverse square law is a fancy term for the idea that the intensity of something per unit area at a range of, say, one meter (e.g. an intensity of 1 watt per square centimeter at one meter's range) will fall off in a geometric fashion. So, at two meters, you'll actually see the intensity drop to 1/4th watt per square centimeter, because while the distance has only doubled, the affected area will quadruple. At three meters it's 1/9th, et cetera. And note that while I used watts above, joules work just as well if you don't need a time factor . . . indeed, the inverse square law applies to all sorts of things, what with it being fairly basic geometry, applied.
The fun part for our purposes, though, is that you can also estimate the strength of the actual source itself if you know this intensity at a given range. This isn't really the inverse square law per se, but it is often discussed in the same breath.
So, in the case of 1 watt per square centimeter at one meter's range, we can work backward. It is 100 centimeters to the point where the intensity is 1 watt per square centimeter, meaning that if we want to work backwards to a 'zero point' all we need to do is get the total wattage for the whole sphere. That's just figuring the whole area of the sphere and multiplying the known wattage on one square centimeter accordingly. So, a 100cm sphere has an area of 4πr^2, or 4π(100cm)^2, or 125,663.7cm^2. Since we know the intensity is 1 watt per square centimeter and we know the whole sphere is 125,663.7 square centimeters, then the point source of energy must be outputting at a rate of 125,663.7 watts, or about 126 kilowatts. (That's a lot, but so is a watt per square centimeter at one meter's range. Per the basic geometry, you'd be looking at 10,000 watts per square centimeter at 1 centimeter range . . . yet the sun only hits us with 0.14 watts per square centimeter, and we only receive about 0.1 watts per square centimeter at the surface.)
- Rocking the Law
So with 150,000,000 meters as the range and two megajoules per square meter as the energy intensity at that range, the total energy of the Death Star blast would be 4π(150,000,000)^2 times two megajoules, or 282,743,338,823,081,391.5m^2 times two megajoules. That gives us about 565,500,000,000,000,000 megajoules, or 565,500 exajoules, or 5.655E23 . . . or, about 135,000 gigatons.
If, however, the estimated range is wrong and it was really, say, 500,000 kilometers (500,000,000 meters), then the final value would be much higher, in the range of 6.3E24 J, or about 1.5 million gigatons.
The sun puts out 4E26J every second, meaning that the Death Star explosion upper limit from the first example would be equivalent to 1/700th the energy output of the sun at the range and with the energy on the surface specified.
Given a range of 500,000km, the comparison comes out rather better, with the Death Star explosion coming in at about 1/64th of the sun's energy every second.
However, given the way the Death Star II played out so very much closer to the surface of Endor, all of these figures seem ridiculously high.
- Addenda
I seemed to recall doing something like this before, so I went to look and found that this is a different technique (insofar as calculating via affecting foliage) than what I used in regards to getting an upper limit for the DS2 explosion in the Superlaser Effect section of my site. See #4 at http://www.st-v-sw.net/STSWsuperl-5.html and you'll see a technique involving evenly heating the atmosphere by fifty degrees Celsius. While I still don't think that was a completely terrible approach, I think the full methodology above is much superior.
Applying it, and using the 3500km value, we get 3E20 J as the upper limit, which is 1/1,333,333 of the sun's output in one second.
Even that is very much too high, given that we know exposed skin was unaffected by the DS2 blast . . . two megajoules per square meter in a short timespan like that would definitely provide a touch of sunburn. Leia didn't even seem to notice the DS2 had gone up until several seconds later, so we can reasonably suppose that she was not burned, blinded, et cetera.
3E20 joules . . . about 73.6 gigatons . . . is thus, to my mind, a very solid upper limit value for the DS2 destruction, and probably applies with caveats to the first Death Star, to boot.
Also, just for bonus points, and to actually apply the main part of the inverse square law a little, if we assume the Falcon and friends were only about 500 kilometers from the second Death Star when it blew, then they would've been getting about fifty times more energy per square meter, or 100 megajoules per square meter. The Falcon would therefore probably have had to deal with, say, 20,000 megajoules . . . 20 gigajoules . . . incident upon her, at worst case. That's a bit less than a five ton bomb.
On a forum it was recently asked just how powerful the Death Star explosion over Yavin IV could've possibly been . . .that is, "what is the most energy released by the Death Star that Yavin IV could have survived?"
Well, that sounded like fun.
A beginning estimate would be to basically figure out a level of energy that would observably damage the moon on the surface. This is useful because of three factors.
1. As the Rebels take off to go fight the Death Star, the red gas giant is visible in the sky. This suggests that the Yavin base ought to have been within line of sight of the Death Star blast.
2. After the battle, we get two useful scenes.
A. We see a brief shot of the planet from a great distance as the surviving Rebels fly home.
B. We see a close shot of the jungle near the Rebel base with its impenetrably dense stone before the medal ceremony.
So, as a result, we know that the condition of the base and nearby jungle was basically the same before and after the Death Star was destroyed. With that fact in mind, we can estimate an energy intensity that would have produced observable effects, then rock the inverse square law back out to the range of the Death Star. Since the planet and jungle are not observably affected, this would serve as a back-of-the-envelope upper limit.
- Atmosphere
Now, in reality, there would be a number of other factors to consider, such as atmospheric attenuation (especially in the sense of what parts of the spectrum make it through), cloud cover more specifically, and so on. So while we could use surface effects and ignore the atmosphere, technically the actual energy would need to be greater. For sunlight on Earth, it is said that about 50 percent makes it to the surface to be absorbed, 30 percent gets reflected at the surface, and 20 percent is absorbed by the atmosphere. That 20 percent is what we need to zoom in on at the moment, because I presume the 30 percent figure is a general value including such things as oceans (what with their being 75 percent of the planet and all), which are not relevant to our inquiry.
Just in case, though, let's assume that a third of the radiative emission of an exploding Death Star would be absorbed by the atmosphere . . . that means we need to increase our relevant values below by 33 percent.
- Range
As for the range to the Death Star from Yavin, most estimates are in the 150,000km range, which I believe is true for both the SFJ pages and Wong's SDN. (Also, some nincompoop on SDN claimed I am the source for a very specific figure from Michael January's pages, but that's just a funny thing I've been wanting to share and is not relevant right now.)
I haven't personally studied the matter, but we'll roll with this figure for now as it is not really inconsistent with the ~77,000km range of the Alderaan shot.
- Damaging the Surface
For ease, we can use estimates from http://glasstone.blogspot.com/2006/04/ignition-of-fires-by-thermal-radiation.html . . . that for a nuclear explosion, the energy required to ignite assorted things like dry and dark rotted wood falls somewhere in the range of 15-25 calories per square centimeter (my own selected range from the wide assortment of figures he reports). Now, Yavin IV didn't exactly look dry, so we could crank this up further to 50 just to try to be sure, and to try to cover the 33% atmosphere absorption mentioned above.
I don't see where a precise time factor is provided, but we can presume for the purpose of this that the nuclear weapon and the Death Star blast would radiate for about the same amount of time.
Translating to joules per square centimeter, then, with a calorie being 4.184 joules, that's 209 J, which in square meters would be 2,090,000 . . . so, two megajoules per square meter. (The total energy incident on Yavin would be interesting but not relevant here.)
- The Inverse Square Law
The inverse square law is a fancy term for the idea that the intensity of something per unit area at a range of, say, one meter (e.g. an intensity of 1 watt per square centimeter at one meter's range) will fall off in a geometric fashion. So, at two meters, you'll actually see the intensity drop to 1/4th watt per square centimeter, because while the distance has only doubled, the affected area will quadruple. At three meters it's 1/9th, et cetera. And note that while I used watts above, joules work just as well if you don't need a time factor . . . indeed, the inverse square law applies to all sorts of things, what with it being fairly basic geometry, applied.
The fun part for our purposes, though, is that you can also estimate the strength of the actual source itself if you know this intensity at a given range. This isn't really the inverse square law per se, but it is often discussed in the same breath.
So, in the case of 1 watt per square centimeter at one meter's range, we can work backward. It is 100 centimeters to the point where the intensity is 1 watt per square centimeter, meaning that if we want to work backwards to a 'zero point' all we need to do is get the total wattage for the whole sphere. That's just figuring the whole area of the sphere and multiplying the known wattage on one square centimeter accordingly. So, a 100cm sphere has an area of 4πr^2, or 4π(100cm)^2, or 125,663.7cm^2. Since we know the intensity is 1 watt per square centimeter and we know the whole sphere is 125,663.7 square centimeters, then the point source of energy must be outputting at a rate of 125,663.7 watts, or about 126 kilowatts. (That's a lot, but so is a watt per square centimeter at one meter's range. Per the basic geometry, you'd be looking at 10,000 watts per square centimeter at 1 centimeter range . . . yet the sun only hits us with 0.14 watts per square centimeter, and we only receive about 0.1 watts per square centimeter at the surface.)
- Rocking the Law
So with 150,000,000 meters as the range and two megajoules per square meter as the energy intensity at that range, the total energy of the Death Star blast would be 4π(150,000,000)^2 times two megajoules, or 282,743,338,823,081,391.5m^2 times two megajoules. That gives us about 565,500,000,000,000,000 megajoules, or 565,500 exajoules, or 5.655E23 . . . or, about 135,000 gigatons.
If, however, the estimated range is wrong and it was really, say, 500,000 kilometers (500,000,000 meters), then the final value would be much higher, in the range of 6.3E24 J, or about 1.5 million gigatons.
The sun puts out 4E26J every second, meaning that the Death Star explosion upper limit from the first example would be equivalent to 1/700th the energy output of the sun at the range and with the energy on the surface specified.
Given a range of 500,000km, the comparison comes out rather better, with the Death Star explosion coming in at about 1/64th of the sun's energy every second.
However, given the way the Death Star II played out so very much closer to the surface of Endor, all of these figures seem ridiculously high.
- Addenda
I seemed to recall doing something like this before, so I went to look and found that this is a different technique (insofar as calculating via affecting foliage) than what I used in regards to getting an upper limit for the DS2 explosion in the Superlaser Effect section of my site. See #4 at http://www.st-v-sw.net/STSWsuperl-5.html and you'll see a technique involving evenly heating the atmosphere by fifty degrees Celsius. While I still don't think that was a completely terrible approach, I think the full methodology above is much superior.
Applying it, and using the 3500km value, we get 3E20 J as the upper limit, which is 1/1,333,333 of the sun's output in one second.
Even that is very much too high, given that we know exposed skin was unaffected by the DS2 blast . . . two megajoules per square meter in a short timespan like that would definitely provide a touch of sunburn. Leia didn't even seem to notice the DS2 had gone up until several seconds later, so we can reasonably suppose that she was not burned, blinded, et cetera.
3E20 joules . . . about 73.6 gigatons . . . is thus, to my mind, a very solid upper limit value for the DS2 destruction, and probably applies with caveats to the first Death Star, to boot.
Also, just for bonus points, and to actually apply the main part of the inverse square law a little, if we assume the Falcon and friends were only about 500 kilometers from the second Death Star when it blew, then they would've been getting about fifty times more energy per square meter, or 100 megajoules per square meter. The Falcon would therefore probably have had to deal with, say, 20,000 megajoules . . . 20 gigajoules . . . incident upon her, at worst case. That's a bit less than a five ton bomb.
2014-01-07
Quick Review: Mythbusters AT-ST Log Smash
So, did the Mythbusters get it right?
In broad strokes, yes. As reader Vorus noted, my old 2002 page on the AT-ST log smash held up well compared to their scrutiny.
In short, they build a rig with telephone poles and hang logs beneath them, pulling them back and letting them smash a test vehicle and then a full-on (and nicely painted) armored truck of the sort you might expect to haul bank loot.
There were some great moments beyond the eminently television-friendly Kari's various attire, such as Grant Imahara describing "back of the envelope" calculations suggesting that the energy involved per log was going to be about a megajoule, followed by a quick flash of assorted scenes from the show of megajoule scale, like car crashes and so on.
The total for both logs of two megajoules falls in line nicely with the notations on my AT-ST log smash analysis page in which, as always, I heap the benefit of the doubt upon Star Wars and basically double the likely figures. The Mythbusters team, unencumbered by claims of bias from a group of Star Wars fans bent on maximizing Star Wars technology at every turn, felt perfectly at liberty to simply follow the facts and not inflate the figures in favor of Star Wars.
But, there were also some oddities.
"Aren't you a little short for a chicken walker?" - The AT-ST height is given as 20 feet (a rather low figure given that the full-scale mock-up in the film was over 8 meters, or 26 feet), and for some reason they felt it necessary to then hang a Ford Econoline test vehicle at that height for smashification. Not only was the altitude completely unnecessary, but it put undue stresses on their rig to move the logs about at that height. Fortunately this was abandoned for final testing, otherwise they would've had to rebuild it.
"Aren't you a little short for an Ewok log?" - Also, despite having a fake log of proper length to width ratio at the beginning of the episode, it is somehow decided that the logs in the film are about ten feet long and five feet wide, which makes them entirely too short. In the scene below, for instance, the logs would not even extend all the way to the edge of the frame.
On my page I came out with 1.5 meters by nine meters, or about five feet in diameter and 30 feet long. And, since their targets were quite a bit shorter than the AT-ST cab section, the five foot logs looked a bit monstrous.
"Aren't you a little short for a log nose?" - The logs themselves were unfinished, in the sense that they were straight or nearly-straight cuts and had no pointed/rounded tips. In the first hit against the armored truck, this seemingly resulted in the left side (as viewed from the front) not actually penetrating the side of the vehicle since it hit the roofline broadside.
"Aren't you a little far back for a logsmash?" - The first strikes were against the flat sides of a van and then the flat side of an armored truck. In the case of the truck the right side simply caved in, which makes sense given that there is likely very little actual vehicle structure back there. The second hit, however, was aimed more forward, and literally tore the roof off of the vehicle and smashed the cabin. There is likely a pillar just behind the driver's door, making that more of a fair test. After all, armored trucks of that nature are mostly intended to be bulletproof rather than structurally strong.
Of course, comparing that to the actual AT-ST's structural failure is interesting, since the first hit, as noted, only poked a hole in the truck. The general structure appeared to remain fairly intact. Compare that to the logsmash itself, in which the AT-ST literally crumples. I guess in an age of blasters, the notion of kinetic impact may seem altogether quite foreign.
"Aren't you a little hard for a log? Wait, don't answer that." - For some unexplained reason, eucalyptus logs were selected explicitly for their hardness. Presumably this was intended to allow repeated strikes without fear of log failure . . . the rig, after all, would've been rather difficult to 'reload'. Nevertheless, it seemed odd when watching initially.
This does bring me back to my page, however. I did assorted calculation via forces and whatnot, but never really got into the actual pondering of hardness of the AT-ST hull. Given the malleability against logs it would be nice to review that ancient page and see what I can come up with in that regard. Presumably I dismissed it back then as relatively unimportant, given that I had already concluded that AT-STs were more or less bulletproof.
In any case, I'd say the Mythbusters team did a great and very entertaining job, and it was great fun to see things get smashed and have Grant Imahara generally confirm my old figures.
2014-01-05
Shuttlecraft Terminal Velocity
Number-crunchy-stuff, ahoy!
Well, *that* needs fixing, so we're fixing to fix that right now.
Applying Google-fu, I found: http://www.wikihow.com/Calculate-Terminal-Velocity . There, we are told that terminal velocity is the square root of the following: 2 times the mass and times gravitational acceleration, divided by the density (of air) times the area of the object times the drag coefficient.
Easy enough, with caveats.
First, we have a mass of the Type-6 shuttle that is an estimate from my Volumetrics pages, scaled down from the density of full-size ships. That said, the Delta Flyer is believed to be of similar density to Voyager, and so we're gonna roll with a similar density, providing a mass of 29 tonnes . . . that's only about half the mass of an M1A1 Abrams tank, so I'm cool with it.
Second, the reason I didn't think I'd be able to calculate it before is because I had no method of determining the drag coefficient. However, I am now willing to make scientific-sounding wild-ass guesses. For instance, a handy-dandy list of various drag coefficients is linked to from the WikiHow page. For example, a cube is said to have a drag coefficient of 0.8, a flat square plate facing the wind provides 1.17. For our purposes, we'll want to look at something akin to a minivan or SUV and then bump it up a notch to factor in the nacelles. Doing so, we find that an SUV commonly falls somewhere in the neighborhood of 0.45 or less, while a bus is listed as 0.6 to 0.8. So, I'm going to say a shuttlecraft should be in the 0.55 range, but your coefficients may vary.
A final problem is finding the frontal area of the shuttlecraft . . . as in, were it to fly through paper and cut it perfectly rather than tear it, thus making a neat shuttle-sized hole, what would the area of that hole be? Fortunately, this is easy enough with Google Trimble SketchUp, at which point we find a frontal area of 7.55 square meters, give or take.
So, now we have basically all the parts we need. For the density of the air we will just rough it as an average of .9 kg/m^3 . . . at sea level it can be 1.2, but at 50,000ft (15km) you're gonna be looking at a much lower value, somewhere in the .3 range. .9 is the pressure at about 7500 feet (~2300m).
The terminal velocity of a shuttlecraft is thus the square root of: 2 times 29,000kg times 9.8 m/s^2, divided by 0.9 kg/m^3 times 7.55 m^2 times 0.55.
Check my math, but I believe that works out to the square root of 568400/3.73725, or the square root of about 152090, which yields about 390 m/s as the terminal velocity.
Interestingly enough, that's greater than Mach 1 at the altitude indicated by our "average" pressure. If we calculate instead based on sea level pressure of 1.2 kg/m^3, then the terminal velocity drops to 338 m/s, which is so close to Mach 1 at sea level STP (340 m/s) that, given the back-of-the-envelope wild-ass nature of our exploration, it might as well be close enough. (And hey, wouldn't it be cool to see the transsonic condensation clouds around a shuttle as it is about to hit the ground?)
So, terminal velocity of a shuttlecraft is approximately 340 m/s (Mach 1) at sea level, and greater than Mach 1 at higher altitudes.
Now, of course, it's worth noting that this is not especially useful for the page in question, since the shuttle from "Rise"[VOY3] was not actually dropping like a brick, but was instead engaged in powered descending flight and then suddenly lost power. The result would be more like a ballistic trajectory than a simple fall. But hey, it was fun anyway, wasn't it?
2013-12-29
Did the Enterprise See Herself?
While writing on a page, a thought crossed my mind.
You see, it seems very odd that the Federation would have authorized the intentional time travel mission from "Assignment: Earth". Even if you assume that the timeline is pretty robust and resilient, it doesn't seem wise to plop an entire bright-white 289 meter Constitution Class starship in 1968 Earth orbit when the same job could be achieved by a specialized low-observability vessel or microsatellite or somesuch.
However, it also occurs to me that "Tomorrow is Yesterday" did feature an accidental dropping of the same Constitution Class starship into Earth's atmosphere. On that occasion the Enterprise escaped via some sort of unusually slow time-warp maneuver. Her sensors were capable of recording information at the time, which is how they were able to do the beaming maneuver as they slowly returned to their time.
So, as they travelled forward in time with sensors recording information, did the Enterprise see herself in 1968, which led Starfleet's eggheads who were later poring through the sensor logs to send them back there? Was her assignment to 1968 Earth thus a predestination paradox?
Just a random thought.
You see, it seems very odd that the Federation would have authorized the intentional time travel mission from "Assignment: Earth". Even if you assume that the timeline is pretty robust and resilient, it doesn't seem wise to plop an entire bright-white 289 meter Constitution Class starship in 1968 Earth orbit when the same job could be achieved by a specialized low-observability vessel or microsatellite or somesuch.
However, it also occurs to me that "Tomorrow is Yesterday" did feature an accidental dropping of the same Constitution Class starship into Earth's atmosphere. On that occasion the Enterprise escaped via some sort of unusually slow time-warp maneuver. Her sensors were capable of recording information at the time, which is how they were able to do the beaming maneuver as they slowly returned to their time.
So, as they travelled forward in time with sensors recording information, did the Enterprise see herself in 1968, which led Starfleet's eggheads who were later poring through the sensor logs to send them back there? Was her assignment to 1968 Earth thus a predestination paradox?
Just a random thought.
2013-12-28
Ben Sisko, Gunslinger
Imagine if you are sent back in time by accident.
You awake in a strange land of oddly dressed humans . . . they speak English but it sounds quite odd to your ears. Seeing your strange clothing they assume you are some sort of spy for their foes and detain you, along with some others.
You soon come to realize that you are in the colony town of Lexington, Massachusetts in mid-April 1775. You manage to escape, but in so doing you inadvertently cause the death of Solomon Brown, who (we shall pretend for the purpose of this gedankenexperiment) fired the first shot against the British redcoats from the area of a tavern on April 19, 1775. Aware that your action has potentially screwed with history, and knowing you have only minutes to make history right, you take on his identity, picking up a weapon of the era and firing the shot at the appointed time.
Okay, so this is all well and good, but there's a problem. Unless you are especially well studied in the arms of the era and not just the historical specifics, you really ought not have much idea of what to do with that weapon.
Sure, just from watching the History Channel or somesuch, you'd know that you're probably dealing with a muzzle-loading weapon, hopefully with a rifled barrel. You'd probably try putting some powder down the barrel, maybe a piece of fabric or paper or somesuch, then a ball. You'd tamp it down and then try to figure out how to deal with actually firing it. Is there something to light which then makes contact? Is there a flint striker? Do you need to put a little powder down some other hole?
You may have been hanging out in the 1770s for a day or two and seen other weapons, but unless you've had a chance to observe them used you may have no idea how to actually make the gun go boom. And if you fail, the American Revolution may happen completely differently.
This brings us to 2024.
In "Past Tense", Ben Sisko is thrown back in time from the 24th Century. Placed in a concentration camp for the poor, his knowledge of the era's history leads him to realize that he is at the precise location which, in mere days, will be the epicenter of the Bell Riots, a watershed event of the 21st Century. While trying to survive, an event occurs which causes the death of Gabriel Bell, hero of the event. Later, as the riots break out, Sisko grabs a shotgun, immediately racks and fires it, and is then able to take control of the situation along with Gabriel Bell's identity. His action gets him shot later, but keeps history correct.
The problem with that story is pretty much identical, except in his case a Remington 870 is dropped into his hands and boom, he's running the thing smoothly.
While I'm hardly a firearms expert by any stretch, I have owned, handled, and fired a variety of pistols and rifles. I am at least basically familiar with the operation of such weapons from .22 revolvers to 1911-style semi-automatic pistols to bolt-action magazine-fed rifles to semi-automatic "assault" rifles. I even shot a friend's 12 gauge shotgun a few times over ten years ago, though as I recall he probably did the loading. I was also recently helping a friend who was having issues with his 12-gauge semi-automatic shotgun, aiding in the disassembly and helping assess what was going wrong with it on the trigger group mechanically, and even watching him fire it.
But when roaming around in a sporting goods store a few months ago, I spotted a particularly interesting pump-action shotgun and asked to take a look at it. The kid behind the counter handed it over and I was giving it a look-see. I had the mind to rack it, pulled back on the fore-end and . . . nothing happened. It wiggled but didn't otherwise budge. Trying to look nonchalant, I fiddled with the safety and messed around otherwise, surreptitiously trying to rack it again, to no avail. As I had no real interest in purchasing the shotgun I didn't just decide to stop and ask for directions from the kid as I normally would be happy to do when lost . . . I simply handed it back and thanked him.
The thing that either eluded my attention or which I otherwise failed to try to operate correctly was the part of a pump-action shotgun most commonly called the action release or slide release (the latter is what I mostly hear, despite it being potentially confusing considering pistols feature that same term for a different purpose altogether). Now, in fairness to myself, my most recent shotgun experience was with a semi-auto that had no slide and no release . . . it operated much as any semi-auto rifle. And the experience a decade prior with a pump probably didn't feature a lot of me working the gun. I can't even recall what kind of pump-action shotgun it was, though I do remember how sore my shoulder was.
I tell that story not to embarrass myself in regards to my ignorance, stupidity, and/or forgetfulness, though I'm sure I've succeeded in doing so. Nor do I tell the story to regale you with overly inflated tales of my manly potency with firearms (lol). I simply tell it to make the point that I am familiar with the general principles of many weapons of the late 20th and early 21st Centuries, and yet properly working a pump-action shotgun, cold and with no preparation, was a mystery to me. Had yet another leftist shooter or some band of terrorists stormed the place at that moment and the kid behind the counter tossed me shotgun shells as his dying act, the most I could've done is try to use the shotgun as a baseball bat to knock the shells in the general direction of the badguys.
Meanwhile, you have Starfleet Captain Commander Benjamin Lafayette Sisko, a relative stranger to these parts temporally-speaking, pick up a period firearm and operate it just as smoothly as a Hollywood actor who'd been shown how to do it shortly before the scenes started filming. And more than that, he had correctly assessed that he needed to do it in order to fire a shot mere seconds after picking it up to get the nearby crowd under control. (In other words, he snatched a weapon without a shell in the chamber, so he had to rack it to load a shell in the chamber and cock the internal hammer, and if the safety was on at the time he also had to disengage that.)
How the hell did that happen? That man should most likely have virtually zero experience with 20th/21st Century projectile weapons, and certainly not a pump-action shotgun. And after reviewing the episode, there's no indication that he had opportunity at any point to carefully observe one being manipulated before he himself had to do so.
And it isn't as if the Remington 870 slide release is the sort of thing you could work by accident, either. Located well ahead of the trigger on the very front end of the long trigger guard's left side, some people find that they cannot operate the slender slide release with a trigger-ready grip, That said, it is worth noting that when Sisko first gets his hands on the shotgun, he works it like a left-handed shooter . . . that is, his left hand is the trigger hand, which would've made it easier to work by accident. Later, he's usually running it right-handed.
Besides, it isn't like phaser rifles have pump slide-action releases that he could've practiced with, and given Sisko's general disdain for much of the 20th Century it seems unlikely that he'd have been futzing around in a Dixon Hill sort of setting on a holodeck. He did have an interest in early 21st Century history, but learning about history is, at least in the modern era, a very different thing than living in the moment and playing with the local doodads.
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Then again . . .
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Then again . . .
After writing the above I have performed a more thorough review of the second part of that episode. The issue I'm seeing is that folks appear to be racking the shotguns without recourse to the slide release at all. To be honest, I'm not sure I can even see the release at the resolution available. And of course, since no shell emerges when they rack it, then it cannot have had one in the chamber at the time.
As a rule, racking without the slide release would only be possible under a particular sequence of events. For instance, if the weapon was already cocked and the trigger pulled with no shell in the chamber, then you would have successfully unlocked the bolt. This allows free movement of the fore-end, meaning that if you have done this you could theoretically then rack the shotgun to load a shell into the chamber with a shell from the magazine without using the slide release.
That doesn't really make a great deal of sense unless you are expressly wanting to have to rack it to prepare to fire (as a roundabout safety precaution or for psychological terror purposes) and want to make that racking as easy as possible. Given the user-unfriendly slide release location on the 870 I can actually kind of see this being a plausible desire, but I'm not aware of this being a big trend in the 870-owner community.
So again, that's a complete series of events that is quite unlikely for most people to engage in, and given that we also see the main "ghost" theater-major thug doing it . . . a guy who seemed intent on shooting someone . . . it becomes even more unlikely.
One possibility is that the slide release somehow was removed from the design or otherwise became obsolete, but this would be incredibly odd. The slide release is there because the gun, when ready to fire, is locked into battery. If it weren't, it would be possible to fire it with the bolt partially open ("out of battery"), meaning you could have hot gas heading toward the shooter. Without a release, the only way to unlock it is to pull the trigger.
I suppose it's possible future shotguns could be made with some sort of electronic slide release . . . that is, if there's a hand on the fore-end (or maybe a squeezing hand) and it is pulling back on the fore-end, it might 'know' that you're trying to rack it and let you do it if your finger's off the trigger. That seems like a terribly bad idea to me and one that would most likely result in a lot of accidents, but still. And worse, of course, is that Sisko's effortless picking up of a shotgun and perfect operation thereof is even harder to believe if you imagine he had to know how to work a shotgun totally different in operation from today's models.
So let's abandon our imagined magic slide release. Besides, in any case, we see multiple occasions where the characters rack the shotgun (with no working of the slide release) to prepare to fire, and then for whatever reason they don't fire, only to repeat the maneuver a few scenes later. This means that they have loaded a shell into the chamber but no longer intend to use it. A few scenes later, they again rack it (without the slide release) to prepare to fire. Since no shell comes out, they must have removed the earlier scene's shell from the chamber. This is not something you can generally do with a finger . . . the most common technique, as seen in this video of a fellow with his 870, is to simply hit the slide release and rack the weapon over and over again so it loads from the magazine to the chamber and then from the chamber unloads out of the ejection port.
So after completely unloading the shotgun in that fashion, they would then have to pull the trigger on an empty chamber to unlock the action and allow the slide release to live free, and then reload the magazine with shells. A practiced shotty-lover can probably do this pretty quickly if for whatever strange reason they had the desire to do so, but most people would probably just flick the safety on.
Of course, if we assume instead that these early 21st Century shotgun models have no safety, then all the racking makes a little more sense as being the only way to safe the shotgun (i.e. not having one in the chamber). But again, I rather doubt that the theater-major thug would have been acting like a responsible gun owner and keeping it as safe as possible . . . he would've been more likely to leave it ready to fire as soon as his finger met the trigger, even if it meant it was possible to have an accidental discharge. It isn't like there was anyone else in the room (or even out of it) that he actually cared about.
Then again, maybe Sisko convinced him to keep it safe in that way that Sisko seemed good at in this episode.
In short, other than the out-of-universe truth that the shotguns weren't loaded at all and were probably otherwise modified for endless racking, I don't have any good answer as to what exactly was going on, though it seems most likely that the characters engaged in a whole lot of unloading, dry firing, and loading off-camera, generally for no apparent reason. Maybe if DS9 gets the HD remastering treatment we'll be able to see if the 870s are missing externally-visible bits like safeties or slide releases.
In-universe, the only good answer I can figure is that maybe everyone spends more time on the holodeck running amok in earlier centuries than we've ever suspected, and that working primitive projectile weapons is something that comes up rather more frequently than we ever would've suspected.
As a rule, racking without the slide release would only be possible under a particular sequence of events. For instance, if the weapon was already cocked and the trigger pulled with no shell in the chamber, then you would have successfully unlocked the bolt. This allows free movement of the fore-end, meaning that if you have done this you could theoretically then rack the shotgun to load a shell into the chamber with a shell from the magazine without using the slide release.
That doesn't really make a great deal of sense unless you are expressly wanting to have to rack it to prepare to fire (as a roundabout safety precaution or for psychological terror purposes) and want to make that racking as easy as possible. Given the user-unfriendly slide release location on the 870 I can actually kind of see this being a plausible desire, but I'm not aware of this being a big trend in the 870-owner community.
So again, that's a complete series of events that is quite unlikely for most people to engage in, and given that we also see the main "ghost" theater-major thug doing it . . . a guy who seemed intent on shooting someone . . . it becomes even more unlikely.
One possibility is that the slide release somehow was removed from the design or otherwise became obsolete, but this would be incredibly odd. The slide release is there because the gun, when ready to fire, is locked into battery. If it weren't, it would be possible to fire it with the bolt partially open ("out of battery"), meaning you could have hot gas heading toward the shooter. Without a release, the only way to unlock it is to pull the trigger.
I suppose it's possible future shotguns could be made with some sort of electronic slide release . . . that is, if there's a hand on the fore-end (or maybe a squeezing hand) and it is pulling back on the fore-end, it might 'know' that you're trying to rack it and let you do it if your finger's off the trigger. That seems like a terribly bad idea to me and one that would most likely result in a lot of accidents, but still. And worse, of course, is that Sisko's effortless picking up of a shotgun and perfect operation thereof is even harder to believe if you imagine he had to know how to work a shotgun totally different in operation from today's models.
So let's abandon our imagined magic slide release. Besides, in any case, we see multiple occasions where the characters rack the shotgun (with no working of the slide release) to prepare to fire, and then for whatever reason they don't fire, only to repeat the maneuver a few scenes later. This means that they have loaded a shell into the chamber but no longer intend to use it. A few scenes later, they again rack it (without the slide release) to prepare to fire. Since no shell comes out, they must have removed the earlier scene's shell from the chamber. This is not something you can generally do with a finger . . . the most common technique, as seen in this video of a fellow with his 870, is to simply hit the slide release and rack the weapon over and over again so it loads from the magazine to the chamber and then from the chamber unloads out of the ejection port.
So after completely unloading the shotgun in that fashion, they would then have to pull the trigger on an empty chamber to unlock the action and allow the slide release to live free, and then reload the magazine with shells. A practiced shotty-lover can probably do this pretty quickly if for whatever strange reason they had the desire to do so, but most people would probably just flick the safety on.
Of course, if we assume instead that these early 21st Century shotgun models have no safety, then all the racking makes a little more sense as being the only way to safe the shotgun (i.e. not having one in the chamber). But again, I rather doubt that the theater-major thug would have been acting like a responsible gun owner and keeping it as safe as possible . . . he would've been more likely to leave it ready to fire as soon as his finger met the trigger, even if it meant it was possible to have an accidental discharge. It isn't like there was anyone else in the room (or even out of it) that he actually cared about.
Then again, maybe Sisko convinced him to keep it safe in that way that Sisko seemed good at in this episode.
In short, other than the out-of-universe truth that the shotguns weren't loaded at all and were probably otherwise modified for endless racking, I don't have any good answer as to what exactly was going on, though it seems most likely that the characters engaged in a whole lot of unloading, dry firing, and loading off-camera, generally for no apparent reason. Maybe if DS9 gets the HD remastering treatment we'll be able to see if the 870s are missing externally-visible bits like safeties or slide releases.
In-universe, the only good answer I can figure is that maybe everyone spends more time on the holodeck running amok in earlier centuries than we've ever suspected, and that working primitive projectile weapons is something that comes up rather more frequently than we ever would've suspected.
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