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.


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.



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.


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.


Shuttlecraft Terminal Velocity

Number-crunchy-stuff, ahoy!

Perusing a page on my site, I saw where I'd said I wasn't sure what the terminal velocity of a Federation shuttlecraft might be.

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?