Archive for March, 2015

Fantasy Round 34: Fortress in the Eye of Time

2015/03/26 Leave a comment

Having been visiting used book stores lately, I’ve re-read one of my old favorites: Fortress in the Eye of Time, the first book in the Fortress series by C.J. Cherryh.

It’s still a favorite of mine (although Michael is not so much of a fan), so let’s discuss it a bit.

Also, here be some spoilers.

I'm not sure which fortress is supposed to be in the eye of time, but the one at Ynefel (pictured on the cover) is probably the best bet.

I’m not sure which fortress is supposed to be in the eye of time, but the one at Ynefel (pictured on the cover) is probably the best bet.

Read more…

Projectiles That Go In Circles

2015/03/01 1 comment

This is a post inspired by the fact that I watched House of Flying Daggers recently, have heard of Wanted, and have been exposed to Batman for ages.  (Do you really need a link for Batman?)

The today’s topic: projectiles that come back, or go in circles, and stuff like that.  There’s ways that can happen in reality… but in fiction, it often goes way too far.

So, let’s talk about that.

This is a batarang.  I think this one is from the Dark Knight. (Image from here.)

This is a batarang. I think this one is from the Dark Knight.
(Image from here.)

The Batarang

The Batarang is, of course, based on the boomerang.  The boomerang could be used to hunt birds, and its main advantage was that if it missed, it would come back to you.  (That said, those were used primarily for play or as decoys; according to Wikipedia, non-returning boomerangs were more likely to be used for hunting.)

If you want to know way more about boomerangs than I have bothered to thoroughly read, check out this paper by John Vassberg.  The bottom line is that the turning of the boomerang results from the spin and the effect of air forces on the boomerang’s airfoil shape.

Based on skimming that paper, a typical boomerang may fly at a speed of 23 m/s (or about 50 mph).  Assuming they made sure their final simulations are reasonable, I’d estimate a turning radius of about 3 m near the sharper point.  This implies a centripetal acceleration of about 176 m/s^2.  For a device that weighs half a kilogram, that’s 88 Newtons — or, roughly 18 g’s, or 18 times the strength of Earth’s gravity.

That’s some pretty sweet acceleration.

However, getting good behavior out of a boomerang requires careful tuning.  Making a scalloped sharp bat-thing go in a circle in the same way is pretty implausible.  If bat-wing-blades worked that well, I would expect airplanes to be using that shape to get better lift out of their wings.  I suspect that Batman’s stuff only works because, you know… he’s Batman.

I do note that the batarang is not entirely inconsistent with common sense: the exploding batarangs don’t come back.

Flying Daggers

The House of Flying Daggers includes a bunch of… well, flying daggers.  Among the highly implausible knife-work that appears in the film, at one point, a single thrown dagger is used to slice a circle of bamboo poles being used to hold two people captive.

How much momentum should the dagger lose each time it cuts through a bamboo pole?

The main contributor is going to be the friction forces of the pieces of bamboo acting against the motion of the blade.

Let’s assume the bamboo is something like wood, so I can get a coefficient of friction from this site.  The static coefficient of friction between metal and wood is given a range of 0.25-0.5.  This essentially tells how “sticky” the two materials are when they are sitting on top of one another.  Since I’ll actually need an estimate for the kinetic coefficient (for when the two materials are moving against each other) let’s go with the lower end of that range.  (Kinetic is generally less than static.)

Further, let’s assume that the mass of the upper portion of each bamboo pole is about 5 kg.

In that case, each pole feels a downwards gravitational force of about 50 Newtons.  This force pushes down on the knife as it goes through, for a typical frictional force of about (50 Newtons)*(0.25) = 12.5 N.  Technically, I should double this number to account for both surfaces of the knife, so I’ll double this number, to 25 N.  Since the knife won’t always be feeling the full weight of the bamboo (i.e., before it has cut through the entire pole), but I should also account for the forces holding the bamboo itself together in addition to gravity,  I’ll just assume those two effects cancel each other out, roughly.

Lastly, we need to know how fast the knife is moving to start with, its mass, and how thick a piece of bamboo is.  Let’s say the bamboo has a thickness of 10 cm (about 4 inches), and that the knife has a mass of 0.5 kg.  And, to be maximally friendly, let’s say its initial velocity is as high as the boomerang in our previous example — 50 mph, or (to make the numbers nice) about 20 m/s.

So, the knife starts at 20 m/s.  How many bamboo poles can it go through, given that it feels a frictional force of 5 N for as long as it is going through each pole?  After each pole, the knife will lose (25 N)*(0.1 m) = 2.5 J of kinetic energy.  To start with, the knife has kinetic energy of (1/2) (0.5 kg) (20 m/s)^2 = 100 J.  It could go through 40 poles before it loses all its energy — which is roughly the number of poles that the knife goes through in the film.

There are, meanwhile, two other details that I’ve been ignoring for this calculation.  The first is gravity, which should be rapidly drawing the flying knife downward.  The second is the part where it goes in a circle — it would need to, somehow, interact with each pole in such a way to get the angular momentum it needs, and always spin in such a way to present its sharp side towards the next pole.  The flaws and variations in the material of the bamboo poles is something that the knife-thrower couldn’t possibly anticipate.  Finally, when the knife is going too slowly, it may stop having enough oomph to get started with the slicing.  Especially if it gets dull from slicing all those poles.

So, no, you couldn’t really do this in real life, even if the energy requirements aren’t too outrageous.

Of course, there are worse offenders…

Magic Bullets

There’s a scene in the movie Wanted where somebody shoots a bullet, has it go in a circle through the skulls of six or seven other people, and then hit the original shooter.

Quoting Michael: “There’s only three problems with that.  I call them energy, momentum, and angular momentum.”

A typical handgun bullet doesn’t have the energy to go through that many people without stopping — which a very hefty gun with armor-piercing bullets might do, but not certainly regular ones; similarly, it somehow keeps transferring momentum to the victims without losing its own; and there’s just no way for a human being to supply angular momentum to a bullet to make it go in a circle like that.  (Consider: if you have an object on a string, and whirl it around, it only goes in a circle so long as you hold the string.  Let go, and it goes off in a straight line.  Making it spin faster requires pulling harder on the string.  Bullets don’t come with strings, or the airfoils that make boomerangs turn, so…)

Rather than discuss how utterly ridiculous this is in more detail, let’s consider the only good reason for a projectile to go in a circle: gravity.  (Airplanes and boomeranges and so forth don’t really experience projectile motion, since air has a major “lift” effect on them.)

Thus: how massive an object would be required to make a bullet go in a circle like in that late scene in Wanted?  And is it going to be a black hole?

From watching the scene, let’s say that the radius of the bullet’s path is about 10 m.  A typical .357 Magnum bullet will have a muzzle velocity of about 400 m/s, which I will assume is not actually reduced by air resistance or going through lots of’ braincases.

In that case, the bullet will need a centripetal acceleration of about 16,000 m/s^2, which is about 1,600 times the acceleration from Earth’s gravity.  How large a mass will be needed for such an immense force?  About 2.4×10^16 kg.  Thankfully, this is far less than the mass of Earth (about 6×10^24 kg).  It is roughly the mass of the asteroid Ida (which has a tiny moon named Dactyl).  Ida itself has a radius of around 16 km.

Fortunately, we don’t need to make the massive object our bullet is orbiting into a black hole — a black hole with this mass would be less than a nanometer in diameter.  So, we can totally fit this in.

On the down side, it’s likely there would be some issues with the disposition of objects in the room.  Humans aren’t really set up to handle 1600 g’s of acceleration…