I think anyone reading this blog probably knows all of this but what the hell, I put several hours into this rant so I might as well put it in a convenient place. Who knows, maybe someone will find it useful to smash some ignoramus over the head.

**VISIBILITY**

First issue, visibility of destinations and obstacles. In the story, this carrier ship is supposed to be 8 kilometers long. If the fighter is traveling at lightspeed then one second away from the carrier, it's 1 light-second away. Well,

arctan((8 km) / (1 light-second)) in degrees = 0.00152894519 degrees

Do you know what that is? From http://en.wikipedia.org/wiki/Naked_eye

> Angular resolution: 1-2' (about 0.02°-0.03°), which corresponds to 30–60 cm at a 1 km distance

0.02 / 0.00152894519 = 13.

In other words, 1 second away from the carrier at lightspeed is still

**thirteen times too small**to see with the naked eye.

Even assuming that Talking Is A Free Action and the protagonist was able to point out and describe the carrier to his passenger instantaneously, the passenger would

**still have been unable to see it**.

In fact, since the human eye sees at about 18 frames per second, the passenger would

*just barely*be able to see a tiny speck of white if he stared at the exact spot the carrier was going to pop up at in advance.

So first he would see nothing, then immediately after he would see a speck, then immediately after that he would see the

*entire ship*.

**KINETIC ENERGY**

Going on, the author doesn't seem to have heard of this thing called "kinetic energy", whether the relativistic or the newtonian kind. Behold,

(((1 / 2) * ((0.9 * c)^2)) * 1 gram) / (1180 (kilojoules per gram)) = 30.8471057 tonnes

The top half of the left hand side of that equation is the kinetic energy of 1 gram of matter (say, space dust) travelling at a velocity of 90% the speed of light. The bottom half of the left hand side of the equation is the heat necessary to fuse lead starting from 0 degrees Kelvin (absolute zero). The right hand side is

**how much of your ship is going to melt away**from a collision with 1 gram of space dust. Assuming nothing heats up that doesn't melt and blah blah blah.

(((1 / 2) * ((0.9 * c)^2)) * 1 gram) / (4637 (kilojoules per gram)) = 7.8498134 tonnes

This is the same equation but including heating from 0 degrees kelvin to fusion, past fusion to boiling, past boiling to ionization. In other words, this is how much of your starship is going to turn into

**pure plasma**from a collision with 1 gram of space dust. It's straightforward to come up with similar equations for iron (steel) and uranium, but unnecessary.

The fact that air going at 90% of the speed of light has the

**oomph**of a thermonuclear explosion is really useful when designing

**real**starships because it means an air shield misted in front of your starship in its direction of travel can reduce asteroids down to free nuclei and anything smaller than that into subatomic particles.

Starship design hint: you can make your starship as long as you want without increasing its cross-section. So it can store

**any**finite amount of air you might possibly need on your interstellar and/or intergalactic trip to vaporize everything in its way. Once vaporized (well, ionized actually) it's easy to sweep the debris out of the way with your magshield. Particle accelerators do that every day.

And lest you think that I'm exaggerating when talking about thermonuclear explosions,

(((1 / 2) * ((0.9 * c)^2)) * 1 kilogram) / (4.2 petajoules) = 8.66656779

a 1 kilogram mass going at 90% lightspeed has a yield of a cool 8.67 megatons.

(((1 / 2) * ((0.99 * c)^2)) * 1 kilogram) / (4.2 petajoules) = 10.486547

at 99% lightspeed, it clocks 10 megatons.

It took until 1954 for the United States to have a nuclear device with a yield more powerful than that.

The most powerful thermonuclear weapon ever tested in human history was the Tsar Bomba, which clocks at the equivalent of 5 kilograms of mass at 99% lightspeed.

Are you starting to get the picture? Hitting anything at lightspeed results in a

**big boom**.

**REACTION TIME - NAIVE**

From http://pseudoastro.wordpress.com/2008/10/26/asteroid-belts-proximity-of-rocks-and-why-navigation-is-not-dangerous-sorry-han-solo/

> Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.

Using the formula for the volume of a sphere, and given that half of (half a meter) diameter gives a radius of 1/4 meter, we have

(4 / 3) * pi * ((1 / 4)^3) = 0.0654498469 cc's.

Assuming it has the density of water, that's 0.686342898 megatons or 686 kilotons. More than 30 times the explosion at Nagasaki. In other words, that's some dangerous fucking asteroids. And they're an average of

sqrt(500 (km^2)) = 7.45871992 × 10-5 light-seconds apart.

In other words, the hotshot pilot protagonist has

**one ten thousandth of a second**to dodge them.

The speed of nerve impulses in the human body is "up to 100 meters / second". In other words, the nerve impulses in the pilot's body have the time to travel a whole

((100 (meters per second)) / 7) * (10E-5 seconds) = 1.42857143 millimeters

yeah,

**one millimeter**in the time he has to dodge an asteroid. Brilliant, are you feeling the gritty realism yet? I know I am!

**DISTRIBUTION OF ASTEROIDS**

Now you may have noticed that the sqrt(500 km^2) neglects the vertical dimension. But 65 cc's is still an enormously dangerous asteroid, and if you go down to 1 cc then you still have to worry about explosions on the scale of 10 kilotonnes. For comparison, the Hiroshima explosion was 15 kiloton. Now, 65 cc's is conveniently close to 64 cc's

64* 1 cc = 2^6 * 1 cc

which means that asteroids of that size are

8^6 = 262 144 times as numerous,

given that it's a power law of degree 3 (every time you 1/2 the size, you increase numbers by 2^3 = 8). And

(8^6) * sqrt(500 (km^2)) = 5 861 718.04 kilometers

That's the height of the asteroid belt that's compensated for by the fact we're neglecting any asteroids less than 65 cc's.

(8^2) * (8^6) * sqrt(500 (km^2)) = 375 149 954 kilometers

That's the height of the asteroid belt that's compensated for by the fact we're neglecting any asteroids with yields of less than about 1 kiloton when travelling at 99% the speed of light.

And we can stop there because 375 million kilometers is almost the RADIUS of the asteroid belt. And of course, explosions of less than 1 kiloton aren't a danger at all to a tiny two-man fighter.

**CHANCE OF COLLISION**

So yeah, anyway. Now that we've established we can use the 500 km^2 number as a simplifying assumption, this is the proportion of the area swept by a 2 meter wide spaceship going through 500 km^2

(2 m) / sqrt(500 (km^2)) = 8.94427191 × 10^-5

which is the same as the chance of hitting that speck of rock somewhere inside it. And

*this*,

(1 million km) / sqrt(500 (km^2)) = 44 721.3595

is the number of such areas you travel through when going 1 million km (less than 1% of the radius of the asteroid belt) inside the asteroid belt. So logically, this

(1 - (8.94427191 × 10E-5))^44721 = 4.17433136 × 10^-18

is the probability that you will cross 44721 such areas without hitting anything. In

**three seconds**going at light-speed.

Do you understand what 10^-18 is? It's one tenth of one millionth of one billionth of one percent. It's one billion billion to 1 odds against. I think your chances of surviving a point blank gunshot to the forehead are higher. How do you like them apples?

**REACTION TIME - REVISITED**

Now that we have a probability of a collision, some playing around with google shows that,

(1 - (8.94427191 × (10^(-5))))^7 749 = 0.500012276

and of course,

7749 * sqrt(500 (km^2)) = 0.577976206 light seconds

So every half-second, there's a 50/50 chance that you'll have to dodge something. If you can convert this to a Mean Time Between Failures (or tell me how to) then you'll know the average reaction time you have between any two dodge-or-die events. Whatever that number actually is, it's going to be close to half a second. And of course, your reaction time had better be

**much**better than this if you want to live through a breezy five minute jaunt.

So yeah anyways. Dodging asteroids at lightspeed? You fail physics forever.

See also,

Atomic Rocket - http://www.projectrho.com/rocket/index.html -- includes technical info on everything up to and including time travel (more plausible than you might think, infinitely more plausible than FTL)

Bad Astronomy and Universe Today Forum - http://www.bautforum.com/ -- if there's any name in starship design I'd recommend it's Isaac Kuo - he does it with today's technology, not hypothetical future technology

Hint: starships don't travel through the solar system, let alone something as dirty as the asteroid belt. They avoid collisions because they're in

**space**.

This review was sponsored by Google TM search engine and Google TM calculator.

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