Outer space is big—really big—and almost entirely empty. One thing done in a lot of sci-fi movies is having the characters fly through an asteroid field, dodging asteroids like road hazards. This is silly. You would be lucky to find one asteroid in the Asteroid Belt flying straight through it, yet alone having to worry about dodging such things. Everything in our solar system is separated by distances much larger then anything we can normally work with.

In my story, ships are regularly required to travel to the outer edge of the solar system. Pluto's outer edge of it's orbit is about 50 AU, or 7.3 billion km, or 4.5 billion miles from the sun. So how long would it take to get there?

If you could get a space craft to travel 500,000 mph (not too crazy for space travel), it would take about 1 3/4 years to travel the distance. It you would like to do that in one month, you would have to travel about 10 million miles an hour.

Rocket engines are a source of force. Force in physics is usually measured in Newtons, which is kg·m/s².

Assuming the force exerted by a rocket engine is constant, we can determine it's acceleration (m/s²) on some given mass (kg). Our mass is the space ship, so knowing it's mass, we can determine acceleration.

What's nice about acceleration is that it keeps increasing velocity. The standard physics equation x = x0 + v0 t + 1/2 a t² will determine a position (x) after some time (t) with an initial position (x0) and initial velocity (v0). We we start of position 0 (x0 = 0) at a complete stop (v0=0), the equation simplifies to x = 1/2 a t². And we can solve for t, so t = sqrt( 2 * x / a ). We know x—that is the distance we want to travel. So what is the optimal acceleration? Acceleration due to gravity naturally. This way, we have linear acceleration and perfect gravity. So acceleration is 9.81 m/s². Thus, if you had the ability to provide a constant force with this rocket engine, it would take about 14 days to reach the outer orbit of Pluto from the Sun. That's not bad considering the distance.

But in working this out, I was faced with one obvious problem. The ship I designed is rotating. It already has gravity. The gravity introduced by linear acceleration will be in the wrong direction. If the rotation provides 1 g in one direction, and linear acceleration provides 1 g in an other direction, people are going to be standing sideways just to stay upright. This isn't going to work.

It was in the car ride back to Wisconsin talking this over with Mikala that a solution was presented. Mikala pointed out what should have been obvious: why not just have the spheres rotate? As the ship accelerates, turn the spheres such that the center of gravity at a 90 degree angle to the floor. In fact, one wouldn't even need to turn the spheres. The bottoms are heavier then the tops, because the tops are empty. If allowed to rotate, the spheres would naturally correct for the center of gravity.

So as the ship accelerates, the rotation is slowed. Since the ship is likely to be large, we will have to gradually accelerate to avoid jarring everything. A change in acceleration is known as jerk. So some constant jerk factor would have to be obtained—likely from that rate at which the rotation could be slowed to a stop.

A couple of weeks ago, I started contemplating an idea of a large transport space ship. On occasion I like to write a little sci-fi, but I'm usually far more interested in the technology then the story. I started to wonder about how a large transport vessel might be laid out.

Not much of a fiction reader, most of my exposure to sci-fi is from movies. And most movies don't try all that hard to create images of accurate space traveling vessels. Most somehow assume there is artificial gravity so they never have to address the issue. But how do you get gravity?

There are two methods that are rather easy: rotation and linear acceleration. Rotation works well if the object is stationary, such as with a space station. Linear acceleration can work if you have the ability to accelerate at 9.8 m/s²—the same acceleration as gravity. That's an acceleration of 79,000 mph²—meaning if you start for a dead stop, after 1 hour would be traveling 79,000 mph—so a fairly hefty acceleration. So I started with rotation.

Rotation provides centripetal force. Most of us as a kid have either tried or have seen someone spin a bucket of water over their head. The centripetal force from the rotation causes the water not to spill out even as the bucket is inverted. One can apply the same principals in space.

Again, most movies get a rotation as a means of artificial gravity wrong. Gravity is a function of the radius of the rotating path, and the speed of rotation. g = R( pi r / 30 )²/9.81 where g is gravity with respect to Earth (so 1.0 is Earth's gravity), R is the radius of rotation and r is rounds/minute of rotation. Two things should stand out about this function: a faster rotation means much higher gravity because the RPM is squared, and gravity depends on the radius. If you picture a circle, the one would be standing on the inside edge. One's feet would be touching the inner edge, and one's head would extend toward the center of the circle. The head, being closer to the center of the circle, would experience less gravity then their feet, because the radius has been reduced by their body length. This effect would be significant for small circles. For example, we could use a radius of 10 meters (so 20 meters, or about 66 feet in diameter). To get gravity at the circles edge would require a rotational speed of about 9.5 RPM. A person who is 5'8" (1.7 meters) tall would only have 82% gravity at the top of their head. Someone lifting 10 pounds would to their head would find it weighting only 8 pounds when level with their head. This may not be terrible. However, there is an other problem. Humans experience the Coriolis effect (i.e. knowing that we are spinning) at speeds above 2 RPM. So someone in such a station would get dizzy.

To get the speed down, it is necessary to increase the radius. One concept studies back in 1975 referred to as the "Stanford torus" was a torus (i.e. the doughnut shape) 900 meters in radius (about 1 mile in diameter). This would require about 1 RPM in order to hold gravity. The "O'Neill cylinder - Island Three" had an outer radius of 16 km (about 20 miles) and would rotate at one round ever 4 1/4 minutes.

After a lot of reading (mostly on Wikipedia), I starting thinking about how to create my own living area. The "Island Three" concept was really neat because it is an open cylinder. You could look up and see the other side—a city scape above your head but not falling on top of you. But this concept was designed to be stationary—orbiting Earth for example. They used large mirrors to reflect in sunlight, which was used for lighting and plant life. My vessel had to move, and the further away from the sun, the less light you get.

An other problem I had with this design was the possibility for catastrophic disasters. The entire atmosphere was contained in a single section. One hull breach, and you risk destroying the entire station. I wanted something more compartmentalized—something that worked in sections that were able to be isolated for other sections.

In other stories I've worked on, I thought up a concept using domes to grow crops. A center light source would shine on a reflective ceiling, and fields would be cultivated by a rotating gantry crane. I wondered if this concept couldn't be expanded to the space port. Rather then just use a dome, I tried using a sphere. It was easier to draw, but after some thought, made more sense. A dome is just a sphere cut in half. So the upper half of this sphere would be the dome, and the lower half could be support equipment. The flat disk in the middle is the plane for which people could live. This idea seemed to work pretty well in my head, so I started to come up with numbers—which is really where I have my fun!

My ring would be 30 km in radius (or 37 miles in diameter). At this size, it would rotate once every 5 3/4 minutes (0.17 RPM). Each dome would be 1 km in radius (1 1/4 miles in diameter), providing 776 acres of land (31 km²). The whole ring could fit 90 such domes for a total of almost 70,000 acres—109 sq. miles (2,800 km²). Because of the large size, even a building 100 meters tall (328 feet, or around 32 stories) will still have 99.67% gravity—so a 150 pound person would weight 149.5 pounds on the top floor.

So far, everything was going well for my space port. But then I remember this thing had to move...

   Yesterday I completed my basic bench layout, and for the most part, everything fits.  Part of the reason is that I used Google Sketchup to do a drawing for placement first.
   The sleeping box has been redesigned.  At first I considered cutting the box down to bench height.  Our new place is quite small.  I instead to use the top of the box as a bunk for our roommate James.  So this leaves the box at a height of 4 feet.  The biggest change is the door.  Previously, the box opened by hinges in the front, done mainly to keep from having to make additional cuts when the box was first assembled.  This was not a good design.  It leaked a lot of light, and was quite heavy.  The sliding door of my box at the Garage works much better.  Since Jai sleeps in the box with me, I decided to recycle the sliding door, but add a second one.  We both go to bed at different times, and this allows her to get into the box whenever she likes without having to crawl over me.

   Having owned and operated a domain for over 5 years I'm use to people trying to sell me junk because the idiots shysters at Network Solutions won't keep my e-mail address private unless I pay for that "privilege".  I've received e-mails in the past from people wanting to sell me a domain name similar to mine.  I ignore them by changing my e-mail address.  So when I received an e-mail the other day, I followed standard procedure.  To my surprise, I received a second e-mail hour latter to the updated e-mail address.  Seems the domain squatters have become more aggressive.  To humor myself, I clicked on the "unsubscribe" link.  Naturally, I received an e-mail today, this time from a different company.
   Let's have a look at the e-mails.  The first one is from 2/20/2010 9:19 AM
The next from 2/22/2010 7:13 AM
I received this e-mail at 2/23/2010 7:06 AM
Now I change my e-mail address.  The next e-mail is to the new address at 2/23/2010 6:01 PM
And the last e-mail from 2/26/2010 11:26 AM
    Some interesting things can be learned.  First, the use of various domain names to push their products.  They use initrustdomain.net, bestinitrustdomains.biz, flexmediadomains.com and easydomainrecovery.com.  Both initrustdomain.net and bestinitrustdomains.biz are owned by "Moniker Online Services LLC" who according to "SPAM Trackers" wiki have a long history of criminal activity.  The other two domains are "owned" by "DOMAINBULLIES, LLC" and "ENOM, INC."  There isn't much information about them.
   What gets me is that the Internet is a major source of commerce.  Isn't the FTC suppose to be doing something about this?  Why then do places like SPAM Tracker talk about groups like Moniker Online Services and how they have operated for years?  The more I deal with law and law enforcement, the more less I believe it functions.  It is both incapibal of protecting people, and incapibal keeping law enforcement from harassing the people it aims to protect.

    Everyone once and awhile I want to quick put a 3D drawing together, and I'll use Google Sketchup. Being the math nut that I am, I love it when I need to do a bit of geometry.
   Today, I happened to need to place an octagon (8=sides parallelogram) such that northern, and eastern edges were also the edges of a square, and the start of the north west corner (X), and south east corner (Z) of the octagon began at the these corners on the square. When you draw a polygon in sketchup, you need to have a center point (I) and an end point at one of the angles (such as A). So if I wanted an 8-sided polygon (octagon) to calculate where the center points, and some end point. If you happen to have Sketchup, try and see if you can eyeball this--you'll find it isn't easy. A geometric solution is needed. Luckily, it's easy.
  Every angle in an octagon is 45 degrees, which is 360 / 8. To find the center of an octagon, pick any edge, and measure 67.5 degrees (45 / 1.5) toward the center. An example in this drawing is line AB or BC. Where the angles meet is the center, point I. Note also that when drawing this polygon, it will be done from point I to one of the point A through H.
   In my requirement description, I want this octagon (call it octagon I) to meet up with square WXYZ such that vertex A meets vertex X, and D meets Z. Sides AB to XY, and DE to ZY and parallel, and vertex A and the same as X, D the same as Z. Now that we know the 2 verities, and two edges, we can find the center. Measure an angle starting at X along XY that measure 67.5 degrees clockwise. And 67.5 counter clockwise starting Z lone ZY. This is the center point.
   Draw your octagon starting at this location I, going to either vertex X or Z. That's it! :)

So my last article on prime numbers led me to ask the question "How many prime numbers are there between 2 and 2³²". I know the quick answer to this question is "a lot". 2³² is over four billion—and that's a fair bit of space to check. It would be easy to make a for-loop and brute force the entire number range, but this is the age of multi-core CPUs—and single threaded is so 2000. Besides, my computer hasn't broken a sweat in awhile.

I decided to do a straight C implementation. I've been doing C and C++ at work, and C++ has been irritating me lately because I tried to do a "quick and dirty" implementation without too much thought into the classes I'd need. That's a recipe for disaster, and how you end up with things like microsoft's windows ME. So vanilla C it was.

Threads and semaphores I used the POSIX standards, and I've tried to demonstrate how to make a dispatcher foreground task with one or more worker threads. The key to this is the dispatch semaphore. Normally, semaphores are used to for mutually exclusive lock, or to send signals between threads. However, they can also be used as a count. Think of them like a library that has a set number (say 5) of books. That means 5 people can check that book out, but if a 6th person wants to check that book out, they will have to wait for one of the first 5 to return their copy. The semaphore does this by keeping a count. When that count reaches zero, the next task that tries to pend the semaphore will have to wait. So the dispatcher is a loop that starts by pending the semaphore. When it gets the semaphore, it will create a task. The task will do some work, then post the semaphore. The semaphore count determines how many tasks can run at the same time. If we limit the number of tasks that can run at once to the number of CPU cores, then we can utilize the CPU to 100%.

First, the dispatcher skeleton code:

Then, the skeleton of the worker thread:

Now just launching a thread to check one number is a bit of a waste—it takes some overhead to start and stop the task. So we give each thread some range of numbers to check. A pthread can be passed parameters, and we pass the starting number, along with the amount of numbers to check. The parameter block is also used for storage of the results. So here is the full worker thread:

The dispatcher has to accumulate the results. Our could try using a global variable, but you can run into issues with non-atomic access. Adding 1 might seem a basic operation, but (at least in a RISC system) it takes 3 instructions to add one to a variable in memory: load the data to a register, add the register by 1, and store the register. You can not (without a semaphore) guarantee that those instructions will not be interrupted by an other thread. So, the accumulation happens in just one place—the dispatch loop.

The dispatcher will assign all the work, but it is finished once all the work has been assigned. We still need to wait for the work to finish. So the last part of the process looks like this:

Now we can print the results to the screen, and we're done.

I knew this code would tax the CPU rather hard, so I wanted to measure how long the process would take. "time.h" is a C standard unit. Unfortunately, "time_spec", which has a high-resolution method for checking the time, it's part of the C99 standard. But we can get time in the resolution of seconds, and that's good enough. To do this, we simply mark the time when the process begins, and mark it again at the end. The difference (for which there is a function to compute) is the total number of seconds that has elapsed.

The answer: There are 203,280,221 primes between 2 and 4,294,967,295, and it took 6,054 seconds (1 hour, 41 minutes) on a quad-core clocked at 2.5 GHz.

The full source code is here.

So I mentioned yesterday that I wrote a simple program to check to see if some numbers in a list were prime or not. I didn't post it because I wanted to spend a little more time with it—giving it an article on it's own.

Finding prime numbers goes back a long time. The easiest way to see if a number is prime is to simply check and see if the number is divisible by all the prime numbers up to the square root of the number in question.

A number (call it x) is prime if there are no two number (call them a and b) such that a * b = x. All non-prime numbers can be expressed as the product of two or more prime numbers. For example, 125 can be made from 5 * 25, but 25 can be made from 5 * 5. So 5 * 5 * 5 = 125, and represents the most factored version of 125. This is true of any number. Since it takes at least two prime numbers to create a factor, we only need to check the primes up to √x (or the square root) of the number. This is because the if x = a * a, then √x = a. If x = a * b, and b is greater a, then a must be less then √x. Thus, we only need to check primes up to √x.

This makes for a nice brute-force test method, and it's pretty simple to implement. If we're going to test a number, we need all the primes up to the square root of that number. For what I needed, the number was capped at 32-bits—or 2³². √2³² = (2³²)^1/2 = 2¹⁶ = 65536. So I need a list of all the prime numbers up to 2¹⁶. We can make the list fairly quick. Have a look at this function:

We have an array to hold our prime numbers. We set the first entry in the array to two—the first prime number (1 is prime, but who cares). From this, we can test to check the rest of the numbers. We loop from 3 to 2¹⁶. For each number, we see if any of the primes already in the list will divide evenly into the number under question. If any of them do, they are not prime.

It happens that there are 6542 prime numbers between 2 and 2¹⁶. Knowing that, we can cap the array size of our lookup table at this value—the number of primes in this range will never change.

We now have enough prime numbers to check any number up to 2³². Unlike the function that generates the prime lookup table, we can throw in one more speed improvement: we only need to check up to the square root of the number in question. So here are the guts of the prime test function:

Note that can add two short cuts. First, checking up to the root of the number like we discussed. And second, we test to see if the number is even—2 is the only even number that is prime.

That's it. Now we have a simple to implement prime number test that works on numbers up to 2³². Here is the full source code to a command line version of the test. Compile it, pass it a number (or several), and the program will tell you if the number is prime or not.

 

(thanks Erica for the correction)