## 2/28/10

### A Perfect Time For Budding Aerospace Engineers

## 2/27/10

### I'm Waiting on Pins and Needles!

I still don't know where I'm going to college next year. The reason is that I haven't heard back from MIT yet. It's extremely likely they'll say "no, we don't want you." And they would have good reason. My ACT was only 29 and their interquartile range is from 31 to 35 or something rediculous like that. Only 6% of their accepted applicants from last year had an ACT of 26-29, which isn't totally forbiding, especially since I'm on the upper end of the 6% and I have a great GPA to back me up, but... good grief, I can't wait until the middle of March to decide where I'm going to go next year!

My next two top contenders are Embry-Riddle Aeronautical University in Florida and Iowa State University. I would be totally fine with going to these schools, I just can't wait to *know* where I'm going.

### A Slick Paint Job

We got to use the resources of the tech-ed department to paint the rocket. We only spent about three hours total on it, which wasn't enough priming and sanding to remove the spiral from the phenolic airframe, but the end result was still pretty good. We used black and gold, the school colors. Its black with a double gold spiral from nose to aft. Silly me, though. I seem to have left my camera at school! Come back on Monday and I'll have some great pictures.

Here's the solution to #4 on my pop quiz:

Sound intensity I = P/(4*pi*r^2), P is power and r is the sound distance from the source. In this case, r = d.

Apparently the actual sound intensity didn't matter because they gave us I and 1/3I. So I plugged some random number like 1 in for I. I_{1} = 1 W/m^2 and I_{2} = .333 W/m^2. Rearanging the intensity equation for power is P = 4*pi*d^2*I. d_{1} = 562 meters. P = 4*pi*562^2*1 = 3.97x10^6 W. I then solved the intensity equation for d so I could find d_{2}. d = (P/(4*pi*I))^(1/2). Plugging 3.97x10^6 in for P and .333 in for I you get 973.4 meters. Once you have this, you can use the kinematic equation d = 1/2at^2 to find t_{1} and t_{2}. Solving for t you get t = (2d/a)^(1/2). t_{1} = 5.79 seconds and t_{2} = 4.40 seconds. The change in time from the first sound emmission to the second is t = t_{1} - t_{2} = 1.39 seconds. But this isn't the final answer, because the question asks for the time between each of the measurements below at the ground-based monitoring station. Sound takes time to travel through the air, so you have to find the amount of time each sound wave took to go from the altitude they were emitted at and the station. t = d/v, so t_{3} = 562m/343m/s = 1.64 seconds, t_{4} = 973m/343m/s = 2.84 seconds. So to find the time between the two measurements you just have to sit there and think about it for a while, and finally you come up with t = 2.84 sec + 1.39 sec - 1.64 sec = **2.59 seconds**

I thought that was a fun problem.

## 2/25/10

### How'd We Miss That?

I recently looked at the *official* rules for the TARC contest, and it turns out we are supposed to be using a streamer for recovery. That changes things a bit. *Quite* a bit actually. At first it didn't seem like our design could slow down enought under a streamer, but it turns out there's more to streamer recovery than meets the eye. There is an entire science behind it. For instance, most people seem to think that you want a really really really long streamer. Not so. For the maximum drag coefficient you want a length to width ratio of 10:1. According to the simulations, a 9 by 90 inch streamer will bring our rocket down in 50 seconds, which is more than enough time.

I'm just glad I caught this as early as I did. We still have at least a month before the deadline.

Today we did some painting in the Tech Ed department. My ears are still ringing from all the noise that goes on in there. Its going to look pretty good, but I entrusted too much of the painting to the freshmen and we got some nasty globs and runs. Its actually not too bad, I just like to give them a hard time.

## 2/23/10

### Tidbit of the Week

If nothing can travel faster than the speed of light, what if you had a spaceship traveling left at 99.99999% of the speed of light and had a second spaceship traveling right at 99.99999% of the speed of light? Relative to each other, wouldn't they be passing each other at 199.99998% of the speed of light?

A few years ago I didn't quite get this, but as I learn more about relativity, the clearer it becomes. In actuality, the spaceships would be passing each other at 99.99999999999999999% of the speed of light. Time actually changes to make sure that they never exceed the speed of light. The speed of light is the most constant thing in the universe, and even time conforms around it.

v = d/t

So if v is constant, time has to travel faster for the pilots of the spaceships than for an outside observer.

## 2/22/10

### Problems

Since school is almost all I have time for these days, I thought I'd bring it to my blog now and then.

In calculus we are doing sequences, L'Hopital's rule, and improper integrals. Pretty straight forward.

Pop quiz:

1. Is 0^0 zero or one? Or neither?

2. Limit as x approaches 1 of x^[1/(x-1)] = ?

3. The third term of an arithmetic sequence is 5 and the seventh is 19. What is the first term?

4. This is a fun problem. Brownie-points to anyone who gets it right:

A rocket, starting from rest, travels straight up with an acceleration of 58.0 m/s^2. When the rocket is at a height of 562 m, it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity *I*. Later, the station measures an intensity 1/3*I*. Assuming that the speed of sound is 343 m/s, find the time that has elapsed between the two measurements.

## 2/16/10

### I'm Back on the Web!

A friend of mine was able to fix my computer, and apparently it wasn't a hard drive failure. All my files are still here, but I had to reinstall all my programs, which is only a minor inconvenience.

We're going for our second launch tomorrow. Supposedly the weather is going to be superb.

## 2/13/10

### Hey, I Know How High It Went!

I figured there's got to be a way to calculate the altitude effectively. I thought it might have something to do with the descent, since the factors are more controlled than on the ascent (variable thrust hence variable velocity hence variable drag hence calculus). The answer was terminal velocity. Solve the drag equation for velocity, and call the drag mass x gravity since the rocket is in equilibrium. The result is the following equation:

v = [(2mg)/(pACd)]^(1/2) (I hate typing math. It always looks like someone just puked jibberish)

where v = velocity, m = mass of the rocket (0.550 kg), g = gravitational acceleration (9.8 m/s/s), p = the atmospheric density (1.2 kg/m^3 at the time), A = the area of the parachute (0.164 m^2), and Cd = the drag coefficient for our parachute (0.75 for a flat sheet). Plugging it all in yields v = 8.54 m/s.

I then went back to the video and timed the descent, which turned out to be 21 seconds. Using d = vt I was able to figure the rocket fell from 179 m, or 574 ft.

It is important to realize that this wasn't the maximum altitude the rocket reached, it was simply the point where the parachute popped out. The rocket may have fallen from a hundred or so feet beyond that.

If anyone finds a problem with any of my thinking, please let me know.

### First Test Launch!

We just conducted our first two test launches of our TARC entry. We have been really pushing ourselves to get ahead of schedule so that we can fit in as many test launches as possible before the official ones this spring. The snow on the ground complicates things just a little, but it actually saved our rocket. The first launch went well on the ascent (the easy part), but when the 24-inch plastic parachute came out it ripped right off and drifted away. The rocket came tumbling down in fifteen seconds and buried itself in the snow. Had there been no snow, the collision would have resulted in some serious damage.

So we ran back to the physics room and grabbed our spare chute. This was a heavy-duty ripstop nylon one, whose shroud lines were stitched right into the canopy. It wasn't about to fail anytime soon, but it was considerably smaller, hence our flight time was still only thirty seconds--ten short of the forty-second parameter. However, the flight was absolutely spectacular.

There isn't any altitude data yet since we are still waiting on our altimeter. Some people tried measuring an angle, but a hundred feet isn't much of a baseline...

Suprisingly our egg survived! After its harrowing first flight we took it out to examine it. Just a few hairline fractures in the shell, but it was still alive! We placed it back in its well-padded cockpit for the next flight, which went beautifully. Our egg survived a flight with fractures in its shell!

Here it is!!