Monday, September 26, 2011

Whats up with those wings?

Hey blog, first off I would like to appologize for the delay in posting a blog. Now to the meat of the blog. So two weekends ago I flew to Cincinati with my girlfriend and on the flight up had a question pop into my head. I was flying on one of the smaller regional jets an Embraer 170 and I was next to the wing. What I noticeced on this plane and I have noticed on other smaller planes before it the 90 degree turn of the wing right at the end.

My question for all of you is why do we find this turn up in the wings? Is its just to look cool or does it actually serve a purpose?

Sunday, September 25, 2011

Mapping g on the moon (and earth!)

http://www.npr.org/2011/09/10/140361610/nasa-launches-probes-to-study-moon describes a recently launched unmanned NASA mission to the moon to map out the moon's gravitational field. What is the gravitational field? It's quite simply the acceleration due to gravity ("g" as we've been calling it in class) as a function of position around the moon!

On the surface of the earth, g is a pretty consistent 9.8 m/s^2, but it does vary depending on your position on the planet, since Earth is not a perfect sphere. And once you start to get out into space, g begins to diminish drastically, dropping off like 1/r^2, where r is the distance between you and the center of the earth!

The moon's acceleration due to gravity behaves much the same way. On the surface of the moon (at least the parts we've been to!), it's about 1/6 of our g on earth (so about 1.7 m/s^2, give or take), and also drops off like 1/r^2 (where r is the distance between you and the center of the moon) as you leave the surface.

These probes will measure these variations in the moon's g as they orbit on opposite sides of the moon! By the way, the GRACE mission (http://www.csr.utexas.edu/grace/) did the same thing on earth! Here's a map of the results, depicting the difference between the local g and the average g: http://www.csr.utexas.edu/grace/gallery/gravity/03_07_GRACE.html, where the red regions represent a higher value of g and the blue regions represent a lower value of g (measured in units of "milligals," which are named after Galileo; 1 gal = 1 cm/s^2).

Throwing a Baseball

So, we all understand the concept behind throwing a baseball. Palm the ball with a firm, but not too hard grip, then throw it with as much force possible at the catcher. In proffesional games, and even on the collegiate and little league levels, pitchers have become so efficient at their sport that they have developed different pitching techniques to alter the trajectory of the baseball.


At the base of these various techniques is the curveball. By altering one's grip, the baseball leaves the pitchers hand spinning in the direction of the pitchers choosing. The different grips the pitcher utalizes all offer different types of spins that the ball will move during flight. These various pitches have been named over the years, often according to the type of trajectory the ball will follow.

This link offers a few examples of the various types of pitches.
http://www.thecompletepitcher.com/different_baseball_pitches.htm


So, why does putting a spin on the baseball change the path it will take from the pitcher to the catcher? Interestingly enough, it manipulates the same laws of physics that an airplane does to gain flight. The airplanes wings are created so that the air on top of the wing has to move a greater distance in the same amount of time as the air on the bottom of the wing. This creates lower air pressure on the top of the wing, giving the plane it's lift.

Now let's inspect the trajectory of a curveball. When it leaves the pitchers hands, it is given a spin. Imagine that the ball is traveling from the right to the left. If you don't understand, then perhaps this awesome little text/image/thing will help.


Catcher <-----Ball----- Pitcher

Because of the spin given by the pitcher, the ball travels with a clockwise rotation the whole time it is in flight.


-------->

.........

.. ..
.. Poorly ..

.. Drawn ..

.. Baseball ..

.. ..

.........

<--------


This rotation, in combination the trajectory of the ball from the pitcher to the catcher, makes it so that the air on the bottom of the ball has to move faster than the air on the top. This means the ball will curve downwards as it travels.


This curve can be altered based on the positioning of the pitcher's grip.


Can you think of any other objects who's path through the air are altered because of its spin or rotation?

Wednesday, September 21, 2011

Results of the 21Sept Brain Dump!

Here are your responses from today's question: "What's the most important thing you've learned thus far in this course?" Enjoy!

Monday, September 12, 2011

Preflight #6 responses

I’ve finished reading your great questions on Preflight #6. I apologize it took me so long to process them, but I think you’ll find the discussion worth the wait! Below are anonymous responses from students, each of which is followed by my commentary.
“I keep getting confused on how the powers like feet/second squared eliminates”
That’s a great topic! The goal is to keep track of what is in the numerator and what is in the denominator. For example, when we calculate velocity as distance / time, we have meters in the numerator and seconds in the denominator. Thus, the units come out to m/s. If we take a velocity and multiply it by a time, we’ll have units of (m/s) * s. The first “s” is in the denominator. The second “s” is in the numerator. If you were rearrange it, you’d have (m*s) / s. Rearrange it a little more, and you’d have m * (s/s), and s/s = 1, so you’re left with meters.
Another way of thinking of it is in terms of negative exponents. Remember that dividing by a quantity is the same thing as multiplying by that quantity raised to the negative one power. So, our first example of calculating velocity as distance / time is equivalent to distance * (time-1). Thinking about the units, we have m * (s-1), which is the same thing as m/s. Our second example of calculating distance as velocity * time then has units of m * (s-1) * s. You can think of the “s” as s1, so that we have m * s-1 + 1 = m * s0 = m * 1 = m.
The negative exponent method also comes in handy when you think about the units for acceleration. Acceleration is defined as change in velocity divided by the change in time, which would have units of (m * s-1) / s, which, again, you can think of as (m * s-1) * (s-1). The units thus change into m * s-1 + -1 = m * s-2 which is the same thing as m / s2.
Ultimately, this is a skill that requires practice, so keep track of your units on every calculation you make (no matter how mundane), and you’ll have it down in no time!
“I only have a little bit of issues when I use Cos, Sin, and Tan to solve for problems.”
I think the most important thing to remember is the flow of information in the trig functions. The trig functions (sin, cos, tan) take in an angle and produce a number. That number is the ratio associated with that trig function (opp/hyp, adj/hyp, and opp/adj, respectively). Just remember that “sin(whatever)”, “cos(whatever)” and “tan(whatever)” are numbers, pure and simple. They’re just wearing a costume until you plug it into your calculator.
The inverse trig functions, on the other hand, take in a number and produce an angle. That number, again, is the ratio associated with that trig function.
“The only thing that I have trouble when it comes to any math is knowing when to use a formula. We learn a lot of formulas, and I can do them pretty good when were on the topic, but when we go back to it a while after i have trouble picking out which formula the question is asking for.”
Choosing an equation is a very important skill! I think that if you practice good bookkeeping at the beginning of the problem (writing down what you know and what you’re looking for), it can help the selection process.
I do notice you highlight that you have trouble remembering after “a while.” Might I suggest trying out problems immediately (or at least within a few hours) after class?
“It is taking multiple attempts for me to get [the problems] correct.”
That’s actually a very typical experience! Succeeding at physics requires practice—lots of it—which is why I think the Practice Flights are so valuable. Be sure to leave yourself enough time to try them again and again until you get the correct answer! (You might even need to plan to spend a few different sessions trying out the problems. For example, half an hour on Friday, Saturday, and Sunday each would be more valuable than 1.5 hours only on Sunday.)

Sunday, September 11, 2011

Physics in Roller Coasters

So we all know what a roller coaster is and, unless you have been living under a rock your whole life, you have probably had the pleasure of riding one. In fact, roller coasters have become so common in the entertainment industry that we rarely consider the forces being manipulated "in the service of a great ride".
roller coaster
The entire concept behind a roller coaster is that it utilizes gravity to make massive amounts of potential energy before releasing them in the form of kinetic energy. In order to create the first and greatest reservoir of potential energy, the cars make their way up the first hill. This ascent is usually initialized with the chain lift, which acts as a conveyor belt to bring the coaster cars to the peak of the hill. At the top, the car's potential energy is as high as it will be for the rest of the ride.
After the chain lift disengages, gravity takes over and acts as the source of energy acting upon the coaster. As the cars descend, their potential energy is lost and in return they gain kinetic energy. Upon reaching the bottom of the hill, almost all of the potential energy is gone. However, on it's way over the next hill, the coaster loses some of its kinetic energy and gains some of its previous potential energy. This continues till it reaches the crest of the second hill
at-which-point the process begins again.

This link has an interactive image/video thing that can help to better explain kinetic and potential energy: http://science.howstuffworks.com/engineering/structural/roller-coaster3.htm

Because an object in motion tends to stay in motion (Newton's first law of motion), the coaster will maintain a forward velocity even as it climbs up the hills, working against gravity. This makes it possible for there to be any variation in the slope of the track, such as the swerves, hills, and loops.

Here is a cool picture that displays forces that I'm not going to talk about.
Due to friction between the train and the track, as well as the train and the air, the coaster continually loses kinetic and potential energy. For this reason, the hills decrease in size as the cars make their way through the ride. Eventually, the ride slows down to the point that it is unable to move forward any longer. By now the tracks have lead back to the beginning of the ride and the brakes are activated.

Sources:
  • science.howstuffworks.com, "How Roller Coasters Work", Tom Harris
  • Common sense and experience


Saturday, September 10, 2011

Mapping g on the moon

http://www.npr.org/2011/09/10/140361610/nasa-launches-probes-to-study-moon describes a recently launched unmanned NASA mission to the moon to map out the moon's gravitational field. What is the gravitational field? It's quite simply the acceleration due to gravity ("g" as we've been calling it in class) as a function of position around the moon!

On the surface of the earth, g is a pretty consistent 9.8 m/s^2, but it does vary depending on your position on the planet, since Earth is not a perfect sphere. And once you start to get out into space, g begins to diminish drastically, dropping off like 1/r^2, where r is the distance between you and the center of the earth!

The moon's acceleration due to gravity behaves much the same way. On the surface of the moon (at least the parts we've been to!), it's about 1/6 of our g on earth (so about 1.7 m/s^2, give or take), and also drops off like 1/r^2 (where r is the distance between you and the center of the moon) as you leave the surface.

These probes will measure these variations in the moon's g as they orbit on opposite sides of the moon! By the way, the GRACE mission (http://www.csr.utexas.edu/grace/) did the same thing on earth! Here's a map of the results, depicting the difference between the local g and the average g: http://www.csr.utexas.edu/grace/gallery/gravity/03_07_GRACE.html, where the red regions represent a higher value of g and the blue regions represent a lower value of g (measured in units of "milligals," which are named after Galileo; 1 gal = 1 cm/s^2).

Wednesday, September 7, 2011

The Spruce Goose?

In 1942 the United States Government contacted Henry Kaiser a shipbuilder, to build a new airplane to carry troops across the Atlantic Ocean. When faced with this challenge Kaiser contacted famous plane designer Howard Hughes to help build this creation. Throughout the project the government kept their support but poked fun at his project. At one point the Senate even called this plane the "flying lumberyard" showing how immense this project was.

This plane was designed to be 400,000 lbs with a 320 foot wing span. THAT IS AS BIG AS A FOOT FIELD!!!! So you might wonder how this plane even would take off. It used eight P&W 4360 which are the same engines that are on the B-50's. However slapping eight huge engines on a pile of wood doesn't do much.
However, why did they call it the "Spruce Goose"? Most people who do not know about the plane think that it was made out of spruce. One of the biggest surprises that people might find out is that most of the plane is made out of birch not spruce. One reason the builders could have decided to use birch is because it is more dense and adds more structural integrity to the plane. Compared to spruce which is a little but of a softer wood.
My question for everybody is with all of the modern technology what is one thing that the designers of this plane could have done differently to make it fly more effectively?

Friday, September 2, 2011

Week 2 In-Class Problem - Monorail!

(Here's why I add the exclamation mark:

)

Below are pictures of your solutions to the monorail problem(s) from today!






What questions do you have for each other about your solutions? Fellow blog authors: What questions do you have for the aviation students about their solutions?