Friday May 10

At this point, I feel like the only thing I have left to do is to fully relate my topic to the activity of skydiving. I’ve talked about it a little bit in terms of terminal velocity, but I haven’t fully explained all the real life applications of the various part of the drag equation for skydivers.

If we go back to that equation, we’ll see that there are four variables: the object’s velocity, coefficient of drag, and reference area, and the fluid’s density. When increased, all four of these things increase the drag force as well. Similarly, a decrease in any of those will decrease the drag force.

This is extremely important for a skydiver, whose terminal velocity is entirely dependent on the drag force exerted on him. The greater the drag force, the lower the terminal velocity. The smaller the drag force, the faster the diver can go before topping out.

Because all four of those variables are part of the larger drag force, this means that skydivers can affect the drag force in many ways, many of them equal. If two equally dressed skydivers found a way to double the reference area for one person and to double the drag coefficient for the other, they would experience equal drag force at equal conditions.

Skydivers can affect these variables in many ways. Their velocity and the density of their surrounding air are largely out of their hands, but they can change their coefficient of drag by changing the material they wear. A baggy t-shirt would have a higher drag coefficient than a scientifically designed skydiving suit (hence why they exist..). Similarly, skydivers can change their body position to change their reference area. When divers are chest-down with their limbs extended, they will experience a much greater drag force than when the are faced head-down with arms at their sides because of this change in reference area. If you’re skydiving, this means that you will experience less drag force and be able to reach a higher terminal velocity in a head down position than with a ‘flat’ or sprawled out position.

This is also why the parachute works to slow down the diver once they have reach the point at which they need to deploy it. When the parachute is deployed, the reference area (and possible the drag coefficient as well) skyrocket. This results in a large increase in the drag force, so that the drag force is actually greater than the skydiver’s weight. This results in deceleration as long as there is an imbalance of forces. This deceleration (obviously) causes the diver to slow down, until he or she has reached a new terminal velocity where the drag force once again equals the skydiver’s weight. At this point, the skydiver simply drifts down towards the earth with a much more reasonable/slow velocity, acceptable and safe for landing!

Seeing as how this is my last blog post, I thought I would revisit my learning objects and give a progress indicator and short explanation for each:

1. Understand the relationships between air resistance, terminal velocity,  skydiving, kinematics, and Newton’s laws.

100%. I have a really good understanding of how these things fit together.

2. Learn to calculate air resistance based on the formula D = (1/2)*p*u^2*Cd*A where D is the drag force (like air friction), p is the density of the fluid, u is the object’s velocity, Cd is the object’s drag coefficient, and A is the object’s cross-sectional area.

100%. It’s pretty much just plugging numbers into the equation; simple.

3. Solve physics problems involving air resistance as it pertains to kinematics and to Newton’s laws.

100%. I solved a problem in my last blog post, and it went well.

4. Understand how the positioning of skydivers affects their velocity, drag force, etc.

100%. Position doesn’t directly affect velocity, though – it affects drag, which then affects velocity.

5. Reach a solid conceptual understanding of the physics behind skydiving, and of why skydivers feel so safe jumping out of planes.

90%. I understand now that parachutes increase reference area so much that a skydiver is just as physically inclined to slow down as he was to speed up due to gravity. Physics don’t fail. I still don’t know if I trust the cords on the parachute, though…

Thursday May 9

I have to first clear up a couple things about my topic:

1. In the drag equation, v is not the absolute velocity of the object, or the object’s velocity compared to earth, but rather the velocity of the object relative to the fluid in which it is immersed.

2. I had a brief moment where I wondered (incorrectly) why two objects or two skydivers with equal area but different forms would have the same drag force, even though I didn’t think they should. For example, why would an object traveling to right like so:

))))))))))

have the same drag force as the same object traveling to the right like so:

(((((((((

I thought that surely the second way would be much less aerodynamic, but I couldn’t figure out why in terms of the drag equation. But then I realized that all shapes and objects have different drag coefficients, which, when plugged into the equation, will affect the drag force accordingly. So yes, the first object would experience a much lower drag force, because its drag coefficient would be much lower.

Now that I think I’ve gotten all of the conceptual information down, I wanted to try some problems. Here’s one:

1. Determine the drag coefficient of a 75 kg skydiver with a projected area of 0.33 m² and a terminal velocity of 60 m/s.
2. By how much would the skydiver need to reduce his project area so as to double his terminal velocity? 

a)

D=(1/2)*Cd*p*u^2*A

(2D)/(p*u^2*A)=Cd

p= density of air,  1.225 kg/m³

u = skydiver’s velocity, 60 m/s

A = reference area, 0.33 m²

D = ??

Cd = ??

We want to find Cd, the drag coefficient, but to do this, we first need to figure out what the drag force is. We can figure this out with Newton’s second law, because the net force on the skydiver is 0 (he has reached terminal velocity, or zero acceleration).

ΣF = W-D = ma = 0 m/s^2

W = D = mg or (9.81 x 75)

so…

(2mg)/(p*u^2*A)=Cd

Plug in numbers, and we get that Cd = 1.011, which is pretty standard for a skydiver (whose drag coefficient will typically be somewhere in between 1.0 and 1.4).

b)

Using the equation we derived for terminal velocity on Tuesday, we know that terminal velocity equals:

V = ((2D)/(p*Cd*A))^(1/2)

If we now set V equal to 2V, and leave everything as a constant, we can solve for q, a random coefficient we will put in front of A, which will show us how the reference area must be affected to double the skydiver’s terminal velocity.

2V = ((2D)/(p*Cd*qA))^(1/2)

4*V^2 = (2D)/(p*Cd*qA)

4(V^2*p*Cd)/(2D) = 1/(qA)

We can now see that whatever happens to A must result in the left side of the equation being multiplied by four. Because qA is in the denominator, it is inversely proportional to the other side, so if we make q equal 1/4 or .25, it will result in the other side being divided by 1/4 as well, or multiplied by four. This mean that to double the skydiver’s velocity, his reference area must be quartered (multiplied by 1/4 or .25).

Sorry if that math was a little hard to follow; WordPress isn’t exactly the best way to show equations. Come Tuesday’s presentations, I’ll have a much neater approach prepared. I think I am now nearly complete with my learning objectives. I understand the concepts well and I am getting better at solving problems now. Tomorrow, I’ll take a deeper look at the different positions skydivers use, when they use them, and why they use them. I think that should tie everything together well.

Sources:

http://physics.info/drag/

Tuesday May 7th

After understanding the drag equation, I wanted to figure out the history behind drag forces and how they were discovered. The equation was conceived by Lord Rayleigh, and has uses in many fields across the world. Car designers, scientists, top professors and astronauts alike all rely on the drag equation for applicable circumstances. I realized that so far, I have failed to define exactly what the drag force is. So: the drag force is a resistive force on an object by a fluid in the direction opposite to the object’s movement. A skydiver, for example, will feel the drag force in the opposite direction of his downward velocity (upwards). Some of the time this is hard to observe, but with lighter objects, we see it every day. Something light like a whiffle ball or a beach ball will slow down quickly due to drag forces. The reason lighter objects are more affected is because the drag equation does not take mass into account. Two equally shaped objects with the same drag coefficient, velocity, and reference area but with different masses will each experience the exact same drag force. Because F = ma, or a = F/m, a lighter object will have a much greater acceleration (or rather, deceleration) with the same force. This is why a beach ball doesn’t get very far, but a baseball does.

With that said, now I can go back to where and why the drag equation is used. Car designers use it to design the most aerodynamic cars possible – a drop in the drag coefficient of a vehicle by just 0.01 can boost the fuel economy of the car by up to 0.2 MPG. Certain cars, like the Toyota Prius, take advantage of this and produce much better mileage than cars with larger drag coeffcients (truck, SUVs, Hummers, etc). Someone at NASA might use the equation to determine the forces on a rocket as it leaves the earth’s atmosphere into space, or a physicist might use the equation in an experiment to discover the most aerodynamic surfaces and shapes. The equation has many, many uses.

Now I’d liketo return to skydiving, and how it is impacted by drag forces. When we think of an object or person in free-fall, what we might think of is someoneor something falling and accerelating for a long, long time. In a vacuum with infinite depth, this would be true, but in real life, it is not. Because we live on earth, an object in free-fall will eventually crash into the earth and stop moving due to the normal force exerted by the earth onto the object. And because we don’t live in a vacuum, we have air surrounding us at all times. This presence of air (a fluid) causes a drag force on any object moving through it. Because of this, objects free-falling in the earth’s atmosphere will not accelerate and accelerate until they hit the ground. Instead, they will reach a terminal velocity, where their acceleration is zero because the net force acting on them is zero as well.

But the skydiver was accelerating when he hopped out of the plane! What happened? Well, even though the drag force is independent of ths skydiver’s mass, it is entirely dependant on the skydiver’s velocity. As the diver picks up more and more speed, the drag force rises and rises. Eventually, the skydiver has enough velocity so that the drag force is equal to the force of gravity. This means that the sum of the forces acting on the diver is zero, and that the skydiver’s acceleration is zero as well. When an object or skydiver has reached this point, it has reached its terminal velocity. See this picture for further exaplanation:

Now you might ask: how do you calculate the terminal velocity of an object? It’s pretty simple really, using Newton’s 2nd law and the drag equation:

Anyways, that’s all I’m going to look at for today. I’ve really cemented my understanding of drag forces, the drag equation, and terminal velocity, and I think I’m going to start looking into solving some simple problems tomorrow. I’m getting close to completing the conceptual aspects of my learning objectives, but I’m not quite there with the quantitative stuff yet.

Sources:

http://auto.howstuffworks.com/fuel-efficiency/fuel-economy/aerodynamics2.htm

http://cnx.org/content/m42080/latest/?collection=col11406/latest

Monday May 6th

The first thing I decided to research was simply the definition of cross-sectional area. There’s only one equation for calculating drag forces, so I felt it would be necessary to fully understand each of the equation’s components to fully understand the concept as a whole. Originally, I thought that the cross-sectional area would be the area of the object when compressed into a two-dimensional shape along the movement axis. For example, if a soda can was falling straight down, the cross-sectional area would be the area of the can if you squished it down into one flat circle. However, this is not necessarily true. The cross-sectional area can actually be any area, depending on how you look at the equation. Because the drag coefficient is based on real-life tests (it is an intrinsic and self-defined number based on the area used), it will reflect the area chosen for the cross-sectional area, or reference area. Typically, people will use the area of a slice perpendicular to the axis of movement as in my example with the soda can. However, you could also consider the area to be the entire surface area of the object, and in real life tests the drag coefficient will reflect that.

I also wanted to fully understand what exactly the drag coefficient is, and in researching cross-sectional area, I think I have accomplished that. It’s similar to the coefficient of friction in that it is affecting the frictional/drag force in a constant way, but it is dissimilar in that it is a much less concretely defined value, which surprised me. It is entirely based on the reference area used, so when scientists discuss drag coefficients and cross-sectional areas, they must specify exactly which they have chosen for their tests.

I think I’ve made good progress on my learning objectives; I understand things conceptually a lot better now, and I’ve made solid progress towards understanding the drag force equation and what it can be used for. I haven’t fully related all of this to skydiving yet, but I’m getting there.

Sources:

http://exploration.grc.nasa.gov/education/rocket/sized.html

Friday May 3rd

For this month’s project, I’ve chosen to learn, write, and eventually present about the physics behind air resistance and how it affects skydiving. Skydiving is something that simultaneously looks both horrifying and incredible. For people to quite literally jump out of a plane from thousands of feet up in the air with nothing but a parachute, I would assume (and hope) that the physics behind the activity are pretty sound. I chose this topic because I want to figure out exactly what those physics are, and why skydivers feel so comfortable and safe free-falling for thousands of feet.

Here’s a quick, simple video on the basics of skydiving in terms we can all understand – force diagrams and Newton’s laws. It doesn’t go into much depth about air resistance itself, but it gives a general overview of what happens to a skydiver’s velocity and acceleration while free-falling based on the forces acting on the diver.

Additionally, here is a photograph of an indoor skydiver. It works much the same way as regular skydiving – the force of air friction equals the man’s weight due to gravity, so the net force acting on him (and therefore his acceleration as well) equals zero. The main difference is that in an indoor facility, the air friction is due to a fan blowing air really quickly rather than the skydiver moving quickly through the air. This means that in an indoor facility, the diver is actually stationary, whereas in actual skydiving, the diver is (obviously) not.

I am not partnering with anyone, so my learning objectives are for me only. They are:

1. Understand the relationships between air resistance, terminal velocity,  skydiving, kinematics, and Newton’s laws.

2. Learn to calculate air resistance based on the formula D = (1/2)*p*u^2*Cd*A where D is the drag force (like air friction), p is the density of the fluid, u is the object’s velocity, Cd is the object’s drag coefficient, and A is the object’s cross-sectional area.

3. Solve physics problems involving air resistance as it pertains to kinematics and to Newton’s laws.

4. Understand how the positioning of skydivers affects their velocity, drag force, etc.

5. Reach a solid conceptual understanding of the physics behind skydiving, and of why skydivers feel so safe jumping out of planes.