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:
- 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.
- 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:
Physics of Skydiving 
Awesome post. Seeing an example of the equation in use is really helpful to supplement the conceptual information. This isn’t a big deal, but if you are having trouble with the equations in WordPress you could always type them up in word and copy them in, I think that would work.
I think you’ve done a really good job explaining your topic. I was interested in your discoveries about terminal velocity because I’ve never understood it before. I also liked how you included a couple of problems. I wonder how different air pressures affect drag and sky diving. Also, does the shape of the diver change their path downwards (ie if a person made their body into a certain shape, would they start spinning or something)?
Due to our time restraints for this project, I chose not to delve too much into the rotational aspect of skydiving. I felt it would perhaps be infringing on some other people’s topics like Galen’s. I do know, though, that a skydiver’s position will greatly affect his or her cross sectional area, which is directly proportional to the drag force. So, a head-down down diver will experience much less drag than a diver with his or her limbs spread out.
Ha, I really enjoy Katrina’s question about spinning. I never thought I’d say this in reference to a blog, but yay, someone is using math! This post was refreshing because most people (including myself, unfortunately) have been sticking to conceptual stuff. Not only did you reference an equation, you applied it. And for some reason that makes me really happy. As Charlie said, using Word would be easier to follow. I don’t know if Word equations copy well into WordPress… if all else fails, you could always screenshot and paste the pictures of the equation into the blog. Regardless, congrats on the math.
Hi Mr. Paige, I really enjoy your blog and I find your revelations to be very interesting. My only issue is at times it can be a little bit difficult to follow your math. This may be a product of wordpress’ limiting factors with its lack of equation creators, I’m not sure but maybe you could take it a little slower in between steps so I can keep up. But other than that great job.