# Air Resistance

## Sidarth Rajan Fall 2020

## Main Idea

Based on what we've learned about Newton’s Second Law and its affiliation to gravity and acceleration, we assume that an object in freefall will fall at a constant acceleration without any other forces acting upon it; however, this assumption is wrong. As an object moves through a medium(whether it be gas or liquid), forces that oppose the motion of the object come into play such as viscosity, drag, and air resistance; moreover, these principles form the basis of the field of physics centered around fluid dynamics, which examines this topic in great detail. Air Resistance is the force we see when we throw an object in the air and it is falling down, if we were to measure the acceleration at which an object is falling, we can see that the magnitude of the acceleration is decreasing due to a force acting in the opposite direction, known as air resistance. Air resistance is particularly important when examining objects in flight as to infer data on the motion of an object, we must properly be able to calculate the magnitude of the drag force in the form of air resistance.

### Formulation

Air Resistance is built upon a relationship of a few variables all pertinent towards the motion of an object falling. These include

- [math]\displaystyle{ \rho = }[/math] a measurement of the density of the medium
- [math]\displaystyle{ v = }[/math] the velocity of the object
- [math]\displaystyle{ A = }[/math] the cross-sectional area(looking for the area coming into contact with the air)
- [math]\displaystyle{ C_{D} = }[/math] which is a non-dimensional constant that determines a relative drag depending on the shape of the object.

### Mathematical Model

A typical drag force can be described by:

- [math]\displaystyle{ F_{D} = \frac{1}{2}C_{D} A \rho v^2 }[/math]

#### Cross Sectional Area

We have encountered the term of 'cross-sectional area' in many cases: for example, when you need to calculate the flux of a liquid flowing through a pipe, you need the cross-sectional area to determine the volume of the liquid flowing through. However, the 'cross-sectional area' here is different. We'd better see it as the 'frontal area': the area facing the direction of the object's movement.

It is easy to understand: we can relate this idea to the choice of the system and moving objects. For example, when a plane is flying forward with a speed [math]\displaystyle{ v }[/math] in the [math]\displaystyle{ +x }[/math] direction and the air is static, only the surface of the plane pointing in the [math]\displaystyle{ +x }[/math] direction is subject to the air resistance since most air particles are colliding with that area with relatively high speed in the opposite direction.

Essentially, what we are describing buy the cross-sectional area is the portion of the surface area that is facing the direction of the object's motion, as this will be the area that has the most in contact with external mediums such as air. When we examine air resistance, we look at the area of the object that is facing the air. Conceptually speaking, we can see from the mathematical model that as the cross-sectional area increases the magnitude of the drag force also increases, this is due to the fact that there is more area of the object coming into contact with the specific medium.

##### Skydiver/Parachute Example

When we view skydivers, we can see that when they move downwards headfirst they tend to travel much faster than when they move stomach first. Why is this?

When a skydiver is traveling headfirst, the cross-sectional area coming into contact with the air is simply the area on the skull, which is substantially smaller than the surface area of the anterior side of your body. Therefore since less area is coming into contact with the air, the drag force is much less, which causes a greater acceleration as opposed to one with a greater surface area.

This also explains why parachutes open up with a wide surface area on the inside, as the entirety of the inside is coming into contact with the air, allowing for a large drag force, which causes the acceleration to fall greatly until one reaches terminal speed.

#### Drag Coefficient

The drag coefficient is a dimensionless number that engineers use to model all of the complex dependencies of shape and flow conditions on motion resistance (drag). It is usually determined by experiment, since the factors are very complex. The drag coefficient is a function of several parameters like shape of the body, Reynolds Number for the flow, Froude number, Mach Number and Roughness of the Surface. In short, different materials, substrates, and solvents have different drag coefficients.

### Computational Model

- [[1]]

## Examples

### Simple

You are standing at the top of a [math]\displaystyle{ 20 }[/math] meter high building, and throw a ball with an initial speed of [math]\displaystyle{ v_{0} = 10 \frac{m}{s} }[/math] in the [math]\displaystyle{ +x }[/math] direction.

**a) If you neglect air resistance, where would you expect the ball to hit on the flat surface below?**

- Without air resistance, the acceleration of the ball will be about a constant [math]\displaystyle{ g = 9.81 \ \frac{m}{s^2} }[/math]

- We can use the kinematic equations under constant acceleration:

- [math]\displaystyle{ v_{f} = v_{0} + at \ \ \mathbf{\text{(1)}} \quad \And \quad \Delta x = v_{0}t + \frac{1}{2}at^2 \ \ \mathbf{\text{(2)}} \quad \And \quad {v_f}^2 = {v_0}^2 + 2a \Delta x \ \ \mathbf{\text{(3)}} }[/math]

- We will break this into two parts:

*The y-direction (vertical)**The x-direction (horizontal)*

*The y-direction:*

- From the problem statement, we are told the ball is thrown horizontally, so there is no initial velocity along the y-direction:

- [math]\displaystyle{ v_{y_{0}} = 0 }[/math]

- Also, the only acceleration affecting the system, the acceleration due to gravity is along the negative y-direction:

- [math]\displaystyle{ a_{y} = -g = -9.81 }[/math]

- Also, we are told the initial height:

- [math]\displaystyle{ y_{0} = 20 }[/math]

- Using equation
**3**, we can find the final velocity along the y-direction of the ball:

- Using equation

- [math]\displaystyle{ v_{y_{f}} = - \sqrt{{v_{y_{0}}}^2 + 2a_{y} \Delta y} = -\sqrt{0 + 2 \times (-9.81) \times (0 - 20)} = -\sqrt{392.4} = -19.81 \ \frac{m}{s} }[/math]

- Now, we can plug this final velocity into
**1**to find the time of flight for the ball:

- Now, we can plug this final velocity into

- [math]\displaystyle{ v_{y_{f}} = v_{y_{0}} + a_{y}t }[/math]

- [math]\displaystyle{ t = \frac{\Delta v_{y}}{a_{y}} = \frac{-19.81}{-9.81} = 2.02 \ s }[/math]

- Knowing the time of flight will allow us to solve for the distance traveled horizontally.

*The x-direction:*

- We know the initial velocity in this direction:

- [math]\displaystyle{ v_{x_{0}} = 10 }[/math]

- Also, there is no acceleration along the x-direction:

- [math]\displaystyle{ a_{x} = 0 }[/math]

- Using equation
**2**, we can find how far the ball goes horizontally:

- Using equation

- [math]\displaystyle{ \Delta x = v_{x_{0}}t + \frac{1}{2}a_{x}t^2 = v_{x_{0}} t = 10 \times 2.02 = 20.2 \ m }[/math]

- Therefore, the ball will land a distance of 20.2 meters away horizontally and 20 meters below the roof of the building.

**b) Do you think your prediction without air resistance is too large or too small? Explain.**

- If air resistance was taken into consideration, it would impede the motion of the ball, slowing it down. Thus, the previously predicted distance would be further than the ball would really go.

### Middling

John is going sky diving for the first time. His mass is 70 kg and his terminal speed is 38 m/s.

**a) What is the magnitude of the drag force on John?**

- Since John has reached terminal speed, he is not accelerating. Equivalently, this means his acceleration is [math]\displaystyle{ 0 \ \frac{m}{s^2} }[/math], his change in momentum is [math]\displaystyle{ 0 \ \frac{kg \cdot m}{s} }[/math], and the net force on him is [math]\displaystyle{ 0 \ \text{N} }[/math]. Thus, we know the two forces acting on him, the gravitational force and the drag force, are balanced:

- [math]\displaystyle{ F_{net} = F_{D} - F_{g} = 0 }[/math]

- [math]\displaystyle{ F_{D} = F_{g} }[/math]

- Since John is close to the surface of Earth, we will use [math]\displaystyle{ mg }[/math] as the gravitational force:

- [math]\displaystyle{ F_{D} = mg = 70 \times 9.81 = 686.7 \ \text{N} }[/math]

- Therefore, the drag force on John is about [math]\displaystyle{ 686.7 }[/math] Newtons pointing away from the Earth.

**b) What is an expression for the value of the cross-sectional area of John?**

- Using our drag force equation and setting it equal to the gravitational force, we see:

- [math]\displaystyle{ F_{D} = \frac{1}{2}\rho v^2 A C_{D} }[/math]

- [math]\displaystyle{ \frac{1}{2}\rho v^2 A C_{D} = mg }[/math]

- Solving this for [math]\displaystyle{ A }[/math] gives:

- [math]\displaystyle{ A = \frac{2mg}{\rho v^2 C_{D}} }[/math]

- If we knew the density of the air and the drag coefficient, we could give a value for the cross-sectional are of John.

### Difficult

Sarah is doing an air resistance experiment in class. The experiment requires Sarah to drop a coffee filter from a height of 2 meters. Let's say that the mass of the coffee filter was 2.0 grams, the radius of its base was 6 cm, and it reached the ground with a speed of 1.0 m/s.

**a) How much energy did the air gain as the coffee filter floated down?**

- This is a conservation of energy problem. We can describe the coffee filter as the system and the air as the surroundings to a good approximation. Thus, by the Energy Principle:

- [math]\displaystyle{ \Delta E_{c} + \Delta E_{a} = 0 }[/math]

- [math]\displaystyle{ \Delta E_{c} = - \Delta E_{a} }[/math]

- The initial energy of the coffee filter can be described as a combination of the initial Thermal, Gravitational Potential, and Kinetic energies:

- [math]\displaystyle{ E_{c_{0}} = U_{c_{0}} + K_{c_{0}} + Q_{c_{0}} }[/math]

- The initial kinetic energy will be [math]\displaystyle{ 0 }[/math] since the coffee is dropped from rest:

- [math]\displaystyle{ E_{c_{0}} = U_{c_{0}} + Q_{c_{0}} }[/math]

- The final energy of the coffee filter can be described as a combination of the final Thermal, Gravitational Potential, and Kinetic energies:

- [math]\displaystyle{ E_{c_{f}} = U_{c_{f}} + K_{c_{f}} + Q_{c_{f}} }[/math]

- Subtracting the final energy from the initial energy will give us the change in energy of the coffee filter:

- [math]\displaystyle{ \Delta E_{c} = E_{c_{f}} - E_{c_{0}} = (U_{c_{f}} + K_{c_{f}} + Q_{c_{f}}) - (U_{c_{0}} + Q_{c_{0}}) = (U_{c_{f}} - U_{c_{0}}) + K_{c_{f}} + (Q_{c_{f}} - Q_{c_{0}}) = \Delta U_{c} + K_{c_{f}} + \Delta Q_{c} }[/math]

- Using common expressions for Gravitational Potential energy and Kinetic energy, we see:

- [math]\displaystyle{ \Delta E_{c} = mg(h_{f} - h_{0}) + \left(\frac{1}{2}\right)m{v_f}^2 + \Delta Q_{c} }[/math]

- We will assume the change in Thermal energy is negligible:

- [math]\displaystyle{ \Delta E_{c} = mg(h_{f} - h_{0}) + \left(\frac{1}{2}\right)m{v_f}^2 }[/math]

- We have all the needed values for this:

- [math]\displaystyle{ \Delta E_{c} = 0.002 \times 9.81 \times (0 - 2) + \left(\frac{1}{2}\right) \times 0.002 \times (1.0)^2 = -0.03924 + 0.001 = -0.03824 \ J }[/math]

- Using:

- [math]\displaystyle{ \Delta E_{c} = - \Delta E_{a} }[/math]

- We see:

- [math]\displaystyle{ - \Delta E_{c} = \Delta E_{a} }[/math]

- [math]\displaystyle{ \Delta E_{a} = -(-0.03824) = 0.03824 \ J }[/math]

- We see the coffee filter lost [math]\displaystyle{ 0.03824 \ J }[/math] of energy, and the air in contact with the coffee filter gained it.

**b) What is the drag force at the end of the object's motion? Assume the drag constant is 2.3 and air has a density of 1.22 [math]\displaystyle{ \frac{kg}{m^3} }[/math]**

- [math]\displaystyle{ \rho = }[/math] 1.22
- [math]\displaystyle{ v = }[/math] 1.0
- [math]\displaystyle{ A = (0.06)^2 \pi }[/math]
- [math]\displaystyle{ C_{D} = }[/math] 2.3

- We now have everything we need to calculate the drag force

- [math]\displaystyle{ F_{D} = \frac{1}{2}C_{D} A \rho v^2 }[/math]
- [math]\displaystyle{ F_{D} = \frac{1}{2}(2.3)((0.06)^2 \pi)(1.22)(1)^2 }[/math]
- [math]\displaystyle{ F_{D} = \frac{1}{2}(2.3)(0.01131)(1.22)(1)^2 }[/math]

- After calculating everything we can see that

- [math]\displaystyle{ F_{D} = 0.01587 }[/math] N

## Connectedness

**How is this topic connected to something that you are interested in?**

- Air resistance plays a big factor in sports, and physical activity. We can see that during sports such as football, rugby, and soccer when the ball is kicked upwards, it travels downwards with an amount of drag, which is important to take into consideration when playing these sports. Furthermore, the sport of ultimate frisbee is entirely based upon the principle of air resistance, as really good players know how to manipulate the air to cause the best throws and motion of the frisbee.

- We can also see that the concept of air resistance in specific its relation to the overall concept of drag, is present throughout the universe in many different forms, one of them being the drag you feel against water. We know that it is much harder to move underwater than it is in air, and that is due to the fluid mechanics behind the motion between both mediums, and the increased drag in water.

**How is it connected to your major?**

- I am a computer science major who is interested in speciation in artificial intelligence and computer vision. Dealing with forces involved in a drone's or vehicle's motion is exactly what I will be having to deal with in the future. Learning how drag force is dependent on the speed proves that it is constantly changing, and must constantly be updated and factored into calculations regardless of whether the motion is vertical or lateral.

**Is there an interesting industrial application?**

- Air resistance has a lot of applications throughout academia, entertainment, security, and many other things. Through academia, we can examine how air resistance impacts the motion of objects in freefall and we can see the magnitude that it has on an object's net speed. Moreoever, through the sports world, we can see how knowing how to control the amount of drag that will be on a ball is imperative towards reaching higher athletic goals.

## History

Aristotle was the first to write about air resistance in the 4th century BC. In the 15th century, Leonardo da Vinci published the Codex Leicester, in which he rejected Aristotle's theory and attempted to prove that the only effect of air on a thrown object was to resist its motion. The first equation for air resistance was:

- [math]\displaystyle{ F_{D} = \rho S V^2 \text{sin}^2(\theta) }[/math]

This equation overestimates drag in most cases, and was often used in the 19th century to argue the impossibility of human flight.

Louis Charles Breguet's paper in 1922 began efforts to reduce drag by streamlining. A further major call for streamlining was made by Sir Melvill Jones who provided the theoretical concepts to demonstrate emphatically the importance of streamlining in aircraft design. The aspect of Jones’s paper that most shocked the designers of the time was his plot of the horse power required versus velocity, for an actual and an ideal plane.

## See also

### Further reading

- http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Free-Fall-and-Air-Resistance
- http://www.forbes.com/sites/chadorzel/2015/09/29/the-annoying-physics-of-air-resistance/[2]
- https://www.youtube.com/watch?v=mP1OTzdB0oI