Can the relationship between drag and velocity, R=bv^2

I think that the relationship between radius and time will be proportional. When a graph between drag and velocity is drawn, it will be more plausible to find the constant from the exponential graph. This is because the original equation is \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv^2\) meaning that the relationship between drag and velocity is \(R\ \propto\ v^2\).

Here, I will attempt to predict the relationships between the variables in my experiment. As I mentioned before, my results are split into two sections. So, I’m going to work backwards from the drag equation, \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv{^2}\), to predict the relationships for radius against time. I have already predicted the relationship between drag and velocity, so there will be no need to do this again.

 

Starting with our drag equation.

 

\(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv{^2}\)

 

Rearranged,

 

\(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dA\)

 

Simplified to surface area and velocity,

 

\(A\propto\frac{1}{v^2}\)

 

The velocity equation,

 

\(v=\ ^d{\mskip -5mu/\mskip -3mu}_t\)

 

\(d\ = distance\ travelled\ \ t= times\ taken\)

 

Substituted the two equations,

 

\(A\propto\frac{t^2}{d^2}\)

 

Simplified to surface area and time,

 

\(A \propto t^2\)

 

The surface area equation,

 

\(A=\pi{r}^2\)

 

\(\pi=pi\ \ r=radius\)

 

Substituted the two equations,

 

\(\pi{r}^2\propto\ t^2\)

 

Simplified to radius and time,

 

\(r^2\propto\ t^2\)

 

Square root on both sides,

 

\(r\ \propto\ t\)

 

This relationship shows that radius is proportional to time. This is easier to visualize because the bigger the surface area you have, the longer it will take to reach the ground. Another way to look at this is through the drag equation, \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv{^2}\) where surface area is proportional to drag, hence increasing the time taken.

 

Making the Parachute

 

• Garbage bags (enough to make parachutes with a radius of 25cm)
• Strings (30cm)
• A mass (10g)
  • Initially, I used a mass of 40g however after some trials it showed to drag the parachute down too much. Affecting how I measured the time, increasing my uncertainties. So, I decided to decrease the mass to 10g, so it was more manageable
• Ruler (1m)

 

Experiment

 

• Parachutes
• Stopwatch to measure time precise to 0.1s

 

Our 2 equations,

 

\(P=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{v}^2\)

 

\(P=\ ^R{\mskip -5mu/\mskip -3mu}_A\)

 

 

Bernoulli’s Principle can now be substituted into the new pressure equation,

 

\(^R{\mskip -5mu/\mskip -3mu}_A=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dv^2\)

 

\(C_d= Coefficient\ of\ drag\)

 

 

Rearranged in terms of drag,

\(R=\ ^1{\mskip -5mu/\mskip -3mu}{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv^2\) (Elert, n.d.)

 

 

This is our final equation.

 

You'll see that \(C_d\) has been inserted as a new symbol "which is absent from the first two equations. This is because shape and texture both affect drag (Elert, n.d.). Therefore, \(C_d\), the coefficient of drag "Is essential to include. We don't have this value, which is a significant flaw in this calculation. I did, however, discover a figure.

 

\(C_d=0.75-0.80\) (European Space Agency, 2018)

 

 

The last variable we need to measure is the density of air.

 

\(\rho=1.204\) (Princeton, n.d.)

 

This is more reliable because, unlike the coefficient of drag, the density of air won't change significantly. There won't be a need for additional estimations because other variables can be measured in the experiment.

 

We can then simplify the equation into one that assumes that drag is proportional to velocity.

 

\(R=bv\)

 

\(b\ =\ Constant\ of\ Proportionality\)

 

It is important to note that this equation does not fit our equation, \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv^2\) properly. This is because \(b=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dA\), and from the equation, it shows that the equation should be\(R=bv^2\) However, I thought it would be interesting to graph these constants and see if experimental results could prove that the true equation is \(R= bv^{2}\)

 

Therefore, this is the simplified equation that assumes drag is proportional to velocity squared.

 

\(R=bv^2\)

 

In my experiment, b will be the constant of proportionality that I will be attempting to find experimentally. 2 constants will be drawn, one for \(R=bv\) and another for \(R=bv{^2}\). Then from these results, I will determine which constant of proportionality is closer to the true value.

 

The calculation for the uncertainty in time.

 

\(\triangle{t}=\frac{Max\ t\ -\ Min\ t}{2}+0.2\)

The second part of our derivation is Bernoulli’s principle,

 

\(P=\ ^1{\mskip -5mu/\mskip -3mu}_2\rho{v}^2\)

 

\(\rho\ = Density\ \ v= Velocity\)

 

Bernoulli’s principle shows how fluid density and velocity affect the pressure. This is important because it shows us that high pressures can lead to low velocities and vice versa. In my experiment, one of the main variables is velocity. So, factors affecting this need to be highlighted.

 

• Make the parachutes; figure 4 shows one of the parachutes I used in my experiment.
• From a safe height, drop the parachute and start the stopwatch
• Stop it when the parachute hits the ground.
• Drop the other parachutes and repeat the experiment.

 

Range 10cm – 25cm

 

Before, I was going to do a range of 5cm – 15cm. However, the parachutes with 5cm and 7.5cm had many problems catching the air and having a proper flight. This resulted in 2 anomalies out of 5, which would cause many problems in analysis. So, I decided to increase the minimum radius. I also saw an overall lack of data since I couldn’t identify a relationship, so I increased the overall range.

 

Number increment of 2.5cm

 

No changes were made to the number since the problems were rooted mainly in the range and provided sufficient data.

 

Repeats 5

 

Although, my range and number showed sufficient data. Experiments like these have many external factors acting on them, thus increasing the random uncertainty. One way to avoid this problem is to average over repeats. This was I could get more precise results and sufficient data.

When doing some research, I found this equation \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\rho{C}_dAv^2\)  (Elert, n.d.). After looking at the derivation, I wondered if it would be possible to find a constant of proportionality between drag and velocity. Even though, it could be argued that the constant is \(^1{\mskip -5mu/\mskip -3mu}_2\rho{C}_dA\). I want to see if this was true experimentally or if perhaps there are other factors involved. Learning about the different effects on parachutes is very important for the design process.

 

First, we need to derive the equation from simple concepts to help us understand this better. 2 equations, the pressure equation \((P=\ ^F{\mskip -5mu/\mskip -3mu}_A)\) and Bernoulli’s principle \((P=\ ^1{\mskip -5mu/\mskip -3mu}_2\rho{v}^2\), are substituted to get our drag equation \((R=\ ^1{\mskip -5mu/\mskip -3mu}_2\rho{C}_dAv^2)\).

 

 

The first part of our derivation is the pressure equation,

 

\(P=\ ^F{\mskip -5mu/\mskip -3mu}_A\)

 

\(P\ = Pressure\ F\ = Force\ A=\ Surface\ Area\)

 

 

Our force here is drag therefore \(F=R\)

 

\(P=\ ^R{\mskip -5mu/\mskip -3mu}_A\)

 

\(R=Drag\)

 

A graphic can make this equation easier to understand if a larger-surface area parachute is deployed. As a result of increased air resistance, the drag will also increase. Skydiving is a good example of this because as the parachute is dropped, the velocity lowers due to drag, making a safe ground landing easier.

To find the constant of proportionality, many steps will need to be taken to find it. Following the equation, \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv^2\) We need to measure the surface area and the velocity. Note that we already have the air density and the coefficient of drag. There will be one experiment which will change the radius, and measure the time and the height will remain the same. From the radius, we can find the surface area and, from the height and time, we can find the velocity. Using the original equation, \(R=\ ^1{\mskip -5mu/\mskip -3mu}_2\ \rho{C}_dAv^2\) , we can find the drag. Then we can graph the results to find the constant. 

 

My results be set out into two sections: radius against time and drag against velocity. The first section shows the problems I had with my experiment and how it effects the final results in my second section.

Although there are many uses for parachutes, moving people or things is the most frequent. Numerous tests were carried out to examine the aerodynamics and other factors influencing flying in greater detail. The surface area was the easiest variable to adjust, but it has a lot of effects, including on velocity, drag, and drop time.

The calculation for the percentage uncertainty.

 

\(\triangle{t}\%=\ \triangle{t}\ {\mskip -5mu/\mskip -3mu}_t\times100\%\)

Independent Variable

 

The variable that I will be changing is the radius measured in m.

 

Dependent Variable

 

The variable that I will be measuring is time measured in s.

 

Control Variables

 

• Length of the suspension lines – changing the length of the suspension lines could cause a change in drop time since, it is changing the structure of the parachute. In response, all suspension lines will be 30cm long

 

• Mass of the suspended object – an increase in mass could cause an increase in force according to Newton’s second law, 𝐹 = 𝑚𝑎. This means that keeping the mass the same will be essential for collecting precise results.

 

• Height from which the parachute is launched – since velocity depends on the height. Changing the height will essentially change the drag, which will make the experiment less precise. The parachutes will be launched from the same height every time.