Recall the parachute problem from long ago
By Zimo Wang, 6th Feb 2023
In 2020, I had posted a series of three videos about parachute experiments in 8th grade, detailing the experimental report and the parachute. Two years ago it did feel like I was pretty good and felt like I had come up with some very complicated wrong formula then said the rest was going to use complex knowledge of fluid dynamics. Today we will further recap this problem, come back and see what I was stupidly doing years before.
1. Quick Recap
Just what was the lab report doing at that time? The research aim was: to investigate the relationship between the area of the parachute and its landing time. According to our common sense, we all know that the larger the parachute's surface area, the longer the landing time, due to the larger air resistance. This is certainly correct. But two years ago, I was trying to solve this problem quantitatively, namely by studying for:
where t is the parachute's landing time and A is the area of the parachute. The following file is the theoretical deduction in the report, which is all an 8th-grade kid could do. It is incorrect, but not all incorrect, as we will see later.
However, the final result I got as shown in the graph is , where x and y are constants. You may think this is correct, but from a practical view, y is not a constant because it involves v, the instantaneous velocity of the parachute. The velocity of the parachute would change all the time as it falls, so v must not be a constant. So that's where the problem lies.
2. What's wrong? What's correct?
So now the question is, which parts of the above deduction we can carry over and which parts are wrong? Let us look at the first equation:
You can't question it anyway, because you're questioning Newton's second law. The left part is the force on the parachute: gravity that orients downwards, air resistance orienting upwards (it's called "resistance" only when it's opposite to the direction of motion), and on the right is mass times acceleration. Therefore, let's look at another formula that we've used:
This is the equation of the motion with constant acceleration we learned in high school. The question is: does the parachute possess a constant acceleration during the landing process? The answer is definitely not. So the mistake comes from the usage of this formula. In fact, the deduction process of the above formula is derived from , which means that the acceleration is constant. Wait, you say how to derive this? In fact, just use the acceleration as the distance to the second-order derivative of time, that is , to solve the differential equation and then substitute the initial conditions. Therefore, this formula cannot be directly applied to the case of a parachute.
3. So how we do?
Then we must use a reliable formula, that is necessarily the result of our fantastic Mr Newton, . We know that weight is W = mg; f (we call the air resistance for f, not fair too annoying!!) is the air resistance, and the lab report also gives a seemingly reliable formula: . But why "seemingly reliable"? In fact, this formula is not suitable for all objects. But in this case, we assume that this formula is correct (and the fact is that this approximation is not very problematic). So we can write Newton's second law as follows:
Because acceleration a is the derivative of the velocity v. We also know that the parachute will definitely have a terminal velocity during landing, where gravity = air resistance (this is mentioned in the previous video, not much to say here). We call this terminal velocity , which is also the equilibrium point of the above differential equation. At this time, v is:
So substituting this value changes the complex equation of air resistance to . Thus the previous differential equation can be directly written as
Then we can solve this equation, and all the processes are mathematical stuff, so we will omit some of them. Separating the variables and then integrating both sides:
After substituting the initial conditions, we can solve for
In this case, we have solved the relationship between time and velocity. We will then use the time to express the velocity, i.e.
Recall the purpose we want: to find the relationship between the parachute's area and its landing time. The first question you may ask is where is area A in the equation!? In fact, area A is in . After finishing all the deductions, we will then substitute it back. The second question is that there is a v in the above formula, and we know that v will change as the parachute landing. But let's think about it, though the velocity will change, the distance that the parachute needs to travel will not change at all. The experiment is done by throwing a parachute down from a certain height (5m), so the distance the parachute travels is a fixed value. So we just need to integrate two sides here:
Until now, all the integration is done! We then just need to put t to the left side of the equation (since we are finding t = ???A), and then we get
Finally, substituting our previous substitution , we get
Perfect! That's the solution we want - the exact relationship between t and A. Until now, let's review all the parameters. m is the mass of the parachute. ρ is the density of the air. Cd is the drag coefficient. A is the area of the parachute. So what does such a function probably look like? We can sketch it roughly as follows.
where x-axis is the area A, and y-axis is the time t. You can see that A is actually positively correlated with t, and the slope is gradually decreasing. This means that if we keep increasing the area of the parachute, its landing time will increase, but it increases more and more slowly.
So when we go back to this screenshot from the video, we can see that the slope of the function derived from the previous formula (8th-grade one) is increasing (left figure), but it does not match the final experimental results (right figure). If we apply the function formula we derived today, we can find that there is consistency between the trend of the data from the theoretical deduction and the experimental result!