Last week, reader James from Melbourne wrote in:

Hi James,

As I'm sure you know, in a vacuum they'd take the same amount of time to fall...but air resistance changes the game.

The amount of deceleration caused by the air you encounter as you fall depends on three main things: your velocity, your surface area, and your "drag coefficient", a number that depends on factors like the shape of an object and the texture of its surface. This number is usually fiendishly difficult to calculate, so it's most often obtained experimentally. At first, I thought that we'd have to make some simplifying assumptions about the drag coefficient of an elephant—you'd need a very large wind tunnel, and a very patient elephant to find it—but lo and behold, the good people in the University of Wisconsin's zoology department have done extensive simulations and experimental testing, using a scale-model elephant made of aluminum. Publishing their findings in the

Specifics aside, the important thing to understand here is what goes into determining an object's

Since you specified such a great height, we know that both the elephant and the brick are going to have time to reach their terminal velocity, which simplifies things—we just need to figure out which one's is greater, which brings us to the crux of your question: is the elephant's greater surface area enough to balance out its increased weight, relative to the brick?

Taking a look at the formula for terminal velocity, we can see that the mass and the surface area are given equal importance, and everything in this equation is multiplied together, so an object's terminal velocity will remain the same as long as mass and surface area increase proportionally. But even before we do any calculating, I have a hunch about our answer, based on a bit of math that goes all the way back to Galileo: the square-cube law.

The square-cube law simply states that, as an object's size increases, its volume grows faster than its surface area. Consider that the formula for the surface area of a sphere is 4*pi*r

Obviously, the difference between the r

However, as any physicist knows, intuition is no substitute for real math—so even though our calculations won't be perfect, we can use that terminal velocity formula to get a rough idea of how accurate this hypothesis is. This is what we call a back-of-the-envelope calculation, which will reveal if one or the other is likely to be a clear winner, e.g. if one has a greater terminal velocity by a factor of ten. To make sure our estimate is conservative, we'll want to use a heavy brick, and a light elephant, so let's go with an African forest elephant, the smallest kind. We'll also have the brick falling lengthwise, with its smallest face pointing downward, and the elephant falling on its back: the "projected area" in the formula above means the surface area of the side facing downward, into the "wind".

A large brick weighs 3.5kg, and its smallest face is 10.6cm by 7.3cm—about 4 inches by 3, for a total area of 77.4 cm

Using the "long cylinder" approximation for the brick's drag coefficient (0.82), we can now calculate:

Well I'll be—it's actually a tight race! For the record, 80 m/s is roughly 200 miles an hour. Redoing the calculation with the brick's largest face pointing downward, though, we find that the elephant's terminal velocity is greater, and the same is true if the elephant is falling headfirst, rather than on its back or side. In reality, both objects are likely to tumble, but asymmetrical objects tend to end up falling with their smaller ends facing downward.

If I had to put money on it, mine would be on the elephant reaching the ground first, but you never know—the calculation of an elephant's drag coefficient was done using ordinary wind speeds and an aluminum scale model. In reality, the elephant's wrinkly skin and large ears, flapping furiously in relative winds of 200 miles per hour, might change its drag coefficient enough to swing things in the brick's favor.

All things considered, the surprising answer is that it's likely to depend on the size of your elephant. In physics, they say, experiment is king—but hopefully, we'll never see this one put to the test; a 2700 kg elephant moving at 80 m/s has about as much kinetic energy as you'd find in eight or nine sticks of dynamite. This "race" might turn out to be a photo-finish...but I wouldn't want to be waiting around at the finish line.

—

I was having a discussion with a colleague about what would hit the ground first if it fell from a plane (let’s say 15,000 ft). An elephant (let’s say African) or a standard brick. Curious to know your thoughts. Thanks!

Hi James,

As I'm sure you know, in a vacuum they'd take the same amount of time to fall...but air resistance changes the game.

The amount of deceleration caused by the air you encounter as you fall depends on three main things: your velocity, your surface area, and your "drag coefficient", a number that depends on factors like the shape of an object and the texture of its surface. This number is usually fiendishly difficult to calculate, so it's most often obtained experimentally. At first, I thought that we'd have to make some simplifying assumptions about the drag coefficient of an elephant—you'd need a very large wind tunnel, and a very patient elephant to find it—but lo and behold, the good people in the University of Wisconsin's zoology department have done extensive simulations and experimental testing, using a scale-model elephant made of aluminum. Publishing their findings in the

*Journal of Experimental Zoology*, they reveal that the drag coefficient is roughly 0.65 for a headwind, 0.8 for a crosswind.Other highlights from the paper include this absolute gem of an image. The title of the paper itself is a sly reference to an old physics joke, the "Spherical Cow"Image Credit: P.N. Dudley, R. Bonazza, W.P. Porter. |

*terminal velocity*—the maximum speed a body can reach in a fall. Air resistance, which slows that fall, depends on the body's speed—the faster you go, the more each air molecule you encounter slows you down. In a vacuum, you'd accelerate indefinitely, moving toward the earth 9.8 m/s faster for each second that you'd fallen, until you hit the ground. But in air, there's a certain speed where the pushback of air resistance is enough to balance out the pull of gravity, and you stop accelerating. How much that pushback slows you down depends on your momentum, which depends on your mass—so between two otherwise-identical objects of different masses, the heavier one will have a greater terminal velocity. For two objects of the same mass, where one has a greater surface area, the one with greater surface area will have a lower terminal velocity.Since you specified such a great height, we know that both the elephant and the brick are going to have time to reach their terminal velocity, which simplifies things—we just need to figure out which one's is greater, which brings us to the crux of your question: is the elephant's greater surface area enough to balance out its increased weight, relative to the brick?

Taking a look at the formula for terminal velocity, we can see that the mass and the surface area are given equal importance, and everything in this equation is multiplied together, so an object's terminal velocity will remain the same as long as mass and surface area increase proportionally. But even before we do any calculating, I have a hunch about our answer, based on a bit of math that goes all the way back to Galileo: the square-cube law.

The square-cube law simply states that, as an object's size increases, its volume grows faster than its surface area. Consider that the formula for the surface area of a sphere is 4*pi*r

^{2}, where r is the sphere's radius, while the formula for volume is 4/3*pi*r^{3}—hence the name "square-cube law".Obviously, the difference between the r

^{2}in surface area and the r^{3}in volume can be enormous, especially when talking about a difference in size like the difference between a brick and an elephant. Since an object's mass depends on its volume, and thus increases with the cube of its size, while surface area increases as the square, it's generally safe to say that big things of roughly equal density fall faster and have greater terminal velocities than small ones, with the exception of shapes like thin sheets. Here, I'd expect we don't even need to take into account the different densities of the elephant and the brick, or their geometries—the square-cube law ought to wash away those relatively minor differences.However, as any physicist knows, intuition is no substitute for real math—so even though our calculations won't be perfect, we can use that terminal velocity formula to get a rough idea of how accurate this hypothesis is. This is what we call a back-of-the-envelope calculation, which will reveal if one or the other is likely to be a clear winner, e.g. if one has a greater terminal velocity by a factor of ten. To make sure our estimate is conservative, we'll want to use a heavy brick, and a light elephant, so let's go with an African forest elephant, the smallest kind. We'll also have the brick falling lengthwise, with its smallest face pointing downward, and the elephant falling on its back: the "projected area" in the formula above means the surface area of the side facing downward, into the "wind".

A large brick weighs 3.5kg, and its smallest face is 10.6cm by 7.3cm—about 4 inches by 3, for a total area of 77.4 cm

^{2}—or 0.00774 m^{2}. A small elephant, on the other hand, is about 4 meters long, 2.1 meters wide (giving it 8.4 m^{2}of surface area on its back), and weighs 2700 kg.Using the "long cylinder" approximation for the brick's drag coefficient (0.82), we can now calculate:

versus...

Well I'll be—it's actually a tight race! For the record, 80 m/s is roughly 200 miles an hour. Redoing the calculation with the brick's largest face pointing downward, though, we find that the elephant's terminal velocity is greater, and the same is true if the elephant is falling headfirst, rather than on its back or side. In reality, both objects are likely to tumble, but asymmetrical objects tend to end up falling with their smaller ends facing downward.

If I had to put money on it, mine would be on the elephant reaching the ground first, but you never know—the calculation of an elephant's drag coefficient was done using ordinary wind speeds and an aluminum scale model. In reality, the elephant's wrinkly skin and large ears, flapping furiously in relative winds of 200 miles per hour, might change its drag coefficient enough to swing things in the brick's favor.

All things considered, the surprising answer is that it's likely to depend on the size of your elephant. In physics, they say, experiment is king—but hopefully, we'll never see this one put to the test; a 2700 kg elephant moving at 80 m/s has about as much kinetic energy as you'd find in eight or nine sticks of dynamite. This "race" might turn out to be a photo-finish...but I wouldn't want to be waiting around at the finish line.

—

**Stephen Skolnick**

Sorry for that mental image; let's just pretend this is how things would go instead.Image Credit: Walt Disney Company, via Giphy |

While I don't know how it would apply in this case, another factor to consider in falling bodies is buoyancy.

ReplyDeleteA balloon and a bowling ball have roughly the same drag co-efficient yet the balloon will fall much slower due to the buoyancy force acting on it.

Which equivalent means that density matters.

DeleteI thought a brick was more dense than water