Cyclopian Cyclones

For the past 2 weeks, I've been struggling with a question from Papou, as well as struggling with a stomach bug which probably didn't help, but I finally have an answer! Papou wondered, How does the Coriolis affect hurricanes in the northern hemisphere and what is the percentage centrifugal velocity increase/decrease at the eight points on the compass assuming the hurricane is moving directly East to West?

Coriolis force is one of the fictitious forces that appear in a rotating reference frame, along with the centrifugal and Euler forces. If you look at articles on why hurricanes spin the way they do, they'll give explanations about how the speed of the Earth's rotation varies depending on how close to the pole you are, which causes the Northern edge of the hurricane to be pushed one direction, and the Southern edge pushed the other way. This explanation makes sense, but it didn't fit with my algebraic understanding of the Coriolis force.

If you look up the equation for the Coriolis force, you'll see it stated as

where Ω is the angular velocity of the rotating frame (for Earth, 2π/day), and v is the velocity of an object in the rotating frame. You may notice, there's no mention of radius here, which is why I was confused by the articles I was reading.

I decided to follow the default route for Physics: Simplify until things make sense! Let's imagine a hurricane directly over the North Pole. A key feature of hurricanes is the eye, which is a region of low pressure at the center. Low pressure in the middle means air will tend to flow from the outside in. As that air accelerates though, it will have a velocity perpendicular to the angular velocity, which is pointing straight up in this case. That will cause the Coriolis force above to push the air in a counter-clockwise direction, exactly what we're expecting.

I tried to put together a full simulation of the pressure difference driving the air inward, but then being balanced by the Coriolis force spinning it in a circle. Unfortunately, I couldn't come up with a model of the pressure that didn't vastly overpower the Coriolis force, so I ended up simplifying things a bit more: We assume the pressure difference causes a constant force inward, and then see how that changes under the effect of the Earth's rotation vector, projected into the local plane of Florida (27.7° N latitude):

For this simulation, I used a diameter of 300 miles and a Category 3 windspeed of 120 mph, but again, I had to fudge a bit to get things to work.

As for the second part of the question, centrifugal force doesn't really come into things here. For most places on the Earth, centrifugal force points upward, so its only effect on the hurricane would be a vertical circulation of the air. That may have an effect in reality, but I'm only considering a 2D model here, so it doesn't really apply.

I always have a hard time picturing these 3D systems, with interacting vectors and different reference frames, but I hope I've given a reasonable explanation for how these types of weather events form. I'm betting there's a way to connect my understanding of Coriolis forces to the one commonly given, but I can't puzzle it out. Thanks for another great question, Papou!

On the rEvolution of Doorways

Dipping once again this week into my list of topics, I chose a particularly old one: revolving doors. Ever since my stay at Mass. General Hospital, I've been curious how efficient revolving doors are at keeping heat in or out, compared with sliding doors. Last time, aside from the disadvantage of chemo effects, I attempted to do a detailed simulation of the motion of particles through the doorway, which never panned out. This time, I took a much simpler approach using an approximation for the rate of heat flow between two reservoirs:

where Q is the heat energy transferred in a time Δt through a surface of area A. The temperature difference ΔT between the two reservoirs is spread over a distance Δx, and the thermal conductivity k is a property of the air, which we can look up. The idea with a revolving door is that the inside and outside are never in direct contact: The air moves from outside, to a segment of the door, then to the inside. I wasn't sure how to get the distance, since it will change with time as the air moves, but I went with a guess of 20 cm. For the dimensions of the door, I found an architecture page that gave some example measurements. I went for a 3-segment door. For each person who enters, the door will turn 120°, so we can try a few different rates.

Along with the heat transferred, we need to know the current temperature in each section of the door. That will be a simple scale factor, the heat capacity. Since the change in energy/temperature is proportional to the current energy/temperature, we'll get an exponential relationship in time, where the door section will approach the in/outdoor temperature, but never quite make it.

I put together a simulation with some values more suited to my current Florida environment: 20°C (68°F) inside and 35°C (95°F) outside. First, we can look at the temperature in each section of the door as it rotates with 1 person every 30 seconds:

I was surprised how consistent the temperature stayed – In total it's only about half a degree C in variation. To get a visualization of what was going on, I made an animation from the same run:

I find it really interesting that the oscillations are consistent enough that the door segments return to their original uniform temperature about every 1.5 minutes. I did not expect such a clean result.

Turning to the comparison with the sliding door, we can consider having direct contact between the inside and outside temperatures for the same time it takes for a person to go through the revolving door. We can add up the energy transfer over time for different rates of entry for both the sliding and revolving door. Below I've plotted the results on a log-log scale:

According to this model, the revolving door is only more efficient if more than 120 people per hour are going in/out. However, I suspect this could change drastically with the choice of the temperature spreading distance I mentioned earlier. A more reliable result may be that all the revolving door cases lose nearly the same amount of energy. I've often scoffed at airports that have motorized continuously rotating doors, on the belief that they were wasting power pumping air between the inside and outside, but perhaps I owe those architects an apology!

Moonshotgun

I got another lunar question from Papou this week: If the same gun was fired on the Earth and Moon, proportionally, how much further would the bullet travel and would it have a greater velocity?

As I thought about this, there were a couple different effects that could make the bullet go further, so let's see how much each one contributes:

Drag

The most obvious difference between the Moon and Earth is the lack of atmosphere. That will come into a couple of these points, but this is the one most people might think of. When a bullet travels through the air, it needs to push the molecules out of the way. According to Newton, the air will push back on the bullet, creating a drag force:

where ρ is the air density, u is the speed of the bullet, A is the cross-sectional area of the bullet, and cd is the drag coefficient. The drag coefficient depends on the shape of the object, since some shapes are more aerodynamic than others. However, for bullets and other projectiles, people instead use the ballistic coefficient, defined as

where m is the mass of the bullet. We can combine these to write

where ad is the drag acceleration. The bullet is slowed down by this drag, meaning it won't get as far before hitting the ground. We can find the ballistic coefficient with tools like this one.

Gravity

Another difference that immediately comes to mind is the difference in gravity: The Moon has about 1/6 the gravity of Earth, which will make the bullet fall more slowly. The bullet's range is limited by how long it takes to hit the ground, so the Moon will allow the bullet to travel further.

Expansion of Gas

Bullets are propelled down the barrel by expanding gas from the gunpowder. Similar to the drag force above, on Earth air will get in the way. In this case though, rather than drag I think a better model is an opposing pressure on the other side of the bullet. If we assume the bullet has constant acceleration a down the length of the barrel L, then the muzzle velocity is 

On Earth, the acceleration will be given by

where P is the pressure from the gunpowder, and P0 is the ambient pressure. Putting everything together, we can get an equation for the muzzle velocity in vacuum based on the one from air:

Plugging in some typical values shows that the speed increase is pretty negligible.

Curvature of Body

Since the Earth and Moon are both (approximately) spherical, as the bullet travels, the ground will begin to slope away, increasing the time it takes for the bullet to hit the ground. When I first started thinking about this question, one thing that came to mind was Newton's Cannonball:

Wikipedia

As the muzzle velocity increases, the projectile has further to fall, until it reaches orbital velocity and continues falling without hitting the body. For short distances, we can approximate this as 

Putting It All Together

I put together some code to run a simulation of the bullet's flight on Earth and on the Moon, but I ran into some problems with the drag calculation using the ballistic coefficient. I ended up approximating the drag using the drag coefficient for a hemisphere. For the bullet/rifle parameters, I used the numbers from Wikipedia's .22 rifle page, although I also tried cranking up the speed to see the differences. By far, the greatest effect on the range is the decrease in gravity, but drag has some influence, particularly for the higher speed. The curvature has no discernible impact.

In case you're wondering whether a gun would even work on the Moon, Mythbusters tested a pistol in a vacuum chamber, and found that the gunpowder does contain all the necessary chemicals to ignite. In fact, during the Space Race, Russian cosmonauts carried the TP-82 Survival Pistol, reportedly to fend off wildlife in case of crashes in the Siberian wilderness, but you can imagine it was advertised in case American astronauts got any ideas. Thanks for a great question, Papou!