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Sunday, August 2, 2020

Not So Oblious

Marika and I have been getting ready to head down to Florida this coming week, and while paying a farewell visit to her grandparents, Papou posed another of his fascinating questions: Given the Earth is an oblate spheroid (flattened at the poles and bulging in the middle), what direction does gravity point relative to the surface?

If we take a cross-section of the Earth through the poles, we get an ellipse, which is given by the equation
where a and b give the half-width and half-height of the ellipse respectively. We want to try varying these two values to see what happens to the gravity. If we assume a constant density, we can keep the total mass constant by fixing the area of the ellipse:
We can also consider the eccentricity of the ellipse, given by
Using these, we can rewrite a and b as

Newton's Law of Gravity says the force between two masses is
where m1 and m2 are the two masses, r is the distance between them, and G is Newton's constant. For our elliptical planet, we can add up little bits of mass throughout the volume to make a complicated integral:
where ρ is the mass density and A is the space covered by the ellipse. Initially, I thought I could avoid evaluating this nasty object by using Gauss's Law, which would suggest that the pull was always toward the center, but the ellipse doesn't offer the necessary symmetry. I decided to prove this to myself by throwing the integral at Python to solve for me:
The red line goes to the origin, and the black arrow points in the direction of gravity, as found from the integral above. You can see that for large eccentricities, the arrow begins to diverge. We can plot the angular difference:
We get up to around 12° difference, which suggests that an assumption of radial gravity won't cut it. Unfortunately, the numerical integration is both slow, and has inaccuracies for certain points. That leads to some uninformative plots:
Here, theta is the angle from the x-axis, and the horizontal gravity is in arbitrary units. The integrator has problems when it gets to the edges of the ellipse, which leads to some of the force arrows going nuts. If we disregard the deviation discussed above, and assume the gravity really does point toward the center, we can get a much smoother graph, which roughly matches the true solution:
As a check on this, it's often useful in Physics to consider the most extreme case (e.g. zero or infinite mass, length, energy). For this system, we could imagine a planet that was flattened into a disk. The mass ends up concentrated in the center, tapering off toward the edges. If you stand in the center, gravity pulls straight down, but moving away, it would begin to pull back toward the center. If you were standing on the edge of the disk, the center would be straight down, so there's no horizontal force. This leads to the 4 zeros on the plot above: the centers of the top and bottom, and the right and left edges. Thanks for another great question, Papou!

Saturday, July 18, 2020

Vroom!

This morning for breakfast, we got some pastries from a favorite bakery. On the way home, another car honked its horn as it passed, and the changing pitch reminded be of something I've often wanted to try: I wondered whether I could simulate the sound of a vehicle moving by an observer, by calculating the Doppler shift at each point.

The Doppler shifted frequency is given by
where v is the velocity of the vehicle, c the speed of sound, and f0 the original frequency. The dot product with r means we only take the part of the velocity that is along the line-of-sight between the vehicle and the observer. That dot product is what leads to the changing pitch: As the vehicle approaches the observer, the angle decreases, lowering the pitch.

I decided to put together another HTML5 doodad to play around with (make sure you have your sound on):

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True Frequency


Velocity (mph)


Sunday, July 12, 2020

Anna, Mom-Eater!

While watching the morning news recently, I noticed their graphic of a wind gauge during the weather segment:
Wikipedia
Initially, I imagined the cups were moving at the same speed as the air, but then I started wondering about how the backs of the cups would slow the rotation – The wind is pushing on both sides, so how much gets cancelled?

The key is that the cup's drag coefficient changes as it rotates. This coefficient is defined as
where F is the force applied to the cup, ρ the density of air, u the speed of the air, and A the area of the cup facing the wind. Wikipedia gives the drag coefficient for the outside as 0.38, and the inside as 1.42. To get the area of the cup we need, we can use the same type of stereographic projection from an earlier post. In theory, the drag coefficient would change continuously as the cup rotates, but we can get an approximation by assigning the inside/outside coefficients to the inside/outside area projections. Putting everything together
This is just for a single cup, so now we turn to the full anemometer design. Initially, they were built with 4 cups at right-angles, but these have mostly been replaced by 3 cups at 120° angles. I decided to try both:
The Wikipedia article I linked to above says, "The three-cup anemometer also had a more constant torque and responded more quickly to gusts than the four-cup anemometer." My model doesn't show this, with the 4-cup design (red) giving less variation in force, and a larger mean force. I imagine this is due to the drag approximation I mentioned, which leads to the sharp knees in the plot above.