Let Them Heat Cake

[Programming Note: I've noticed looking through older posts that the equations have picked up extra + signs, due to my impolite hotlinking.  I plan to go through and fix those when I have a chance, but until then, don't take them at face-value.]

Another great question from a reader, this time my own mother, Sally: I’m making a carrot cheesecake for Steve’s birthday. The cheesecake layer is supposed to go into an 9” springform pan but I only have 8”[...] I assumed I’d just make the cheesecake layer thicker, but how to determine baking time? The volume is the same but the height is 25% greater. Thicker tends to mean longer baking. But how do i determine that? What role does oven temp play? What role does moisture level of cheesecake play?

I've talked about heat transfer before, in this post about making granita, but I'd like to take a different approach this time, using the more general heat equation:

This says that the rate of temperature change at a point in the cake is proportional to the variation in temperature nearby, and the thermal diffusivity, α.  This quantity depends on the substance we're interested in, but I don't think it's been tabulated for cheesecake batter, so we'll assume it's about the same as water, 0.143 × 10⁻⁶ m²/s [this is the role the moisture level plays].

Technically, this is a 3-dimensional problem, but thanks to the cylindrical symmetry of the cake, we can just consider a 2d cross-section through the center.  We can assume the outer surface of the cake is fixed at the oven temperature, 325°F.  Then we can use a numerical solver to find the temperature throughout the cake over time.  As it happens, I wrote a solver for this equation a few years ago for a Computational Physics class.  After a few adjustments, it was ready to go: bake.py

First, we need to find the internal temperature that the cake reaches after the prescribed 45 minutes of baking.

The final temperature after 45 minutes is 206°F.  This is a little close to the boiling point of water, 212°F, where I expect things to get a bit off from the approximations I'm making, but we'll go with it.  Then we can start again with a narrower cake of the same volume, and find how long it takes to get to that temperature.

The final time is about 70 minutes, which isn't completely unreasonable, but I take no responsibility for any charred cakes this calculation results in.  Thanks for a great question, Sally!

[Edit: One first posting, I mistakenly used 8/9" as radii, rather than diameter.  It doesn't actually make a difference in the final results, since the thickness dominates.]

Precipitous Power

We've been getting a lot of rain here in Ann Arbor the last few days, and it reminded me of a question I've often wondered after being hit in the head by a particularly large drop: How much power could you generate from rainfall?

The kinetic energy of each drop is

where m is the mass and v is the velocity when it hits the ground.  Clearly we need some statistics about raindrop size; a quick search turned up this paper, from a collaboration between Brookhaven National Laboratory and a group of Chinese institutions.  The important info we need is in Fig. 5:

The y-axis, volume-mean diameter, refers to the diameter of a perfect sphere with the same volume/surface area ratio as the drop.  Coincidentally, this is also known as the Sauter mean diameter.  Unfortunately, the paper doesn't provide a fitting function for these curves, but we can take a guess, estimating values from pixels:

x	y
1.149825784	0.8509578544
3.066202091	0.9689655172
4.982578397	1.078544061
7.857142857	1.137547893
13.79790941	1.230268199
27.21254355	1.348275862
48.67595819	1.390421456

I chose the convective points for their larger range of rates.  Based on those, I asked R for a log fit and got this, with an R-squared of 0.992:

The relationship between mass and diameter is

where ρ is the density of water.  We still need to find v, but if we assume the drops hit terminal velocity for a sphere, we can use eqn 3 from this paper:

where g is the acceleration due to gravity, and I've substituted in the drag coefficient for a sphere.  Putting these together, the energy becomes

This is the energy of each drop, so we need to multiply by drops per time to get power.  The drops per time is related to the rain rate from above:

where A is the collection area.  Putting everything together, we have

If we plug in values, this comes to

At the highest rain rate from the paper, this comes to 0.081776 Watts per square meter, while solar cells can put out a couple hundred Watts per square meter, so maybe not the solution to our energy problems!

This one wound up being a bit dense, sorry.  I'll try to pick a more accessible topic next time!

Lunar Loop-the-Loop

This question comes from my girlfriend Marika's grandfather, who studied physics in his youth: What shape does the moon trace as the Earth orbits the Sun?

The idea is, the Moon is going in (approximately) a circle around the Earth, while the Earth is going in (approximately) a circle around the Sun.  This is a surprisingly simple situation to model using a parametric equation:

The 'R's are the radii of the Earth's and Moon's orbits, and the 'T's are the time each takes to complete a circuit.  This type of curve is called an epitrochoid, and can take on some interesting shapes:

 

The general form for this type of curve is

We can relate this to the earlier equations with

The site I linked above states that there will be loops if b > a/c, so let's plug in some values (from Wikipedia):

Nowhere close, a result borne out by the plot:

If you look carefully, you can see that's not quite circular, but it's awfully close.  Thanks for a great question, George!

[Edit: If you're curious about the b > a/c condition, you can think of it this way – There will be a loop anytime the Moon is moving backwards faster than the Earth is moving forwards.  The speed of each is given by 2πR/T, and if you plug in the R and T relations for a, b, and c, you'll see that's exactly what b > a/c means.]