Checkerboard Cake Volume Ratios

I.  Checkerboard Cake Volume Ratios

Have you ever wondered how to construct a circular checkerboard cake, but you wanted it explained in one second with a dry textbook photo?  Well, you've come to the right place.  Just imagine stacking these cakes on top of each other and cutting a typical slice, also known as a circular sector.  The cross section would be a 3x3 checkerboard.
Pretty neat, huh?

Three Layers of a Checkerboard Cake

Now, I know what you're thinking.  "But Lan, the ratios!  How would I know how much of each batter to make?"*  Before you look it up, I can assure you that the internet says, "Make too much batter and use what's left for cupcakes.  We don't know how many cupcakes, and if you overfill your cake pans, that's on you."  Apparently, people who are better at math don't make checkerboard cakes, and people who are better at baking don't mind making the wrong amounts of batter.  So being mediocre in all ways, I had to be the hero.

For a 9-inch circular cake pan, a classic checkerboard mold will create 3 cakes, each with an outer 9-inch diameter circle, a concentric middle circle with a 6-inch diameter, and a concentric inner circle with a 3-inch diameter.  The radii of these 3 circles are 4.5, 3, and 1.5 inches respectively, and the length of a side of each square is 1.5 inches.  The cakes are made with a dominant batter, chocolate, and a secondary batter, vanilla.  We'll call 1.5 inches r for radius, just for all you 8-inch-checkerboard-cake-making fools.

Review:
The area of a circle = π×r², where r is the radius, and
The volume of a cylinder = (π×r²)(height).
The height for each of the 3 cakes is the same, and we'll call it "height."
In the best of all possible worlds, notice that height = r!

If

V_i = the volume of one inner ring of batter,

V_m = the volume of one middle ring of batter, and

V_o = the volume of one outer ring of batter,

Then

V_i = (π×r²)(height),

V_m = (π×(2r)²)(height) - V_i, and

V_o = (π×(3r)²)(height) - V_m - V_i.

This means

V_i = (π×r²)(height),

V_m = (π×(4r²)(height) - (π×r²)(height), and

V_o = (π×9r²)(height) - (π×(4r²)(height) + (π×r²)(height) - (π×r²)(height),

And

V_i = (π×r²)(height),

V_m = 3(π×r²)(height), and

V_o = 5(π×r²)(height).

To see how these relate, we can divide everything by (π×r²)(height), so

V_i = 1,

V_m = 3, and

V_o = 5.

The ratio of all three rings is 5:3:1.

An outer ring is 5/9 of one cake, a middle ring is 3/9, and an inner ring is 1/9.

5:3:1

From this picture, you can see that you will need 15 parts dominant batter and 12 parts secondary batter to make an entire cake, or

(2V_o + V_m + 2V_i) : (V_o + 2V_m +V_i) = 15:12 = 5:4.

So the dominant batter is 5/9 of the total batter, and the secondary batter is 4/9.

II.  Making 5:4 Obvious

Now, I know what you're thinking again.  "But Lan, I won't remember that!"*

Oh, but you will!  I have pictures!  First, let's talk about that 5:4 dominant to secondary batter ratio.  This, of course, involves all 3 layers of cake.  Imagine cutting a sector, bisecting that sector, rotating it, and attaching it back together.  The birds eye view of such a rearrangement would look like this.

5:4

It's pretty easy to imagine that the ending result is somewhat close to a square prism.  Whether you're the kind of person that likes to imagine decreasing the arc length and watching what happens as it approaches 0, increasing the subdivisions and rotating every other tiny sector, or simply squishing up your cake a bit, you can turn the figure on the right into a real honest-to-goodness square prism.

Top Layer, Middle Layer, & Bottom Layer 

And because 3x3 checkerboards are symmetrical about the axis we rotated that piece of cake, which is the axis extending from this screen to your eyes, our new square of cake looks exactly like a checkerboard, through and through.  Count the sections with me.  Five chocolate squares.  Four vanilla.

III.  Making 5:3:1 Obvious

Let's get back to that 5:3:1.  In a nutshell, look down at the figure on the right and count the triangles.  Let them burn into your memory.  Find the 13 isosceles eagerly awaiting your attention.  They will convince the majority of you.

5:3:1

But for every hundred people that will read this post in a casual yet enthusiastic manner, there will always be one that says, "But, Lan!  That picture with the 13 (16 if you counted the ones in the central figure) eager isosceles wasn't good enough!"*
To which I would say, "I hadn't wanted to get into inscribed polygons."  But then those people, who would have remained unconvinced that the differences between the figure on the left and the figure in the center maintain a ratio of 5:3:1 wouldn't have been able to remember the diagram at all, because it would have been useless.  Furthermore, we all know that those kinds of people are not the kinds of people who would be well equipped to make this cake without further discussion.
In other words, some people reading this post currently believe that the figure on the left has a ratio of 5:3:1, and they also believe the figure on the right has the same.  But they may not find the figure on the right to be a good enough representation of the figure on the left, which means there's no reason they'd remember that V_i's, V_m's, and V_o 's bear a 5:3:1 relationship.  I would know - I used to be one of them.

Before we begin, let's agree to divide everything by height and work in two dimensions; after all, we're only looking at one cake.  Now that that's been sorted, we need to define the concept of what I'll call a sliver.  Simply put, a sliver is the curvy part of a sector that's been cut off.  The curvy little chocolate piece on the left of the 5:3:1 diagram minus the little chocolate isosceles is the smallest of the three slivers we're interested in.  With a streak of unbridled creativity, let's call this tiny piece S_i.  I'll let you decide what to call the middle sliver and the outer sliver.

Red Velvet Slivers

Review:
The area of a sector = (π×r²)(a/360), where r is the radius, and a is the angle of the sector.
The area of a triangle = bh/2, where b is the triangle's base and h is the triangle's height.
sin(θ) = (opposite side)/(hypotenuse), where θ is an angle in a right triangle.
cos(θ) = (adjacent side)/(hypotenuse), where θ is an angle in a right triangle.

Our little chocolate isosceles can be bisected into two right triangles, each having a top angle of half of a.  Let's call (a/2) θ from now on.

To calculate the area of S_i, we first need to compute the area of the smallest sector and then subtract the isosceles.  Because a/2 = θ, the area of the smallest sector is (π×r²)(θ/180).  The area of the isosceles is (half the base)(height), which is also (the side opposite θ)(the side adjacent θ).

Enlarged Top Isosceles & Right Triangles

Because the hypotenuse of the right triangle is r,
sin(θ) = (half base isosceles)/r,
cos(θ) = (height isosceles)/r, and
The area of the isosceles = [sin(θ)][cos(θ)]×r².

Now we know

S_i = (π×r²)(θ/180) - {[sin(θ)][cos(θ)]×r²},

S_m = (π×(2r)²)(θ/180) - {[sin(θ)][cos(θ)]×(2r)²}, and

S_o = (π×(3r)²)(θ/180) - {[sin(θ)][cos(θ)]×(3r)²}.

Once again, because we're only interested in the relationships between these areas, we can divide everything by (π*r²)(θ/180) - {[sin(θ)][cos(θ)]×r²}, and we're left with
S_i = 1,
S_m = 4, and
S_o = 9.

Comparing Differences

To transform the inner chocolate sector from the left figure
to the inner chocolate isosceles in the central figure,
we must subtract S_i from the bottom of the curve, which is a change of -1.

To transform the middle curvy vanilla piece from the left figure
to the vanilla trapezoid in the middle of the central figure,
we must subtract S_m from the bottom and add S_i to the top, which is -4 +1 = -3.

To transform the outer curvy chocolate piece from the left figure
to the chocolate trapezoid at the bottom of the central figure,
we must subtract S_o from the bottom and add S_m to the top, which is -9 + 4 = -5.

Of course, we had already discovered that the ratios of the areas of the circles are 5:3:1, and this relationship would also be true for any circular sector.  Furthermore, the ratios of the areas of the triangles show themselves to be 5:3:1, so of course the differences between the two, namely the various additions and subtractions of slivers, would do the same.  With any luck, however, every single one of us has come to the conclusion that the 13 eager isosceles are not only an acceptable representation of the real deal, but also more memorable than ever.

IV.  Time to Bake

Now that everybody can remember the ratios forever and ever, we can bake.  If you want your squares to resemble squares even a little bit, you will need a very thick cake batter, a steady hand, and a perfectly level oven.  Making one full batch of thick batter cake yields about 2215 g, and when 104 g of chocolate are added, the total weight is theoretically 2319 g.  However, some of our ingredients will be lost from sticking to bowls and spatulas, so we'll plan on a little over 2250.

With a total of at least 2250 g of batter, slightly over 1000 g should be reserved as the secondary batter.  The remainder, when combined with the melted chocolate, should yield slightly over 1250.  That's our 5:4.  Using our 5:3:1, each cake before baking should weigh 750 g, which means each V_o = 417, V_m = 250, and V_i = 83.

If you've made it this far, and your reaction is, "But Lan, it's easier to bake whole cakes first and then cut them up to get cleaner results.  Don't you know that's how people do it?"*
I would say, "Yes, I do, but some of us have our standards."
And if you nodded your head to that, then cheers.

Link to Full Recipe
*narrated in Liên Kratzke's voice because she's the only person who ever says, "But Lan."