The Mercator Saga (part 2): Finding the Scale

Mercator projection of the World (also a handout students used). Basic grid spacing is 15°, but it includes the tropics of Cancer and Capricorn so don’t be deceived.

You really don’t want to read all the details. But among the crucial steps in getting the lesson outlined last time to work right, the crucialest might be the place where the students figure out that the y-scale goes as the secant.

How do we help them figure that out? (And what do we mean by that?) That’s what we’ll talk about today.

First of all: that y-scale is the number of centimeters per degree in the y-direction, and that depends on latitude. And when you look at a Mercator-projected Earth (like the one in the figure) you can see that this scale increases with latitude. How do we know? Because the lines of latitude get farther apart. So more centimeters per degree.

Coping with Converging Longitudes

To help students get a handle on this, I gave them some questions for homework between the two sessions. One was to figure out how many km there were per degree of latitude, given that the Earth’s circumference is (about) 40,000 km. [ We had had a discussion of the historical meaning of the meter as part of the globe/map literacy phase. ] That answer was easy for calculus students: about 110 km.

We noted that this was the same for a degree of longitude at the equator but that a degree of longitude got smaller (in km) as you traveled towards the poles. The homework question was, “why.” This they understood as well: the lines of longitude converge.

Then I asked, at what latitude is the number of km in a degree half of what it is at the equator. Only one student in about 30 got it in the homework. That was OK, though I had hoped for a few more kids to get it on their own. Zoya knew it would be this way; they had virtually no experience with spheres.

So how to help them figure it out? There was a lot of give-and-take here. These students quickly understood that the circumference of the relevant circle was half the circumference of the equator. They even said the two circles were similar. They painstakingly calculated the radii and discovered (sigh) that the radius of the small circle was also half the radius of the big one.

So I drew them a diagram of a hemisphere, pulled right out of my astronomy heritage. It goes with the question, “We measure latitude in degrees. That sounds like an angle. Where is the angle?”

Shows the relevant triangle in the 3D-ish hemisphere.

At this point, with some kicking and screaming, and a flat, 2D cross-sectional diagram, the students were able to tell me that theta was 60°.

Comparing the World to the Map

Now I could introduce a chart I had made for myself, to help me understand what was going on. The left side compares the shapes of 1° grid “squares” on the globe. Starting out, it looks like this:

Chart Phase One

That is, even though the box is 1° by 1° at both latitudes, at 60°, it’s not square when you measure it in km. It has a 2:1 aspect ratio. We can now look at the globe to see that this is indeed kinda the case.

At 60° North, the red box is roughly 10° by 10°. Notice the 2:1 aspect ratio.

“But on the Mercator map,” I say, “the lines of longitude are parallel. They stay the same distance apart.” So we add to our chart, assuming a scale (at the equator) of 1 cm = 110 km:

Helper chart, phase 2. How many cm tall must the top box be?

Clearly, the question mark must be 2 cm in order to make the rectangles the same shape. (Sleight-of-hand note: on the fly, I decided not to make any big deal about this. In retrospect, this was wise. Or lucky.)

Then the question is, suppose it’s not at the equator and not at 60°? Then what’s the shape of the rectangle? Back on the white board, we enhance the chart:

The chart, phase three. What happens at an arbitrary angle?

It is still not obvious to the students because they mostly used 30-60-90 properties to get the answer for 60° rather than trig. But another trip to the cross-section and they see that the width on the globe is $110 \cos \theta$, and then that the height on the map is $1/\cos \theta$. A couple checks (e.g., with 0° and 60°) convince us that this is correct: The scale of a Mercator map—the number of centimeters in one degree of latitude—is proportional to the secant of the latitude.

Onward

Soon: how we got from there to integration, how we did the integral (which was way cool), and some reflection.