1919 Mercator Charts - Scale — Page 313, Lesson 283

Let’s pick this up right where the worked example left off. We’ve just finished Example 4, where we found the scale of a Mercator chart at 30°S by two different routes — first as a pure scale/departure problem, then by finding the scale at the Equator as a normal representative fraction and converting it with the secant of latitude. That second method gave us the same answer, 1:12 831 000, and it’s the one we’re going to lean on now. Now I want to walk you through Example 5, because it looks more complicated than it really is. Here’s the setup: at 40°N, the scale of a Mercator chart is 1:10 000 000. The question asks: what is the distance in centimetres between the 160°E and 160°W meridians at 20°S? At first glance, that seems like a trap. The scale is given at 40°N, but the distance between the two meridians has to be calculated at 20°S. A student might immediately reach for the scale conversion formula we just used — converting the scale from 40°N down to 20°S with the secant rule. But here’s the key insight: that is unnecessary. Think about what a Mercator chart actually looks like. The meridians are drawn as parallel straight lines, evenly spaced. Because they’re parallel, the distance between any two meridians on the chart is the same at every latitude. The chart distance between 160°E and 160°W will be identical at 40°N, at 20°S, and at the Equator. So the whole problem can be solved completely at 40°N, where the scale is given. We don’t need to touch the latitude conversion at all. So let’s do that. The distance between 160°E and 160°W is 40° of longitude — from 160°E to 180° is 20°, and from 180° to 160°W is another 20°, so 40° total. Now, the scale at 40°N is 1:10 000 000. That means 1 centimetre on the chart represents 10 000 000 centimetres on the Earth’s surface. But we need the Earth distance in the same units. The Earth distance corresponding to 40° of longitude at 40°N is the departure — the east–west distance along that parallel. And departure is found by multiplying the change of longitude in degrees by 60 nautical miles, then by the cosine of the latitude. So at 40°N, 40° of ch.long times 60 NM gives 2400 NM, and then we multiply by the cosine of 40° to get the actual departure at that parallel. Now, the scale formula is: Chart Length divided by Earth Distance equals the representative fraction. We know the RF is 1:10 000 000, and we know the Earth distance in nautical miles. We convert that Earth distance into centimetres — 1 nautical mile is 1852 metres, and 1 metre is 100 centimetres, so 1 NM is 1852 × 100 centimetres. Multiply the departure in NM by that factor to get the Earth distance in centimetres. Then we set up the proportion: Chart Length over Earth Distance equals 1 over 10 000 000. Cross-multiply, and the chart length in centimetres falls out directly. That’s the whole trick of Example 5. The moment you recognise that the meridians are parallel on a Mercator chart, the distance between them is latitude-independent, and you can solve entirely at the latitude where the scale is given. No secant conversion needed. The scale conversion formula is only for when you’re moving the scale itself from one latitude to another — not for measuring a distance that’s constant across latitudes. So the takeaway from this pair of examples: on a Mercator chart, the meridians are parallel, so the chart distance between two meridians is the same at all latitudes. When a problem gives you a scale at one latitude and asks for a distance at another, check whether you’re measuring between meridians. If you are, you can ignore the latitude difference and work at the latitude where the scale is known. That’s the clean, professional shortcut — and it’s exactly what the examiner is testing here.