SME
Cartography
Map projections
Use this for “On a Mercator map of a 10 cm globe, how stretched is 60° N, and how high above the equator is it drawn?”
Diagram
k
=
1
cos 60
°
=
2;
areas
grow
×
k
2
=
4.
y
=
R
ln
tan(45°
+
φ
2
)
=
10
×
ln
3.73
=
13.17
cm.
Conformal: each circle stays a circle, growing toward the poles, which never fit.
This is an example. Type your own values to start a new problem.
k
=
1
cos(
°)
R
Globe radius
example
a
Area factor
calculated
y
Height of the parallel
calculated
Formulas
k
=
1
cos(
φ
°)
2
=
1
cos(60°)
a
=
k
2
4
=
2
2
y
=
R
×
ln(tan((45
+
φ
2
)°))
13.1696
=
10
×
ln(tan((45
+
60
2
)°))
Assumptions
•
The Earth is a sphere, shrunk to a globe of radius R before it is unrolled onto the cylinder.
•
Mercator is conformal: small shapes keep their angles while areas grow k² times.
•
The poles never fit: y grows without end as φ nears 90°, so the map stops near 85°.
Step-by-step
WE KNOW
φ = 60°, R = 10 cm
FIND
scale factor (k), area factor (a), height of the parallel (y)
Step 1 · Find scale factor (k)
Use k = 1 ÷ cos(φ°)
Each
parallel
is
drawn
as
long
as
the
equator,
though
on
the
globe
it
is
only
cos
φ
as
long.
k
=
1
cos(60°)
k
=
1
0.5
k
=
2
Step 2 · Find area factor (a)
Use a = k²
Stretched
k
times
east–west
and
k
times
north–south,
so
areas
grow
k
2
times.
a
=
2
2
a
=
4
Step 3 · Find height of the parallel (y)
Use y = R × ln(tan((45 + φ ÷ 2)°))
Mercator spaces the parallels farther apart toward the poles, just enough to keep shapes true.
y
=
10
×
ln(tan((45
+
60
2
)°))
tan((45
+
60
2
)°)
=
3.732
ln(3.732) = 1.317
y
=
13.1696
cm
Check
Put the numbers back in. Both sides should match.
2
=
1
cos(60°)
✓
4
=
2
2
✓
13.1696
=
10
×
ln(tan((45
+
60
2
)°))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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