SME
Cartography · Problem type
Cylindrical equal-area map
Use this for “On a cylindrical equal-area map of a 10 cm globe, how high is 60° N drawn, and how are shapes stretched there?”
Diagram
y
=
R
sin
φ
=
10
×
sin
60°
=
8.66
cm.
kE
=
1
cos φ
=
2,
kN
=
cos
φ
=
0.5:
kE
×
kN
=
1,
so
areas
are
true
and
shapes
squashed.
This is an example. Type your own values to start a new problem.
y (cm)
=
R (cm)
sin(
φ
°)
kE
East–west scale
calculated
kN
North–south scale
calculated
a
Area factor
worked out (not typed)
Formulas
y
=
R
×
sin(
φ
°)
8.6603 = 10 × sin(60°)
kE
=
1
cos(
φ
°)
2
=
1
cos(60°)
kN
=
cos(
φ
°)
0.5 = cos(60°)
a
=
kE
×
kN
1 = 2 × 0.5
Assumptions
•
The Earth is a sphere, shrunk to a globe of radius R; each parallel is carried straight across onto the cylinder.
•
East–west the map stretches by kE; north–south it squeezes by kN.
•
The area factor is kE × kN; it stays 1 everywhere, so areas are true while shapes are squashed.
Step-by-step
WE KNOW
φ = 60°, R = 10 cm
FIND
height of the parallel (y), east–west scale (kE), north–south scale (kN), area factor (a)
Step 1 · Find height of the parallel (y)
Use y = R × sin(φ°)
Straight
across
from
the
globe
onto
the
cylinder:
the
parallel
is
R
sin
φ
above
the
equator.
y
=
10
×
sin(60°)
y
=
10
×
0.866
y
=
8.6603
cm
Step 2 · Find east–west scale (kE)
Use kE = 1 ÷ cos(φ°)
Every
parallel
is
drawn
as
long
as
the
equator,
though
on
the
globe
it
is
only
cos
φ
as
long.
kE
=
1
cos(60°)
kE
=
1
0.5
kE = 2
Step 3 · Find north–south scale (kN)
Use kN = cos(φ°)
The
parallels
crowd
together
toward
the
poles,
squeezing
north–south
by
cos
φ
.
kN = cos(60°)
kN = 0.5
Step 4 · Find area factor (a)
Use a = kE × kN
The area factor is the two stretches multiplied.
a
=
2
×
0.5
a
=
1
Check
Put the numbers back in. Both sides should match.
8.6603 = 10 × sin(60°) ✓
2
=
1
cos(60°)
✓
0.5 = cos(60°) ✓
1 = 2 × 0.5 ✓
Topic check
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