A cartographer is projecting a cylindrical map of longitude and latitude using the Mercator projection. If the true width of a region between 0° and 6° east longitude is 420 kilometers, and the map scale at the equator is 1:10,000,000, what is the width of this region on the map in centimeters?

["Projecting Longitude with Mercator: Calculating Cylindrical Map Width at 0°–6° East", "When cartographers project the Earth’s surface onto a cylindrical map using the Mercator projection, accurate representation of distances—especially along lines of constant longitude—remains a key challenge due to the distortion inherent in projecting a spherical globe onto a flat cylinder. One critical measurement involves determining how wide a region appears on a cylindrical Mercator map, particularly for east-west extents along specific latitudes.", "Consider a region between 0° and 6° east longitude. On Earth, 1 degree of longitude at the equator spans approximately 111 kilometers. Therefore, 6° east corresponds to:", "[\n6^\circ \ imes 111~\ ext{km/degree} = 666~\ ext{km}\n]", "However, the Mercator projection preserves angles but distorts scale—especially at higher latitudes and along meridians. While the equator scale is 1:10,000,000 (meaning 1 cm on the map = 10,000 km in reality), this scale applies uniformly only along the equator. East-west distortion increases with latitude due to convergence of meridians.", "Despite this non-uniform distortion, Mercator maps maintain conformality (angle accuracy), meaning straight lines on the map represent rhumb lines (paths of constant bearing). For small east-west extents near the equator, the longitudinal width on the map closely approximates true width, adjusted only by minor scale variation.", "Given the equatorial scale of 1:10,000,000, 1 km corresponds to:", "[\n\frac{1~\ ext{cm}}{10,!000,!000~\ ext{cm}} = 10^{-7}~\ ext{cm per km} = 1~\ ext{metre per centimeter} \ imes 10^{-5} \quad (\ ext{wrong unit handling})\n]", "Correctly:\n1 cm on map = 10,000 km = 10,000,000 cm = 100,000 meters\nSo:\n[\n666~\ ext{km} = 666,000~\ ext{m} = 666,000 \ imes 100 = 66,600,000~\ ext{cm}\n]", "Thus, on the map, the width is:", "[\n666,!000~\ ext{cm (true width)} \div 10,!000,!000 = 0.0666~\ ext{cm} = 0.666~\ ext{mm}\n]", "But this ignores projection geometry.", "In Mercator, the projected width ( w_{\ ext{map}} ) for longitude span from ( \lambda_1 ) to ( \lambda_2 ) is:", "[\nw_{\ ext{map}} = \ ext{scale factor at equator} \ imes \Delta\lambda \ imes 111~\ ext{km/degree}\n]", "The Mercator scale factor at latitude ( \phi ) is:", "[\nk(\phi) = \frac{1}{\cos\phi}\n]", "But at the equator (( \phi = 0^\circ )), ( \cos\phi = 1 ), so the scale remains 1:10,000,000 across the equator — effectively constant for small longitudinal intervals near the equator.", "Thus, even with convergence, a small 6° stretch at the equator is projected with scale preserved:", "[\n\Delta\lambda = 6^\circ, \quad \ ext{True width} = 6 \ imes 111 = 666~\ ext{km} = 66,600,000~\ ext{cm}\n]", "Scale: 1 cm = 10,000 km = 100,000,000 cm?\nNo:\n1:10,000,000 → 1 cm = 10 km? Wait — correction.", "Clarify units:\n1:10,000,000 scale means 1 cm = 10,000,000 cm = 100 km? No.", "Wait: 10,000,000 cm = 100,000 meters = 100 km? No:", "100,000 cm = 1 km → So 10,000,000 cm = 10,000,000 ÷ 100,000 = 100 km? No.", "10,000,000 cm = 10,000,000 ÷ 100,000 = 100 meters? No.", "100,000 cm = 1 km → So:", "[\n1~\ ext{cm on map} = 10,!000,!000~\ ext{cm} = 100,!000~\ ext{m} = 100~\ ext{km}\n]", "Yes — scale is 1 cm = 100 km.", "But the problem states: “the map scale at the equator is 1:10,000,000” — this is ambiguous.\nTypically, 1:10,000,000 means 1 cm = 10,000,000 cm = 100 km.", "However, many cartographic standards use 1 cm = 100 km for large-scale cylindrical projections near equator.", "But 6° longitude = 666 km → ( \frac{666}{100} = 6.66 ) cm — only if scale preserved.", "Given the Mercator projection’s equatorial scale is stated as 1:10,000,000, and assuming 1 cm = 100 km along equator (common in such cylindrical systems), the projected width of 6° longitude is:", "[\n\frac{666~\ ext{km}}{100~\ ext{km/cm}} = 6.66~\ ext{cm}\n]", "But this contradicts scale distortion?", "No — on Mercator, longitudinal scale does not change with latitude at small east-west scales near the equator — distortion from pure scale increase is avoided via the standard cylindrical projection that restores distance along parallels.", "In the equirectangular projection, distances along parallels increase with ( \cos\phi ), but in the Mercator projection, that distortion is eliminated, and the scale factor becomes ( \sec\phi ), increasing toward poles. However, at the equator, ( \cos 0^\circ = 1 ), so the scale factor is exactly 1 — no distortion.", "Therefore, across latitudes near 0°, the Mercator projection accurately represents east-west distances.", "Hence, a region from 0° to 6° east spans ( 6^\circ \ imes 111.32~\ ext{km/degree} \approx 667~\ ext{km} )", "With equator scale 1:10,000,000 → 1 cm = 100 km → so:", "[\n\frac{667}{100} = 6.67~\ ext{cm}\n]", "But the scale 1:10,000,000 is likely meant as 1 cm = 10 km? No, that’s too small.", "Most standard systems use 1:10,000,000 to mean 1 cm ≈ 100 km (since 1:10,000,000 = 1 cm = 100 km). This is consistent with large-scale urban or regional mapping.", "Therefore, the width on the map is:", "[\n\frac{666~\ ext{km}}{100~\ ext{km/cm}} = 6.66~\ ext{cm}\n]", "But to respect precision:\n( \cos 0^\circ = 1 ), so Mercator preserves scale exactly at equator → no convergence effect.", "Thus, the width is simply:", "[\n\Delta\lambda \ imes \left( \ ext{width at equator per degree} \right) \div \ ext{scale factor}\n]", "( \Delta\lambda = 6^\circ ), true width = ( 6 \ imes 111.32~\ ext{km} \approx 667.92~\ ext{km} )", "Scale: 1 cm = 100 km → even at equator, standard efficiency assumes 1 cm = 100 km, so:", "[\n\frac{667.92}{100} = 6.6792~\ ext{cm}\n]", "But the problem says the scale at equator is 1:10,000,000 — so interpreting as 1 cm = 10,000,000 cm = 100 km — yes, consistent.", "So final answer:", "[\n\boxed{6.68~\ ext{cm}}\n]", "However, for exactness, since 6° × 111 km = 666 km, and 1 cm = 100 km on map:", "[\n\frac{666}{100} = 6.66~\ ext{cm}\n]", "Rounded to two decimal places:", "[\n\boxed{6.66~\ ext{cm}}\n]", "But in alignment with geographic projection standards and the precise equatorial scale, the width of the region on the Mercator map at the equator between 0° and 6° east is best calculated as:", "[\n\ ext{Width on map} = \Delta\lambda \ imes \frac{111~\ ext{km}}{100~\ ext{km/cm}} = 6 \ imes \frac{111}{100} = \frac{666}{100} = 6.66~\ ext{cm}\n]", "Thus, the cylindrical Mercator-projected width is 6.66 centimeters.", "---", "### Key Takeaway:\nDespite the Mercator projection’s known scale distortion at higher latitudes, at the equator where ( \cos 0^\circ = 1 ), the projection preserves east-west scale accuracy. Therefore, a longitude band from 0° to 6° spans exactly ( 6 \ imes 111 = 666 ) km, which maps to 6.66 cm on a 1:10,000,000 scale equatorial cylindrical map.", "This precise projection enables reliable navigation and spatial analysis, critical in applications like environmental monitoring and infrastructure planning — where accurate representation of east-west distances near the equator is essential."]