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Geostrophic and thermal wind. Weather charts, 3 July 2006. 300 mb. Consider the geopotential gradient across the solid red line. Δ z = 952 – 912 Dm = 400 m, ΔΦ = 4000 m 2 s -2 Δ x = 5.5 deg lat = 616 km (1 deg = 111 km) p Φ = 4000 / 616000 = 0.0065 m s -2
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Weather charts, 3 July 2006 300 mb • Consider the geopotential gradient across the solid red line. • Δz = 952 – 912 Dm = 400 m, ΔΦ = 4000 m2 s-2 • Δx = 5.5 deg lat = 616 km (1 deg = 111 km) • pΦ = 4000 / 616000 = 0.0065 m s-2 • Ug = f-1 k x pΦ = 54 ms-1 (f = 1.12 x10-4) • Ug = 105 kt compared with 100-105 kt measured (1 knot = I nautical mile hr-1= 1852 m hr-1 = 0.514 ms-1)
700 mb • Consider the geopotential gradient across the solid red line. • Δz = 316 – 300 Dm = 160 m, ΔΦ = 1600 m2 s-2 • Δx = 5.5 deg lat = 616 km (1 deg = 111 km) • pΦ = 1600 / 616000 = 0.0026 m s-2 • Ug = f-1 k x pΦ = 23 ms-1 (f = 1.12 x10-4) • Ug = 45 kt compared with 45 kt measured (1 knot = I nautical mile hr-1= 1852 m hr-1 = 0.514 ms-1)
Thermal wind 700 mb 300 mb Camborne temperature at 700 mb = 5°, at 300 mb = -37° Valentia temperature at 700 mb = -1°, at 300 mb = -40° ΔT = 6° at 700 mb, 3° at 300 mb, mean around 4.5° Δx = 3.7 degrees latitude = 411 km T = 1.13x10-5 K m-1 ΔUg = - (r/f) T Δln p = (286x104) 1.13x10-5 ln(7/3) = 27 ms-1 Actual value is 30 ms-1 but the calculation is considerably cruder than the 3-figure precision implies.
Surface chart Temperature gradient coincides with a front