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The geostrophic wind relation

The geostrophic wind relation. V g = (1/f ) k x  H p z = (g/f) k x  H z p . Northern Hemisphere:. p =1016 mb. (-1/ )  H p z or -g H z p. v g. p =1032 mb. 90 . 90 . Z = 900 dam. -f k x v g. Z = 912 dam. Computation of geostrophic winds.

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The geostrophic wind relation

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  1. The geostrophic wind relation Vg = (1/f)k x Hpz = (g/f) k x Hzp

  2. Northern Hemisphere: p=1016 mb (-1/)Hpz or -gHzp vg p=1032 mb 90 90 Z = 900 dam -fk x vg Z = 912 dam

  3. Computation of geostrophic winds • Draw a line normal to the isobars or height contours at the point in question. This will be the n-axis • Lay off a distance n on this axis, centered on the point in question • The value of n can be chosen for convenience by observing that the speed of the geostrophic wind, in finite-difference approximation, is given by • Vg=(g/f)(z/ n)

  4. Now, we compute n (the magic distance!) • n = (g/f)(z/Vg) • Now choose z/Vg = 60 meters/20 knots, so that a difference of 60 m in the distance of n yields a speed of 20 knots. (sea-level isobars at 8-mg intervals can be converted to 6-dam contours of the 1000-mb surface)

  5. Therefore: • n (degrees of latitude) =(5.15/(104f)

  6. Latitude (deg.) n (deg. Latitude)90 3.5385 3.5580 3.5775 3.6570 3.7665 3.9060 4.0955 4.3350 4.6045 5.0040 5.4835 6.1330 7.0525 8.3120 10.315 13.5510 20.605 39.610 

  7. In practice, one simply lays off the appropriate distance n and inspects how many 6 - dam channels are contained within it. Each channel is worth 20 knots. If contours are at intervals of 12 dam, each channel is worth 40 knots.

  8. The thermal wind can also be found using the same technique: vT = (g/f) k x (z2 - z1)

  9. Once the direction and speed of the geostrophic wind have been found: The components may be found: Ug= - (geostrophic wind speed) sin (wind direction) Vg = - (geostrophic wind speed) cos (wind direction)

  10. I like Montreal because the magic distance is 5 deg. Latitude!

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