Solution to Problems #1. I. Information Given or Otherwise Known. = 28 m/s. Heading of the ultralight aircraft!! h

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1 METR 520: Atmospheric Dynamics II Dr. Dave Dempsey Dept. of Geosciences, SFSU Spring 2012 Solution to Problems 1 Problem 1 An ultralight aircraft is flying. I. Information Given or Otherwise Known Horizontal air speed of an ultralight aircraft c a = 28 m/s. Heading of the ultralight aircraft h = 60 clockwise from north = 30 math angle The ultralight s heading is unchanging. Horizontal ground speed of the ultralight aircraft = 40 m/s Direction of motion of the ultralight aircraft c = 0 math angle The ultralight is flying on a constant pressure surface. Coriolis parameter f = 10-4 s -1 The pressure field is steady. The wind is geostrophic. Acceleration of gravity g =9.806 m/s 2 II. Information Desired Slope of the trajectory of the ultralight aircraft in m/km z ul s s [where z ul sis the altitude of the ultralight and s is the distance along the projection of its trajectory onto a horizontal plane]. III. Relations Needed 1 = + c a [Relation between horizontal velocity of an object relative to the ground,, the horizontal wind velocity,, and the horizontal velocity of the object relative to the air, c a.]

2 III. Relations Needed cont d 2 A B = ABcos A B [The dot product between two vectors, A and B, expressed in terms of the scalar components of the vectors in polar coordinates, where A and B are the directions of the two vectors defined in the plane containing both vectors.] 3 0 = g p Z f ˆk g [Statement of geostrophic balance in isobaric coordinates, assuming a hydrostatic atmosphere. g is the geostrophic wind velocity, and Z is the geopotential height of an isobaric surface. ˆk is a vertical unit vector.] 4 A B = AB sin A B [The magnitude of the cross product between two vectors, A and B, expressed in terms of the scalar components of the vectors in polar coordinates, where A and B are the directions of the two vectors defined in the plane containing both vectors. I. Solution A We are given that the ultralight flies on an isobaric surface and that the pressure is steady, so we know that the isobaric surface isn t rising or sinking. If we ignore the small difference between the altitude of the ultralight and the geopotential height of the same spot, then under these conditions it follows that: z ul s Z s, p s s where Z s, p is the geopotential height along the ultralight s trajectory. But Z s, p s is just the component of the geopotential height gradient in the direction of the horizontal projection of the ultralight s trajectory, which is also the direction of c g. Any component of a vector in the direction of second vector can be found simply by taking the dot product of the first vector with a unit vector in the direction of the second vector. We can construct a horizontal unit vector in the direction of the horizontal projection

3 of the ultralight s trajectory simply by dividing by its magnitude. Hence, it follows that: 5 z ul s s Z s, p = s pz B Solve 3 for p Z, substitute into 5, and invoke the information given that the wind is geostrophic so that = g : 6 z ul s s * f g ˆk + C Solve 1 for and substitute into 6: 7 z ul s s z ul s s c g c g * f g ˆk + c g c a = f, g c g *. ˆk + cg -. f g ˆk * ca Note that ˆk is perpendicular to, so D Apply 2 to 7, noting that = 1: z ul s s 8 z ul s s f g ˆk ca cos g * ˆk ca f g ˆk c a cos g * ˆk ca * ˆk + ca ˆk cg / 1 01 = 0 above.

4 E If we apply 4 to ˆk c a and note that 1 ˆk is vertical while c a is horizontal, so that the angle between them is 90 and sin 90 = 1; and 2 ˆk c a is rotated 90 counterclockwise horizontally from c a but is the same length as c a because ˆk = 1, then we can write ˆk c a ˆk ca = h + 90 and substitute into 8: = c a and 9 F Substitute values and units into 9: z ul s s 10 4 s m s 2 = 28 cos m s cos *104 s 1 m s = *10 4 m m = 14 m *104 m * 1000m km m s 2

5 . Check Solution Although the ultralight is pointed toward the east-northeast, it is actually moving eastward. This implies that the wind must be blowing at some angle to the southeast, as in the diagram below: Note that c g > c a is given, and the directions of g and h are also given and are roughly as shown here. Since the wind is geostrophic and f > 0 implies that the location is in the Northern Hemisphere, if follows that the geopotential height gradient and the horizontal pressure gradient must be oriented to the southwest at some angle so that the horizontal PGF pushes to the northeast at some angle and is opposed by the Coriolis force acting to the right of the wind, toward the southwest at some angle. This means that the ultralight is moving at an angle down the height gradient, toward lower heights, so the slope of its trajectory should be negative, as in fact it is. The magnitude of 0.14 m/km = 140 m/1000 km is very representative of height gradients at midlatitudes at 500 mb above the polar front, as we learned when scaled the horizontal equations of motion in isobaric coordinates. It would produce a geostrophic wind of g = g f 10m s2 pz = 10 4 s m m = 14m s, which is physically quite reasonable and fits well with the relative sizes of the vectors in the diagram above recalling that = 40 m/s and c a = 28 m/s.

6 Problem 2: Suppose that the actual horizontal wind is directed. I. Information Given or Otherwise Known Difference in directions of the geostrophic and actual horizontal wind g = 30 math angle where the wind is to the right, or clockwise looking down, of the geostrophic wind, and hence at a smaller angle math angle, and hence the positive sign for Geostrophic wind speed g = 20m s Coriolis parameter f = 10-4 s -1 ertical component of the wind velocity w = 0 Coriolis and pressure-gradient forces are the only significant forces acting horizontally II. Information Desired Rate of change of horizontal wind speed D D III. Relations Needed 1 D = 1 H p f ˆk [The horizontal velocity equation, neglecting all forces except the horizontal pressure-gradient force and the Coriolis force arising from the horizontal velocity only since w = 0 is given.] 2 0 = 1 H p f ˆk g [Geostrophic wind equation.] 3 A B = ABcos A B [The dot product between two vectors, A and B, expressed in terms of the scalar components of the vectors in polar coordinates, where A and B are the directions of the two vectors defined in the plane containing both vectors.]

7 4 D = ŝ D = D [The rate of change of horizontal wind speed, D, is the component of the vector acceleration, D, in the direction of the horizontal wind. This component can be obtained by taking the dot product of the acceleration with a unit vector, ŝ, in the direction of the wind velocity, which can be created by dividing the wind velocity by its own magnitude,. Note that this relation is really just an application of Eqn. 3, so one could argue that Eq.3 should be invoked to extract this relation during the solution procedure, Section I. below, rather than listing it as a separate, independent relation here. ] 5 A B = AB sin A B [The magnitude of the cross product between two vectors, A and B, expressed in terms of the scalar components of the vectors in polar coordinates, where A and B are the directions of the two vectors defined in the plane containing both vectors.] I. Solution A Solve 2 for 1 H p and substitute into 1: 6 D ˆk g f ˆk B Substitute 6 into 4: D 7 D = f ˆk g f ˆk ˆk g f ˆk ˆk g Note that ˆk is perpendicular to, so ˆk = 0 above.

8 C Apply 3 to 7 and note that = 1 and cos = cos: D 8 D ˆk g cos ˆk g ˆk g cos ˆk g ˆk g cos ˆk g D If we apply 5 to ˆk g in 8 and note the following: a ˆk is vertical while g is horizontal, so that the angle between them is 90 and sin 90 = 1; b ˆk g is rotated 90 counterclockwise horizontally from g ; and c ˆk = 1, then we can write ˆk g = g and ˆk g = g + 90 in 8: D g cos g

9 E Substitute values and units into 9: D = 10 4 s 1 20m s cos = 20 cos s 1 m s = s 1 m s. Check Solution The magnitude of D is on the same order as synoptic-scale PGF/mass and Coriolis force/mass, which seems reasonable, given that the actual wind is more than 10 different from the geostrophic wind in this example as it would have to be, given the 30 difference in their directions. In the Northern Hemisphere f > 0 given, higher pressure lies to the right of the geostrophic wind. Since the actual wind lies to the right of the geostrophic wind, it follows that the actual wind is blowing in a direction at an angle toward higher pressure, which means that there is a component of the horizontal pressure gradient that opposes the wind, causing it to slow down. Hence, the minus sign looks correct.