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Atmospheric Science 4320 /
7320
Lab Portion / Anthony R. Lupo
Lab 1 - Coriolis
 “Thursday is Lab Day”
 Lab 1: Real and Apparent Forces, The
Coriolis Force
 Read Ch. 1 from Holton p 14-19
 Newton’s Second Law:


dV

 F  ma  m dt
Lab 1 - Coriolis
 Then the summation of F ( SF ) involves
several forces. Thus,

SF = PGF + CO + Gravity + Friction
+ Viscous forces
 where gravity =
absolute gravity + centrifugal force
Lab 1 - Coriolis
 Other forces such as electrostatic forces or
magnetic forces are negligible for typical
scales of atmospheric motions and are thus
neglected!
 Real forces: PGF, Gravity, Friction, and
Viscous forces!
 Must exist in both inertial (non
accelerating) and non-intertial coordinate
systems.
Lab 1 - Coriolis
 Apparent forces: Coriolis force 
 
 2 V
 Coriolis Force is due to the fact that
the coordinate system we use is on a
rotating earth, which is of course,
NOT an inertial coordinate system. (V
!= 0).
Lab 1 - Coriolis
 A “derivation”:
 First, let’s define our “position vector”

r  xiˆ  yˆj  zkˆ
 And,
(1)
 dr Dr

V

  r
dt Dt
 or 
 

V  V    r
Lab 1 - Coriolis
 Now the same for the acceleration in a
moving system: (2)
 Then put (1) into (2):



 dV DV 
a

  V
dt
Dt




dV d V 
dr 


     V       r
dt
dt
dt

Coriolis
Centrifugal
Lab 1 - Coriolis


 r
(recall cross product
– the resultant has to be
mutually perpendicular to all three!)



dV d V 
2


 2  V   r sin rˆc
dt
dt

Coriolis
Centrifugal Acc.
(points to or away from axis of
rotation)
Lab 1 - Coriolis
 Then, substitute above expression
into the Equation of Motion.
 The Coriolis Force: ( -2xV )
  0iˆ   cos ˆj   sin kˆ
Lab 1 - Coriolis
 When the vertical velocity is small
compared to the horizontal motions,
the horizontal component of the
Coriolis force is:

f = 2sin()
 “” is your latitude.
Lab 1 - Coriolis
 Coriolis force deflects moving objects on sufficient
time and space scales to the right in the Northern
Hemisphere and to the left in the Southern
Hemisphere.
 For horizontal motions:
 This “sine” relationship (cross product) assures that
when the rotation vector is perpendicular to the
motion vector, but in the same plane as V ( = 0), and
thus  x V (at the equator) is perpendicular to the
horizontal plane, f = 0. The Coriolis force is all in the
vertical!
Lab 1 - Coriolis
 f = 0  No horizontal comp!!!
 Then when the rotation vector is
perpendicular to the motion vector
(angle  = 90, or p/2), thus  x V
perpendicular to the vector V and lies
in the same plane, f = 2 coriolis
force, or is at a maximum.
Lab 1 - Coriolis
 f is the coriolis parameter of
“planetary vorticity”. Recall vorticity is
the curl of the velocity vector!