Hail Encounters

What is the probability of an airplane being in intense hail 
continuously for a given period of time?  For purposes of the
analysis, make the following assumptions:

    (a) Every hail storm is circular with radius R and moves 
        in a straight line with velocity u.  Also, every hail 
        storm exists for a length of time T.
    (b) The initial positions of storms are randomly and
        uniformly distributed over an infinite flat plane
        and throughout time.  Also, the directions of motion
        of a storm is randomly and uniformly distributed
        from 0 to 2 pi.
    (c) Storms occur at an average rate such that a stationary
        observer has an expected value of N_0 storms per unit
        time, and every stationary point has an equal probability
        of being in a storm at any randomly selected instant.
    (d) An airplane is modeled as a moving observer travelling
        in a straight line with velocity v.  The direction of 
        motion is uncorrelated with the location or direction
        of motion of any storm.

First we will determine the probability that a moving observer will
encounter a storm at least once during a given time interval.  Then
we will impose the requirement that the duration of the encounter 
must be at least t.

Since the probability of a storm at any given instant is independent
of position, it follows that every observer (moving or stationary)
has the same expected total duration of time-in-storm per unit time.
However, the expected number of encounters will depend on the relative
motion of sotrm and observer, and on the spatial and temporal extent
of storms.  Letting N_v denote the expected number of storms encounters
per unit time for an observer moving with velocity v, and letting D_v
denote the average duration of each encounter, we have

                       (N_v)(D_v) = (N0)(D0)                      (1)

where D_0 is the average duration of a storm for a stationary observer.
The value of D_0 can be deduced from the diagram below.



The total swept area of a storm is (pi) R^2 + 2 R u T, and the portion 
of that region in a storm at any given time is (pi)R^2.  The total 
(area)(time) of the storm is pi R^2 T, and this must equal the
product of the total swept area times the average duration for 
which each encountered point is in the storm.  That is

              pi R^2 T  =  (pi R^2  +  2 R u T) D_0

so we have
                        pi R T                1
          D_0  =    --------------  =   --------------
                    pi R  +  2 u T        1       2 u
                                         ---  +   ----
                                          T       pi R

If an observer has velocity v (relative to the ground) at an angle q
relative to the storm's direction of motion, then the relative velocity 
between storm and observer is the vector sum of magnitude
                   ______________________________
          u_r  =  / u^2  +  2 u v cos(q)  +  v^2

Therefore, we have
                                 pi R T
                  D_v[w]  =  ----------------
                             pi R  +  2 T u_r

Integrating this over q from 0 to 2 pi and then dividing the result by 
2 pi gives the average duration of encounter for observers moving 
with velocity v and arbitrary direction q

                             pi R T
           D_v  =  ------------------------------
                    pi R  +  2 T sqrt(u^2 + v^2)

which, of course, reduces to D_0 when v=0.  We can now solve (1) for 
N_v, which gives
                                           __________                               
                             pi R  +  2 T / u^2 + v^2
               N_v  =  N_0  --------------------------
                                 pi R  +  2 u T

This is the expected number of encounters per unit time for a moving
observer.  Therefore, letting n_v denote the ACTUAL number of
encounters per unit time for an observer moving weith a velocity v,
Poisson's distribution gives the probability of one or more encounters
as
                                                _________
                           /       pi R  + 2 T /u^2 + v^2 \
  Pr{n_v >0}  =  1  -  exp( -N_0 ------------------------- )       (2)
                           \           pi R  +  2 u T     /

To determine the probability of at least one encounter that lasts for
a duration at least equal to some specified time t, we must find the 
fraction of all encounters having duration greater than or equal to
t.  This can be deduced from the diagram below.



The shadded area is the locus of points that are in the storm for at 
least the time t.  This region has an area of

      a  =  pi (w/2)^2  +  w(u_r T  +  2 R  -  2 u_r T  - w)

where
                          ____________________
                   w  =  / 4 R^2  -  (u_r t)^2

Obviously these equations assume the inequalities

                    t < T    and    t < 2 T/u_r

If either of these is false, then no observer will be in the storm
for a time greater than t.

Now, if the storm velocity u can be considered negligible compared
to the observer's velocity, and if T/t is greater than about 30,
then the ratio of the toal number of encounters to those that last
for at least a time t is given by
                             ________________
                            /       / v t \ 2
                  f_t  =   /  1  - ( ----- )
                         \/         \ r R /

Using these same assumptions, the adjustment to the expected number
of encounters due to the motion of the observer given in equation
(2) reduces to
                               2 T v
                f_v  =  1  +  -------
                               pi R

and the equation for the probability of at least one encounter
lasting for at least a time t becomes

      Pr{n_v >0 : dt > t}   =   1  -  exp(-N_0 f_v f_t)