Difference between dates

Me again....

Ok, I thought I had this question answered, but appreantly not. Here's what I'm trying to do:

Read a command line argument as a:
3-letter month string (eg. Jan)
day number (eg. 5)
4-digit year (eg. 1940).

Next I need to calculate the number of days from Jan 1, 1940 to that date.
**No dates before Jan 1, 1940 or after the current year
**Needs to take leap years into account
**Cannot use the date command
Example:
      $ ./cal_dif Jan 5 1940
      4 days

Here's what I have so far:
#! /bin/sh
# ./cal_dif MMM DD YYYY

# Checks for correct arguements
[ $# -ne 2 -a $# -ne 3 ] && (echo "usage: $0 mon dd [yyyy]" && exit 1 )

# Converts month to lower case
tr '[A-Z]' '[a-z]' < $1

# Converts alpha month to interger
case $1 in

      jan) mon=1;;
      feb) mon=2;;
      mar) mon=3;;
      apr) mon=4;;
      may) mon=5;;
      jun) mon=6;;
      jul) mon=7;;
      aug) mon=8;;
      sep) mon=9;;
      oct) mon=10;;
      nov) mon=11;;
      dec) mon=12;;
      *) echo "usage: $0 mon dd [yyyy]" && exit 1;;
esac

year=$3

# Add the number of days appropriate to the month.
  case $mon in
    1|3|5|7|8|10|12) day=31;;
    4|6|9|11) day=30;;
    2)
      if [ `expr $year % 4` -eq 0 ]; then
        if [ `expr $year % 400` -eq 0 ]; then
          day=29
        elif [ `expr $year % 100` -eq 0 ]; then
          day=28
        else
          day=29
        fi
      else
        day=28
      fi
    ;;
  esac

# If days entered > day then $2 invalid
if [ $2 > day ]; then
  echo "Days in month incorrect" && exit 1
  ;;
fi

Now that I have an intereger month, day and year I think I need to convert it to epoch then check that the epoch value is not less than Jan 1, 1940. How do I do this though?