I can fudge it by flying over the beacon, turning 180 degrees, waiting 'a bit', then turning outbound. Just wondered if there was a rule of thumb for what 'a bit' is, or if there's another way.
So assuming (a) I've got my trigonometry right and (b) mathematically exact turns, no wind, etc. then I reckon the 'bit' should be ((1-cos(a))/sin(a)).(60/pi) seconds, where 'a' is the angle between the hold inbound course and the course being followed when joining - for direct entry joins on the turn-greater-than-180-degrees side.
This gives 13 seconds for the extreme at 70 degrees. 30, 40, 50 and 60 degrees would be 5, 7, 9, 11 seconds respectively. So course difference divied by 10 (drop the rightmost digit), double, subtract 1.
This method also gets you on the outbound leg early by the same amount.
MC