January 15, 2007
And the Winner is...

Congratulations to Ian Tosh from Rule Broadcast Systems and former employee (now freelance sound mixer) Bryan Dembinski for correctly answering last month’s riddle, reproduced below.

The correct answer, as supplied by Charles Parra of Denecke, is as follows:
  • Atomic Clock: 01:00:00
  • GR#1(30ND): Display = 01:00:00:00;
    Actual elapsed time (no frames): 01:00:00.0
  • GR#2 (30D): Display = 01:00:03:18;
    Actual elapsed time (no frames): 01:00:03.6
  • GR#3 (29.97ND): Display = 00:59:56:12;
    Actual elapsed time (no frames): 00:59:56.4
  • GR#4 (29.97D): Display = 01:00:00:00;
    Actual elapsed time (no frames): 01:00:00.0035982

The offset with drop frame comes from the fact that it is just a way of "fudging the books". The two frames being dropped per minute don't add up to a "whole number". Therefore, there is always a slight offset.

The riddle:

Suppose I have 4 Denecke GR-1's and an Atomic Clock accurate to within a few picoseconds.

  • I set GR-1 #1 to 30 ND.
  • I set GR-1 #2 to 30 DROP
  • I set GR-1 #3 to 29.97 ND
  • I set GR-1 #4 to 29.97 DROP

Each GR-1 is set to 00:00:00:00. When the Atomic Clock strikes midnight, it instantaneously triggers the GR-1 boxes to begin counting forward.

The Question: When the Atomic Clock reaches 1:00am, what is the time code displayed by each GR-1, and what is the corresponding elapsed time of that time code?


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