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:
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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? |
[PRINTER FRIENDLY VERSION]
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