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Hrm, those aren't working either. I wonder if my initial calculation of the tidal height is off.
I double-checked the math on that by plotting out a graph, and 4.88 seems to be fine, so 7.8 should? work. (I can't get much more precise without trying to find a function that matches the curve, and that seems way too complex for what I'm thinking this should be.)
I also tried running my final numbers with 7.8m, and rounding way down to 7m, and none of that did it either.
I also tried running my final numbers with 7.8m, and rounding way down to 7m, and none of that did it either.
Not something about the excessively precise timing somehow, is it?
The precise timing is the landing at 9 hours and 6 minutes, which is needed for getting the height of the water. The tidal height is 5m at 9, 3.8m at 10, so it's slightly different at 9.1 hours. I got it worked out to 4.88m.
I wonder if "when the tide is falling, round down" refers not to the tidal height, but the carrying capacity of the ship?
Congrats on solving this symbol! The answer was "14021 Tonnes".
For those of you wondering how this answer could be arrived at, here's a quick explanation.
An actual navy guy came up with this puzzle. Here's what he had to say about it:
Apparently this is a VERY simplified version of figuring out how much your ship can carry.
For those of you wondering how this answer could be arrived at, here's a quick explanation.
An actual navy guy came up with this puzzle. Here's what he had to say about it:
Quote:
So the difference between hour 9 and hour 10/100 X 0.1 (because 6 minutes is .1 of an hour)
Subtract from 5m
Theres no indication where the tide is being measured so its a static +3m over the bar, even at the lowest tide
so 7.8m
The next part is a case of exactly the same maths but referencing a weight to a draught
The formula is (Max weight of cargo at Max Displacement/(Max Draught [Full]-Max Draught [Empty])) x available depth of 7.8m -Empty Max Daught))
Subtract from 5m
Theres no indication where the tide is being measured so its a static +3m over the bar, even at the lowest tide
so 7.8m
The next part is a case of exactly the same maths but referencing a weight to a draught
The formula is (Max weight of cargo at Max Displacement/(Max Draught [Full]-Max Draught [Empty])) x available depth of 7.8m -Empty Max Daught))
Apparently this is a VERY simplified version of figuring out how much your ship can carry.
Formatting, my nemesis.
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