Forums » Epic Week 2023 » Puzzle 5 Symbol (SOLVED)

This topic is for discussing clues and potential answers to the Puzzle 5 Symbol in Epic Week 2023. Please stay on topic. Remember, these puzzles are designed to be challenging and require a group of people brainstorming together; don't feel intimidated if you don't know the answer right away. Every little idea can help. In puzzles, all information you will need to solve it is contained within Epic Week materials.

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I tried

9:06
Six minutes past the ninth hour
6 past 9

so it's probably not about that specific bit.

How, though? There's been monsters in the way since this opened up.

...Did I try on the wrong one? 🤔 I thought there were no monsters in the way at the same time as there were none on a different one. Like two crystals free at once, I mean. I guess I'll retry those here when it's free ^^' in case I had the wrong symbol

Based on what I've looked through so far and what has been revealed without the third clue, "land" and "timing" would connect with the following:

maybe this
or
this

perhaps:
9:06
6 minutes
6 minutes past
before dawn
round down
round down for safety
3 meters
3 groups
3
6
9

I guess that could be so if it happened in the itty bitty gap between puzzle 2 being solved, the solve spawn being put down, and the hourly spawn. I didn't see a gap between them but I absolutely could have missed it.

wow i do be popping off here

okay so weight as the newest factor.

"We will have to choose the ship that undertakes the solo mission carefully, as well as what we send with it. We will have a great deal of cargo to haul to have everything that will be needed to establish the camp and its defenses. But if we weigh it down too much, it won't have enough clearance. I am still calculating the exact cargo capacity that will be possible, but if I can make the math work, we will be better positioned from day one."

alright these are where I'm looking at

Maybe it's the amount of walruses/cargo, related to the sketches of ships' hulls? Helpful links, thank you... Perhaps it's 20000, like the number of maximum loading on the last figure.

I don't really know how to read water level maps (is that the tide or just the regular waveline?) but maybe the clearance has something to do with that. Seems kinda complicated so maybe not.

i think ive figured out where we need to be looking actually.

i think the answer is numeric and has to do with some variation of math in reference to: the time, (9:06), and finding which vessel out of the 6 pictured on the last piece of research has the ability to contain the most cargo, while still being able to clear with draught.

HMMM

alright ive guessed:

hull 1-6
every number including first decimal from 0.1-14
all of the numbers on every sheet (last 3 i think)
every max cargo load
weight, clearance, draught, cargo
sobbing.

this map

this graph

this graph

there is some kinda math going on here and i cannot.

edit: the crulest thing kim has ever done is have a bunch of writers try and do math /s

Good thing some of us studied math, then.

Alright, lemme write this out as I go so I have some notes -

They're arriving by hour 9, but need to wait 6 minutes for the proper lighting conditions.

Hour 9 5
Hour 10 3.8

So at 9:06 you'd be looking at about 4.88 height - the graph there looks pretty linear, hasn't yet started to curve out for the bottom - but... there's a note that says "when tide is falling always round down for safety", so the tide is falling so that's either just straight 4 or 4.8, if we assume that a tenth position is okay since that's what the rest of the numbers are.

Here I am interrupted by a goblin, so this'll have to wait to get further

Other helpful notes: "Low water clearance: 3 meters

Tidal range not the depth of water

Remember to add the clearance"

Alright, so - I ended up with an answer of 6400 T, but that doesn't seem to do it. So let's put down my math and see if people can massage the hole out of it.

So each boat hull has different draughts (the amount of space the boat takes up underneath from the top of the water). They all also have different max tonnage they can hold - and each boat has a different amount of tonnage per meter of displacement. We obviously want to maximize the amount of cargo that can be taken while still being able to clear the sandbar and make landing at the appointed hour.

So, I'm calculating the tonnage per meter as follows: Max tonnage / (Max draught fully loaded -- max draught empty)

Which works out to:
Hull 1: 4600 / (4.2 - 2.5) = 2705.88 T
Hull 2: 8000 / (5.3 - 2.8) = 3200 T
Hull 3: 12000 / (6.8 - 3) = 3157.89 T
Hull 4: 15000 / (8.1 - 3.5) = 3,260.87 T
Hull 5: 18000 / (9 - 5.1) = 4615.38 T
Hull 6: 20000 / (10.2 - 7.6) = 7,692.31 T

The bigger boats can take way more stuff, but they also start out too tall for the 4.8 m height we got from earlier math. So let's just scratch 5 and 6 right off the list.

From the remaining boats, Hull 1 could go in fully loaded, but the others would have to only be partially loaded, and that's what we need to figure out, I think.

So, I arrived at the answer I got by taking our 4.8 m, subtracting the empty weight displacement, and then multiplying that by the tonnage per meter of displacement. That gives us in order:

Hull 1: 4600 T (Max weight)
Hull 2: 3200 * 2 = 6400 T
Hull 3: 3157.89 * 1.8 = 5684.2 T
Hull 4: 3260.87 * 1.3 = 4239.1 T

So it looks like Hull 2 is the ship Sylviornis wanted, and that it could take 6400 T of assorted cargo. But apparently my math has veered off into a ditch somewhere, so please correct me.

I'm wondering if it's not the "Tidal range not the depth of water" note that's causing the issue, but I'm not entirely sure where to get the actual depth from, if that's the case.

I am learning way more about calculating moorage than I ever wanted to.

(I legit have a fear of falling into water, especially deep water, LOL. This will never ever come in handy for me.)

Ok starting this by saying maths is not my strong point, but here goes.

Looking over the research, I think maybe we need to find the "Exact cargo capacity that will be possible".

Not sure I interpreted everything right, but hopefully maybe some points I've gone over might help with brainstorming the real answers...

I think the answers could be:
- 12,000 walruses
- 18000T
- 18,000,000kg
- Hull 5
OR
- 8,000 walruses
- 12000T
- 12,000,000kg
- Hull 3

Some of my reasoning/interpretations that led to this:
- "Solo ship should arrive by hour 9 but may need to wait 6 minutes to make landing occur exactly half an hour before dawn"
- "Always round down when tide is falling for safety"
- Walruses are 1500kg.

- If we assume "rounding down" was just meaning rounding down decimal points involved in the meters...
- Hour 9 height is 5m, plus ("Remember to add the clearance") the 3m clearance is 8m total.
- Hull 5 has 5.1m draught when empty, 9m draught when fully loaded, and the max weight is 18000T.
- 18000 T is 18,000,000kg.
- 18,000,000 kg % 1500 kg = 12,000 walruses

- Does rounding down intend for us to round down to the next hour? Normally that would mean rounding down from the decimals but... if so the secondary working could be:
-- Hour 10 height is 3.8m, plus ("Remember to add the clearance") the 3m clearance is 6.8mm total.
-- Hull 3 has 3m draught when empty, 6.8m draught when fully loaded, and the max weight is 12000T.
-- 12000 T is 12,000,000kg.
-- 12,000,000 kg % 1500 kg = 8,000 walruses

I tried all the answers from above, and it was a valiant go, but none of them worked.

I'm going to go back to my initial math from above, now that we've got that bonus clue - add the clearance to tidal height that was calculated earlier, and we've got 7.8 m.

Which means any of the first three ships would work fully loaded. I think hull 6 would still be too big so I'm not gonna bother with it. The others...

Hull 4 = 14021.74 T
Hull 5 = 12461.53 T