Mass/weight of a triere

[P. Lilius Frugius Simius]

Jasper,

10-12 L of water a day per rower??? That's even more weight that I expected (twice as much as I used on the last row of calculations)... we are starting to get closer to the 70 tons of weight for a "full rigged", full manned trireme...

Yes, the ramming speed is the key to all this dance of numbers... I am 100% positively sure that the speed needed to perforate a trireme hull wasn't very high at all... With all that inertia, an attacking ship would surely need only a small speed to crash a hole on your lower hull and sink you.

Therefore the speed was only used to get there, and not to allow you to escape. Once the collision was inevitable, I am sure the speed would have been lowered as much as possible...

We can make another exercise, it will be especulative, but it will help us to have a better idea of the dimensions of what we are talking about in here...

If we take a trireme of 60 total tons at 7 knots speed v0 and stop rowing, in ideal conditions of calmed water and no wind, we can try to approximate how long it will take the ship to stop herself, and how much space does she needs. Unfortunately the formulae will need a lot of approximations, and its confidence, therefore, low. But as a 1st approximatio it will have to do... :-P P

Drag force for an object on water is proportional to the speed, -bv, where b is a constant that depends on the water density and object's shape, and v the speed.

Therefore, F = m a (approximating zeros...) => v = v0 exp(-bt/m) and the distance to stop will be x = -bv0/m

Now, the trouble here is finding a good enough value of b... Let's consider the trireme a huge kayak (OK, this may be a bit too far-fetched, but...). Experimental measures thrown a varying distance to stop on quiet waters and low winds for a K-4 kayak of 50-100m from cruise speed (about 10 knots) down to "zero"... Considering the K4 has a ICF max. length of 11 m (and a min. displacement of 30 Kg), thus we find a "penetration distance" of 5 to 9 times it's length.

If we get b from here, it gets: b = (40 + 4*75) kg * 8 * L m / 5.15 m/s ~ 5281 kg s

Thus, we get a squeashy value of "penetration distance" at 7 knots for our trireme of x = b * v0 / m = 0.32 m (this is, it would crash against the water like if it were concrete). However, this means that the semi-time (the time to reduce the speed by 1/e) is 11.36 s, this is it would be moving at less than 1 knot in about 25 seconds (which is almost completely stopped) and it would take about the whole length of the ship (35 m) to stop...

This actually means nothing, but we can conclude that a trireme had an enormous drag, and that the length of the prow ram was significantly smaller than what the trireme could ram by its own inertia: slowing down a bit makes perfect sense, specially considering the difficulty to get the ship moving from full stop (my kayaking sources point to the most difficult thing of moving the craft)...

OK, doubtly useful, but fun and illustrative, nevertheless... :-) )

laters!

PS- data from Olympias are also posted here: http://www.mlahanas.de/Greeks/war/Trireme.htm and displacement is about 45,000 kg...