ship dimensions (TL1)

[DanHoward]

Originally Posted by Polydamas

Thanks for taking the time to look up some stats, I could not spare it! It seems like their reconstruction is heavier than we are estimating the Kyrenia ship was.

Their first reconstruction of the Ulu Burun ship was a lot lighter but after examining a wider range of data, they decided that it was unrealistic and created a new model. The result is pretty good considering the limitations involved. IMO the modelling of the Kyrena ship is looking like it is too light since they used many of the same assumptions that the Ulu Burun team initially used.

[whswhs]

Originally Posted by Agemegos

Therefore it seems to me from linearity and similarity considerations that if you have an ellipsoidal hull with half-length a, half-length b, and half-height c floating to draught h in the intended orientation (with two principle axes parallel to the liquid surface and one perpendicular to it), that the hull must displace

V = ab/c² ∙ πh²(3c-h)/3

= π abh²(3c-h)/3c²

Translating to length A and beam B, and with c the height from keel to gunwale, I get

V = π ABh²(3c-h)/12c²

I take it that ab/c^2 is a proportionality constant and that the rest is just the original equation expressed in terms of the vertical principal axis?

[Agemegos]

Originally Posted by whswhs

I take it that ab/c^2 is a proportionality constant and that the rest is just the original equation expressed in terms of the vertical principal axis?

Exactly. abc/c³ is the ratio of the volume of the ellipsoid to the volume of a sphere of radius c.

[whswhs]

Originally Posted by Agemegos

Exactly. abc/c³ is the ratio of the volume of the ellipsoid to the volume of a sphere of radius c.

It looks to me as if it's possible to simplify this, by normalizing to the volume of the body (equated to 1) and by doing the analysis for a sphere, as the proportionate depth should be the same for a sphere and a general ellipsoid. As a further step, working with a hemisphere rather than a sphere gets us a suitable expression for most watercraft. So we have

π abh²(3c-h)/3c² / π ab(2c)/3c² = h²(3-h)/2. And that lets us get

h V
0.1 0.0145
0.2 0.056
0.3 0.1215
0.4 0.208
0.5 0.3125
0.6 0.432
0.7 0.5635
0.8 0.704
0.9 0.8505
1.0 1.0000

which lets us estimate the fraction of a ship's height that is submerged ("draft") if its burden is a given fraction of its flotation rating, for the case of a ship that is an ellipsoid with one of its principal axes vertical. At least, I think I've done the algebra right, and the numbers don't look too implausible.