Generally speaking the high levels of cover are things like bunkers and bulkheads. So assuming a 3 foot by 4 foot shield (about 90cm x 120cm) with a faily conservative foot thick (about 30cm) and mild steel being 7.85g per cm^3. That comes to 324,000 cm^3 and 2543kg in weight, or about 2.5 tonnes.
"mild steel is 7.85g per cm³"... ok I don't know about that but I believe you and it's ok for me too agree upon "plasteel probably is about as dense as mild steel"
so a 90cm*120cm shield with a weight of 9kg should be about 9500/7.85/(90*120) => 0.11205473 cm thick (if my formula is right) and those 0.112 cm provide 8 AP
a guardshield with the same material weighting only 1,5kg should be (1 500 / 7.85) / (90 * 120) = 0.0176928521 thich thus 0.018 cm provide 6 AP
if we try to create a function x(thickness)=AP with x(0)=0, x(0.018)=6, x(0.112)=8 should it be a logarithmic function? should it be an exponential function? and what would be the required thickness for 16 or 32 AP ?
I think we can see that either my formula is wrong, or the density is somewhat wrong or it is just really ridiculus to approach it this way...
I agree that the naval shield should not count as "total concealment" but I think it shows that fluff text is just not enough for shield rules
and about flexible weapons yes they can't be parried but that's not a problem of shields, shields are not superior or inferior to anything else here
but another question... what does craftsmanship do to shields? as it clearly is a weapon it does not receive the plus 1 armour and as I don't get the +5/+10 for blocking... what's it good for? (except for the mono acreage buckler which is a ranged weapon if thrown and therefore does never jam ;P )