Ok, think 17 story building...You'd want to multiply the draught by 2.5 or even 3.
As you can (hopefully) see, she's got a lot more height to the flight deck than just equal to her draught.
90 feet from the the flight deck to the water....
Ok, think 17 story building...You'd want to multiply the draught by 2.5 or even 3.
As you can (hopefully) see, she's got a lot more height to the flight deck than just equal to her draught.
No. In my post before the one you quoted I provided the actual volume of the Ford-class carriers determined from the GRT registered with the USCG. That was 323,288m^3 (23,092 dTons) which is slightly lower than my calculation of 2 x length x beam (waterline) x draft = 331,068 m^3 (23,686 dTons)You'd want to multiply the draught by 2.5 or even 3.
She does indeed. However, multiplying the draft by 2x takes into account the non-box shape of the hull.As you can (hopefully) see, she's got a lot more height to the flight deck than just equal to her dr
Not quite - one GRT = approx 2.8 m^3, so you need to multiply the GRT by 5 to get dTons; your formula would divide the GRT by 5 so you'd get 1/25 of the actual dTons.(GRTx2.8)/14=Traveller displacement tonnage for any ship of known GRT
You're right. I clearly hadn't had enough coffee when I wrote that.100,000x2.8 = 280000
280000/14 = 20,000
1 GRT is 100 cubic feet or 2.8 cubic meters.
(GRT x100) gets you volume in cubic feet
Divide by 500 to get Traveller displacement tonnage
(GRTx2.8) gets you volume in cubic metres
Divide by 14 to get volume in Traveller displacement tonnage.
Your suggestion 100,000 x 5 gives 500,000.
It won't really help.Inertial compensation...
Fair enough, but you didn't mention that, and that makes it a non-universal method, and some ships (most steel-hulled cargo ships and big 20th century warships) are full in cross section, whilst others (sailing ships, for example) are not.She does indeed. However, multiplying the draft by 2x takes into account the non-box shape of the hull.
That doesn't mean that an outside force (such as the engines) that's accelerating the whole ship doesn't care - inertia still exists, as does mass. This is especially the case if you take it that inertial compensation only functions in areas with emitter plates or the like, and isn't a field that encompasses the whole ship. If you do have it as a field, you could rule that the effective centre of mass is in the field's centre (or some other arbitrary point), which would be interesting (and what if the field goes off?), but I don't.Inertial compensation deals with all that.
Inside the ship you can't tell if you are accelerating or not moving at all.
Sounds like this might be a plot complication to me.I'm sure the designers of ships in the TU have to consider thrust lines and getting them lined up so they go through a ship's centre of mass, and that they need to make sure that cargo holds are placed such that cargos within the ship's designed mass limits won't move the CoM too much. I'm also sure that proper, balanced placement of high-mass cargo is part of the cargomaster's job, along with making sure that cargo is placed so that the first stuff to come out isn't buried in the back corners, that any cargo passengers are entitled to access in-flight is actually accessible, and so on.
Not according to RAWGiven most of us are using some flavor of grav lift, do starships have weight limit for grav systems, more buoyancy then thrust, or need to be properly trimmed/balanced/ballasted?