HMS Victory stats question

[admin]

Hello everyone

 

Assuming base Victory has the following weaponry:

• 12lb
• 12lb
• 32lb
• 42lb

What would be the effects on speed and waterline position if we take off all those cannons from the ship

 

if base speed is 100% what would be the new speed of the ship without cannons

if base waterline is 0 what would be the new waterline in meters?

[Poyraz]

Just playing with numbers here,

 

According to http://en.wikipedia.org/wiki/HMS_Victory:

 

- Total displacement of the ship is 3.500 tonnes.

- Total gun weight is around 240 tonnes.

 

So, the guns without ammuniton are 7% of the ships total mass which actually looks like way less than I expected.

 

With basic assumption that the force taken from the wind via the masts is F = m x a and it is constant since nothing changes but the mass. Therefore, we will end up having a 8% increased acceleration for a 7% drop in ship's mass.

 

The maximum limit velocity on the other hand won't be affected due to increased acceleration in my opinion. The limiting parameter here would be the ships' volume/area under the waterline which changes the drag/friction coefficent for the ship.

 

Regarding the ships volume under the waterline, it will decrease proportional to 7% change in displacement volume. But, here is more dimensions of the ship size at the waterline needed.

[maturin]

Removing the guns will also lower the ship's center of gravity considerably. This will make the ship stiffer, meaning that she will heel and roll less, giving you an additional speed boost. (But it will also make the ship a lot less comfortable to be on, because her movement will be jerky.)

[BungeeLemming]

and dont forget to mention all the additional equipment wich is used for shooting.

All tools to sight the guns and then powder, ammunition etc.. theres more to the armament than the pure mass of gthe cannons

[maturin]

Not to mention that if you take the guns out of a ship, you're also not going to have the crew to shoot them, either. 900 men weigh plenty.

[BrutishVulgarian]

That Poyraz 7% displacement thingy seems like a good way to determine Wetted Surface Area, as long as it looks good. Wetted Surface Area is a speed killer, that's why the monohull snobs at the America's Cup were forced to sail catamarans.

"Performance

Any surface that's in contact with the water causes drag; frictional resistance depends on how much surface area is in contact with the water. Less wetted surface area generally means less frictional drag.

In a sailboat, frictional drag usually dominates at low speeds, so minimal wetted surface is important to light-air performance. (This is better understood in terms of the ratio of sail area to wetted surface.)" http://marine.marsh-design.com/content/wetted-surface-area

Where's Ryan21 when we need him?

[Poyraz]

I used to toy with Unit3D some time ago and as far as I remember, you can define the movement of the objects with defined forces or transform position and defined speeds. Using the forces on the objects results in more realistic behaviour and Naval Action looks like using those forces.

 

I have no idea on the coded mechanics of the game but in overall the concept looks like the "sum of applied forces" model.

• On vertical (z - axis) it is buoyancy and gravity
• On the horizontal (x - axis, ship's sailing direction) it is:

F_Sails = Sum of all forces on sails and mast (vectoral)

F_Drag = Sum of all underwater drag forces

F_Total= Total applied force on the ship

 

F_Total = F_Sails - F_Drag

 

m = Total mass of the ship

x"  = Acceleration of the ship

ρ_Air = Density of air

ρ_Water =Density of water

S = Sail surface

A = Projection are under waterline

C = Drag coefficient of the ship

v_Wind = Wind speed (assuming it is relative bigger than ship speed)

v_Ship = Ship speed

 

m . x"  =  (1/2 . ρ_Air . S . v²_Wind )  -  (1/2 . ρ_Water . C . A . v²_Ship)

F_Total                    F_Sails                                   F_Drag

 

When the ship is at her maximum speed F_Sails is equal to F_Drag  so, that the total force on the ships is zero which leads the ship sailing at constant speed.

 

v²_Ship = ( ρ_Air . S . v²_Wind ) / ( ρ_Water . C . A )

                                                                                                                      ___

According to this equation, we can see that ship speed is proportional to √1/A  and given that the 7% mass reduction leads to 7% less area underwater we will end up having a speed increase of 3,7%.

[akd]

Please clarify: are you looking at a theoretical unarmed Victory or a Victory that has pushed her guns overboard?

Interesting tidbit from Forester's The Age of Fighting Sail: starting ten tons of Constitution's water over the side would make an inch difference in her draft.

[Poyraz]

As for the height change over waterline:

 

Using x,y,z coordinates; x showing sail direction, y horizontal and z height and assumig the ship dimensions are uniform in all x, y and z directions like in the picture below.(which is not in reality for HMS Victory)

 

 

A mass reduction of 7% percent will lead to a draught change of 7% for this geometry. And assuming that HMS Victory has a draught of 9 meters under water, dropping guns will lead to a raise of 0,63 meters in z direction over the loaded waterline.

 

For a more precise result, the integral of the hull under waterline has to be calculated though.

 

 

[Ink]

I think this is the formula we need.

 

roughly, I got q=9.22, so for a total gunweight (only guns, tho) of 293 tonnes it's about 32 cm.

 

Did anyone else meet the formula before? What do you think guys?