GCS Vs 12" Cannons

[Captain Seafort]

If he reported trouble after two hits. That could have been the 400GW. So then if the E-D was hit six times then the total would end at 1200GW.

You're getting energy and power mixed up - while the former can be simply added up, the latter can't, as it's time-averaged.

[m52nickerson]

And if I shone a torch in your face and then punched you, would you doubt that was a KE attack? We have an explicit statement from Worf as to the nature of the impact - the light show was probably some kind of containment field.

Only if your fist was glowing! Even if it was a containment field that shot should have impacted the shields and continued or dispursted, instead it wraped around the shields much like the Breen energy weapon did in DS9.

Besides Worf could have been wrong. It was you that has said that the visuals should be held higher then the Dialoge.

You're getting energy and power mixed up - while the former can be simply added up, the latter can't, as it's time-averaged.

Did you not add the wattage of the shells from the ship together at the begining of the thread?

I don't see why you could not add them up, each hit will reduce the overall streight of the shields, past a certain threshhold.

[Praeothmin]

The ship's sensors also registered that none of their previous shots were having any effects, yet the enemy ship was then destroyed by one torpedo at the end.

So?

So their sensors were fooled by the Douwd in the case of the ship's resistance, twice.

However, the damage to the E-D, including the collapse of the shields, would have been registered by internal sensors, rather than being a Douwd creation. Those sensors registered a 400 GW hit, which knocked out the shields.

Makes sense, especially when you factor in the very real injuries suffered by the crew, which were not Douwd induced.

I went over the numbers for fun.

A 1 Watt= 1 Joule / 1 second.

If the 400GW is kinetic energy, then it is most likely "Translational kinetic Energy", because the ship rocks when it is hit.

If so, then here goes:
400GW = 400 000 000 000 billion Watts.
If this is delivered in 1 second (which we'll assume for this calculation), then we're faced with 400GJ.

How fast would a 1000kg car need to go to have that kind of energy?
Let's see.
"Translational kinetic Energy"'s formula is as follows:
K.E. (kinetic energy) in joules = (m.v*v)/2, where "m" is the mass in kg, and "v" is the speed in meters/second.
v*v means "v" squared.

So, we have 400GJ = (1000 . v*v) / 2
That gives us 400GJ x 2 = (1000.v*v)
800GJ / 1000 = v*v m/s
The square root of 0.8G, or the square root of 800 000 000 = 28 284.3 m/s, or 101 823.48 km/h.

So 400GW is like being hit by a 1000kg car shot at 101 823.48 km/h.

[Captain Seafort]

I ran a few calculations earlier in the thread - 400GW is, more or less, the power of a broadside from Hipper class heavy cruiser. That, more or less, is where the title came from when this was split.

[Sionnach Glic]

Out of curiousity, is that just the kinetic energy from the impact of the shells themselves, or does that also include any explosives in the shells?

[Praeothmin]

I ran a few calculations earlier in the thread - 400GW is, more or less, the power of a broadside from Hipper class heavy cruiser. That, more or less, is where the title came from when this was split.

The only nitpick I have is that the Kinetic Energy, since you're using "muzzle velocity", is the enrgy at the gun's muzzle end.
Once we factor in atmospheric drag, and distance, this amount will drop.
So, if the E-D was at, say, 2 km, its shields might be able to survive the first volley...

Out of curiousity, is that just the kinetic energy from the impact of the shells themselves, or does that also include any explosives in the shells?

Since we are using muzzle velocity, then this value is simple kinetic energy, and does not factor in the explosive effects...

[Aaron]

The only nitpick I have is that the Kinetic Energy, since you're using "muzzle velocity", is the enrgy at the gun's muzzle end.
Once we factor in atmospheric drag, and distance, this amount will drop.
So, if the E-D was at, say, 2 km, its shields might be able to survive the first volley...

Uh...yeah, if the E-D was cruising around in atmosphere.

Since we are using muzzle velocity, then this value is simple kinetic energy, and does not factor in the explosive effects...

I'd be curious to see what the energy of an HE shell for those guns are.

[Captain Seafort]

Out of curiousity, is that just the kinetic energy from the impact of the shells themselves, or does that also include any explosives in the shells?

Just the power of the impact, based on muzzle velocity and the mass and length of the projectile.

[Praeothmin]

Uh...yeah, if the E-D was cruising around in atmosphere.

That was a joke...
Of course, if the E-D wasn't cruising around in the atmosphere, then we've have to wonder what an earth seaship was doing in space... Unless it was in an anime...

Just the power of the impact, based on muzzle velocity and the mass and length of the projectile.

That's pretty much what I said:

Since we are using muzzle velocity, then this value is simple kinetic energy, and does not factor in the explosive effects...

[Mikey]

Unless it was in an anime...