The danger of the Death Star

[Jaded Masses]

Many apologies if this has been bought up before.

Known facts:
The death star destroyed a planet.
It used DET
It got its energy from internal reactions, ie, it "made" its own power.
The reactor was <160km in size (much less)
It was an expansion and refinement of technology, not something completely new.

My question:
If the reactor was that small and the power to destroy a planet came from it, then wouldn't it be possible to simply use death star sized (or many smaller ones) to power planetary shields with strength in excess of of the death stars weapon? This would render it all but impotent in attacking the core worlds (likely he only ones who could afford such defenses, and though most planets would still be destroyable by it, it would create havens where people would go to feel safe, defeating the purpose of rule though fear.)

I have several ideas to account for this.
shields are far more inefficient then weapons, making it so you need more energy for a given amount of defense compared to that amount of destruction.

There is some physical limit as to how strong a shield can be.

The energy you put into your shield is spread though out the entire surface
surface and a weapon is focus to one spot, locally defeating the shield. I'd say this was a given but it begs the question why can't shields be focused as well. Are the computational speeds needed to focus the shield not doable by Star Wars standards? This seems highly unlikely, given that today I'm pretty sure reactive armor is activated on the contact of the weapon. Is it simply not possible to focus shields? This I also doubt because I think we have seen shields change shape, ad we know that their strength goes up and down.

Core worlds shields were more then up o the task but these would presumably be the worlds where the Empire has the most influence and that knowledge was kept under wraps. This means people will still think the death star is un stoppable.

Some where or another I saw a quote that the Death Star used a 'trick' to get around the shield, if not destroy the planet. Is this bunk? Is it related to the Galaxy guns method of shield penetration? (whatever that is)

Thats all I can think of.

Edit: fixed my greater-then to less-then sign

[Rogue 9]

Actually, given the real size of the Death Star (not the one 120 km that's even more screwed up than the 8 km SSD figure) that reactor may well be 120 km across.

[Spanky The Dolphin]

Death Star I was 160 km in diameter.
Death Star II was about 900 km in diameter.

[Jaded Masses]

Death Star I was 160 km in diameter.
Death Star II was about 900 km in diameter.

That doesn't change my question much....

I'm aware of the size difference, I was using the first DS as my example, but 900km still fits comfortable on a planet several times.

[Kazuaki Shimazaki]

Many apologies if this has been bought up before.

Known facts:
The death star destroyed a planet.
It used DET
It got its energy from internal reactions, ie, it "made" its own power.
The reactor was >160km in size (much less)
It was an expansion and refinement of technology, not something completely new.

It should be <160km. The reactor cannot be bigger than the whole DS

The reactor size IIRC was more like 16km.

I have several ideas to account for this.
shields are far more inefficient then weapons, making it so you need more energy for a given amount of defense compared to that amount of destruction.

Actually, it is probably a matter of limits in dissipation. The energy has to be shunted away somehow, and to safely shunt away 1E38J would take some doing.

The Acclamator's power output is 2E23W. If it follows the 25% power to shield convention suggested in the SWSB (I think it was the SWSB,) 5E22W of that goes to shields, and the Acclamator shield dissipation is 7E22W.

There is some physical limit as to how strong a shield can be.

Definitely. With energies of that class being moved around, it has real momentum (lots of it.) After I realized what kind of momentum is involvd in moving that kind of energy around, I'm surprised the Death Star doesn't kill itself just moving those doses of energy around. The Death Star, however, is fully artificial, while a planetary shield ultimately rests on a very natural (with realistic limits) planet.

The energy you put into your shield is spread though out the entire surface surface and a weapon is focus to one spot, locally defeating the shield. I'd say this was a given but it begs the question why can't shields be focused as well. Are the computational speeds needed to focus the shield not doable by Star Wars standards?

Think about the momentum of the energy you would have to shift. Sure you would be faster than the already focused beam? It is acting, you are REACTING.

This seems highly unlikely, given that today I'm pretty sure reactive armor is activated on the contact of the weapon. Is it simply not possible to focus shields? This I also doubt because I think we have seen shields change shape, ad we know that their strength goes up and down.

ERA uses a different principle. It uses the explosive already right under the (generally HEAT) impactor to disrupt the jet. Nothing is really shifted. IMagine if ERA meant the explosive used has to be shifted into place from the time of launch detection to where it is needed, and remember that rockets are faster than the time of the AT-3 Sagger (100m/s.)

Some where or another I saw a quote that the Death Star used a 'trick' to get around the shield, if not destroy the planet. Is this bunk? Is it related to the Galaxy guns method of shield penetration? (whatever that is)[/quote]

The DE Sourcebook suggests some kind of neutrino beam. It never specifically says the beam gets a free (or nearly free bypass.) Instead, the RPG section below it says the beam could be blocked to a great degree (never mind the whole RPG roll concept kinds of falls down with high power disparities - it gets to the point where it says the equivalent of "When a normal Terran ship takes a direct nuke hit, roll to see if it takes any damage.")

A neutrino beam also has EXTREMELY low interactivity with matter. If you want your beam to interact efficiently with a planet, the beam is a lousy choice.

It is contradicted by the BTM CD, which links the superlaser to a big compound TL anyway.

[Jaded Masses]

So basically having x amount of avaliable energy does not equal having x amount of shield defense, whereas with weapons it pretty much 1 to 1, therefore shields could never practically stand up to a turbo laser if given equal amounts of energy. This makes perfect sense I guess, but if the death star devoted 25% of its power to shields, then a planet using 2 death star sized reactors for shields only it will have 2.7 times as much shielding as the DS as main weapon (assuming all that 75% goes towards the main gun). Is the disparity between firepower and defense that large?

[Solauren]

You have to remember something

The Death Star fired enough power to kill a planet onto a SINGLE POINT

Shield power is spread out over the entire surface area.

You'd have to put the equal power of a Death Star superlaser at EVERY POINT in the shield to be able to stop it without damage to the planetary surface.

Case in point, take a battering ram slamming into a wall.
It hits with a impact force of say, 100 pounds in a 10 x 10 foot area. That's 1 pound per foot. The wall takes it, as it can take 200 pound of force in a square foot.

Now, you fire a gun into the wall, the gun hits with 10 pounds of force. The bullet hits a 1/10 inch by 1/10 inch area, or 0.01 square feet. That's the equal of 1000 pounds hittin that one point in the wall. The bullet tears into the wall (of through it, dependin on a variety of factors, like is it a wood wall vs glass wall vs stone wall, etc)
The bullet goes through, and delievers 800 points of damage to whatever is beyond the wall.

In any event, the bullet punches through the wall, continues on, and damages something the wall was protecting

In this case, the Death Star beam hits the planetary shield, punches a whole in it (a vulnerability in Star Wars shields exploted by Torpedoe Spheres, the ability to punch a hole in a single point with enough firepower), goes on thorugh, and the Death Star superlaser hits the protected planet.

If the shield is dissipating enough power theplanet does blow up, it's kill gonna be heavily damaged (i.e mantle collapsing). Then again, just turn the juice to the Superlaser up 50%, and watch it punch the shield and blow the planet.

That was the idea of the Death Star. Overpower Planetary shields

[JME2]

We need to get DW in on this thread; he'll have a expert's POV on energy factors and such.

[Howedar]

There is no reason to assume that 1J of shielding takes the same energy input as 1J of turbolaser. Nor is there any reason to assume that shields are infinitely scalable.