, however I feel that a more in-depth analysis at the issue needs to be done.
First some facts. The E-D was brought to 20 million kilometers as stated in dialog:
PICARD 2: Helm take us in to twenty million kilometres.
WESLEY: Aye, sir.
RIKER: Mister Worf, divert enough power to the shields to offset the increased radiation and magnetic fields.
DATA: Sir, at twenty million kilometres, our shields will only be effective for eighteen minutes.
PICARD 2: Noted, Mister Data.
The starship was also exposed to the emissions beam as seen here:
Data's statement of "our shields will only be effective for eighteen minutes" is rather interesting in and of itself as the ship was not shaking or taking any internal damage during all of this. Perhaps he means the shields will hold well enough before the ship and crew are endangered and take damage? Hard to say, but let's assume for the sake of being conservative that he means the shields will totally and completely collapse and the ship will take severe damage instantly and the crew likely killed right then as well.
As noted in the other thread, pulsars, like the Crab Nebula Pulsar, can be as much as 100,000 times more energetic in output than our own Sol, which puts out around 3.8e+26 watts.
For the sake of being conservative, I'll also leave out the emission beam from the pulsar sweeping down over the E-D, which would concentrate a lot of energy at the ship and assume the pulsar is radiating evenly in all directions. I will use the following to calculate the surface area of the "shell" at which the E-D was at at 20 million km radius:
r = radius
A = surface area
C = circumference
π = pi = 3.14159
Thus:
r = 20,000,000,000 m
A = 5.0265482457437e+21 m^2
So now we have a surface area for the imaginary shell, now let us divide the Sun's energy into that area:
3.8e+26 W/ 5.03e+21 m^2 = 76,000 W per m^2.
So if the Sun was all that was involved here it would be 76 kilowatts per meter squared on the E-D shields, which is pretty lousy. The shields would only be taking 17.9 gigawatts total and 19.33 TW in 18 minutes. This is what it looks like the SBC person did since they only calculated a ton or so of energy for the E-D's shields. However, if we plug in the fact that pulsars can be up to 100,000 times more powerful than Sol, we get a different picture:
76 KW x 100,000 = 7,600,000 KW or 7.6 GW per m^2
Now that's quite a bit different results.
To be conservative again, let's assume that the E-D was facing directly at the pulsar to present the minimum possible shield aspect to the radiation and magnetic field (which it was not as seen in the pictures), this would be at least an ellipsoidal cross-section of 500 meters x 150 meters x pi = 235,619.45 meters squared. So we then have the following:
235,619 m^2 x 7.6 GW = 1,790,704.4 GW total or 1,790 TW. That's not a ton of TNT, that's about .4 megatons. In 18 minutes the E-D shields would take 496 megatons. Just shy of half a gigaton's worth of energy!
Now given that the E-D was presenting the much larger surface area of her dorsal shields to the Lonka Pulsar, we would have a shield surface area more like 1,099,557 m^2 and a total energy absorbed each second of 8,356.63 TW or slightly less than two megatons. In 18 minutes the shields would then absorb an incredible 9,025,160.4 TJ or 2.15 gigatons.
Unless someone can find some serious fault with my calculations, I think we can lay the Lonka Pulsar issue to rest.
-Mike
