Ok, I was poking at this as a way of seeing if there was some way to derive numbers for range based off other characteristics. So far unsuccessful, but whatever.
Scaling here, I get a width of 1015 meters for an ISD. Fatter then I expected.
Now, in ESB we see the Avenger jump to lightspeed. It goes from a couple of km in front of us to BVR in about a second (I haven't timed it yet)
The human eye can be treated as a optical telescope - that's how they got the 20/20 vision rating in the first place. Here
it says that the angular resolution of an optical telescope for light in the visible range can be simplified to resolution = 138/aperature diameter in mm. The human pupil is 5 mm wide. So we get an angular resolution of 27.6 arcseconds.
27.6/3600 = 0.007667 degrees.
Now we do some simple trig.
D = (360*L)/(2*pi*angle)
At 1015 meters in width, the maximum distance a human eye could see the star destroyer would be 7,585 km away. (Not accounting for things like engine luminosity, background light, etc. I'm treating it just like we see in the film)
Now if this takes place over 1 second (like I said I haven't timed it) then it is moving at 7,585,463 m/s or 2.53% of lightspeed. This is not out of line, as the X-wings had to get up to 10% of lightspeed to reach the Death Star in the battle of Yavin.
Now the Venator, with a power of 3.6*10^24 watts and an acceleration of 3000 Gs, would mass 4*10^11 kgs. Using the cube law to scale up, the Imperial class would mass 1.14*10^12 kgs.
KE = 0.5*m*v^2
KE = 3.27*10^25 watts.
So there you go - as a rough calc, an Imperial class Star Destroyer must consume at least 3.27*10^25 watts to jump to hyperspace. I'd like to refine this further in the future, as it makes a pretty solid case to use against those "movie canon only" fools.
Any comments would be appreciated.