Hyperspace energy requirements

[Ender]

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.

Image
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.

[Ender]

Also, for the acclamator and Venator using this method one gets 1.04*10^23 and 3.35*10^24 respectively.

[Vehrec]

My only critique is that if it accelerated from a dead stop to lightspeed, and traveled the distances you say it did in 1 second, then wouldn't its average speed be 7,585,463 m/s and it's final speed be closer to 5.06%C? After all, if you accelerate at a constant speed down a mile long track in exactly one minute, your speed at the end of the track will be 120 MPH, not 60.

My only other suggestion might be a more in-depth volume analysis of the Imperator and Venator classes. Those notches in the sides, and the massive through hanger might be a complicating factor.

[Fingolfin_Noldor]

Are we assuming that this is raw acceleration and not some fudgy science that doesn't exist yet?

[Solauren]

There's also the possiblity the illusion of that level of acceleration is created by the fact the ships are already approaching lightspeed, and what we see is the human eye's interuptation of it.

[Fingolfin_Noldor]

There's also the possiblity the illusion of that level of acceleration is created by the fact the ships are already approaching lightspeed, and what we see is the human eye's interuptation of it.

Here's the thing, light travels at a fixed c, so perhaps, the ship, as you say, has already traveled faster than the speed of light, and what we see is an extreme case of just the instantaneous image of the starship which arrived at our current position, but the image is dated even before it arrives.

[Surlethe]

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.

The Avenger is, IIRC, accelerating, as others have pointed out. If we assume a to be constant through the jump. Since d = (at^2)/2, a = 1.52E7 m/s/s and after 1 s, v = 1.52E7 m/s. This is, of course, 0.0506c.

The force of the engines multiplied by the distance the ships traveled will give the work done. If you're correct (and not circular) in your derivation of mass, W = Fd = (ma)d = (1.14E12 kg)(1.52E7 m/s/s)(7.585E6 m) = 1.3E26 J. Over one second, that is of course 1.3E26 W.

You'll get the same numbers if you plug into (mv^2)/2 (the units of which are J, not W, by the way).

Are we assuming that this is raw acceleration and not some fudgy science that doesn't exist yet?

I'm not really sure that it matters. Even if it's some sort of "energy effect", we still saw the ship do so much work on itself to push it that far in the blink of an eye.

Here's the thing, light travels at a fixed c, so perhaps, the ship, as you say, has already traveled faster than the speed of light, and what we see is an extreme case of just the instantaneous image of the starship which arrived at our current position, but the image is dated even before it arrives.

If the ISD had already gone FTL, you'd expect it to redshift while it disappeared, wouldn't you?

[Fingolfin_Noldor]

If the ISD had already gone FTL, you'd expect it to redshift while it disappeared, wouldn't you?

Why would it redshift? The redshift effect largely comes from the stretching of space itself. Unless of course, the space around the ISD actually gets warped sufficiently that the light does get redshifted.

[Vehrec]

It should be redshifting, because the Doppler effect works on light just as well as it does on sound. In other words, the waves get stretched because the source is receding from the observer. Common sense really. if you are dropping balls onto a converyer belt, they will be a certain distance apart. And if you run while dropping them, that changes the distance.

[Wyrm]

Why would it redshift? The redshift effect largely comes from the stretching of space itself. Unless of course, the space around the ISD actually gets warped sufficiently that the light does get redshifted.

Any relative speed produces redshift, son. It doesn't matter if the relative speed is produced by stretching space or good ol' fashioned thrust, it's the same redshift.

[Fingolfin_Noldor]

Why would it redshift? The redshift effect largely comes from the stretching of space itself. Unless of course, the space around the ISD actually gets warped sufficiently that the light does get redshifted.

Any relative speed produces redshift, son. It doesn't matter if the relative speed is produced by stretching space or good ol' fashioned thrust, it's the same redshift.

Aw... frak... I forgot.... Darn...

[evillejedi]

Based on models that match the scale and major contours of the Venator and ISD (which can be provided, I am using a volume measure tool for a solid CAD model)

The venator has an internal volume of ~ 13 million m^3 with the notches. It is a really tiny ship because of the hull geometry, not to mention that at least 2 million m^3 is hangers) the largest spherical reactor it can fit is around 1 million m^3 assuming that it protrudes into the superstructure as shown in the ICS.

by contrast the ISD can house a main reactor sphere with a volume of 10 million m^3, though the ICS shows a reactor that is smaller. the ISD hull and superstructure is around 100 million m^3 with the notches.

Even accounting for differences in the scaling of individual elements (the models are not dead on accurate, but are quite close) the volume situation doesn't change much, the ISD is still over 7.5 times larger in volume and can house a reactor that is 10x larger.

[NecronLord]

Are we assuming that this is raw acceleration and not some fudgy science that doesn't exist yet?

To be honest, it fits quite well with Dr Saxton's ~1e25W estimate of its output. It could build up the power for that acelleration in under a minute, assuming it has storage capacity, so we don't need to invoke any funky subspace stuff.

[Ender]

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.

The Avenger is, IIRC, accelerating, as others have pointed out. If we assume a to be constant through the jump. Since d = (at^2)/2, a = 1.52E7 m/s/s and after 1 s, v = 1.52E7 m/s. This is, of course, 0.0506c.

The force of the engines multiplied by the distance the ships traveled will give the work done. If you're correct (and not circular) in your derivation of mass, W = Fd = (ma)d = (1.14E12 kg)(1.52E7 m/s/s)(7.585E6 m) = 1.3E26 J. Over one second, that is of course 1.3E26 W.

Ah, danke. I was doing BOTE and figured I missed something.

You'll get the same numbers if you plug into (mv^2)/2 (the units of which are J, not W, by the way).

Yeah, I made the leap of dividing again without noting it in my post.

Are we assuming that this is raw acceleration and not some fudgy science that doesn't exist yet?

I'm not really sure that it matters. Even if it's some sort of "energy effect", we still saw the ship do so much work on itself to push it that far in the blink of an eye. [/quote]I'm open to the possibility of some kind of space warping happening - we do see the rotational wake after it jumps.

I'm very much hazy on how this would follow then, but as I understand it we would need to measure the work done to accelerate it within it's frame of reference, and warping space would give the effect of moving at vastly faster speeds because it's frame itself would be shifted, ala the real life warp drive. For this same reason this could not be applied to Trek ship doing the jump to warp. If I'm wrong (or not even wrong ) please correct me.

[Ender]

Based on models that match the scale and major contours of the Venator and ISD (which can be provided, I am using a volume measure tool for a solid CAD model)

The venator has an internal volume of ~ 13 million m^3 with the notches. It is a really tiny ship because of the hull geometry, not to mention that at least 2 million m^3 is hangers) the largest spherical reactor it can fit is around 1 million m^3 assuming that it protrudes into the superstructure as shown in the ICS.

by contrast the ISD can house a main reactor sphere with a volume of 10 million m^3, though the ICS shows a reactor that is smaller. the ISD hull and superstructure is around 100 million m^3 with the notches.

Even accounting for differences in the scaling of individual elements (the models are not dead on accurate, but are quite close) the volume situation doesn't change much, the ISD is still over 7.5 times larger in volume and can house a reactor that is 10x larger.

What program are you using? Because I have a long list of things I want to measure.

[Ender]

Also, props to Surlethe for providing the original idea here.

(we desperately need a *throwing up the horns, rock out* emote for situations like this)

[Surlethe]

Ah, danke. I was doing BOTE and figured I missed something.

A minor ass-covering note: the Lorentz factor for 0.056c is 1.0015, so since the calcs are only to two or three significant digits anyway, I don't see any need to take it into account.

Are we assuming that this is raw acceleration and not some fudgy science that doesn't exist yet?

I'm not really sure that it matters. Even if it's some sort of "energy effect", we still saw the ship do so much work on itself to push it that far in the blink of an eye.

I'm open to the possibility of some kind of space warping happening - we do see the rotational wake after it jumps.

I'm very much hazy on how this would follow then, but as I understand it we would need to measure the work done to accelerate it within it's frame of reference, and warping space would give the effect of moving at vastly faster speeds because it's frame itself would be shifted, ala the real life warp drive. For this same reason this could not be applied to Trek ship doing the jump to warp. If I'm wrong (or not even wrong ) please correct me.[/quote]

To be honest, I've really no idea how to approach this. If you incorporate space-time warping, it's a general relativistic problem, and I've no experience whatsoever in that area of physics. Also, we don't know what the mechanism for the jump to lightspeed is. We can guess, but since it's impossible and therefore unscientific, we won't be deriving any real numbers from it.

Fun story: I once used some bullshit about "quantum effects" and the jump to lightspeed to confuse some trektards on another board. I postulated that the transition to hyperspace occurred when the ship's length to an at-rest observer was the planck length, and integrated from 0 to like 0.999999999999999999c to get the proper length dilation factor. Of course, you can get pretty much any energy level you like since the curve is horizontally asymptotic to c.

Also, props to Surlethe for providing the original idea here.

(we desperately need a *throwing up the horns, rock out* emote for situations like this)

\m/

[Vehrec]

Me personally, I'd like to measure the Executor, so we can get some idea of what kind of power output a star dreadnought has. These calculations seem fairly accurate for the destroyer scale craft, and I see no reason why we couldn't scale them up. On the other hand, some kind of space-time warping to achieve hyperspace seems likely. We are talking about a FTL drive here, so either relativity or causality just went out the window. That's going to do a number on any physics you can think of.

[evillejedi]

I can export to any standard 3d format, but it is in 3ds max currently.

Executor with dimensions of 17600 x 4900 x 1150 m is ~ 14.5 billion m^3 (let me know if you want a different scaling ratio)

largest spherical reactor is 275 million m ^3, largest practical cylindrical reactor would be ~ 650 million m^3 and it could house many of these

for reference a wingless MC-80 has a volume of ~30-35 million m^3 and could mount around 6 million m^3 of reactor space if it had multiple main reactors. (since we don't have a cutaway reference like the ISD from the ICS)

[Ender]

I can export to any standard 3d format, but it is in 3ds max currently.

Executor with dimensions of 17600 x 4900 x 1150 m is ~ 14.5 billion m^3 (let me know if you want a different scaling ratio)

largest spherical reactor is 275 million m ^3, largest practical cylindrical reactor would be ~ 650 million m^3 and it could house many of these

for reference a wingless MC-80 has a volume of ~30-35 million m^3 and could mount around 6 million m^3 of reactor space if it had multiple main reactors. (since we don't have a cutaway reference like the ISD from the ICS)

Heh, you just opened yourself up to a long list. To start with, if you can find them

Viscount (17 km)
Home 1 (3800 m and at 3200 m)
Liberty type (1500 m)
Unnamed wingless type (the liberty sans wings, at 1500 and 1600)
MC80 (1200)
MC80a (1300)
MC80b (1200)
MC 90 (1255)
Victory (900)
Acclamator (752)
Strike cruiser (the retconned dagger shape if possible 450)
dreadnaught (600)
Nebulaon B frigate
Trade Federation Cruiser (3170)
Invisible Hand (1088)
Executor (19 km)
Munificent (825)
Recusant (1187)
Any and all starfighters you can get your hands on.

Any of this that you can get I'd appreciate. Just the volumes, though the reactor stuff is potentially useful as well. Thanks.

[Ender]

Me personally, I'd like to measure the Executor, so we can get some idea of what kind of power output a star dreadnought has. These calculations seem fairly accurate for the destroyer scale craft, and I see no reason why we couldn't scale them up. On the other hand, some kind of space-time warping to achieve hyperspace seems likely. We are talking about a FTL drive here, so either relativity or causality just went out the window. That's going to do a number on any physics you can think of.

existing estimates place it between 1.7*10^27 and 3*10^27 watts.

[evillejedi]

Viscount (17 km)
do not currently have a model

Home 1 (3800 m and at 3200 m)
3800m is 1e9 m^3, 3200m is .56 e9 m^3

Liberty type (1500 m)
35 e6 m^3 the wings are narrow and don't contribute much

Unnamed wingless type (the liberty sans wings, at 1500 and 1600)
34 e6 m^3 the wings are replaced by some additional bulges that give it more volume, the tail fan is also shorter so I pegged this ship at 1400 m and the liberty at 1500m, the 1600 m scale would produce 65 e6 m^3

mC40 (530 m)
6.4e6

mediator est (2350m)
185 e6

BAC(850)
12.6 e6

Republic destroyer est
21 e6

Defender (SOTG version)
37 e6

Defender (CTD version)
15e6

MC80 (1200)
30 e6

MC80a (1300)
assuming this is the 5 engined ship from tie fighter, 60 e6

MC80b (1200)
30e6
at 2500m (no real evidence to support this, but it makes more sense going up against an SSD) 295 e6

MC 90 (1255)
45e6

Victory (900)
25 e6 and my model is slightly taller than it has been represented, max reactor 1.6 e6

Acclamator (752)
7.2 e6 .5e6 on the reactor

Strike cruiser (the retconned dagger shape if possible 450)
not sure what the retconned shape is
1.5e6

dreadnaught (600)
2.8e6 .5e6 on the reactor


Nebulaon B frigate
.327e 6 @ 300m


Trade Federation Cruiser (3170)
2e9 m^3

Invisible Hand (1088)
18 e6

Executor (19 km)
19e9

Munificent (825)
4.2 e6 though this is closed geometry

Recusant (1187)
5e6 though this is a closed geometry

carrack
.68 e6

enforcer/vindicator/interdictor hull
3.8 e6

corellian gunship
.035 e6


star fighters are probably a lot less accurate, the models may not have the exact scale so the variation from the actual may be a lot higher given the smaller numbers
tie defender
23m^3
tie interceptor
9.6 m^3 (cockpit+spars)
tie fighter
9.6 m^3 (cockpit+spars)
tiebomber
25m^3
tie advanced (tie fighter)
25m^3
assault gunboat
65 m^3

xwing
31 m^3
awing
21m^3
bwing
84 m^3
ywing
65 m^3
ewing
23 m^3

[Surlethe]

The Executor is 5 times longer than the Home One but only 19 times as voluminous?

[CmdrWilkens]

The Executor is 5 times longer than the Home One but only 19 times as voluminous?

Makes a degree of sense as Home One is essentially an ellipsoid for its length while the Executor has a much narrower blade-like profile so the additional length, much like the wings on the Liberty-types, doesn't add as much proportional volume compared against an equal length on Home One.

[Ender]

Thank you very much EJ, I don't suppose I could press you to grab as many for thigns that are in the AOTC ICS and ROTS ICS as possible, could I please? What you have provided so far is very useful, but I would like to try and have models that I can compare results on with published info to verify my methods. Thanks