Fox2! wrote:kzt wrote:I don’t dispute this is a dead horse, but the reasons it comes ip as viable are different than you state.
The elements that govern beam divergence in lasers (Gaussian beams)are the wavelength and the beam waist diameter. When the wavelength is gamma rays and the diameter is in meters the beam maintains coherency at light hours range.
And while it is difficult to predict where a warship will be in an bour, predicting where say a 200km long orbital platform is a whole different animal. You can look it ip and probably get it to plus or minus a millimeter.
So why doesn’t it work? Reasons. Next topic.
The inverse square law says that you lose energy at a non-linear rate with increasing range. Double the range, you only get a quarter of the energy delivered. Regardless of frequency/wavelength or beam diameter.
Inverse square law discussion...wrong. The energy per unit area falls. The energy *does not* fall. For a laser, the angular divergence at the exit is diffraction limited to lambda/R, wavelength/beam diameter. Said differently, you can formally treat the divergence as inverse square, but the formal origin point for the divergence is *not required to be inside the laser*. Even with a 1960s He-Ne laser, the divergence point might be many kilometers rearward of the laser cavity.
This point is discussed at length in my novel MinuteGirls, where the effective maximum range of a xraser (xray laser) was determined by the size of the enemy ship and its maximum jerk (time-rate-of-change of the acceleration). In the words of Grand Commodore Pyotr Eustasovitch Kalinin, States of Lincoln Planetary Self Defense Fleet:
"...In the end, the effectiveness of all xraser-type weapons is determined by range and the maximum acceleration and jerk of your target. Once upon a time, scientifiction authors gave radiation weapons a maximum range. However, as is immediately apparent from Physics 3, a xraser beam is close to diffraction limited, so its divergence is roughly lambda/r. A battlecruiser xraser port is 3 yards across, while a typical xraser output wavelength is 10-10 yards -- you can do this in American Scientific Units if you want. That's a homework assignment. The range of angles on beam output is, oh, 10-10 radians, give or take, so the beam diameter doubles in 1010 yards, or a bit under 107 miles. Even against screened targets, the effective range is much larger, though pointing finally is challenging. The effective range of a xraser against unscreened soft targets like planets is a large multiple of this, which is why Mercury, Venus... all get planetary defense screens.
"...However, to hurt your target, you must hit your target. An enemy ship displaces as x = 0.5 a t2, x being displacement, a being acceleration, and t being time. Time? A lidar pulse is reflected back to you; you retarget to hit the target. At one light-second range, your targeting is two seconds late. Homework: Why not three seconds, since the lidar pulses must first go from you to the target?
"Your target has some radius R. If during t your enemy moves less than its ship radius R, then you always get to hit him. If during t your enemy moves more than R, life becomes interesting. An enemy who keeps constant acceleration might as well have parked -- constant acceleration is predictable, and that which can be predicted can be targeted. An enemy who always shoots at your start point, if you always move more that R during his t, always misses.
"If your enemy keeps changing acceleration, he gets harder to hit. He can't see your xraser beams incoming -- he can still dodge. For example, suppose during t your enemy has randomly displaced on average through 4R. 1/16 of that 4R circle is him. The rest is empty. If you cannot predict his position 15/16 of your shots miss. There is a fine argument: if you are dodging, should you go random after each time you've displaced through your own diameter? Or is patience better? Your materials course is doing polymer properties -- look up 'wormlike chain'. Oh dear, I just gave away an exam one question, didn't I? And how does that depend on what your enemy thinks you will do?
"After all, x goes as t2, right? No. Once you go random, you are doing Brownian motion, which is why every command officer must understand stochastic processes. For Brownian motion, x grows as the square root of t, meaning your chance to hit a distant enemy falls linearly with time and range. Double the range means double the tracking time delay means half the chance of hitting.
"But Mister a comes in, too. The larger your a, the closer you can be to the enemy before he can hit you. Ships with larger maximum accelerations are harder to hit - unless they are so badly designed that they can't change their accelerations quickly. Constant acceleration is worthless as a defense in battle.
"Except, except, acceleration straight at or away from your enemy doesn't affect his aim. Your motion along that axis is straight line, not random evasion, so closing or retreating, ship-on-ship, is indeed x proportional to t-squared. However, a ship that is englobed -- targeted from multiple directions -- loses this option, and in order to dodge must randomize acceleration along all three axes. Thus, being englobed is bad news, especially for an American officer whose ships are slower than his foes to start out with.
"Let's put in some realistic numbers. A typical American ship can pull 30 gravities. An FEU battleship pulls 100. Also, R for a Villiers class is 200 yards. R for a Marco Polo Bridge class like the Large Battleship Death-to-the-Imperialist-Warmongers we are touring next week is 400 yards. The FEU ship can close to within one light-second of the American before the American's chance to hit approaches 100%, while the American can be as far out as 2-3 lightseconds while the FEU ship is still fairly certain to hit. In round numbers, for that pair of ships we must be within 50,000 leagues to be sure of hitting, while an FEU ship only needs to be within 150,000 leagues -- closer against smaller ships. At larger distances, it is harder to hit. At the 3 lightseconds at which an FEU ship is sure of hitting a Marco Polo Bridge class, our chances of hitting fall to one in three or less, so we need three times the firepower to even things up. Furthermore, FEU beams and screens are much better than ours, so in the end we want around ten times the firepower -- meaning ten times the hull weight -- to engage on even or better terms."
The Virginia Squadron, thought Kalinin, is at pointblank range of the FEU Demon-class ship, ranges so short that both sides are firing missile volleys "