Jonathan_S wrote:I suspect if a GT is going to make a snap shot as it overflies a target it's going to need to stop acceleration a little while before and pivot its entire body so it coasts past the target with its nose already pointed at the target's most probable location. That way it can start firing the moment it clears the wedge's edge; and try to correct onto target as it goes. (Though depending on its closing speed it may get only fractions of a second to react, so it might have no more chance to correct for a bad guess than a laser-head would. At high relativistic closing speeds it could clear one wedge, and then be blocked by the other in less time that it takes to see the target and react)
That's something we can calculate.
Perls of Weber: Wedge Geometry lists for an SD:
- throat: 190 km
- kilt: 40 km
- width: 300 km
If the ship is dead centre in the wedge and the torpedo passes it exactly atop and perpendicular to the wedge, we'll get average values. That is to say: the ship will not be dead centre and the torpedoes will not pass exactly perpendicular, but since the throat is wider than the after aspect, we can calculate what missiles will see, on average, passing by an arbitrary SD.
From the centre of the wedge, the distance from floor to ceiling is (190 + 300) / 2 = 245 km. That means from the ship, the floor or ceiling is 122.5 km away in the altitude direction and the edge of that wedge is another 150 km away in the lateral direction. We don't need to calculate the angle (it's simple trigonometry, though). We can just use triangle similarity.
If the torpedo is passing at 30,000 km away, it's at 200x the distance from the ship as the edge of the wedge is from the ship. That means it a clear shot for a transversal distance of 200 x 245 = 49,000 km. If that missile is moving at a third the speed of light, it's going to have a clear shot for 0.49 seconds. From this, you need to subtract 0.1 seconds light-speed lag for the reaction of the missile once it has cleared the wedge, seen the ship and reoriented. Minus a bit more because the reaction isn't instantaneous.
At the same speed, if it wants to have a shot for 3 seconds, it needs to be 6x further away, so 180,000 km, but that also increases the time-lag reaction to 0.6 seconds, making the shot on target last for 2.4 seconds.
Or, at the same distance, it would need to be moving at one sixth the speed, or a mere 12,500 km/s, to be able to have 3 seconds of opening and only 0.1 second time-lag.
It's simple math to calculate where it needs to be and what speed so it has a full 3 seconds firing time including the time lag to orient itself. Exercise is left for the reader.
But either scenario presents its own problem. On one, the torpedo is far enough away that its energy is attenuated considerably. At 200,000 km, can a graser penetrate an SD sidewall? Plus, the ship will be jinking all the way, so the 0.6 second time-lag is significant to react to the evasions. It's unlikely to be able to maintain the full 3 seconds on target. In fact, it'll be challenged to even hit the ship: this is why multiple lasing rods have an advantage, since they can bracket all the possible locations the ship can be in the time it'll take for the laser to arrive and ensure that at least some beams strike.
On the other scenario, the torpedo is moving very slowly compared to the ship and it's extremely close. More than that, it's already been within 50,000 km from its target for about 0.8 seconds before it cleared the wedge and was able to engage. A Keyhole II-equipped ship will likely be firing at it before it can engage. The ship can patiently wait for the missile to get closer because it can't fire until it's cleared the wedge, so it can spend those 0.8 seconds refining the target and burning through its stealth.
As for the rate of rotation, if my trigonometry didn't fail me, it's 0.45 rad/s ≅ 26 deg/s ≅ 4.3 rpm.