Sound speed propagation and detection

USS Moose · 02-07-15, 07:55 PM

From my basic understanding of physics and SSP and sub positioning in this particular scenario, time would be negligible (or the same, depending on how you look at it). Looking at the graphs of common SSP types, velocity is more or less linear over any given amount of time in a function of depth over temp, therefore we can assume that acceleration is constant. By way of analogy, lets imagine two cars:

Car A: initial velocity (Vi) is 20 m/s; final velocity (Vf) is 10 m/s

Car B: Vi is 10 m/s; Vf is 20 m/s

let's say the distance they have to travel is 150 m.

from the kinematic equation (remember acceleration is constant so it's not in this formula), d = [(Vi + Vf)/2] * t, we solve for time giving us

(d * 2)/(Vi + Vf) = t

Both cars have to travel the same amount of distance and since the sum of their initial and final velocities are the same, the answer for time is the same. In this case, it would be 10.0 seconds.

So back to subs, the time it takes for the sound(s) of Sub A to be detected by the sonar of Sub B (and vice versa) would be the same because even though one accelerates and the other decelerates, the EXTENT of their acceleration is the same. Therefore, I believe SPPs are generally used only to determine shadow zones/depths of detectability.

Either my assessment is correct or I just wrote an entire thing of nonsense. Either way I hope that it opens up a new perspective.

02-07-15, 07:55 PM	#2
USS Moose Nub Join Date: Jan 2015 Posts: 4 Downloads: 0 Uploads: 0	From my basic understanding of physics and SSP and sub positioning in this particular scenario, time would be negligible (or the same, depending on how you look at it). Looking at the graphs of common SSP types, velocity is more or less linear over any given amount of time in a function of depth over temp, therefore we can assume that acceleration is constant. By way of analogy, lets imagine two cars: Car A: initial velocity (Vi) is 20 m/s; final velocity (Vf) is 10 m/s Car B: Vi is 10 m/s; Vf is 20 m/s let's say the distance they have to travel is 150 m. from the kinematic equation (remember acceleration is constant so it's not in this formula), d = [(Vi + Vf)/2] * t, we solve for time giving us (d * 2)/(Vi + Vf) = t Both cars have to travel the same amount of distance and since the sum of their initial and final velocities are the same, the answer for time is the same. In this case, it would be 10.0 seconds. So back to subs, the time it takes for the sound(s) of Sub A to be detected by the sonar of Sub B (and vice versa) would be the same because even though one accelerates and the other decelerates, the EXTENT of their acceleration is the same. Therefore, I believe SPPs are generally used only to determine shadow zones/depths of detectability. Either my assessment is correct or I just wrote an entire thing of nonsense. Either way I hope that it opens up a new perspective.