by KD/3rd
Classic setups to make a run on shipping can be lost with a miscalculation of a targets course and speed; and how can it happen? Well for one thing by not having a handle on relative motion.
Basically relative motion is the resultant direction and speed generated by two moving objects. An obvious example of one is when your in your car traveling down a one lane road at 60 while another car is coming at you at 60. Your relative speed is 120.
We really don’t know what that oncoming car’s speed is now, do we. We just took a whack at it to talk about relative speed. We do know it’s really going south cause we see it coming at us. Now lets throw a curve here, what’s the relative direction of travel? Draw a full circle to represent our speed of 60 mph.
Lets say, for example, that road your traveling is due north. Now draw straight a line, representing north, out from the center of the circle to where it touches the 60 mph speed circle. Label that point as “r”. This line is your “course and speed vector” (000 at 60). Here’s some standard abbreviations.
r = your own course and speed
m= targets course and speed
m1= first position of target
m2= second position of target
Let’s say you seen that car coming just as it passes the Jones farm, and you know that farm to be three miles down the road. Now draw another full circle that we will use as distance, and lets use nautical miles cause that’s where were going to end up anyway. Mark that second circle with an “x”, again due north of you (000 at 6000 yards) because that’s where you see the car. Then label it “m1”. Now a minute later you see your target car passing Al’s Garage and you know Al’s is one and a half miles – 3000 yards... from the Jones farm. So draw another full circle representing 3000 yards from the center and place an “x” due north of the center on the circle. Label it “m2”.
Now here’s a little war chant for you. R is to M as M1 is to M2, parallel to and going in the same direction as M1 is to M2. Now here’s the tricky part. From “r” extend a line “parallel to and going in the same direction as m1 is to m2. How far? Well if a car travels 3000 yards in one minute what’s the speed? Here’s another war chant. Distance traveled in 3 minutes divided by 100 equal’s speed (in knots). So, we take this 3000 yards x 3 and come up with 90 kts. We draw the “relative speed vector”. from “r”, due south because that’s parallel to and going in the same direction as m1 is to m2.
Now back to the question. How long is that line. Well it’s a line representing a relative speed of 90 knots. You know how fast your going, 60, but we were not sure about the target. Now we found the “relative speed” is not really 120 but 90. Se we draw a line representing 90 knots from “r”. Looks like ‘m’ is touching the 30 know circle. So our target car is not really going 60 knots, morel like 30.
We demonstrated relative movement using cars now lets move to our playing field, the ocean. I don’t know what type of grid system SHII will have in it, but it would be nice if it was good old Latitude and Longitude, and if they are using real world charts, well why not. Anyway for the sake of demonstrating this segment lets assume some things; we have been told where a four ship convoy is, along with its course and speed.
It’s 0300 and we are patrolling surfaced on a course of 215 at 12 knots. We plot the reported convoy position and find its bearing 315 true at 30 nautical miles from our present position. Its course is 090 at a speed of 12 knots.
Our job is to intercept and destroy those ships.
Lets first plot out the convoys course and speed by drawing a line on a bearing of 090 and scale its length to represent 12 knots. Label that lines starting point as “e” and ending point as “m”(a speed vector). Now select a range scale of your choice and plot out its reported position bearing 315 at 30 miles (60000 yards) from “e” (Don’t worry it will become clearer). Mark that position as “m1” (Obviously the speed scale, and range scale, you select must be used through the entire solution or your likely to be keel hauled in Brazil).
OK so far we have the convoy’s course and speed vector, and its position plotted. Now how do we intercept those four vessels? We create the desired relative movement. Remember the first lesson where we said “r” is to “m” as “m1 is to “m2”? We have “m1” we have “m” so where do we want ‘m2”? Relative motion steps in. When we plot out relative movement we always remain at “e”, the center of our plot. That convoy is not going to help our solution one bit by heading towards us so we have to head towards them. We always remain at “e” so where do we want “m2” to be? You got it, at the center or our plot “e”. We now have another piece of the solution; a line labeled “m1” on one end and “m2” on the other. Our war chant: r is to m as m1 is to m2, parallel to and going in the same direction as m1 to m2. We have m we have m1 and m2. Now all we need is “r” (our course and speed vector). And were really close to finding it.
Let’s copy the m1/m2 line to our speed vector and attach it to “m”. Remember this line must be parallel to m1/m2. If you put a compass on that line it would be running 315/135. The mystery, someplace along that line is our course. The major concern is can we intercept the convoy? Remember the war chant. 1. Do we have the speed? 12 knots is it. 2. Can we do it –intercept? 3. What’s our intercept course at 12 knots? 4. What’s the minimum speed we could intercept at?
