Quote:
Originally Posted by vdr1981
Yes they are marked on the map, but we still can not know the exact uboat position if we are reading bearins only from one NDB (the closest), and that is what we can do with this feature. Maybe I didn't understand what are you trying to say? Can you give some example ?
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Yes sure.
When two radio beacons are in range, our position can be simply plotted on map as the intersection of the bearings relative to the two antennas, as illustrated in the picture below:
This is a valid method we can use with lighthouses, day beacons and landmarks as far as their position is marked on map, but unfortunately it won't work with your radio beacons because the radioman will always report the bearing of the nearest antenna.
Nonetheless just one beacon is enough to get our position relative to it. The method we need to follow is slightly more complicated than the one used when two beacon signals are available, but not that much. What we need to do is stopping our boats when we think that we are within beacon's range and asking our marconist for a bearing to it. Knowing that, we should put ourselves on a ±90 deg route relative to the reported bearing (270 deg in the example below, i.e. 90 deg to port) and we should sail at constant speed for a measured amount of time, so that we can estimate as accurately as possible the distance covered during that lapse of time. The more we move relative to the first bearing, the more accurate our calculations will be. When we are satisfied with the distance covered we stop the boat and we wait for the next radio signal.
At this point, our measurements will describe a right angle whose two catheti are respectively the leg between the two bearings (
a in the figure above) whose length we know, and the distance between our submarine and the antenna when the first bearing was taken (
b). We need to calculate the hypothenusa (
c), i.e. the distance bewteen the antenna and our current position. To do that we need the angle
α (alpha) between the two bearings. This is equal to the absolute value of the current bearing minus the first bearing.
In our example
α = |240-270|= 30 deg
From there, an elementary trigonometric rule can be used for calculating
c as:
a/sin
α
Assuming that in our example
a = 5 km we would have that:
c = 5/sin(30) = 5/0.5 = 10 km
Knowing the bearing and the distance from a fixed point, we basically know our position.
Cool, isn't it?
