The signal strength received from sound, radio and light transmitters reduces as the distance from the transmitter increases. Consider a light bulb in the centre of two transparent concentric spheres. One sphere being twice the diameter of the other. Let us call the strength of the light received at the surface of the inner sphere Pr. Then the strength of the light received at the surface of the outer sphere will be a quarter of Pr. This is because the area of the outer sphere is four times that of the inner sphere*. This effect is called the “Inverse Square Law” as the power level received, Pr at a distance, d, from a transmitter is inversely proportional to the square of the distance. In algebraic form Pr. µ 1/d2, or Pr = k/d2 where k is a constant. Although most transmitters do not radiate energy equally in all directions, but use arrangements to produce beams of energy, such as the reflector in a torch, the energy leaving the transmitter still spreads out in accordance with the inverse square law.

In a radar or sonar system a double inverse square low applies. The energy leaving the transmitter travels a distance, d, to the target, following the first inverse square law, where some of it is reflected**. This reflected energy then travels back the same distance, d, towards the receiver, following the second inverse square law. This assumes that the receiver is CO-located with the transmitter as is normally the case. Some energy has therefore travelled a distance 2d. The sonar operator is therefore relying on an “inverse fourth power law” for his received signal. Thus his power received, Prs µ 1/d4. This compares with the power received at the target, Prt µ 1/d2. When d = 1 unit, Prs = Ľ Prt. /p>

The use of algebra is regretted but it is probably the best way of illustrating the matter, although the diagram shown in the Effective Range page maybe of value, as maybe the following arithmetic notes:

1 nautical mile = 2,027 yards.
3 knots = 6,081 yds/hr.
30 knots = 60,810 yds/hr.
Time for MHV to travel 1,000 yds at 3 knots = 9.9 mins

Time, ‘t1’, for the MHV and weapon to meet from an initial range of 2,000 yds is found from:

‘t1’ = x/6,081 = (2000 – x)/60,810
whence x = 181.8 yds
and ‘t1’ = 0.0299 hrs = 1.8 mins = 108 secs.

‘t2’ = y/6,081 = (1,000 – y)/60,810
whence y = 90.9 yds.
and ‘t2’ = 0.015 hrs = 0.9 mins = 54 secs.

*Area of a sphere = 4pr2
Thus area of a sphere of radius 1 unit = 4p
And the area of a sphere of radius 2 units = 4p22 = 16p.

** The amount of energy reflected from the target depends upon its area, shape and surface characteristics. For Checkmate the area is small, the shape is hemispherical, but the surface is hard. The net effect will be only a small percentage of the incident energy being reflected.