7/8 Rule

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Use this rule when the first bearing to an object is 30 relative, and the second bearing is 60 relative.
What it tells you is that when the first relative bearing is equal to 30, and the second is equal to 60, 0.875 multiplied by the distance run is equal to the distance from the object when abeam. It also tells you that 0.5 multiplied by the time run is equal to the time until the object is abeam, and distance run multiplied by 0.5 is equal to the distance until the object is abeam.
This works because the triangle between the first and second positions and the object is isoceles, meaning that two of the angles are the same and two of the legs are the same length. Therefore distance run is equal to distance to the object at the second position. The pythagorean theorum and sine functions can also show us that in a 30-60-90 triangle, like the one formed between the second position, the position when the object is abeam, and the object, the shorter leg is half the length of the hypotenuse, and the longer leg is 0.875 (or 7/8) the length of the hypotenuse.
sin 60 = opposite / hypotenuse
sin 60 = 0.875
0.875 = opposite / hypotenuse
0.875 x hypotenuse = opposite
cos 60 = adjacent / hypotenuse
cos 60 = 0.5
0.5 = adjacent / hypotenuse
0.5 x hypotenuse = adjacent

Back to Special Case Bearings


Question: You are steering course 090 at 16 knots. At 0800 you spot a mermaid sunbathing naked on a rock bearing 060. At 0845, she has turned over and bears 030. When will the rock be abeam? How far would you have to swim to reach the rock when it is abeam?
-First, find your relative bearings
     090 - 060 = 30
     090 - 030 = 60
-Find your time and distance run
     0845 - 0800 = 45 min
     45 min/ 60 x 16 = 12nm
-Multiply the time run by 0.5 to find the time until abeam
     45 x 0.5 = 22.5 min
-Add the time until abeam to the time of the second observation to find the ETA until abeam
     0845 + 22.5 = 0908
-Multiply the distance run by 0.875 (or 7/8) to find distance to the rock when abeam
     12 x 0.875 = 10.5 nm, that's a long swim