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Re my recent posts:  I’m sure you all spotted a few minor errors but were too well mannered to mention it. 

My 2023 Mar 29, 13:48 – 0700, for “last seriously revised June 1959 for UK Maths 'O' levels.” read: “last seriously revised June 1960 for UK Maths 'O' levels.”,

for “Therefore, distance= height x 360 x 60/2 x 3.14159 x angle in minutes = 3438/angle in minutes.” Read, “Therefore, distance= height x 360 x 60/2 x 3.14159 x angle in minutes = height x 3438/angle in minutes.”

My 2023 Mar 29, 10.02 -0700, I wrote “Let’s also decide that at these sort of distances, we can assume a flat Earth.  Then the angle measured will be the angle between the top of the object and the sea horizon.  The perpendicular will be ‘altitude’ of the top of the object amsl (available from your chart) – 2m (quoted height of eye) – height of tide amsl.”, but we also need to allow for the dip of the sea horizon.  The question is which dip, the flat earth dip at 3nm and 2m ht of eye of 60arctan 2/(3 x 1852) = 1.24’, or the spherical earth value of 1.76root2m=2.5’?

David McN
Before trying you distance-off measurements, bearing in mind that you’ll be bobbing up and down as well a using a certain degree of approximation, it might be worth working out the effect of a 1’ measuring error  at 20 miles distance off.  At 20 nm range from a 350m object you might expect to measure around 60arctan350/20x1852=60arctan0.00945=32.5’.  If you only measured 31.5’, that would correspond to a range of 20.74nm, but if you measured 33.5’, the range would be 19.4nm.  Therefore, 1’ overall measuring error at around 20 miles range is going to correspond to about 0.7nm range error.  Closer to the shore, the effect of 1’ error will be less.  DaveP