Well I never.... did you know ?

[Danny_Labrador]

1.17 times the square root of your height of eye,(in feet) = Distance to the horizon in nautical miles

 

[Goodge]

So where did you get that little gem from ?

 

[Pie]

That'll be that golden ratio 'phi' - I think /forums/images/graemlins/confused.gif from reading that Da Vinci Code book.

 

[Aja]

Cant be right. I cant see the other side of the road let alone the horizon. Bloody grat building in the way.

Donald

 

[Lakesailor]

What if I'm stood in a hole?

 

[catmandoo]

good job you have not got eyes in the bck of your head or have you ?

 

[Rowana]

From my schooldays -

The angle of the dangle is at right angles to the sag of the bag when the throb of the knob is a constant /forums/images/graemlins/blush.gif /forums/images/graemlins/blush.gif /forums/images/graemlins/blush.gif

 

[peterb]

2.2 times, if you measure it in metres. Works for radar sets, too.

 

[Phoenix of Hamble]

And VHF which is also line of sight.......

 

[alant]

This means if your Ht of eye is 2m (I should be so lucky), the horizon is 2.0976 Nm away.
Or 2.569 Nm when @ 3m.
Since this is last one ain't much above the waterline when sitting in a cockpit, how come I can see the beach when known to be at around 4 Nm off (GPS etc position), when sailing across Poole Bay? Or is this an optical illusion.
/forums/images/graemlins/confused.gif

 

[AngusMcDoon]

Because the beach is above sea level, by definition. If it wasn't, it would be seabed.

 

[Beagle]

Isn't the complete calculation bewteen 2 objects on a clear and visible day something like:

2.1 *( SQRheight eyes above water + SQRheight of object)

heights: metric, result in Nm. Works great for lighthouses (or other tall objects) as well

 

[AngusMcDoon]

Yes, something like that. I can't remember it exactly either.

 

[peterb]

[ QUOTE ]
This means if your Ht of eye is 2m (I should be so lucky), the horizon is 2.0976 Nm away.
Or 2.569 Nm when @ 3m.

[/ QUOTE ]

Square root of 2 is about 1.414. Multiply by 2.2 and you get about 3.1. And, as Beagle has said, you need to add in the horizon distance of the object you are looking at.

 

[tugboat]

Re: From my schooldays -

No no Jim, the angle of dangle is inversely proportional to the sag of bag when the throb of knob is constant. Didn't they teach you anything in BP? /forums/images/graemlins/grin.gif

 

[alant]

I've obviously mulitplied HoE in metres by 2.2 & then sq rooted it. /forums/images/graemlins/blush.gif

Using Tables,
2m = 2.9 Nm (3.11 using 2.2xsq rt)
& 3m HoE = 3.6Nm (3.81 using 2.2xsq rt)
2' giving 1.25Nm. (1.655 using 1.17xsq rt)

None of which seem to exactly match these multiples, unless being affected by abnormal refraction. Someone must know why.

 

[Seafort]

Would tables not be for an average Latitude? A bit like the definition of a nautical mile.

With the TTs, charts and distance off tables, plus 2 depth sounders I muddle by... /forums/images/graemlins/grin.gif

Dave.