How to determine distance from an object of know height using a sextant?

[sharkwatcher]

After searching for hours thru the old post to no avail, I am asking for some help. I used to know the formula for calculating the distance from an object of known height using the angle determined using a sextant. Does anyone still do that and remember how to do it and what the formula is? I promise to write it down and stick it in my sextant case this time, LOL.
Thanks in advance,
Mike

[barnakiel]

Use plain trigonometry functions. No tricks.

b.

[seaward 42]

[seaward 42]

[seaward 42]

[sharkwatcher]

Thanks guys, very helpful. I also after looking thru all my nav reference books found table 9 in Bowdich Which does all the math for you! Thanks again. It's good to know that there are still people that do this type of hands on, non electronic navigation.

[gjordan]

Dont forget that you have to correct for height of eye and any error in the sextant itself. I never did use that method, but it is supposed to work well. _____Grant.

[Paul Elliott]

Do note that the first two formulae use an approximation, where sin(x)/x approaches one for small angles. This is entirely valid for reasonable distance cases. These formulae also assume "flat earth" geometry, which is also a reasonable approximation for short distances.

The third formula takes into account the curvature of the earth, but you can't use it for "distance off" unless the object has dipped to the horizon, and in that case no sextant is needed.

[sharkwatcher]

Thanks Guys!!!

[seaward 42]

https://www.youtube.com/watch?v=eooqcaFFz9o

[barnakiel]

I no longer resolve this from the sextant (I simply read my distance off from the gps). I sometimes use the same rule though - for lights popping up on the horizon - same mathematics but the triangle is reversed. It is a quick and dirty method but works fine for morning landfalls.

If you are into it, you can use horizontal angles too in many places - there are countless parts of coasts where two or more charted marks are available.

b.

[Jdege]

For small angles, sin(x) ~= x, so you don't need trig, just ratios.

In the artillery, they measure angles in mills, where 6400 mills = 1 circle. IOW, 2000*pi.

One degree is 17.8 mills, one mill is 3.75 minutes.

And one mill is one meter in elevation at 1000 meter's distance. Our half a yard at 500 yards, etc.

My thumbnail at arms' length is nine mills across. A 30 foot boat would be one mill at 30,000 feet, or nine mills at 30,000/9 ~= 3,300 feet. A 10 meter boat that spans half a thumbnail is approximately 2,000 meters distant.

[Cadence]

Get a cheap slide rule calculator. the tan of your angle =the known height / the distance you are looking for to the object.

[SaltyMetals]

Interesting but hang on a second. Measuring the height of an object like a lighthouse will give a different angle depending on the state of tide. Surely there needs to be a correction that takes tide into account?

[S/V Alchemy]

Originally Posted by barnakiel

I no longer resolve this from the sextant (I simply read my distance off from the gps). I sometimes use the same rule though - for lights popping up on the horizon - same mathematics but the triangle is reversed. It is a quick and dirty method but works fine for morning landfalls.

If you are into it, you can use horizontal angles too in many places - there are countless parts of coasts where two or more charted marks are available.

b.

This method is great for determining distance off, although most people can just use a handheld compass. If you know distance off, you can calculate speed.

You can tell where I stand at the helm, because my eyeline crosses little bits of tape at 45 and 90 degrees relative to me staring straight at the forestay!