MacGregorSailors.com

Hello all,

(Jenna sets down her mineral water, puts feet up on settee.)

I've been wondering, out of curiosity: If I'm sailing on the ocean, and I see something in the distance, how much of it can I see as we approach one another?

Can someone who knows math please give me an equation I can use? I can use it if I learn it.

The height of my vision on my boat above the average sea level at that time (not worrying about swells and waves).
The curvature of the earth (assuming good seeing conditions).

I should be able to determine how much of the object I'm not seeing at what distance.

Thank you

Jen

boaterjen wrote: I've been wondering, out of curiosity: If I'm sailing on the ocean, and I see something in the distance, how much of it can I see as we approach one another?

Can someone who knows math please give me an equation I can use? I can use it if I learn it.

Jen

Hi Jen,

very interesting question ! Don't think that without knowing the height of the object that you are looking at you will be able to tell how much of it you will be able to see in a given distance, as this will determine the height of the object above the sea level. There is a formula that will help you to determine the distance to the horizon. Or practically, the distance to an object that you can just see occuring at the horizon :

The distance in kilometers is approximately 3.57 x the square root of the height of the observer (your eye height in meters) above the water level.

If you want the result in nautical miles, divide by 1.852. This formula follows a simple geometrical model without taking the refraction of light by the atmosphere or the curvature of the horizon into account. As long as you as the observer are close to sea level, ie on a boat and not somewhere high up on a mountain, the latter is not important at all.

Hope this is helping you a bit at least.

The formula for a right angle (isosceles triangle) will determine any of the three sides of a right angle as long as you know enough about the other two angles. Assuming the mast of the ship is perpendicular to the water you have already determined that the mast is a 90 degree angle, but without knowing the height of that mast it gets really hard to determine the final angle.

The one I really find more useful is Chapman Piloting and Seamanship 66th Edition, page 151. To determine relative position in relation to other craft. If you line up with two fixed positions on your boat to the front of the craft in the distance you can tell if the craft will overtake you, follow you, or intercept your course.

I think Jen's question is rather more complicated. What she is asking, if I get it right, is what % of a boat is hidden below the Earth's curvature, at a given distance, or, conversely, at what distance will you see, say, 50% or an object.

Radar folks have worked on this problem for many years, and here's a good site that somewhat explains it:

http://www.rfcafe.com/references/electr ... -sight.htm



...and of course, Wikipedia also covers it, with a very good formula, albeit a complicated one...

http://en.wikipedia.org/wiki/Horizon



Of course, neither explanation completely compensates for Einstein's theory of General Relativity, and more specifically, the Einstein tensor effect.

And don't forget to factor in the undulation of the geoid in calculating the true height of msl above the WGS84 elipsoid. Sorry, I've been trying to figure out why my GPS has so much alttitude error.

~Rich

To heck with using GPS for altitude - NOAA is mapping the gravitational field of the country in order to do accurate surveying by measuring the strenght of gravity.

See details of the GRAV-D program http://www.ngs.noaa.gov/GRAV-D/

While in the Navy many years ago I know for times when the fire control (artillery system) on our boat wasn't working we had a device that you could sight the other ship with and on a scale in the viewfinder mark height from waterline to top of mast. Then refer to a reference book that gave mast heights and you could calculate distance to target. This is not exactly the same thing but along those same lines. I think someone even took this process to the golf course and you can shoot the pin on the green and knowing height can calculate distance to the hole. As said before it's all about triangles.

Just stick to small lakes.

What are you guys doing with your Macs that you need to know the altitude ??? I am always at sea level

I was playing with the portable GPS that I use in the airplane, when the math question came up.



~Rich

Live at 5,390', sail at 8,180', sea level is 1,250 miles away

Bob

About 5 miles

Pick up a Golfers Driving Sight for $100 and when you can focus its calibrated to give the distance.

kmclemore wrote:I think Jen's question is rather more complicated. What she is asking, if I get it right, is what % of a boat is hidden below the Earth's curvature, at a given distance,

Yes. That's it. I'm wondering: I'm sailing along. My eye is maybe 6' above the water. I see an island. How much of the island is not visible to me becaue it's below the curvature of the earth? At what distance do I begin to lose sight of the bottoms of objects? If I know the distance, I should be able to figure out how much of an object at distance I'm not seeing.

I'll look at the information above and see if I can larn dis. Thanks!

Jen

kmclemore wrote:IRadar folks have worked on this problem for many years, and here's a good site that somewhat explains it:

http://www.rfcafe.com/references/electr ... -sight.htm

I took a look at this. This is what I'm looking for. But I went though college without taking one single math course, proper. I took statistics instead and logic, math equivalencies. I've learned since that I could have done it, but at the time, I didn't. So I'm not trained to follow all that.

An equation should be able to be made that I can do that's easy (I know a simple versio of Trig my husband taught me one morning)...

If my boat is at A, eye at 6'. Another boat is at B, 8 nautical miles away. The curvatuer of the earth is in the way--though I'm glad the earth isn't flat, so one can travel 'round it without falling off, which would not mix well with a daily shot of rum. So how much of the bottom of that other boat at B am I not able to see yet? If I view it with a right triangle, I am at a height, and the other boat at B is to the right at the end of the triangle tip. The height of tis right triangle is not 6', as it'd go down to the level of the boat at B, over the arc of the earth....

I'm a liberal arts major, but you see where I"m thinking?

A simple version of somethig would be needed for me, but if I have one, I can use it to rough it out.

Ideas?

Thank yuo

Jen