Deck height & Horizon distance


Arun Vajpey

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I request some accurate information about how far a person standing on the deck of a ship can see in good light and a calm sea.

To hypothesize, let us assume that the person is standing on a ship's deck that is 40 feet above the sea level. Add another 5 feet to the eye level and so he/she is looking out from 45 feet. In that case, how far away will the plain horizon be assuming that there are no mirage or other optical effects to affect the line of vision?
 

Arun Vajpey

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Thanks. But I noted that the picture was that of a rather uneven landscape. Will it be slightly longer - perhaps 8.5 miles or so over a calm, flat sea?

Also, it the hotel room comment in that tool, they talk about looking at a ship from a vantage point. That is what I was getting into.

Applying that logic to my hypothesis and ignoring any drift momentarily, from an eye level of 45 feet, the plain horizon is about 8.2 miles across the sea. But if there was another ship stopped at 8.2 miles with its broadside to the first ship, the original observer would see all of its profile right up to the waterline (perhaps with a bit of curvature distortion). But that ship could be further away from the observer and so partly below the horizon but our viewer might still be able to see part of it, right?

If that second ship was similar in dimensions to the observer's ship and had its uppermost deck at 40 feet, is there a tool to calculate how much farther the second ship can be from the observation point for only its superstructure (part of the ship higher than to topmost deck) to be visible?
 
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Aaron_2016

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I'm sure some of the mariners here will know. My house is about 40 feet above sea level. I photographed this cruise ship which appeared on the horizon 8 miles away. All of her decks can be seen. When I used the binoculars I could even see her bulbous bow breaking the surface just a few inches above the water.

cruise-ship1.png


I also saw this French aircraft carrier. She was around 20 miles away (more than 10 miles beyond the horizon) and only her top decks could be seen with binoculars.

aircraftc.png
 
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Doug Criner

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Here's the formula from Bowditch:

D = 1.17 sqrt h, where:

D = distance to visible horizon, nautical miles

h = height of observer's eye, ft above sea level
 

Arun Vajpey

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I'm sure some of the mariners here will know. My house is about 40 feet above sea level. I photographed this cruise ship which appeared on the horizon 8 miles away. All of her decks can be seen. When I used the binoculars I could even see her bulbous bow breaking the surface just a few inches above the water.

I also saw this French aircraft carrier. She was around 20 miles away (more than 10 miles beyond the horizon) and only her top decks could be seen with binoculars.
Thanks again. I found the highlighted part about the Aircraft Carrier very interesting because it is on those lines that I am looking for answers. Most medium and large ships are also tall and when observed from another ship the superstructures may be visible well beyond the calculated horizon distance.

But there is also another factor - distortion due to the Earth's curvature. If you were seeing the superstructre of that French Carrier from 20 miles away, it would appear as though the ship had a list away from your vantage point. At night, this might create a more confusing picture with the relative positions of masthead lights etc.

There was a British chap named David Hutchings who did a lot of interesting work about horizon distances in relation to Titanic and Californian. I think he might even have been a member of ET at some stage.
 
May 3, 2005
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Here's the formula from Bowditch:

D = 1.17 sqrt h, where:

D = distance to visible horizon, nautical miles

h = height of observer's eye, ft above sea level
I have seen constants of 1.2 or 1.5 used in the formula.
Are there reasons why the diferent constants might be used ?
Say.....Under different conditions ?
Is 1.17 considered the most accurate ?
This is the first time I have seen it quoted.
 
May 3, 2005
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Thanks again. I found the highlighted part about the Aircraft Carrier very interesting because it is on those lines that I am looking for answers. Most medium and large ships are also tall and when observed from another ship the superstructures may be visible well beyond the calculated horizon distance.

But there is also another factor - distortion due to the Earth's curvature. If you were seeing the superstructre of that French Carrier from 20 miles away, it would appear as though the ship had a list away from your vantage point. At night, this might create a more confusing picture with the relative positions of masthead lights etc.

There was a British chap named David Hutchings who did a lot of interesting work about horizon distances in relation to Titanic and Californian. I think he might even have been a member of ET at some stage.
Both the height of the observer and the height of the object being observed must be taken into account.
I believe there is also a formula for this, but I would have to look it up.
 
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Doug Criner

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I have seen constants of 1.2 or 1.5 used in the formula.
Are there reasons why the diferent constants might be used ?
Say.....Under different conditions ?
Is 1.17 considered the most accurate ?
This is the first time I have seen it quoted.
The constants are different for whether the height of the observer is measured in feet or meters. I believe that 1.17 is the accepted constant for height in ft. 2.07 is used for height in m. My source is Bowditch, American Practical Navigator, 2002 (published by U.S. Government) - Article 2202, p. 330. The formula in Bowditch is based on nautical miles, not statute.
 

Arun Vajpey

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Both the height of the observer and the height of the object being observed must be taken into account.
I believe there is also a formula for this, but I would have to look it up.
I would be very grateful if you could post the information when you find it. If we are talking about looking at one ship from the upper deck of another, surely the heights of both ships would have to be considered, in which case the distance of visibility for only the uppermost deck & superstructure would be considerably greater than the calculated horizon distance.
 

Dave Gittins

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In practice, these things are worked out using nautical tables, such as Nories or Inmans. For example, Nories Extreme Range Table tells me that if the observer's eye is 30 metres above the sea an object 10 metres high will just tip the horizon at 18.1 miles. Major libraries should have tables and you can play with them all you like.
 
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Arun Vajpey

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I found this useful calculator on the web. Earth Curve Calculator .

According to the table there, from an eye height of 45 feet looking across an unobstructed flat surface (like a flat calm sea), the hidden height of an object 20 miles away would be about 92 feet. I took that to mean that the first 92 feet of the object above the sea level would be hidden from the observer but anything above it would show.

Applying this principle to the Titanic, can someone tell me the approximate height of the ship from the sea level to the boat deck as it was during its maiden voyage? It would have been approximately 2/3 of its maximum payload, right?
 

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