Wednesday, November 23, 2016

Calculating how long ships are visible cruising over the horizon

An interesting problem which I was faced with is how can I explain boats going over the horizon and disappearing on a flat earth model. I did think on this quite long and did not see anyone give a proper calculated answer. You only see movies of the ships disappearing without the correct theory or real answer behind it. In this case I did find it and have found a way to calculate the relationship between eye-height, distance to visible horizon, cruising speed of the ship and how long you can still see it over the horizon till it disappears in the horizon-mirror.

What is the horizon-mirror?

You cannot see further than the horizon. This is 100% depended on the height of your eye. You line of sight is most important in this. Everyone knows that the higher you are the further you can see. The horizon always levels up at your line of sight (this actually is an important argument for a plane). The horizon-mirror occurs due to the fact that all the lines of sight converge in such a way, that there is to much information. This means to much colour at a very slope angle (think of 0.02 degree or something like that). It results in to much info which results in a mirror. After the visible horizon mirroring occurs which normally results in a mirrored sky. Directly after the horizon the mirroring-effect starts.

Theory on mirroring after horizon

You can argue in two ways. Or your line of sight goes down and meets the horizon, or the plane line of sight goes up. From your point of view this actually is not of importance as the angle between these is very little. But having said that, almost none realises that our line of sight is mirrored from the point of the visible horizon. This means that the mirrored line of sight is going up from the horizon. Object which are located behind that point on a plane, will be cut of from the below up. This type of mirroring is everywhere visible. On the beach, desert or at sea. It doesn't show what is below the mirrored line of sight, but mirrors what is above. This principle is very important to understand the type of calculations in this (see impression below article).

How much can you see of buildings behind horizon

The mirroring-effect results in only one thing. Everything is cut from below and the rest is mirrored below the mirrored line of sight. This principle you can observe in many video's trying to prove or disprove a globe or flat earth. But no-one takes in concern our line of sight. This is most important as it explains almost everything on this subject. What is cut of from the expected picture is in direct relation to the angle of your line of sight. That angle is calculated with tan-1(h/d) with h=height and d=distance. If your eye-height is at 2 m and the horizon is at 5,6 km (this still is somewhat guessing) then the angle would be tan-1(2/5600) = 0,02 degree. From the horizon the mirrored line of sight goes up at an angle of 0,02 degree. Based on this you can calculate how much of buildings (measured from sealevel) have disappeared in the mirror. If the building is at 1 km behind the visible horizon than actually 1*tan(tan-1(2/5600))*1000 = 0,35 m would be invisible. At 10 km that would be 3,5 m. If you look at a globe than that would give 0,078 m (1 km) and 7,8 m (10 km). PS these calculations start from the point of the horizon. It suggests that on a globe the drop will be much quicker in relation to the flat earth.

Calculations on disappearing-time ships over horizon

When looking at a video on ships disappearing over the horizon this relation can be used to calculate how long it will take for the ship to disappear for real over the horizon on a flat earth model. If a ship is cruising at 30 km/hour and in the video the ship is visible 27 min, then how high is that ship up to the top of the shipmast: (30/60*27)*tan(tan-1(2/(56000)))*1000 = 4,8 m. You can turn this around also: How long would the ship be visible if it has a topmast of 7 m then 7/[(30/60)*tan(tan-1(2/5600)]/1000 = 39.2 min. What would be the speed if it is 6 m high and travels 40 min: [6/(tan(tan-1(2/5600))*60/40]/1000 = 25,2 km/hour In this case there are several variables and can only be checked if you know the dimensions of the ship, your eye-height, the cruising speed and the real distance to the visible horizon. But given the video and these results it shows that the relation is most-likely correct and therefor can be used as additional proof for a flat earth.

Comparison to a globe-model

If the forgoing actually would be on a globe earth, then ofcourse the  figures would be different. From the point of the visible horizon the boat gradually would sink behind the wall of water. Every km it would go down (6371^2+1^2)^(1/2)-6371 = 7,8 cm. For the following km this is the same but from our point of view the curvature of the first km needs to be accounted also. So at 2 km the boat would already have disappeared 31 cm. At 10 km this already would be 7,8 m which means that the ship already would have disappeared completely. In the video the author says that the ships is cruising at considerable speed so as stated the ship would probably have a speed of 30 km/hour (about 16 knots) or more. In the mentioned 27 min it would have travelled 13,5 km at which the boat would have disappeared behind 14,3 m of water. This is absolutely not the case as shown in the video, as the ship clearly disappears above the visible line of horizon. Next to that the video shows that the boat disappears on eye-level or our line of view in the so-called mirror. These two facts and the forgoing analysis support the flat earth model and don't support the globe-model.

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