Lighthouse Visual Ranges

Extraordinary claims require extraordinary evidence.

— Carl Sagan

Introduction

Figure 1: Flat Point Lighthouse in Nova Scotia.

Figure 1: Flat Point Lighthouse in Nova Scotia.

It is Easter and I have a terrible cold − I am not going to see family this year. Instead, I am going to lay here and write for a while.

Yesterday, I received a question from a reader who was puzzled by a web page written by a flat earther that presented a seemingly rational argument in favor of the flat earth position. In a nutshell, the flat earther's argument says that to see a lighthouse at long distance on spherical Earth would mean that you would have to be able to see around the horizon, which they claim is not possible. Therefore, the Earth must be flat. I may be mischaracterizing their argument, so you may want to visit web sites that go into the details of the flat earth rationale. Of course, I argue that refraction can and does literally allow you to see "around" the horizon.

The questioner's request was to help him understand the fallacies in the arguments being made. I generally avoid these discussions and simply refer people who write me on this subject to wikis and blogs that are focused on these topics. For example, in the case of the flat earthers, the RationalWiki does a good job of debunking their arguments. However, I am in bed and feeling cranky, so here I go.

While I will not argue about non-falsifiable arguments, I can discuss the parameters critical to computing lighthouse ranges and how nominal lighthouse ranges are computed in a standard table of lighthouse ranges. As a side topic in Appendix A, I review one of flat earther calculations and with my own calculation show that refraction and tidal water level changes can easily explain the ability to see a lighthouse over-the-horizon. You do not need to assume that the Earth is flat.

Background

I have done a fair amount of optical range modeling in the Earth's atmosphere. Truth be told, this is all related to my interest in battleships and optical fire control systems. However, I did write a blog post that referenced lighthouses and the material presented here is based on that post.

Analysis

Formula

Starting Point

Equation 1 gives the distance one can see from an elevated point (e.g. lighthouse or battleship mast) to a point on the horizon, assuming a nominal level of refraction.

Eq. 1 \displaystyle s\left[ {\text{km}} \right]=3.86\cdot \sqrt{{h\left[ m \right]}}

where

    • s is the distance along the Earth in kilometers.
    • h is the height above the sea in meters.

The weakness of this formula is that the front coefficient (3.86) assumes a specific lapse rate, which describes the atmospheric temperature variation with height. Variations in the lapse rate can produce enormous increases or decreases in the visual range of a lighthouse. Variations in the lapse coefficient can even explain atmospheric ducts that can guide optical signals (and other electromagnetic signals) over very long ranges.

Also, the height of the lighthouse above the sea varies with time (i.e. tides) and weather (e.g. storm surge). Most official tables of lighthouse ranges are based on the mean high water level about the lighthouse.

Model Representation

Figure 2 shows how the visual range of a lighthouse depends on both the height of the lighthouse and the viewer. Figure 2 is actually taken from a post on battleships, but the geometry is identical for both cases. This means that I must apply Equation 1 twice:

  • Compute the range from the lighthouse to the horizon.
  • Compute the range from the viewer to the horizon.
  • Add the ranges together.

Figure 2: Figure from My Battleship Work --- Also True for Viewing Lighthouses.

Lighthouse Formula

Figure 3 shows my lighthouse formula. I also grabbed some data from an old UK publication on nominal lighthouse viewing ranges for comparison.

Figure M: My Lighthouse Formula.

Figure 3: My Lighthouse Formula.

This formula also reveals another source of variation − the height of the observer above the sea. This will vary since the height of the deck of each ship is different. The standard for computation is 15 feet, but that can vary dramatically by the type of ship. For example, the flat earther web page in question here assumed a 24-foot high deck. To duplicate the UK table values, I needed to use 15 feet for the deck height.

Publication from Lighthouse Manufacturer

Results

Figure 4 shows that Equation 1 from my blog post produces the same result as that specified for nominal lighthouse ranges. So I believe that my model is reasonable.

Figure M: My Lighthouse Ranges Versus an Old Specification.

Figure 4: My Lighthouse Ranges Versus an Old Specification.

Conclusion

Reported lighthouse ranges are a strongly dependent of the assumptions made in their calculation.  You do not need to assume the Earth is flat to explain how lighthouse range estimates can vary so much from a pure spherical model. The variation naturally occurs because of changes in water level and temperature.

Appendix A: Worked Flat Earther Example

I will work one of the flat earther examples. My intent is to first show how they performed their calculation. I have highlighted the statement "not an uncommon thing", which is a very imprecise statement. In general, lighthouse ranges are stated in a conservative manner that ensures that they are almost always visible at the stated range. This statement indicates that the ranges quoted here are NOT common.

The distance across St. George's Channel, between Holyhead and Kingstown Harbour, near Dublin, is at least 60 statute miles. It is not an uncommon thing for passengers to notice, when in, and for a considerable distance beyond the centre of the Channel, the Light on Holyhead Pier, and the Poolbeg Light in Dublin Bay. The Lighthouse on Holyhead Pier shows a red light at an elevation of 44 feet above high water; and the Poolbeg Lighthouse exhibits two bright lights at an altitude of 68 feet; so that a vessel in the middle of the Channel would be 30 miles from each light; and allowing the observer to be on deck, and 24 feet above the water, the horizon on a globe would be 6 miles away. Deducting 6 miles from 30, the distance from the horizon to Holyhead, on the one hand, and to Dublin Bay on the other, would be 24 miles. The square of 24, multiplied by 8 inches, shows a declination of 384 feet. The altitude of the lights in Poolbeg Lighthouse is 68 feet; and of the red light on Holyhead Pier, 44 feet. Hence, if the earth were a globe, the former would always be 316 feet and the latter 340 feet below the horizon!

Figure 5 shows how I duplicated the flat earther's calculations. There are some issues with their calculation:

  • purely geometric, no refraction.
  • assumes a 44-foot Hollyhead lighthouse (Wikipedia states its height as 70-feet ) and a 68-foot Poolbeg lighthouse (reference states 66-feet )
  • Lighthouse heights are normally stated with respect to the high-water level, but I am unclear what the flat earthers are using.
Figure M: Worked As a Flat Earther.

Figure 5: Worked As a Flat Earther.

Figure 6 shows the geometry that I will be assuming here.

Figure 6: Basic Ship Geometry.

Figure 6: Basic Ship Geometry.

Figure 7 shows that the lights could easily seen at low tide and with reasonable atmospheric conditions.

Figure M: Refracton Makes the Lighthouses Visible on a Spherical Earth.

Figure 7: Refraction Makes the Lighthouses Visible on a Spherical Earth.

 
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4 Responses to Lighthouse Visual Ranges

  1. Ronan Mandra says:

    I was curious as to what effect on the value would be if I played with the 3.86 km constant in the above formulas. It turns out that 3.85 provides a better match to the Specified Range values. The best fit, ignoring significant digits, was 3.84783759. These values were found by Excel Solver with this criteria:
    1. Minimize Sqrt(Sum Diff^2)
    2. By changing the 3.86 factor

     
    • mathscinotes says:

      The leading coefficient is primarily a function of the lapse rate and everyone assumes a different value. When I do my calculations, I try to use NASA or the Wikipedia (i.e. sources readily obtainable on the web). However, there are numerous different sources and they all have different values. In the case of lighthouses, the sources are old and we may never know what exactly they used. Your approach to determining the coefficients is probably the best way to go today.

      The solver add-in for Excel is an excellent way to optimize coefficients. I use it myself for many things.

      mathscinotes

       
  2. LUDICROUS MARBLE says:

    earth IS flat- employ: a gyroscope after becoming familiar with its rigidity in space; a laser beamed over a still lake after dark; sunlight splayed through patchy clouds from "93MILLION miles away" ( REALLY? common sense doesn't kick in and reason that impossible as rays would ALL BE PARALLEL!!)

     
  3. Flat party pooper says:

    Flat Earther main tactic: Unable to address any specific points raised by an article, so go for some flat Earther goon-approved claims to distract people...
    Same old fallacies we see debunked over and over.

    Gyros are adjusted by schuler tuning for navigation.
    "Lasers beamed" are subject to atmospheric refraction close to a surface like ALL light.
    The sun's rays ARE almost parallel. Remember when you see rays and shadows under clouds, you are looking up at the sun and perspective makes the rays converge at the sun. Evidence: Look for images of the sun's rays see from the side over clouds to see them parallel.
    Evidence: Stand at the base of some skyscrapers for the same effect.
    And really, don't appeal to "common sense" because you didn't use any actual confirmable evidence... Try using evidence.

    And it's simple. How can the sunset light the TOP of any object or clouds and not the base or you on a flat Earth with no land based obstructions.

     

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