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Mercator maps are conformal, which means that over small distances, shapes are preserved, though you do have to correctly adjust the distance scale. So how much does that matter over 76 km, would the bend in the line even be visible?

Yea, admittedly at these distances (<100km) the difference will be very small unless at extreme latitudes.

Looks like the points are at approximately N57°18.280', W3°55.948' -> N56°54.152', W4°56.664' if anyone wants to do the math. My guess is the rhumb line and GC route differ by no more than a few meters.

I was hoping the article would go more into how they searched the data and found the points. An exhaustive search of all points in the U.K. would obviously be computationally prohibitive. Maybe create millions of polygons where the edges are roads, and sort them by area to reduce the search space? From the article's text it looks like he "eyeballed" the map to find empty areas to focus on, which is unsatisfying :)

I get a bit over 200 metres from the following R code. (Edit: I forgot how to format code for HM, sorry.)

    library(oce)
    lat <- c(57+18.280/60, 56+54.152/60)
    lon <- -c(3+55.948/60, 4+56.664/60)
    m <- lm(lat ~ lon)
    # Find points along a great-circle route
    gc <- oce::geodGc(lon, lat, 0.001)
    # 'scale' is distance associated with 1 degree latitude shift
    scale <- geodDist(mean(lon), mean(lat), mean(lon), mean(lat)+1)
    # 'e' is latitude shift from great-circle route to 'straight' line route
    e <- scale\*sapply(seq_along(gc$longitude),
        function(i)
            gc$latitude[i] - predict(m, data.frame(lon=gc$longitude[i])))
    plot(gc$longitude, e, type="l",
         xlab="Longitude [degE]",  ylab="Distance error [km]", col=2)
    max(e)

I’m not familiar with R, but the lm(…) line looks like it’s linearly interpolating the latitude and longitude. However, that’s not what straight lines in a Mercator projection are. You might be confusing it with the cylindrical projection.

The Mercator projection stretches things away from the equator vertically to compensate the horizontal stretching, in order to preserve local shapes. This makes it a bit better for these kinds of local measurements than one might think at first glance.

Yes, `lm()` linearly interpolates. Thanks for pointing out my error.

To be fair, I’m not sure this will actually make a significant difference. :) Even 200m wouldn’t be a huge error.

I tried a computation that I think may be correct. It gives 319 metres of northing separation between rhumb and creat-circle paths. In case it's of any interest, the R code is below.

    # Northing distance between rhumb and great-circle paths.
    
    library(oce)
    lat <- c(57+18.280/60, 56+54.152/60)
    lon <- -c(3+55.948/60, 4+56.664/60)
    
    # Great-circle path, first in longitude-latitude space, and then
    # (for Mercator projection) in easting-northing space.
    gc <- oce::geodGc(lon, lat, 0.01) # same results for 3rd arg any value < 0.01
    gcXY <- lonlat2map(gc$longitude, gc$latitude, projection="+proj=merc")
    
    # Create function, f(easting), that computes northing value on
    # a rhumb line.  We need this to make the two lines share easting
    # values.
    p <- oce::lonlat2map(lon, lat, "+proj=merc")
    m <- lm(y ~ x, data=p)
    X <- seq(p$x[1], p$x[2], length.out=length(gc$longitude))
    Y <- predict(m, data.frame(x=X))
    f <- approxfun(X, Y) # interpolating function
    
    # Compute the northing error.
    northingError <- f(gcXY$x) - gcXY$y
    maxNorthingError <- round(max(abs(northingError)), 1)
    message("maxNorthingError=", maxNorthingError, " [m]")
    plot(gcXY$x, northingError, type="l", xlab="Easting [m]", ylab="Northing error [m]")
    mtext(paste("max northing error ", maxNorthingError, " [m]"))

If it were to get technical enough to differentiate between rhumb and great circle, then maybe it should factor in elevation changes too.

If you include elevation then you have to include tolerance for deviation. I live in an old house, so I can't even walk across the room without a vertical deviation of some sort :)

Edit: My point here is that if you try to plot a literal straight line through 3D space, it will be necessary to allow some amount of deviation in one axis. Else your longest 3D line will be the distance between your first two points.

He does: 75.9km on the map, 77.0km when accounting for elevation. With the caveat: "Well, it's never as much as I'd think and it depends upon the accuracy of your terrain model"

That number surprised me, but because of how much extra it was rather than how little extra -- I always assume it's negligable, but in this case was over 1%

I also agree it seems like a lot. If there was 3000m of elevation gain and 3000m of elevation loss, I would only expect a couple or hundred extra meters of distance walked. 77.23 ≈ sqrt(77^2+(3+3)^2).

Reading total elevation gain/loss from his graph it hard but 3000m up and down seems about right.

The average gradient of Scotland is 19.3%.

Then it becomes a coastline problem, with different answers depending on the resolution of your elevation data.

Great video on this from Matt Parker (focusing on area in this case): https://youtu.be/PtKhbbcc1Rc

Specifically, the projection used is the Ordnance Survey National Grid projection, which is Transverse Mercator.

The OS published a good introductory (but detailed) guide some years ago: https://www.ordnancesurvey.co.uk/documents/resources/guide-c...

> Without crossing a pub road

Or crossing water.

I think given this constraints a line from any non-pub-road, non-water place in GB to the center of the earth is the longest straight line, without considering space bending around mass. :)

The question was "what (and where) is the longest distance you can walk in a straight line..." so that rules out the center of the Earth.