How do I get directionality from the departure() function

I have two sets of coordinates and I need to calculate the distance between them in longitude converted to meters. The departure() function serves this purpose well. However, it returns all positive values for distance regardless of which coordinate is more east/west than the other. Is there a way I can get directionality from this function, or by altering the output in some indexed way?

lat1=[45.73366;42.30074;42.11033;37.69088];
lon1=[-115.85149;-115.54721;-108.03860;-96.52568];
lat2=[45.94414;41.20918;42.64974;36.66341];
lon2=[-115.28569;-116.87984;-108.61347;-95.06520];
coords=table(lat1,lon1,lat2,lon2)

coords = 4×4 table
     lat1      lon1       lat2      lon2  
    ______    _______    ______    _______

    45.734    -115.85    45.944    -115.29
    42.301    -115.55    41.209    -116.88
     42.11    -108.04     42.65    -108.61
    37.691    -96.526    36.663    -95.065

coords.lon_diff_m=departure(coords.lon1,coords.lon2,(coords.lat1+coords.lat2)./2,wgs84Ellipsoid,'degrees') 

coords = 4×5 table
     lat1      lon1       lat2      lon2      lon_diff_m
    ______    _______    ______    _______    __________

    45.734    -115.85    45.944    -115.29         43956
    42.301    -115.55    41.209    -116.88    1.1083e+05
     42.11    -108.04     42.65    -108.61         47344
    37.691    -96.526    36.663    -95.065     1.297e+05

%output in meters since wgs84Ellipsoid uses meters for semi-major axis definition
%average latitudes for longitudinal distance calculation

As you can see, all the distances are positive values. I would like the output to be directional where, if coordinate 1 is west of coordinate 2 the output is negative, and if coordinate 1 is east of coordinate 2 the output is positive.

I suppose I could use an index vector of lat1<lat2 to multiply the rows that qualify by -1.

idx=(coords.lon1<coords.lon2);
coords.lon_diff_m(idx)=coords.lon_diff_m(idx).*-1

coords = 4×5 table
     lat1      lon1       lat2      lon2      lon_diff_m
    ______    _______    ______    _______    __________

    45.734    -115.85    45.944    -115.29        -43956
    42.301    -115.55    41.209    -116.88    1.1083e+05
     42.11    -108.04     42.65    -108.61         47344
    37.691    -96.526    36.663    -95.065    -1.297e+05

I came up with this answer as I am writing this question. However, if anybody has a better way to do this or sees an error in my method, please let me know.

Answer 1

[William Rose]

@Steve K,

Multiply the values from departure by sign(lon1-lon2).

Define points:

lat1=[45.73366;42.30074;42.11033;37.69088];
lon1=[-115.85149;-115.54721;-108.03860;-96.52568];
lat2=[45.94414;41.20918;42.64974;36.66341];
lon2=[-115.28569;-116.87984;-108.61347;-95.06520];

Compute longitude difference with +/-:

lon_diff_m=departure(lon1,lon2,(lat1+lat2)/2,wgs84Ellipsoid,'degrees').*sign(lon1-lon2)

lon_diff_m = 4×1

1.0e+05 * -0.4396 1.1083 0.4734 -1.2970

That works.

[William Rose]

@Steve K, departure() is useful if you are doing latitude sailing, as was common pre-good-chronometers, i.e. pre-1730s. Nowadays one would sail or fly a great circle track. What is the current usefulness of departure()? Surveying? Hiking the Mason Dixon line, or the 49th parallel?