How does Earth’s curvature affect flight times?

Question

I have heard people say that the flight time from Fort Lauderdale to Seattle is the longest possible flight time within the continental United States. However, upon further consideration, I realized that the curvature of the Earth may cause the visible distance on a map to decrease when traveling north (the circumference of a cross section of the Earth is smaller further from the equator). A change in cross sectional circumference as one travels north can affect the true distance between Fort Lauderdale and a destination. This is not accounted for in a 2D map which is what most think of when considering flight times. With this in mind, does the Earth’s curvature affect the apparent distance (on a 2D map) between Fort Lauderdale and Seattle, and if so, is there another location in the continental United States with a longer flight time from Fort Lauderdale? In other words, suppose you were to fly around the globe across the equator. This would take longer than flying around the globe at a point north of the equator (say Seattle). This is due to the curvature of the Earth, so would this curvature also take effect when traveling from Fort Lauderdale to Seattle. The Earth is “wider” across Florida’s latitude than it is at Seattle’s. This means that it must take longer to travel from Florida to San Diego than from Maine to Seattle. My questions is if that difference could account for a change in flight time. Although Seattle is north of Fort Lauderdale and thus farther, there is added distance closer to the equator due to the Earth’s curvature. So may there exist a location south of Seattle that is further simply due to this change in cross sectional circumference?

Answer 1

What a fun question! In short, yes. Based on most projections people use, they might think Seattle is the furthest in great circle distance, but something else could be!

I offer for your education (and mine): https://en.wikipedia.org/wiki/Two-point_equidistant_projection

It is common knowledge that you cannot flatten a sphere onto the plane, so all maps are distorted, but it is less common knowledge that you CAN make a map to answer your specific question regarding Fort Lauderdale in particular! Because you can use the two-point equidistant projection with Fort Lauderdale as a loci to create an accurate map representing great circle distances!

A cursory googling allowed me to locate this handy visualization tool: https://www.jasondavies.com/maps/two-point-equidistant/

which gave me hope that there COULD be a flight path to a large airport in the continental 48 states further than Seattle.

Now based on this other useful tool:

Seattle: https://www.airmilescalculator.com/distance/fll-to-sea/ : 2717 miles

South, in Oregon: https://www.airmilescalculator.com/distance/fll-to-oth/ : 2739 miles

North, in Washington: https://www.airmilescalculator.com/distance/fll-to-clm/ : 2784 miles

Unfortunately, I could not find an airport in California that was further. Eureka cuts it close at 2700 miles.

Something else we can do is consider the spherical law of cosines, and how it changes in response to North-South vs East-West. For latitudes $\phi_{1,2}$ and longitude difference $\Delta \lambda$, the great circle angle is $A = \arccos(\sin{\phi_{1}}\sin{\phi_{2}} + \cos{\phi_{1}}\cos{\phi_{2}}\cos{\Delta \lambda})$. Let's note that $\phi_{2}$ is fixed at Fort Lauderdale.

Then, let's consider the ratio of $\frac{dA}{d\phi_{1}}$ and $\frac{dA}{d\Delta\lambda}$ to see how much additional West we need to compensate for some South. Considering the ratio means we get to ignore the derivative of $\arccos{x}$ and skip to the fun chain rule bit:

$\frac{dA/d\Delta\lambda}{dA/d\phi_{1}} = \frac{-\sin{\Delta\lambda}\cos{\phi_{1}}\cos{\phi_{2}}}{\cos{\phi_{1}}\sin{\phi_{2}} - \sin{\phi_{1}}\cos{\phi_{2}}\cos{\Delta\lambda}}$

I'll assume these coordinates:

Seattle: 47.4480° N, 122.3088° W

Fort Lauderdale: 26.0742° N, 80.1506° W

For which I find a ratio of roughly 2.1: every 2.1 degrees South (closer) can be compensated by 1 degrees West (further), locally around Seattle when considering distance to Fort Lauderdale.

Answer 2

Yes, there is a place south of Seattle, namely Crescent City, and one south of Fort Lauderdale, namely Key Largo, whose airports are slightly further apart than Seattle and Fort Lauderdale airports are, at least according to this online geodesic calculator. Plugging the coordinates for Seattle-Tacoma International Airport and Fort Lauderdale-Hollywood International Airport from google maps into this calculator returns a distance of $4372.134149$ km. Doing the same for Del Norte County Regional Airport, Crescent City and Ocean Reef Club Airport, Key Largo returns $4412.495930$ km, about $40$ km greater.

However, the difference between the distances is attributable mostly to the fact that Crescent City and Key Largo subtend a larger angle at the center of the Earth than Seattle and Fort Lauderdale do, and has very little to do with the change in the curvature of the Earth. The contribution of the latter to the difference is so small as to be barely detectable. The distance between two points on a sphere with the same latitudes and longitudes as Seattle and Fort Lauderdale Airports is $0.685540$ radians ($4367.6$ km on a spherical Earth of radius $6371$ km) and that between two points with the same latitudes and longitudes as Crescent City and Key Largo is $0.691664$ radians ($4406.6$ km on the same sphere), still about $40$ km greater.