Don't feed the troll!! Now we will probably get another sixteen hundred word post.
I don't think they can yet face up to the simple fact that there really are flights to/from Australia and South America, Australia and South Africa, South Africa and South America as well as New Zealand and South America.
And, especially that these flights really are the distances determined on the Globe - very close to Google Earth distance from airport to airport.
I have measured long distances in southern Australia (yes, we have some long straight roads - like a 145 km straight sealed road) and all distances agree perfectly (as near as I could measure) to the Globe distances for km/degree! Here is just one example:
from | Sp | to | Sp | on road distance | Sp | direct distance | Sp | km/degree |
Balladonia (-32.35° 123.62°) | Sp | Eucla (-31.68° 128.88°) | Sp | car oddo 532 km, Garmin Nav 531.5 km | Sp | Garmin Nav 503 km | Sp | Garmin direct 94.5 km/° |
The km/° is taken as at an average latitude of -32.01°.
The km/° at the equator can then be calculated from (km/° at Lat)/cos(Lat) or 94.5/cos(32.01°) = 111.4 km/°.
These figures give a circumference of the earth at -32.01° of 360° x 94.5 km/° | = 34,032 km and |
a circumference at the equator of 360° x 111.4 km/° | = 40,104 km. Look familiar? |
In this I have compared the distance on the car oddo (532 km) with that on the Garmin navigator (531.5 km) mainly to quell the doubts that some might have of GPS and map distances south of the Equator. Now the accuracy of what I have here could be questioned, but I am sure the distances are within 1% of the correct values. Anyone can check the Lat, Long co-ordinates of the Balladonia Roadhouse and Eucla (they are both tiny places).
Now, I know I am calculating the circumference at the equator
assuming the earth is a Globe, but
Now I dare any Flat Earther to calculate what the the circumference at the equator would be if we assumed the earth was a flat disk!
i warn you that you might not like what you find!