Distance between Seymour and 10318
Distance between Seymour and 10318 is 462.4 km. This distance is equal to 287.32 miles, and 249.51 nautical miles.
The distance line on map shows distance from Seymour to 10318 between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Seymour to 10318,
It takes 0.51 hours to arrive.
| Seymour | 38.9592201 | -85.8902547 |
|---|---|---|
| 10318 | 40.9343802 | -81.1153542 |
| Distance | 462.4 km | 287.32 miles |
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