Distance between Seymour and Pleasure Ridge Park
Distance between Seymour and Pleasure Ridge Park is 90.54 km. This distance is equal to 56.26 miles, and 48.85 nautical miles.
The distance line on map shows distance from Seymour to Pleasure Ridge Park between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Seymour to Pleasure Ridge Park,
It takes 0.1 hours to arrive.
Seymour | 38.9592201 | -85.8902547 |
---|---|---|
Pleasure Ridge Park | 38.1453477 | -85.8582969 |
Distance | 90.54 km | 56.26 miles |
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