Distance from Savage, Maryland to Chipley, Florida
Wondering how far is Chipley, Florida from Savage, Maryland? Here's our driving distance calculation from Savage to Chipley.
The driving distance between Savage, Maryland and Chipley, Florida is 911 mi or 1466 km . It takes 13 hr 55 min to reach Chipley from Savage by road. The distance by air route is 759 mi / 1221 km.
There are 2 routes to reach Chipley, Florida from Savage, Maryland.
911 mi / 1466 km - 13 hr 55 min
981 mi / 1579 km - 14 hr 6 min
Savage, Maryland
Time | 16:37:49 |
Latitude | 39.144487 |
Longitude | -76.8317325 |
State | Maryland |
Zip | |
Timezone | America/New_York |
Chipley, Florida
Time | 16:37:49 |
Latitude | 30.7818584 |
Longitude | -85.538541 |
State | Florida |
Zip | 32428 |
Timezone | America/Kentucky/Monticello |
Nearby Cities
South Laurel
7 mi / 11 km
Konterra
8 mi / 13 km
White Oak
16 mi / 26 km
Layhill
18 mi / 29 km
Aspen Hill
20 mi / 32 km
Dundalk
24 mi / 39 km
Rockville
25 mi / 40 km
Rosedale
26 mi / 42 km
Derwood
27 mi / 43 km
Essex
27 mi / 43 km
Savage Distance to Popular Cities
Route | Distance |
---|---|
Savage to South Laurel | 7 mi / 11 km |
Savage to Konterra | 8 mi / 13 km |
Savage to White Oak | 16 mi / 26 km |
Savage to Layhill | 18 mi / 29 km |
Savage to Aspen Hill | 20 mi / 32 km |
Savage to Dundalk | 24 mi / 39 km |
Savage to Rockville | 25 mi / 40 km |
Savage to Rosedale | 26 mi / 42 km |
Savage to Derwood | 27 mi / 43 km |
Savage to Essex | 27 mi / 43 km |
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