Distance from Clayton, New Mexico to Sanibel, Florida
Wondering how far is Sanibel, Florida from Clayton, New Mexico? Here's our driving distance calculation from Clayton to Sanibel.
The driving distance between Clayton, New Mexico and Sanibel, Florida is 1743 mi or 2805 km . It takes 26 hr to reach Sanibel from Clayton by road. The distance by air route is 1421 mi / 2287 km.
There are 3 routes to reach Sanibel, Florida from Clayton, New Mexico.
1743 mi / 2805 km - 26 hr
1735 mi / 2792 km - 26 hr
1807 mi / 2908 km - 26 hr
Clayton, New Mexico
| Time | 01:52:34 |
| Latitude | 36.451693 |
| Longitude | -103.1841039 |
| State | New Mexico |
| Zip | 88415 |
| Timezone | America/Denver |
Sanibel, Florida
| Time | 03:52:34 |
| Latitude | 26.4489695 |
| Longitude | -82.0223137 |
| State | Florida |
| Zip | |
| Timezone | America/Kentucky/Monticello |
Clayton Distance to Popular Cities
| Route | Distance |
|---|---|
| Clayton to Hough | 104 mi / 167 km |
| Clayton to Richfield | 105 mi / 169 km |
| Clayton to Bushland | 119 mi / 192 km |
| Clayton to Adrian | 125 mi / 201 km |
| Clayton to Amarillo | 132 mi / 212 km |
| Clayton to Big Bow | 136 mi / 219 km |
| Clayton to Liberal | 142 mi / 229 km |
| Clayton to Skellytown | 141 mi / 227 km |
| Clayton to Canyon | 142 mi / 229 km |
| Clayton to Palisades | 143 mi / 230 km |
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