Distance from Caney, OK to Swainsboro, GA
Driving distance between Caney, OK and Swainsboro, GA is 1,006.09 miles (or 1,619.14 kilometers). If you drove an average car (using about 3.6 gallons of gas per mile), the gas would cost you around $, since you would need about gallons of fuel.
Leaving from Caney, OK you would reach Swainsboro, GA in about 15 hours 10 mins if you keep an average speed of miles per hour, assuming you don’t make any breaks or get stuck in traffic.
This is a map overview of the best driving route from Caney, OK to Swainsboro, GA in which you can see the origin and destination points marked with A (Caney) and B (Swainsboro) respectively as well as blue line along the road route you can take. You can zoom in to see more details, including road numbers and even street names both in Caney, OK and Swainsboro, GA as well as every other town you may pass by on your way.
Fuel consumption if you drive from Caney, OK to Swainsboro, GA
To cover the 1,006.09 mi that separate Swainsboro from Caney you are going to need between of fuel, considering you drive an average car (30 mpg) and , if you drive an SUV or a heavy car (20 mpg). If you drive a fuel efficient car (45 mpg), your consumption will only be .
Cost of driving from Caney, OK to Swainsboro, GA
Depending on the fuel efficiency level of your car, driving from Caney to Swainsboro will cost you between and . An average car would make the trip for . We calculate the route cost using real time average gas prices for the US and other costs such as tolls are not included in this calculation.