Distance from Romulus, New York to Seneca Falls, New York

Romulus Seneca Falls
Total Distance
12.1 mi
(19.5 km)
Travel Time
16 minutes

The driving distance from Romulus, New York to Seneca Falls, New York is 12.1 mi (19.5 km)

The straight-line (air) distance is approximately 6.9 mi (11.1 km)

Estimated travel time by road is 16 minutes

We found 3 routes between Romulus and Seneca Falls.

Available Routes

Shortest
NY-414 N/Ovid St and Bridge St
12.1 mi
19.4 km
17 minutes
Fastest
NY-414 N/Ovid St
12.1 mi
19.5 km
16 minutes
NY-96 N
14 mi
22.5 km
19 minutes

Route Directions

12.1 mi
16 min

Route Directions

14 mi
19 min

Route Directions

12.1 mi
17 min

Journey Details

Traveling from Romulus, New York to Seneca Falls, New York covers 12.1 miles (19 km) in a north direction. This mixed highways and local roads, moderate route typically takes 17 minutes under normal conditions.

Drivers have 3 route options between these locations (12-14 miles / 19-22 km). The primary route shown here saves 2 minutes compared to alternatives.

The route features scenic views. During night weekday travel, expect minimal traffic congestion with average speeds around 41 mph (66 km/h) .

Journey Waypoints

Romulus

Starting Point

McDuffie Town, New York

16 mi to destination

1 mi
MacDougall, New York

12 mi to destination

2 mi
Kuneytown, New York

Traveled 8 mi from origin

1 mi
Canoga Springs, New York

4 mi to destination

2 mi
Waterloo, New York

Traveled 16 mi from origin

2 mi
Seneca Falls

Destination