The 49ers’ schedule isn’t out yet, but based on their opponents we can already determine that their travel schedule could be a rough one. A chart posted by ESPN’s Adam Schefter shows San Francisco is slated to travel 29,958 miles this season, which is the second-highest number in the NFL behind the Seahawks.
It’s not a huge shock that teams on the coasts tend to travel more miles. For the 49ers though they have an unusually robust travel schedule this year thanks to the schedule rotation and their first-place finish in the NFC West last year.
Here’s a look at the full chart via Schefter:
Seahawks will travel 31,600 miles this season, the most in the NFL, per @billsperos.
Here’s how much each team will travel prior to the NFL’s full schedule release Thursday. pic.twitter.com/GTz6CU3idh
— Adam Schefter (@AdamSchefter) May 10, 2023
On the 49ers’ schedule are the four NFC East teams, of which they’ll travel to Philadelphia and Washington. They also face the AFC North and travel to Cleveland and Pittsburgh. Thanks to their first-place finish they’ll also head to Minnesota (the NFC North winner) and Jacksonville (the AFC South winner).
There’s a strong chance given their history under head coach Kyle Shanahan that the 49ers cut down on this travel number significantly by staying on the East Coast at least once.
During Shanahan’s tenure they’ve stayed in Youngstown, Ohio and the Greenbier in West Virginia to mitigate some of the travel between back-to-back long road trips. Depending on the schedule this year with so many games to the Midwest and beyond we could see the 49ers stay on the road multiple times, which would dramatically reduce some of their heavy travel requirements.