On Mon, 24 Sep 2018 09:25:21 -0700, Jeff Liebermann
wrote:

Total distance
traveled is:
1.0 + 1.4 + 1.0 = 3.4 meters.
At 45 degrees, the minimum road width that this could be accomplished
would be the aforementioned length of the rail crossing, plus twice
the turning radius, which I would guess would be no less than 0.5
meters. Therefore the road width required would be:
(0.707 * 3.4) + (2 * 0.5) = 4.4 meters
With a 3.5 meter wide road, it can't be done without riding into
opposing traffic.

Oops. That should be:
(0.707 * 3.4) + (2 * 0.5) = 3.4 meters
which is still a tight squeeze on a 3.5 meter wide road.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558