Index: src/train_cmd.cpp
===================================================================
--- src/train_cmd.cpp	(revision 19022)
+++ src/train_cmd.cpp	(working copy)
@@ -3193,9 +3193,37 @@
 	uint hash = (y_diff + 7) | (x_diff + 7);
 	if (hash & ~15) return NULL;
 
-	/* Slower check using multiplication */
-	if (x_diff * x_diff + y_diff * y_diff > 25) return NULL;
+	/*
+	 * Slower check using multiplication.
+	 *
+	 * The vehicle 'distance' has to account for a wagon on diagonal tracks
+	 * and one on non diagonal track. In this case the distance on one axis
+	 * is 5 and on the other it's 1. Using Pythogoras that gives a distance
+	 * of 5**2 + 1**2 = 26, which is the minimum safe distance (squared).
+	 * The 5 in this formula coincides with half the distance on straight
+	 * track plus 1.
+	 *
+	 * As we can have shorter vehicles we need to calculate the minimum
+	 * distance for them, which is multiplication expensive, so we first
+	 * do a simple sweep by comparing the distance of the trains. Note that
+	 * this also uses Pythagoras and we 'just' remove the sqrt from both
+	 * sides of the equation.
+	 */
+	int distance_squared = x_diff * x_diff + y_diff * y_diff;
+	/* Check for the simply case to prevent costly calculations. */
+	if (distance_squared > 25) return NULL;
 
+	/* Step 1: calculate the distance between the centers. Wagons are
+	 *         always centered by (x + 1) / 2, so we use that here too. */
+	int center_distance =
+			(Train::From(v)->tcache.cached_veh_length + 1) / 2 +
+			(tcc->v->tcache.cached_veh_length + 1) / 2;
+	/* Step 2: calculate distance (for the longest axis), which is
+	 *         half the distance plus 1. */
+	int min_distance = (center_distance + 1) / 2 + 1;
+	/* Step 3: compare the distance of the wagon with the minimum distance. */
+	if (distance_squared > min_distance * min_distance) return NULL;
+
 	/* Happens when there is a train under bridge next to bridge head */
 	if (abs(v->z_pos - tcc->v->z_pos) > 5) return NULL;