Tour de Suisse 1939
Stage 5
Thun → Luzern
Stage Data
Stage 5 of the 1939 Tour de Suisse, held on August 9, 1939, covered 208.0 km from Thun to Luzern. The stage was completed at an average speed of 38.5 km/h, which was 2.6 km/h faster than the edition average of 35.9 km/h, ranking as the 2nd fastest stage of the race.
Road race208.0 km9 Aug 1939
Thun → Luzern
Stage 5 | Thun - Luzern
Avg Speed
38.5 km/h
Startlist Quality
312
Nearby Stages
2
Grenchen → Morat
195 km
3
Morat → Sierre
192 km
4
Sierre → Thun
174 km
6
Luzern → Lugano
205 km
7
Lugano → Rorschach
313 km
8
Roschach → Zürich
218 km
Frequently Asked Questions
How long was Stage 5 of the 1939 Tour de Suisse?
Stage 5 covered 208.0 km from Thun to Luzern.
What was the route of Stage 5 in the 1939 Tour de Suisse?
Stage 5 ran from Thun to Luzern, a distance of 208.0 km.
What was the average speed of Stage 5 in the 1939 Tour de Suisse?
The average winning speed for Stage 5 was 38.5 km/h.