Battle of Doiran (1916)

From Wikipedia, the free encyclopedia
  (Redirected from Battle of Horseshoe Hill)
Jump to: navigation, search
Battle of Doiran
Part of Macedonian front (World War I)
Date 9 August 1916 - 18 August 1916
Location Lake Dojran, present-day Republic of Macedonia
Result Bulgarian victory
Belligerents
Bulgaria Bulgaria  France
United Kingdom United Kingdom
Commanders and leaders
Dimitar Geshov Unknown
Strength
Second Thracian Infantry Division 3 French divisions
1 British division
45,000 men
400 guns
Casualties and losses
851 3,200

In the beginning of August 1916 three French and one British divisions with 45,000 men and 400 guns launched an offensive against the Bulgarian positions at Lake Dojran, defended by the Second Thracian Infantry Division. The attack began on 9 August with heavy artillery fire on the positions of the 27th Chepino Regiment and 9th Plovdiv Regiment. All four attacks that followed - on 10, 15, 16 and 18 August were repulsed by the Second division and the Allies were forced to retreat to their original positions with heavy casualties.

Other sources state that the French took Tortoise Hill (Tortue) and Doldzeli, in total 30 square km, but at a very high cost.[1][2] The British 7th Battalion of the Oxfordshire & Buckinghamshire Light Infantry took Horseshoe Hill.[3][4]

Notes[edit]