Battle of Leuze

The Battle of Leuze was a minor Cavalry engagement of the Nine Years' War that took place on 18 September 1691 between a detachment of French and a superior Allied force.

Battle of Leuze
Part of the Nine Years' War
Combat de Leuze by Joseph Parrocel
Date 18 September 1691 Location Leuze-en-Hainaut, present-day Belgium Result French victory
Belligerents
France	England  Dutch Republic  Scotland
Commanders and leaders
Duc de Luxembourg	Friedrich of Waldeck
Strength
28 squadrons (about 4,000 men)	72 squadrons (about 12,000 men)
Casualties and losses
400 dead or wounded	1,500 - 2,000 dead or wounded

Marshal Luxembourg had been informed that William III of Orange had left for England, in the supposition that the campaign of 1691 was at its end. He was also informed that Marshal Waldeck, who was left in charge, was preparing to retire into winter quarters.

Luxembourg was near Tournai and sent out a reconnaissance under Marsilly, from whom he learned that the main body of the Allied army was retreating, leaving a rear-guard of cavalry at Leuze. Luxembourg acted immediately. He send a detachment to follow the movements of the main body, and with the squadrons of Villars and Marsilly he attacked the smaller party without warning. The French cavalry charged, only using their swords.

The Allied cavalry was superior in numbers, but could not deploy efficiently because of the limited space they were manoeuvred in.

The Allied infantry, including Mackay's Regiment [1] also took part in the battle but did not see much action.

The battle lasted for only two hours, and the French had the advantage. The surviving Allied cavalry was brought off by troops that Waldeck sent back.

Luxembourg had been in the thick of the action and been forced to beat off an attack, directed at his person.