WW11 major 'Malmedy Massacre' of 84 POW American troops during Battle of the Bulge, Belgium

Comment #4

jarasan — Fri, 27 May 2022 14:40:43 GMT

TY Sir Knight!

Comment #3

rdgrnr — Thu, 26 May 2022 23:16:14 GMT

I knew an old guy who spent half the war as a POW in a German prison camp.<br />He hated those people with a PASSION until his dying day.

Comment #2

eddessaknight — Thu, 26 May 2022 22:44:01 GMT

Above: Pre-Memorial Day Reembrance

Comment #1

eddessaknight — Thu, 26 May 2022 21:26:48 GMT

<br />Billow, who died May 17, was attached to the Army’s 285th Field Artillery Observation Battalion when his unit surrendered and he was taken prisoner by Waffen SS soldiers as German forces launched an offensive in Belgium to try to change the war’s tide in December 1944.<br />R.I.P.

Original Blog Entry: WW11 major 'Malmedy Massacre' of 84 POW American troops during Battle of the Bulge, Belgium

eddessaknight — Thu, 26 May 2022 21:23:53 GMT

December 17, 1944

According to various verified accounts, the Germans opened deadly fire on the unarmed prisoners in a field, a barbarous retaliation killing of more than 84 in what came to be known as the Malmedy Massacre, which was outside of the International Geneva accords

This Malmedy massacre was a German war crime committed by soldiers of the Waffen-SS on 17 December 1944, at the Baugnez crossroads near the city of Malmedy, Belgium, during the Battle of the Bulge. Soldiers of Kampfgruppe Peiper summarily killed eighty-four U.S. Army prisoners of war who had surrendered after a brief battle. The Waffen-SS soldiers had grouped the U.S. POWs in a farmer's field, where they used machine guns to shoot and kill the grouped POWs; the prisoners of war who survived the g

....... :-(

Requiem en pace +

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