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

Published:

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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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Entry #1,610

Comments

Avatar eddessaknight -
#1
MOUNT JOY, Pa. — Harold Billow, 99, the last known survivor of a World War II POW massacre during the Battle of the Bulge, will be laid to rest Thursday in Pennsylvania.

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.
R.I.P.
Avatar eddessaknight -
#2
Above: Pre-Memorial Day Reembrance
Avatar rdgrnr -
#3
Dirty basturds.
I knew an old guy who spent half the war as a POW in a German prison camp.
He hated those people with a PASSION until his dying day.
Avatar jarasan -
#4
TY Sir Knight!

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