Battle of Puketutu

Lieutenant-Colonel William Hulme arrived at the Bay of Islands at the end of April 1845 with 460 soldiers, marines and volunteers. The British destroyed Ōtuihu, the coastal pā of the neutral chief Pōmare, who was suspected of secretly supporting Hōne Heke. Hulme and his force then moved inland to confront Heke at his new pa at Puketutu, beside Lake Ōmāpere.

Skirmishing between Māori forces had delayed the completion of this pā. Strong defences consisting of double or triple palisading were in place on three sides, but the rear of the pā was vulnerable. A British assault party of more than 200 men attacked Puketutu on 8 May. They were surprised by 140 fighters led by Te Ruki Kawiti who had been hiding in the bush. Turning to deal with Kawiti, the British appeared to be gaining the upper hand when Heke led a group from the pā. Fierce fighting ensued. Heke’s party returned to the pā with the British in hot pursuit, then Kawiti’s men regrouped and attacked again.

After forcing Kawiti back a second time, Hulme called off the attack. Lacking artillery, he may have felt that a frontal assault would be unwise. Puketutu had been built away from civilians and crops and had no long-term strategic value for Heke. After the battle it was simply abandoned.

Governor Robert FitzRoy reacted to Hulme’s occupation of an empty pā by reporting to his superiors that the rebels had been ‘beaten and dispersed’. Maori casualties were higher than British – 28 killed compared with 15 – but some reports claimed that up to 200 Māori had been killed. Māori learnt an important lesson at Puketutu: the British were a formidable foe in open battle. This would influence Maōri tactics in future clashes.