Makapuʻu

Makapuʻu Beach

Makapuʻu is the extreme eastern end of the Island of Oʻahu in the Hawaiian Islands, comprising the remnant of a ridge that rises 647 feet (197 m) above the sea. The cliff at Makapuʻu Point forms the eastern tip and is the site of a prominent lighthouse. The place name of this area, meaning "bulging eye" in Hawaiian, is thought to derive from the name of an image said to have been located in a cave here called Keanaokeakuapōloli. The entire area is quite scenic and a panoramic view is presented at the lookout on Kalanianaole Highway (State Rte. 72) where the roadway surmounts the cliff just before turning south towards leeward Oʻahu and Honolulu.

The Makapuʻu area is reached approximately 2 kilometers (1.2 mi) east of Waimānalo Beach on Kalanianaole Highway (State Rte. 72) or from the Honolulu side (south shore; Hawaiʻi Kai) travelling east along the same highway beyond Sandy Beach. The Makapuʻu Point State Wayside Park, a 38-acre (15 ha) roadside park, is about midway up the draw on the right-hand side coming from Hawaiʻi Kai.

Features of special interest in this area include:

Makapuʻu Point lighthouse
Makapuʻu Beach from Makapuʻu Point

References

See also

This article is issued from Wikipedia - version of the 5/1/2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.