Alfred topographic maps
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Alfred
United States > New York > Allegany County > Alfred
The elevation of Alfred is about 1,700 feet (520 m) but rises to a high point of 2,355 feet (718 m) at the summit of Jericho Hill just south of the village. The hills greatly affect the weather in the region, which results in quick changes as well as different conditions in neighboring valleys.
Average elevation: 1,982 ft