A hotel is being billed by neighbors who aren't too happy with the amount of noise coming from the hotel. The Gansevoort Park Avenue in Manhattan, New York is looking at a bill of $21,000 for disturbing the peace in the neighborhood.
New York is known as the city that never sleeps, but residents haven't been able to get a good night's sleep due to loud partying going on around the trendy hotel. The hotel, which has a nightclub and a rooftop pool, has attracted a lot of young guests who like to have a good time, but the neighbors aren't too happy, according to the New York Post.
"It's terrible. It's just so loud - it's like they're having a party out on the street," one neighbor tells CBS New York.
"These kids come 11, 12 at night and they leave at 2, 3 in the morning," says another guest, "the taxis are picking them up and there's a lot of noise going on."
Some of the neighbors had enough of the noise pollution. They pulled together and decided to charge the hotel $1000 per hour of sleep that they lost. The total comes to $21,000, for April 14 alone.
The neighbors don't expect the hotel to actually pay them but they do hope the hotel will get the message and will do something to address to complaints that they've given over the past three years.
"The hotel keeps making promises, says 'yes yes yes, we'll remedy the problem' and then it's not remedied," says one neighbor.
"We've had meetings... Maybe for two weeks, it improves. Then they're having these crazy parties again," resident Mario Messina said.
With the summer coming up, residents are hoping that the warm-weather parties at the hotel won't get too out of hand this season.
Hotel reps would not respond to calls or e-mails for comments on the issue, The Post notes.
This article is copyrighted by Travelers Today, the travel news leader
