DAVID LEE ROTH TO OPEN FOR KISS ON THEIR NORTH AMERICAN “END OF THE ROAD” TOUR DATES
KISS announced on Tuesday (December 17th) that former Van Halen lead singer and solo star David Lee Roth will be their special guest on the final leg of their End Of The Road tour.
“David Lee Roth will bring his exciting stage show featuring a twin guitar big rock sound and a set list of classic hits and future energy,” read a statement about the addition of Diamond Dave.
KISS’s North American dates begin on February 1st in Manchester, New Hampshire, and run through October 2nd in Fort Worth, Texas. The tour will consist of two legs, with gigs in South America and Europe taking place in between. View complete dates, and ticket information, here.
As previously reported, David Lee Roth will be performing his David Lee Roth: Rocks Vegas residency at the Mandalay Bay Resort And Casino. For more information about this special event, please go here.
 that former Van Halen lead singer and solo star David Lee Roth will be their special guest on the final leg of their End Of The Road tour. )
A new Kiss character – DLR as the Joker.
Would not be surprised if other special guest creep out of the woodwork. Maybe Cher can make an appearance.
From David Garibaldi to David Lee baldy….
Well played
I guess I`m odd man out as usual. I like Dave he`s pure entertainment.
I like it when a performer goes out there and has fun and doesn`t just go through the motions.
He obviously doesnt have the voice he wants had but he gets the job done in a way only Dave can do. Even still he`s still a better vocalists then some that people pay to see (vince neil).
.I might go now and leave when Kiss comes on, lol.
Well, Gene has always claimed he “discovered VH,” so I guess a full circle moment, huh? 😉
Doug, I thought it was vh1. Merry kissmas!
While it’s great that Dave is going to Vegas and on the KISS tour to play Van Halen music, the sad aspect of this story that I think people are overlooking is that this is confirmation that Van Halen, the band, has ceased to exist.