Look in the yellow pages under "windshield repair". A glass company can come to your location and repair the break (if it smaller than a one dollar bill) by injected a polymer resin into the break and sealing it. I would call my insurance company first to see if it is covered. Most insurance companies will pay one hundred percent of a repair.
2006-06-23 17:53:26
·
answer #1
·
answered by Beware the fury of a patient man 6
·
0⤊
0⤋
Call a professional. Look in the phone book under Windshield Repair. If you do not you run the risk of the chip turning it a spider-web like crack.
2006-06-23 13:23:47
·
answer #2
·
answered by Leemo 4
·
0⤊
0⤋
Here in the UK if the chip is small enough, it can have an injection to refill and strengthen the glass. If left, the weather will affect it, stress of poor surfaced roads will eventually make the cracks appear, and 1 speed bump too quick, the window will give way under strain. Get it sorted fast, it could save your life.
2006-06-23 13:33:18
·
answer #3
·
answered by Hussydog 4
·
0⤊
0⤋
You will most likely have to get your windshield replaced. This has happened to me on a few occasions. Each time a line forms that continues to spread across the windshield, especially from thruway driving. So, most likely it will spread like mine did and you will need to get it replaced.
2006-06-23 13:21:56
·
answer #4
·
answered by jordanrrr 2
·
0⤊
0⤋
Contact an auto glass company immediately. They can seal it, to prevent the crack from spreading, thus making it difficult to see. This can cost anywhere from 50-100 bucks, unless you have a good hook-up or know how to flirt without giving away all of the goodies . Good Luck :)
2006-06-23 13:39:14
·
answer #5
·
answered by CC65 4
·
0⤊
0⤋
Go to Novus Auto Glass. If you have a bit of mechanical expertise and are careful, you can purchase materials to do this at some auto parts stores.
2006-06-23 13:23:30
·
answer #6
·
answered by expatmt 5
·
0⤊
0⤋
Not really....duct tape.....eventually it will get bigger
2006-06-23 13:22:07
·
answer #7
·
answered by Elle 3
·
0⤊
0⤋