X metal frames

[PitBoss]

L&G's

I find myself having spent a lot of hours on this forum the past couple of weeks and I have yet to find a complete list with pictures of all x metals released.

I was hoping that someone here would take the time to show what all the frames are, top, front and side view as well as coins (if they) came with the shades.

I understand that there is some hundreds of colorcombos but that's irrelevant if the special editions don't vary substantially in design.

anyone up for the task?

regards
an Oakley rookie

 

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[motoGP_fanatic]

Try the site o-review.com. It won't have the different view photo of X-Metals but it will be treasure of info there.

 

[OakleyFrankFMJ]

Csdj, cant see you getting such a detailed request I'm afraid, try this!:
www.oreview.com/database_filter_model.php.table_name=glasses

I concur.

 

[CarGuy]

Yes no need to have someone compile a list for you as Dann has already done that for you over on o-review.

 

[Carrera1963]

You have to copy and paste that O Review link rather than click it directly, or it will just send you straight back here.

 

[Dantes]

Link above is wrong, this is the correct link

http://www.o-review.com/database_filter_model.php?table_name=glasses&family=X-Metal&page=0

Cut and paste to browser.

 

[NCHornet]

You have been correctly advised.

 

[Litos]

Link above is wrong, this is the correct link

http://www.o-review.com/database_filter_model.php?table_name=glasses&family=X-Metal&page=0

Cut and paste to browser.

Why does it take you back to the home page when you click on the link ??

 

[PitBoss]

thanks guys.
this was very helpful.

what great resource

 

[CarGuy]

Why does it take you back to the home page when you click on the link ??

Because it reduces the amount of traffic there not used by forum members there but rather by the forum members here.
Imagine a noob here discovers Ichiro x metals, fails to use the search bar here and instead goes there. Imagine how much traffic goes there from here and doesn't join or contribute there? Bandwidth = $