Fenton "Gone With The Wind" lamp, pitcher & bowl

[Holly_Hobby]

My mom went from an apartment to a nursing home, so she gave me her 25" Fenton lamp, pitcher and bowl from the '60's. Unfortunately, she has over $500 in medical bills that need to be paid, so I want to sell them to pay some of them down.

Is there a way to find out what the lamp is worth? Local antique dealers don't even want to look at them. I admit to being intimidated by EBay, because I've never even been to the site.

Any advice? Does DU have a "For Sale" forum? Thanks.

[Vinca]

1. If you could post pictures and sizes, we could probably give you an estimate.

Fenton is funny because values depend a great deal on colors and patterns. Condition is also extremely important. If the pieces aren't in perfect condition, you're apt to be pretty disappointed. I have a Fenton book and have access to ebay's ended sales, but I need more information.

[Holly_Hobby]

2. Thanks, Vinca

I'll take pics over the weekend. All 3 pieces are in perfect condition. They are a powder blue in color with embossed flowers. Purchased in the '60's.

I have my grandmother's lamp, same style, in the custard color, but I'm not parting with that one.

The local antique dealers aren't buying anything in this economy.

[Whoa_Nelly]

3. On RubyLane-$395

http://search.rubylane.com/search?ss=fenton+gone+wind&hsb=Find+an+Item

Quite a few at tias.com
http://www.tias.com/cgi-bin/search.fcgi?database=%2Faltavista%2Fstores%2Fitems&max=50&noResultPageTemplate=NoResultPage2.html&resultPageTemplate=ResultPage2.html&resultItemTemplate=CatResultItem.txt&resultTableTemplate=CatResultTable2.txt&itemResultPageTemplate=ResultPage.html&itemResultItemTemplate=ResultItem2.txt&itemResultTableTemplate=ResultTable.txt&itemDatabase=%2Faltavista%2Fstores%2Fitems&bothSearch=1&primaryServer=&minPrice=1&matchAll=1&searchText=fenton+gone+wind&search=Find

Some at goantiques.com
http://www.goantiques.com/search/search_results.jsp?newSearch=yes&channel=&category=&keywords=fenton+gone+wind&itemType=&image.x=0&image.y=0