reglazing traditional portholes help required...

yourmomm

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ok. trying to remove the circlip ring on the inside of the portholes which retains the glass. it has two indentations opposite each other, i assume to stick some sort of tool in and rotate it out on a thread? i have no such tool, so am left trying to hammer these indentations anticlockwise in a vice with plenty of wd40. heat doesnt seem to work either. they are stuck fast. any ideas? also, anyone know who can supply toughened glass to fit 6inch portholes? many thanks...
 
Can't help with the circlips, but I was googling polycarbonate sheet today and acryllic sheet. I just typed in poly or acryllic sheet I think.

There was one site quite high up on the search list selling polycarb of marine grade for boats and I guess you could cut that yourself (which of course you can't with toughened glass)

Sorry, can't give you the site as I was using someone elses computer so have no history etc
 
This rings don t usually come out easily and I get them out by drilling through one side so it splits. It then peeles off...Check first you cann get threaded rings. The Westerly ownerts site or Trafalgar might help. You need two flat gaskets one round section gasket plus glass and ring for each port hole. You may need new hinge pins.


J0hn
 
No experience of portholes, but I have undone similar threaded rings on large optical assemblies.

To stand any chance of success, you must apply a turning force via both notches at the same time. You can get pukka tools for this, otherwise make something up that will locate in both notches and give you something to hold on to to turn (opened up needle nose pliers, as has been suggested, if you've any big enough).

Penetrating oil & cleaning the threads are good; after that, it's a matter of repeatedly trying to unscrew the ring - 'jiggling', rather than a steady pressure, trying not to put any sideways force on it. It probably won't be very tight, but a tiny amount of friction will lock the two together if there is any sideways force, at all.

0.02p

Andy
 
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