Thread: Sleeving or filling an oversized pivot hole

I have several questions: 1. What is better, sleeving an oversized pivot or filling it with epoxy? 2. Somewhere in SRP I read something on filling the hole with epoxy and drilling it out but I have been unable to find it agian. Can someone give me the link to where the illustration is? I kind of like the idea of filling it with epoxy because it's less moving parts and seems it would keep for a longer period of time.



Thanks for any feedback!

How oversized is the hole? Are you talking about two holes in the scales or the one in the tang?

Just the one for the pivot point. I'm using 1/16" brass rod but the pivot hole is about 1/8".

Sleeving an oversized pivot - Straight Razor Place Wiki

and


Adjustable Pins aka Microfasteners - Straight Razor Place Wiki

here is what you are looking for... be sure to go all the way down the second link

Originally Posted by kcarlisle

Just the one for the pivot point. I'm using 1/16" brass rod but the pivot hole is about 1/8".

So do you mean you want to make the pivot hole on the scales smaller or you want to make the hole in the blade smaller?

I guess you don't want to but you could just use 1/8" rod.

Originally Posted by 2knives

So do you mean you want to make the pivot hole on the scales smaller or you want to make the hole in the blade smaller?

I guess you don't want to but you could just use 1/8" rod.

the fix is to use a 1/8 " brass tubing, sleeve it on the next size down (OD=0.096"), and then use the regular 1/16 rod.
I have not used the epoxy method so I can't comment.

The blade I'm re-scaling is a George Wostenholm & Son's Pipe razor. The old scale is going BYE, BYE as it was broken in half and I'm putting a new scale on it. The original pin was 1/16" but the pivot hole is about 1/8".

After posting this, I looked at another Wostenholm (still pinned) and was able to wiggle (pull) the blade back and forth. It's obvious the pivot hole in it is also oversized. Years ago, was it a standard practice to pin razors with an approximate 1/8" hole with a 1/16" pin? I'm positive there was nothing that fell out when I un-pinned the one I'm working on right now.

At this point, I'm a little confused as to what to do.

Originally Posted by kcarlisle

The blade I'm re-scaling is a George Wostenholm & Son's Pipe razor. The old scale is going BYE, BYE as it was broken in half and I'm putting a new scale on it. The original pin was 1/16" but the pivot hole is about 1/8".

After posting this, I looked at another Wostenholm (still pinned) and was able to wiggle (pull) the blade back and forth. It's obvious the pivot hole in it is also oversized. Years ago, was it a standard practice to pin razors with an approximate 1/8" hole with a 1/16" pin? I'm positive there was nothing that fell out when I un-pinned the one I'm working on right now.

At this point, I'm a little confused as to what to do.

did you follow the links? they show how to deal with your problem.

Also, if you are making new scales, there is no reason you can't pin with 1/8" rod if it fits how you like.... I've done it before with pleasing results.

Some old razors had odd shaped pivot holes (many that I've seen anyhow), in fact I can't think of one older sheffield blade that didn't have a slightly irregular pivot hole. Long story short, what you are seeing isn't uncommon. I do think that the links will help you out though.

Originally Posted by kcarlisle

Years ago, was it a standard practice to pin razors with an approximate 1/8" hole with a 1/16" pin?

At this point, I'm a little confused as to what to do.

I would say 9 out of 10 of the razor I have come across have a pivot hole in the tang of the blade that is larger than 1/16".

Let's hash this out with good'ole mathematics. Let's say you have circular hole (pivot hole in tang) that has a 1/8" diameter and you put a 1/16" rod into the hole. Now take the area of the hole A=PI(r)^2 -> PI(1/8)^2 = 0.04908" = A1. Now take the orthogonal cross sectional area of the rod PI(1/16)^2 = 0.0123" = A2. So A1-A2 = 0.04908 - 0.0123 = 0.03678... seems like a lot of area left over; but, we must remember that it is a circle... so, When dealing with a circle, the you have a radius and a diameter. The radius of the 1/8" hole is 1/16". The radius of the 1/16" rod is 1/32". So 1/16 = 0.0625 = r1 & 1/32 = 0.03125 = r2 and if you subract r1-r2 = 0.0625-0.03125 = 0.03125 which equals 1/32. If you are lost right now, because it certainly is confusing, what this math indicates is that you are looking at a 1/32" gap from pivot pin to tang hole when the rod is centered... pretty small.

In conclusion, if the razor is pinned right and tight (not too tight though), I don't think you'll have much of a problem with the hole in the tang being 1/8" and the rod being 1/16". If you're a perfectionist then you might want to fill the hole in the tang... If you'd rather spend your time doing something else, then I think you'll be juuuuust fine.

Cheers!

-2knives

Originally Posted by kcarlisle

The blade I'm re-scaling is a George Wostenholm & Son's Pipe razor. The old scale is going BYE, BYE as it was broken in half and I'm putting a new scale on it. The original pin was 1/16" but the pivot hole is about 1/8".

After posting this, I looked at another Wostenholm (still pinned) and was able to wiggle (pull) the blade back and forth. It's obvious the pivot hole in it is also oversized. Years ago, was it a standard practice to pin razors with an approximate 1/8" hole with a 1/16" pin? I'm positive there was nothing that fell out when I un-pinned the one I'm working on right now.

At this point, I'm a little confused as to what to do.

Originally Posted by 2knives

I would say 9 out of 10 of the razor I have come across have a pivot hole in the tang of the blade that is larger than 1/16".

Let's hash this out. Let's say you have circular hole (pivot hole in tang) that has a 1/8" diameter and you put a 1/16" rod into the hole. Now take the area of the hole A=PI(r)^2 -> PI(1/8)^2 = 0.04908" = A1. Now take the orthogonal cross sectional area of the rod PI(1/16)^2 = 0.0123" = A2. So A1-A2 = 0.04908 - 0.0123 = 0.03678... seems like a lot of area left over; but, we must remember that it is a circle... so, When dealing with a circle, the you have a radius and a diameter. The radius of the 1/8" hole is 1/16". The radius of the 1/16" rod is 1/32". So 1/16 = 0.0625 = r1 & 1/32 = 0.03125 = r2 and if you subract r1-r2 = 0.0625-0.03125 = 0.03125 which equals 1/32. If you are lost right now, because it certainly is confusing, what this math indicates is that you are looking at a 1/32" gap from pivot pin to tang hole when the rod is centered... pretty small.

In conclusion, if the razor is pinned right and tight (not too tight though), I don't think you'll have much of a problem with the hole in the tang being 1/8" and the rod being 1/16". If you're a perfectionist then you might want to fill the hole in the tang... If you'd rather spend your time doing something else, then I think you'll be juuuuust fine.

Cheers!

-2knives

I absolutely disagree here about just leaving it, and yes obviously if I am taking the razor apart to fix it I am a perfectionist.... Dave posted the links for you, the fix is so simple is is scary, in factIi probably could have done three in the time it took to do the math, just slip in a sleeve and be done with it, or do the epoxy thing, either works ... My personal preference is the sleeve but that doesn't really matter...