Q on spring length/binding point

[beardyweirdy]

Now I have been spending a while on my practice rim lock (is 'rim' the same as 'cylinder', btw?), I wanted to ask a question or two.

I have been reading on the subject, and that's what prompts these queries.

First off, is there anyway to tell from the cut away of this lock which is the binding pin, or does the binding pin vary from time to time?

My understanding of the literature is that the binding pin HAS to be the one to lift and get in place first, before all others - true?

Secondly, these springs... Am I correct in stating than in THEORY a spring could be compressed way beyond anything that would normally allow the binding pin to get to the shear line? Is it this length of the spring (or what I think is truly called its frequency) that means that it is possible to go hopelessly past the binding point and get all befuddled with what's going on?

Thirdly, this practice lock is spanking new, and the pins seems to travel smoothly and with little force. However, i have a much, MUCH older lock of the same type (different make etc.)and I find the springs MUCH harder to move. It was finding that the practice lock's springs were relatively easy to move that made me realise how hard it was to move the springs in the older lock, and have now a few bent picks.

BW

I haven't dared even ask about spool pins and how to cope with them - I guess that's for another thread which I'll search for.

[EmCee]

Hi beardyweirdy

'Rim' means that the lockcase (which includes the bolt that actually secures the door) is mounted on the surface of the door as opposed to in a hole (mortice) in the body of the door.

'Cylinder' just describes a shape - your photo shows a pin-tumbler cylinder.

A 'rim' lock can be a lever lock or a pin-tumbler cylinder lock. With a rim lever lock all the mechanisms (levers and bolt) are in the lockcase on the surface of the door and small holes are needed through the door for the key and perhaps a spindle for a handle. With a rim cylinder lock the working mechanism of the lock - the actual 'cylinder' as in your photo - would be fitted into a large hole through the door, and a tang on the rear would work the bolt held inside the nightlatch lockcase mounted on the opposite surface of the door.

I can't tell from the photo which is the binding pin (other more expert members of the forum might be able to do so). The reference to binding pin is usually about 'feel'. I don't have any cutaways so I'm not sure, but if you watch closely while applying and releasing tension you might be able to see one of the pins moving a bit as it gets caught at the shear line. The pins in any one lock will always bind in the same order (I suppose this might change if any of pins or chambers wears more than others).

There is not just one binding pin - all of the pins will bind in turn. It is only possible to pick locks because of imperfect tolerances, so yes, you have to feel for the first binding pin, lift it so that the driver pin is trapped at the shear line...at which point the slight rotation of the plug will cause another pin to bind and you have to find and lift that one...then the next...etc..until you have lifted all the pins to the shear line. All of the pins bind, therefore, in turn, but you won't know in advance which order they will bind in.

I suppose in theory a spring could be crushed to the point where it loses its 'spring'. If the lock has the driver pins on top then gravity would still pull them down. I think the length of the spring is more to do with the amount of pressure pushing down on the pin and the amount of space needed in the chamber for the pin (or pin stack - there can be more than one pin). It is certainly possible to push the driver pin too high so that the key pin is then positioned across the shear line. The idea is to find the binding pin, then push up slowly and carefully waiting for the small 'click' and very slight movement of the tensor as the plug rotates a gnat's whisker. If you push the pin up too hard/fast, it's very easy to push the key pin up into the chamber.

Security pins complicate matters because they 'false set'...giving you the same 'click' and tensor feedback you would get if the pin had set correctly.

Cheers...