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Joined: Apr 2011
Posts: 610
Sidelock
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OP
Sidelock
Joined: Apr 2011
Posts: 610 |
I was wondering if anyone had stats on psi exerted by the hook on the pin upon firing.
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Joined: Jan 2003
Posts: 144 Likes: 3
Sidelock
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Sidelock
Joined: Jan 2003
Posts: 144 Likes: 3 |
A lot would depend on how effectively it was jointed (or re-jointed) on the circle and draw. Effective bearing surfaces there will distribute the stress and strain off the pin to some degree. Calling Pugwash ... he could answer better.
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Joined: Jul 2011
Posts: 839
Sidelock
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Sidelock
Joined: Jul 2011
Posts: 839 |
Let's see... how about 10,237 psi on the "hingh pin" ?
Of course it's a lot less outside of Calcutta.
Especially without US foreign aid.
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Joined: Oct 2010
Posts: 969 Likes: 38
Sidelock
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Sidelock
Joined: Oct 2010
Posts: 969 Likes: 38 |
Gough Thomas, who was a highly qualified engineer once wrote that the strain on the pin is about 1/3 of a ton, but did not elaborate on how he calculated it.
Calculating the strain would involve taking into account the Poisson effect of thick walled tubes contraction and expansion of the chambers, the force exerted by the recovery of the action body as well as the forward thrust of the charge going through the forcing cone etc.
The action body flexes back on firing, according to accepted theory. If it flexes back then it must recover and that recovery must by definition involve a rather powerful push on the barrels and through them on the cross pin.
More than calculations I would love to see a high speed film sequence of a double shotgun action during firing. That would give visual proof of what actually happens and not what is commonly thought happens.
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Joined: Jan 2002
Posts: 5,954 Likes: 12
Sidelock
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Sidelock
Joined: Jan 2002
Posts: 5,954 Likes: 12 |
Short and simple store is as follows. The maximum back-thrust on the action face would be akin to the bore area times the maximum pressure. Assume pi X (0.729/2)**2 X 10,000 psi = 0.4172 X 10,000 = 4,172 force pounds. The hook-pin contact on a typical W&S BLE measured about 90% of a 0.500" hinge pin and 0.350" wide hook. Circumference of the pin is pi times diameter = 3.14 X 0.500 = 1.57". The hook bares on 90% of the pin, so the effective bearing length is 1.413". The bearing area then is length times width = 1.413" X 0.350" = 0.495 square inches. Assuming equal pressure distribution (not really valid, but OK for this kind of swag), 4,172#/ 0.495" = 8,428 psi. That is well below the working strength of typical soft, low carbon steel. Nothing of the hook or pin is likely to yield! If a gun is worn off-face, the hook and pin go into a battering mode and that is a different story altogether.
If, the concern is wear, then we have to have a working idea of the action flex as wear requires relative motion. Clearly, there is little motion from firing. However, there is relatively a lot of motion when the barrel is cycled open and closed. Modern high pressure lubes will handle the above pressures nicely. However, grit will infect and taint the lube such that it will act as a grinding agent. Clean modern lube is, IMO, the key to longivity of gun bearing surfaces.
Questions, comments? As always, check my math and logic, please.
DDA
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Joined: Apr 2011
Posts: 610
Sidelock
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OP
Sidelock
Joined: Apr 2011
Posts: 610 |
So, a small pin would be taking more psi vs a larger pin?
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Joined: Jan 2002
Posts: 5,954 Likes: 12
Sidelock
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Sidelock
Joined: Jan 2002
Posts: 5,954 Likes: 12 |
Good point!
Yes, a small pin would have less area (square inches) over which to distribute the force (pounds). A large bearing will carry more load than will a small one for equal strain on the components.
Questions?
DDA
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Joined: Dec 2001
Posts: 775
Sidelock
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Sidelock
Joined: Dec 2001
Posts: 775 |
RM: I think you better re-check your math and reasoning. I have never owned a gun with a hinge pin as large as 0.50". I would think that most of mine are 0.375" or less, and the bearing area cannot be wider than the hook. Also, the hook cannot bear on over half the pin diameter, or you wouldn't be able to remove the barrels without removing the pin. I don't know what other makers used, but the pin in the LC Smith gun is a taper pin, according to the drawings. Also, most guns that I am familiar with have a portion of the barrel lug that bears against the frame and will carry part of the load if properly fitted.
Last edited by Tom Martin; 05/31/13 10:36 PM.
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Joined: Jun 2012
Posts: 72
Sidelock
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Sidelock
Joined: Jun 2012
Posts: 72 |
Clean modern lube is, IMO, the key to longivity of gun bearing surfaces.
Rocketman, would you mind sharing what brand of lube(s) you prefer? Also does anyone have any photos exactly where to lube a box lock, 2 AH Foxes, and an AYA 453 ? TIA
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Joined: Dec 2001
Posts: 12,743
Sidelock
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Sidelock
Joined: Dec 2001
Posts: 12,743 |
I do have an old H&A Single on which the lug has a hole rather than a hook so it is necessary to remove the pin to take it down. Even on this gun the forward half of the pin would bear no load upon firing. Assuming that you were meaning the hook covered 90% of "Half" the pin then that would be 45% of the circumference. Even using the .500" this would give .707 x .350 = .247 sq In for a load of more like 16,686 PSI on the hinge. It would of course vary from this depending upon actual diameter of the pin & width of the lug. A question I have is are we as concerned so much about the actual area involved as to the shear strength of the pin. It would seem to me, that at least as far as the pin is concerned that once an adequate with was achieved making the lug wider would not increase its strength to any significant amount. The force require to deform or eventually shear it at the juncture with the frame sides would remain essentially the same regardless of its width.
Miller/TN I Didn't Say Everything I Said, Yogi Berra
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