inch/lbs ?

[Richard]

Im not sure I understand. What is inch/lbs as relating to lock strength. What does this translate in pounds. For instance, Cold Steel says thier Gunsite holds something like 125 pounds of weight on the handle and the lock holds. Im sure most of us have seen the proof video so we know the weight is on the end of the handle. What would that be in inch/lbs. I hope somebody can put this in english for me. Thanks

 

[GK]

Inch-Lbs is a measue of torque, or force around a pivot point. Typically used in tightening bolts. Imagine a 250 lb weight hanging off a 1 inch long lever. You can get an equivalent torque force with a 62.5 lb weight at 4 inches, i.e. hanging at the tip of a 4 inch blade. Hope this helps...

GK

 

[Sal Glesser]

Rich - Let's say we fix a knife into a fixture in the open position. Pressure is applied directly on the handle 2.5" from the pivot pin. Pressure is increased until the knife lock defeats. Let's say that the graph shows that it defeated at 100 lbs of pressure. 100 lbs of pressure applied at 2.5 inches from the pivot = 100 X 2.5" = 250 inch/lbs of force to defeat the lock. Hope that also helps.
sal

 

[Kelt]

does that mean that if you were to choke up on the handle of a Jot Singh, closer towards the blade that it might take more pressure to cause the lock to fail? or would it take less? i'm trying to think of the lever principle here (damn, should have thought this out before i had my Fosters!), so i thought it would take more pressure as you got closer towards the pin. Sal, Help?!

 

[Richard]

Thanks guys, now I wont feel like such an idiot.

 

[Outlaw_Dogboy]

Kelt34, how about this? It will take more pressure (force), but it will fail at the same torque (in-lbs). 100 lbs at 2.5 inches from the lock = 100lb x 2.5 in. = 250 in-lbs.

OR

250 lbs at 1.0 inches from the lock = 250 lb x 1.0 in. = 250 in-lbs. Same torque, but more pressure (force).

------------------
Work hard, play hard, live long.
Outlaw_Dogboy



[This message has been edited by Outlaw_Dogboy (edited 30 August 1999).]

 

[Kelt]

Outlaw: yeah, d-uh i'm feeling stupid. just switch the numbers around and it makes sense. thanks. guess i should have written it down. never do math under the influence. unless you're the materials acquisitions guy working for the pentagon; then it won't matter anyway.

 

[Outlaw_Dogboy]

Don't feel stupid. It's a matter of terminology as much as anything, IMO. Bottom line, it is better to "choke up" on the knife.

------------------
Work hard, play hard, live long.
Outlaw_Dogboy

 

[Cliff Stamp]

Sal, you are mixing up terms and units here :

Let's say that the graph shows that it defeated at 100 lbs of pressure. 100 lbs of pressure applied at 2.5 inches from the pivot = 100 X 2.5" = 250 inch/lbs of force to defeat the lock.

You apply a force at a distance to produce a torque :

"Let's say that the graph shows that it defeated at 100 lbs of force. 100 lbs of force applied at 2.5 inches from the pivot = 100 X 2.5" = 250 inch-lbs of torque to defeat the lock."

As for the units, lbs measure force, torque is force x lever arm so the unit lbs x inch, commonly wrote as inch-lbs or inch.lbs . Inch/lbs would be inches divided by pounds which is a different thing totally.

Pressure is force / area in the standard system usually wrote psi (pounds per square inch) and is not involved at all here. It does come up a lot in material testing though as it is how tensile stresses are evaluated and similar properties.

As for the CS question, if the weight is on the end of a 5 inch handle that is 625 inch-lbs of torque if the weight is perpendicular. It is less if it isn't and the value depends on the angle. Assuming the weight is 45 degrees off of the perpendicular that gives about 450 inch-lbs of torque.

-Cliff


[This message has been edited by Cliff Stamp (edited 31 August 1999).]