Axis lock: Tested to Failure?

Joined
Nov 26, 2000
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515
I work in a knife shop and my Assistant manager carries a Benchmade Ares Axis. Anyways, he says that he has seen somewhere that the Axis-Lock was tested and failed at 3,000 lbs. of pressure placed on the blade by a robotic arm. I don’t believe that this is the truth, I think that it will probably fail at 300-500 lbs. of force.
Does anybody know what the failure point is for the Axis Lock?

I don’t want to know this because I am looking for the toughest lock, etc. but I just want to know out of curiosity.

Thanks!
 
3000 lbs.??? I highly doubt any folding knife could handle that much force. The pivot pin would simply shear in half. I've heard that BM's Axis lock has tested at 200 lbs. and that seems reasonable. It certainly would seem to be more than adequate for the requirements that should be expected of a folding knife.

BTW, that Ares is an excellent all-around folder.
 
knifenerd- 3,000 lbs:eek: ?
I was like :eek: :confused:
That is like a Hummer being supported by an open blade.
I have heard that it was tested with 200 lbs. without damage to the lock, and that seems a little more realistic. I think the "without damage" part is not truthful either, it seems like there would be quite a bit of blade play.
 
Lets see, on a 710 you have roughly a 4" blade, thats 1/3 of a foot. Granted the arm probably wouldn't put the load right at the point, but it would probably be at least an inch in front of the pivot, 1/12 of a foot.

So 3000lbs of force 1/3ft from the pivot = 1000 ft/lbs of torque. Yeah right.

At 3" out, you have 3000lbs X 1/4'= 750 ft/lbs of torque, again... yeah right.

2" out, 3000X 1/6= 500 ft/lbsd of torque...yeah right.

1" out, 3000 X 1/12'= 250 ft/lbs of torque. Don't think so...
 
Originally posted by Akwacko
@ 2633lbs my arm broke. ...

Wimp. ;)

Actually K Man, if you do a search, you should come up with a really long and tedious thread on this topic. I believe the 3,000 pound number is foot-pounds. If you divide by twelve and multiply by the two inches or so from the pivot to the actual testing point, you probaby get around 500 inch-pounds, which is more in the realm of common sense. Basically, the lock isn't likely to break with you holding onto the knife.
 
Steve Harvey :

I believe the 3,000 pound number is foot-pounds. If you divide by twelve and multiply by the two inches or so from the pivot to the actual testing point, you probaby get around 500 inch-pounds

Foot pounds to inch pounds is converted by a 12 multplier. There are 12 inch lbs in one ft lbs as there are 12 inches in one foot. Second, you would divide by the distance to get the weight applied. Or multiply the weight by the perpendicular distance to get the applied torque.

What generally happens is that a huge number is quoted which is so vague that nothing can be made of it and very frequently the units are all mixed up. Like for example "The lock took 1000 lbs of pressure". This is like saying "The knife has a blade length of 2 feet per second".

Even if the number was quoted in the right units so sense could be made of it (500 in.lbs), it is still pretty meaningless because it is really artificial. It is a perfectly perpendicular static load. In actual usage you can expect side loads and impacts, both of which can cause release or failure under much lower forces than the huge numbers describe.


Basically, the lock isn't likely to break with you holding onto the knife.

You don't have to be able to hold onto the knife to the level of its break point for it to get enough impact to break it, even with the blade being knocked out of your hand, this doesn't absorb all the energy.

Most of this isn't even relevant to use however (break point), because the stability is always the critical factor. What differences does it make if the lock takes 1000 in.lbs + in a vice when you can twist it a few degrees and now it takes 100 in.lbs. A description of the weak points of the locks would be of benefit.

-Cliff
 
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