Is this true???

Here is the reply from SOG:

The test was of lock pressure by lever principle, it measured the force on the blade which was held in a vice. The results where calculated and the calculations revealed the force placed on the blade was equal to 1100 lbs of pressure.

Regards,
Jose Lopez
Retail Sales & Customer Service
SOG Specialty Knives & Tools
6521 212th St SW
Lynnwood, WA 98036


Looks to me that they are say the the lock really did hold under 1000 lbs of load (lever action). I think the 1100 is a typo. "Believe it or not", I guess.

Regards
 
Looks to me that they are say the the lock really did hold under 1000 lbs of load ...

The representative simply doesn't understand what the test measures and how to report it and thus what they are saying is meaningless jibberish. Any folder can take a force of 1000 lbs if you apply it close enough to the pivot and the same folder will fail with a force of 1 lbs if you apply it far enough away from the pivot. Locks of that type will fail under a given torque because you basically have to induce them to rotate to make them break/release. To break a lock by a force you would have to for example pull the blade right out of the handle or drive it back through the handle.

-Cliff
 
The representative simply doesn't understand what the test measures and how to report it and thus what they are saying is meaningless jibberish. Any folder can take a force of 1000 lbs if you apply it close enough to the pivot and the same folder will fail with a force of 1 lbs if you apply it far enough away from the pivot. Locks of that type will fail under a given torque because you basically have to induce them to rotate to make them break/release. To break a lock by a force you would have to for example pull the blade right out of the handle or drive it back through the handle.

-Cliff

I could be wrong but my understanding of the reply was the the weight was "calculated to be 1000 lbs". That could mean just about any weight at just about any distance from the lock, as long as it calculates out the be 1000 lbs of "lever load". Failure could be a number of things: the locks horizontal pin breaks or the handle holding the lock breaks. I will try to get a more specific answer.
 
Re: Mode of failure-
I recall seeing tests in one of the magazines when the Axis lock first came out, and they were talking about how the lock was so strong that the blade would break first.

And then I saw the pictures.

Yeah, the "blade" broke- at the tang in front of the pivot. This seemed to make perfect sense to me, since the design requires the tang to be like half as wide as the blade body. Though I love the Axis design on the whole, if I were to come up with any improvements it would be to get rid of them Omega springs, and change things so that the tang could be made wider. It doesn't make much sense to come up with the ultimate in lock strength if it also requires you to significantly weaken the blade at the same time.
 
... my understanding of the reply was the the weight was "calculated to be 1000 lbs". That could mean just about any weight at just about any distance from the lock, as long as it calculates out the be 1000 lbs of "lever load".

Yes, my point was that you can't actually do that calculation. The language isn't correct. As an example, I can give you a bucket and ask you how many rocks it will hold. This question doesn't have an answer because it depends on the size of the rocks. What Benchmade is doing is telling you the bucket holds a lot of rocks which makes you think it is a very big bucket but unless they tell you the size of the rocks you really don't know anything at all.

For example, lets assume that I take a knife which has a failure point of 500 in.lbs on the lock. This means that I can break it with a force of 100 lbs at five inches, 500 lbs at one inch and 10 000 lbs at about 1/16" of an inch. Thus you can't quote a force as a lock strength because it is indeterminate, just like you can't specify the size of a bucket by the amount of rocks it can hold. What most people do is use really small rocks and not tell people just how small the rocks really are. An in.lbs is a really small torque. A normal man can exert about 300 in.lbs for example with just his writs.

... the "blade" broke- at the tang in front of the pivot.

Did they mention how much torque was required to do this, it takes a lot to crack off steel through its width like that, even if the width is small. It would be similar to bending a steel through a spine of similar thickness.

-Cliff
 
Well, in their defense simply numbers are thrown around by a lot of people. However, it is tale telling that BM states (and that is from their previous webpage) that the axis lock holds 800 lbs, while Spyderco for example states 200 lbs per inch bladelength. Not only is the former incomplete information, it also makes the number appear very large even though at a blade length of 4 inches both indicate the same lockstrength, which of course is a simple sales pitch. As comparison, the Ti ATR by Spyderco held 800 lbs. one inch from the pivot and was subsequently improved. So, it is a strength that a number of well engineer locks achieve.
 
Yes, my point was that you can't actually do that calculation. The language isn't correct. As an example, I can give you a bucket and ask you how many rocks it will hold. This question doesn't have an answer because it depends on the size of the rocks. What Benchmade is doing is telling you the bucket holds a lot of rocks which makes you think it is a very big bucket but unless they tell you the size of the rocks you really don't know anything at all.

For example, lets assume that I take a knife which has a failure point of 500 in.lbs on the lock. This means that I can break it with a force of 100 lbs at five inches, 500 lbs at one inch and 10 000 lbs at about 1/16" of an inch. Thus you can't quote a force as a lock strength because it is indeterminate, just like you can't specify the size of a bucket by the amount of rocks it can hold. What most people do is use really small rocks and not tell people just how small the rocks really are. An in.lbs is a really small torque. A normal man can exert about 300 in.lbs for example with just his writs.

Did they mention how much torque was required to do this, it takes a lot to crack off steel through its width like that, even if the width is small. It would be similar to bending a steel through a spine of similar thickness.

-Cliff

Yes, I know exactly what you are saying (the company is SOG, by the way). This is mostly just a paraphase of what you just said but it's the for of my email to them so what the heck:

That is why I sent them another email asking if it is 1000 foot lbs or 1000 inch lbs (or whatever) of force. They are both 1000 lbs but obviously quite different in the amount of force. One is 1000 lbs at 1 foot from the locking mech and the other is 1000 lbs and only 1 inch from the locking mech. I have not receive a response as yet and I'm not hold out much hope.

Regards
 
That is why I sent them another email asking if it is 1000 foot lbs or 1000 inch lbs ...

It has to be 1000 in.lbs, even the stronger fixed blades will crack at torques of only several hundred ft.lbs.

Well, in their defense simply numbers are thrown around by a lot of people.

Yes, the concept is mangled constantly both due to lack of understanding as well as likely intentional inflation of the numbers.

-Cliff
 
Quote:
Originally Posted by sog View Post
SOG has 1000 lbs of force written down. It is probably inch/pounds. But i can't prove it. Guess you'll have to call them and find out.


"Strength: The Arc-Lock™ was independently lab tested at over 1000 lbs. of pressure without lock failure."

From :

http://www.sogknives.com/Folding/SpecElite/SE18.htm

Inches per pound would be another insensible unit. Torque is the cross product of force and distance so the units are (force)*(length). In the standard system usually ft.lbs or in.lbs. Lock strengths are always quoted in in.lbs because the numbers would be very lot in ft.lbs and thus not visually impressive. For example 600 in.lbs sounds like a lot, however 50 ft.lbs does not. They are however both the same number.

-Cliff

I changed it to pressure in another post and didn't in this one. Pressure and force are very close, but it is a misquote.

So in the previous post about molecules it is accepted that steel has molecules. I don't have an electron microscope or their steel to prove them wrong about the molecules changing shape. So i am not going to put them down if i don't have proof first.

I guess the people manning the phones don't know difference between inch or foot lbs. I'll try CS and maybe i can get an e-mail back from someone that does.
 
Pressure and force are different quantities, pressure equals force divided by area so the same force can produce different pressures. Just like speed and distance are related as speed is distance divided by time.

Forging steel to effect the structure does mainly by changing the grain structure which isn't a molecular event, it has to do with recreating the crystal structure to minimize the grain size by enhancing the number of independent nucleation cites. This is a large scale change, far beyond a molecular size change.

There are also effects on the carbides as forging will break up the as cast carbide networks, but again it doesn't change the shape of the molecules it changes the shape of the structure. This is no different than if you took a piece of snow and squat it to make a shape you are not actually changing the shape of the "snow molecules".

To exert a change on molecules you would do something like hydrogenate a fat, this changes liquid fats like corn oil into solid fats like margerine. This actually changes the nature of the individual fat molecules.

-Cliff
 
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