strength of the axis lock?

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jzmtl
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#21

Post by jzmtl »

dialex wrote:Indeed, this is something that bothers me as well. I mean, according to the statements above, if you have say a 4" blade knife, it will hold 4 x 200 lbs near the tip but only 1 x 200 lbs near the pivot? :confused:
I always thought it simply means this, the inch.lb is the base torque value, and length of blade is the multiplication factor. In case of a 3.5" MBC lock, it means the lock itself would have to stand up to 200 in.lb X 3.5" = 700 in.lb of torque. It makes sense because this way no matter how long the blade in the knife is, locks in the same rating category would be able to stand up to the same amount of pressure applied to tip of blade.

So essentially rating is not purely based on lock strength, but rather whole knife. The longer the blade, the stronger lock needs to be in order to get the same rating.
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The Mentaculous
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#22

Post by The Mentaculous »

Yes, I would think 200 in lbs per inch on a 4" knife means it can withstand 200 in lbs at 4 inches, and 800 in lbs at 1 inch. If the knife was 2" blade instead of 4", it would only have to be 1/2 as strong overall to still qualify--200 in lbs at 2", as opposed to the 400 inch lbs at 2" on a 4" knife. This makes complete sense, just like jzmtl said, because the longer a blade is, the more leverage there is being applied.
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The Deacon
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#23

Post by The Deacon »

dialex wrote:Indeed, this is something that bothers me as well. I mean, according to the statements above, if you have say a 4" blade knife, it will hold 4 x 200 lbs near the tip but only 1 x 200 lbs near the pivot? :confused:
No Alex, it's the other way around.

What Sal's actually saying is that the lock of a 4" bladed knife needs to be four times as strong as that of a 1" bladed knife, and twice as strong as that of a two inch bladed knife, to achieve the same "inch pounds per inch" rating. So a 2" bladed knife could be MBC rated with a (relatively) weaker lock than a 4" bladed one. That makes perfect sense, since the blade of the 2" knife can never exert as much leverage on the pivot and lock as a 4" blade can. So an MBC rated 4" blade would need an 800 in/lb lock (200 lb X 4") while a 2" bladed one would only need a 400 in/lb lock (200 lb X 2").
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Tsujigiri
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#24

Post by Tsujigiri »

defenestrate wrote:MBC = Martial Blade Craft

Not certain about the Bradley (though I'm pretty sure it's one of the stouter linerlocks) but the Manix 2 uses the CBBL which is known to be rather stout. I believe they qualify as MBC rated generally.
Great, thanks!
ghostguy1
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#25

Post by ghostguy1 »

My friend, I don't think that's what it means. Obviously, Sal can better explain what is meant, but I think, based on his definition or specifications, for a martial blade craft knife to be classified as such, it must be able to support or withstand 200 in lbs of force/in of knife blade. Since the greatest torque can be applied end of the blade the calculation would be made with the full length of the blade. So to qualify as a MBC knife a one inch blade should be able to support 200 in lbs torque, a 2 inch knife should be able to support 400 in lbs of torque, a 3 inch knife should be able to support 600 in lbs of torque and so on. In light that when alan elishewitz tested a number of factory knives and one custom knife and found that the average torque supported was only 300 something in lbs. It means that most of the spyderco knives would exceed the strength of the other knives. However, for those who are not impressed by the demonstration of a knife handling 400 lbs 4 inches from the pivot, I don't know what to tell you because I am impressed. It means that knife can handle 1600 in lbs of torque which is more than twice that of the chinook if the numbers above If the numbers are correct above than that knife can support or handle more than twice the torque of the chinook. If I remember my college physics, torque is equal to force X torque arm length. Wt is not truly equal to force because force is equal to mass X acceleration, but it is good enough for this argument since acceleration is acceleration of gravity which is constant in this instance. If the numbers are correct above, the chinook should be able to handle 200 in lbs per inch of blade X 3.5 inches which is equal to 700 in lbs. It also means that the knife should be able to handle 700 lbs one inch from the pivot. I don't know if this makes sense. Hope it does. :)
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