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- Mar 5, 1999
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That one is not going to fail.
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Price is$300$250 ea (shipped within CONUS). If you live outside the US, I will contact you after your order for extra shipping charges.
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It came! Love the BAS
- Thread starter dewingrm
- Start date
For an estimate the force that can be exerted on what you are trying to pry out do a torque balance.
Torque = force (1) x distance (1) = force (2) distance (2)
The force (1) provided by you is the same. Distance (1) is the distance between your grip point and the pivot point, distance (2) is the distance from the pivot point and the blade contact point. Force is the prying force.
Sorry if this is not correct but my physics is a bit rusty.
Will
Torque = force (1) x distance (1) = force (2) distance (2)
The force (1) provided by you is the same. Distance (1) is the distance between your grip point and the pivot point, distance (2) is the distance from the pivot point and the blade contact point. Force is the prying force.
Sorry if this is not correct but my physics is a bit rusty.
Will
- Joined
- Sep 7, 2001
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- 5,946
So, theoretically...
The blade being 1.5" which makes the distance between the two blade lengths 9.2% longer.
9.2% more chopping power? or should we use the length of the arm and blade, which would make the percentage change less. But then we have the added weight.
blah... it's been too long since high school physics. You have both the 16.5 and the 18 don't you Will?
The blade being 1.5" which makes the distance between the two blade lengths 9.2% longer.
9.2% more chopping power? or should we use the length of the arm and blade, which would make the percentage change less. But then we have the added weight.
blah... it's been too long since high school physics. You have both the 16.5 and the 18 don't you Will?
For leveraging, I would take the contact point (with the wood) between your grip and the final point where the blade contacts the wood (near the tip).
How to define chopping power? Other factors such as blade geometry, individual ability, etc. go into determining chopping performance.
I have both the 18" (my is 17.5") and 16.5" WWII. The 18" can go through a 2x4 in about 10 blows, the 16.5 takes around 17 blows.
Will
How to define chopping power? Other factors such as blade geometry, individual ability, etc. go into determining chopping performance.
I have both the 18" (my is 17.5") and 16.5" WWII. The 18" can go through a 2x4 in about 10 blows, the 16.5 takes around 17 blows.
Will
- Joined
- Mar 5, 1999
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"I have both the 18" (my is 17.5") and 16.5" WWII. The 18" can go through a 2x4 in about 10 blows, the 16.5 takes around 17 blows."
That's a significant difference.
That's a significant difference.
- Joined
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Maybe, it's the added torque with the extra length , plus the added mass which adds to the potential kinetic energy blah blah...
If the velocity between the two blades doesn't change much the added mass could be a signifigant factor... I dunno.
Or we can say in general "Big nife chop bettah than little nife".
If the velocity between the two blades doesn't change much the added mass could be a signifigant factor... I dunno.
Or we can say in general "Big nife chop bettah than little nife".
- Joined
- Mar 5, 1999
- Messages
- 34,096
I think that's exactly why the 20 inch AK and GRS outchop everything.