Does thickness matter more than overall volume?

It all depends whether you want to do damage to the log you are splitting or the knife you are using.
And how much of the pie it takes to replenish your strength after the testing. But it could also be best to just save the knife to slice pie and relax and let someone with the proper wisdom worry about the issue.
 
rhino said:
You're tweaking at brain cells that haven't been touched since I was a sophomore in Mechanics of Materials class back in the mid 1980s! I remember we had to calculate Area Moment of Inertia when studying bending. How is the section modulus related to Area Moment of Inertia?
Click to expand...

like you, it's been too long. but with a little google-fu - Area Moment of Inertia is same as Second Moment of Area (SMA) - used in calculating deflection and stress of a beam due to applied moment.
Section Modulus, specifically Elastic Section Modulus (ESM) - applicable up to the yield point, which seems appropriate here, but not quite.

ESM=SMA/y
y=distance from axis

further yield moment=ESM x yield strength of material

but on a Monday morning with less than 1/2 cup of coffee in my system, I'm not sure how this would apply to the strength vs batoning
 
Lesknife said:
It all depends whether you want to do damage to the log you are splitting or the knife you are using.
And how much of the pie it takes to replenish your strength after the testing. But it could also be best to just save the knife to slice pie and relax and let someone with the proper wisdom worry about the issue.
Click to expand...

Nobody told me we're having pie. What's the matter with you guys?
 
l1ranger said:
like you, it's been too long. but with a little google-fu - Area Moment of Inertia is same as Second Moment of Area (SMA) - used in calculating deflection and stress of a beam due to applied moment.
Section Modulus, specifically Elastic Section Modulus (ESM) - applicable up to the yield point, which seems appropriate here, but not quite.

ESM=SMA/y
y=distance from axis

further yield moment=ESM x yield strength of material

but on a Monday morning with less than 1/2 cup of coffee in my system, I'm not sure how this would apply to the strength vs batoning
Click to expand...

Thanks! I branched off into thermodynamics, fluid dynamics, and heat transfer as soon as I could!
 
rhino said:
Thanks! I branched off into thermodynamics, fluid dynamics, and heat transfer as soon as I could!
Click to expand...
somebody has to do it. I stayed in civil, but they kept throwing structural stuff at me.
I do some structural now, but don't have to actively figure out any of these things - thank goodness
 
l1ranger said:
somebody has to do it. I stayed in civil, but they kept throwing structural stuff at me.
I do some structural now, but don't have to actively figure out any of these things - thank goodness
Click to expand...

After grad school, I sustained a back injury that prevented me from working for about 10 years, so my career as a mechanical engineer was lost anyway. Most of what I have forgotten is now of no consequence!
 
Isn't the only difference between Scandy grind and Saber grind is the latter has a micobevel?
 
NeedItCheaper said:
Might be some poor math. If that's the case, hopefully someone who's less mathematically challenged than me could fix this equation.

For years I've liked saber-grinds for batoning. One of my assumptions was thicker blades of the same material/heat treat/etc... were by definition stronger. But then it dawned on me I didn't know the overall volume and a thinner blade with more material might be stronger. The crux of this theory is people often look at thickness of a blade and discount the triangle-ish (I know they all aren't perfectly flat nor symmetrical) shape at the bottom. The thickness argument should only be true if we were dealing with two rectangles of different thicknesses. So I took some of my blades and measured them to come up with two theoretical knives and calculate the volume or rather amount of metal used. Both blades will be 1.25 inches in height. The scandi-grind will be 5/32 thick and the saber-grind will be 3/16 or rather 6/32 thick. Length of blade will be discounted and assumed to be the same length. I've converted some of the common fractions we use to make the equations easier to work with. Please feel comfortable with correcting any mistakes.

Scandi 1.25 inches height (or 40/32)
- 5/32 Thick * 32/32 height = 160/32 .... 32/32 represent 1in worth of flat before the blade begins to taper.
- Blade taper; 5/64 (half base) * 16/64 (height) = 80/64 or 40/32 ... Both my scandi-grinds have about .25in in taper.
- Total = 200/32

Saber 1.25 inches height
- 6/32 (or 3/16) * 12/32 (or 3/8) height = 72/32
- Blade taper; 3/32 (half base) * 28/32 height = 84/32
- Total = 156/32

Assuming this is a fair estimate, there's still a problem. In my testing, thicker blades allow for less strikes in batoning. This means the potential increase in durability from a scandi-grind would need to out weigh the increase in strikes. What are your thoughts on all this?
Click to expand...
NeedItCheaper said:
Yeah, but it's been raining for 3 days now and I had to cut my backpacking trip short because it was too overwhelming. I'm stuck inside and clearly have too much time on my hands.


Your wish is my command

FFG 1.25 inches height
- 6/32 (or 3/16)
- Blade taper; 3/32 (half base) * 40/32 height = 120/32
- Total = 120/32

I wonder if FFG with the same dimensions would literally be 40% less strong. 120/200 = .6

Edit; Yes I edited this immediately after posting.
Click to expand...

When you multiply fractions, you have to multiply the numerators and the denominators.

Scandi 1.25" height (or 40/32)
- 5/32" Thick * 1" height = 5/32 sq.in. (this is where you should've realized the problem with your math...)
- Blade taper; 5/64" (half base) * 1/4" (height) = 5/256 sq.in.
- Total = 45/256 sq.in. (or 0.17578125 sq.in.)

Saber 1.25 inches height
- 3/16" Thick * 3/8" height = 9/128 sq.in.
- Blade taper; 3/32" (half base) * 28/32" height = 21/256 sq.in
- Total = 39/256 sq.in. (or 0.15234375 sq.in.)

FFG 1.25 inches height
- 6/32" Thick (or 3/16")
- Blade taper; 3/32" (half base) * 40/32" height = 120/1024 sq.in.
- Total = 30/256 sq.in. (or 0.1171875 sq.in.) [half the cross-section area of the Scandi example and about 25% less than the Saber]
 
rhino said:
You're tweaking at brain cells that haven't been touched since I was a sophomore in Mechanics of Materials class back in the mid 1980s! I remember we had to calculate Area Moment of Inertia when studying bending. How is the section modulus related to Area Moment of Inertia?
Click to expand...
I'm a structural engineer so I do this kind of thing all the time.
Section modulus is a convenient intermediate calculation but isn't required for what we are talking about.
To calculate stress due to bending, start with
M=amount of bending force applied
c=half of the thickness, i.e. distance from center of the section to extreme fiber
I=simple moment of inertia, also ma measure of the lateral stiffness of the section
Bending stress=M x C / I
(Section modulus, S, = I/C so you can see how it is just an intermediate calculation bending stress = M / S)

Someone mentioned a comparison of 2 blades- a FFG with 1/4" thickness at the spine, and a Scandi that is 1/8" thick for most of its height. The volume or weight of these 2 sections would be virtually the same. I'm sitting on the sofa with my laptop without any other resources handy. If I'm figuring this correctly the 2 blades would have the same lateral strength but the FFG would be twice as stiff.
 
Thick. 😉
F10FEE6A-E8A2-4858-8762-5B582B288C13.jpeg
3CCBD40C-55BB-4E76-9BD6-C504B54EA776.jpeg
 
Last edited:
bdmicarta said:
I'm a structural engineer so I do this kind of thing all the time.
Section modulus is a convenient intermediate calculation but isn't required for what we are talking about.
To calculate stress due to bending, start with
M=amount of bending force applied
c=half of the thickness, i.e. distance from center of the section to extreme fiber
I=simple moment of inertia, also ma measure of the lateral stiffness of the section
Bending stress=M x C / I
(Section modulus, S, = I/C so you can see how it is just an intermediate calculation bending stress = M / S)

Someone mentioned a comparison of 2 blades- a FFG with 1/4" thickness at the spine, and a Scandi that is 1/8" thick for most of its height. The volume or weight of these 2 sections would be virtually the same. I'm sitting on the sofa with my laptop without any other resources handy. If I'm figuring this correctly the 2 blades would have the same lateral strength but the FFG would be twice as stiff.
Click to expand...

NERD ALERT!

Meh, I think I'm bitter because you reminded me that I had a solid A going into my Mechanics of Materials final exam back in 1984 or so . . . and I bombed the biggest problem on the exam and got a B+ for the course. It was an early example of my innate ability to snatch defeat from the jaws of victory!
 
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