# yield and ultimate fail in battle

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Posted (edited)
It's a theory that hard-wood blades are able to excise cortical plate in skull-bone .  My inquiry is about the ultimate failure of 2mm thick wood blade at maximum human arm-speed of 40metre/sec ( assumed to be accelerating). As surgical chisels need an upper limit of 2mm to avoid random bone cracking and damage,  then 2mm wood is given to obtain a smooth surface to the skull-trauma fracture.

A .2cm  slice of wood of .1kg at 40m⋅s2  strikes on  5cm of bone (1  cm 2) = 16MPa. .
Wood density ~ 1ton cu.m.  The wood is impacted perpendicular to 2mm 'sharp' edge. Parallel ultimate failure is 40-70MPa and by analogy the perpendicular fail is ~8 - 14MPa for hardwood. The 60cm blade length adds to angular velocity.

Bone has ultimate fail of ~120 - 200MPa and by Hankinson equation at 45° it fails at ~38 - 63MPa.

It appears that failure of the blade is likely (and certain from a perpendicular strike by an opponent's heavier blade against the blade-width surface.  If the blade stays together at 16MPa it can't at faster speed and slice bone , even perpendicular on bone for  ~26MPa bone fail . myth busted ?
Edited by nameless

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1 hour ago, nameless said:

A .2cm  slice of wood of .1kg at 40m⋅s2  strikes on  5cm of bone (1  cm 2) = 16MPa. .

Can you walk through this calculation?

40m⋅s2? Is that supposed to be 40m/s2? How do you arrive at that number? How do you get 16 MPa? What do 5 cm and 1 cm2 refer to?

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Maximum velocity for baseball / cricket ball throw is ~ 100mph , ~ 40m/sec.  I'm wrong to say 2mm thickness at contact , it should be .02mm sharp edge.

A 5cm length of cut gives .1 sq cm.  So  .1kg. 402  / .1 is 1600 which is 16MPa.  Maybe.

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12 hours ago, nameless said:

Maximum velocity for baseball / cricket ball throw is ~ 100mph , ~ 40m/sec.  I'm wrong to say 2mm thickness at contact , it should be .02mm sharp edge.

A 5cm length of cut gives .1 sq cm.  So  .1kg. 402  / .1 is 1600 which is 16MPa.  Maybe.

That's a mass, multiplied by a speed squared, and divided by the area? That's not the formula for pressure (mv^2 is not a force), and doesn't have units of pressure (units matter)

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OK then given velocity , mass and area of contact could you demonstrate the MPa equation?

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32 minutes ago, nameless said:

OK then given velocity , mass and area of contact could you demonstrate the MPa equation?

No. It’s not enough information. You need the force, which depends on how long the interaction takes to change the momentum.

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Posted (edited)
Back to the drawing-board:

"30 people weigh about 26,000N.  An approximate minimum cross-sectional area of the femur is ( .0003 sq m). We divide the compressive force by the cross-sectional area to find the compressive stress on the bone. Our approximate value for the ultimate strength of bone that would be required to support 30x body weight was 80 MPa, which is actually less than the measured value of 205 MPa ( compressive fail) , so the claim that the femur can support 30x body weight seems  reasonable."

Here we go with 9000N for ball and the same for return strike. The .145kg weight is approximately equal to the hardwood .1kg in my theory question. "  If the ball contact is 5cm by 2cm= .001sq m ( see image) then the bat gets 9MPa.  Perpendicular fail for bat hickory wood is around  8-12MPa.

For the wooden blade on skull, contact may be 5cm by .2cm . Mass  is .1  kg , same acceleration as ball.  So it's .0001sq m , stress 63MPa.  Perpendicular fail is ~ 8- 14MPa and skull-bone fails at around 205MPa. . Bone is OK.
Edited by nameless

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3 hours ago, nameless said:

Our approximate value for the ultimate strength of bone that would be required to support 30x body weight was 80 MPa, which is actually less than the measured value of 205 MPa

To a good approximation; nothing ever fails in straight compression.

You may find this next equation helpful.

The (average) force exerted by an object when it comes to a halt is equal to the (average) force used to accelerate it times the ratio of the distance it was accelerated to the distance over which it stops.

So, if I drop a rock that weighs 1 newton off a table that's a metre high and it comes to a halt when it hits the carpet  which is 1 cm  thick, the average force on the floor is 100N.

It's also important to recognise that the "strength" (however you measure it) of the object which strikes a target isn't as important as people think.

You can cut steel with water if the water is moving fast enough.

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I am at a complete loss to understand what you are trying to do here.

This has not been helped by you evident struggle with mechanics in general and stress analysis in particular.

John Cuthber is noted for incisive comments in simple plain English and he has provided another one for you here. +1 John.

Please explain in plain English what you are trying to achieve, without mixing imperial and metric measurements or inappropriate formulae from mechanics as you have done up to now.

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4 hours ago, nameless said:
Back to the drawing-board:

"30 people weigh about 26,000N.  An approximate minimum cross-sectional area of the femur is ( .0003 sq m). We divide the compressive force by the cross-sectional area to find the compressive stress on the bone. Our approximate value for the ultimate strength of bone that would be required to support 30x body weight was 80 MPa, which is actually less than the measured value of 205 MPa ( compressive fail) , so the claim that the femur can support 30x body weight seems  reasonable."

You have a force given in this example

Quote
Here we go with 9000N for ball and the same for return strike. The .145kg weight is approximately equal to the hardwood .1kg in my theory question. "  If the ball contact is 5cm by 2cm= .001sq m ( see image) then the bat gets 9MPa.  Perpendicular fail for bat hickory wood is around  8-12MPa.

For the wooden blade on skull, contact may be 5cm by .2cm . Mass  is .1  kg , same acceleration as ball.  So it's .0001sq m , stress 63MPa.  Perpendicular fail is ~ 8- 14MPa and skull-bone fails at around 205MPa. . Bone is OK.

In this example, the contact time is used to calculate the force. That's missing from your scenario, as I already pointed out. Why would you use 9000N when the example you cite says the average force (not peak force) is twice that? Baseballs are ~7.5 cm in diameter, and bats can be up to 7.0 cm, so your estimations are probably off there, too.

Baseball bats are typically made of ash.

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Posted (edited)

John,

>The (average) force exerted by an object when it comes to a halt is equal to the (average) force used to accelerate it times the ratio of the distance it was accelerated to the distance over which it stops.

In the 30 people situation, the forces are given without time but a is G.  The baseball and wooden blade reach acceleration at point of delivery and deccelerate on contact ,  with probably equal time. So I used the examples as close duplicates with their force values.

>It's also important to recognise that the "strength" (however you measure it) of the object which strikes a target isn't as important as people think.

My inquiry is about ultimate fail of the 2mm wood as well as non-failure of the bone. The strength of each is not used to describe the other's result.

studiot,

The issue is in my first post.  I'm struggling with what I learnt in 1964 but we're getting there.

swansont,

The bat  F of 18,000N is divided between stopping and then repelling the ball., with about equal velocity . Then given the similarity in actions , the blade had roughly that F , delivered immediately on accelerating to stop on the bone.

The image shows the deformation of the baseball  and "woods used to manufacture custom pro wood bats are maple, ash, and birch. Old Hickory Bat Company" but it's not really relevant .  A skull is given with cm scale attached and the blade is required to be 2mm thick.

image of skull

Edited by nameless

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9 hours ago, nameless said:

studiot,

The issue is in my first post.  I'm struggling with what I learnt in 1964 but we're getting there. ﻿

swansont,

The bat  F of 18,000N is divided between stopping and then repelling the ball., with about equal velocity . Then given the similarity in actions , the blade had roughly that F , delivered immediately on accelerating to stop on the bone.

The image shows the deformation of the baseball  and "woods used to manufacture custom pro wood bats are maple, ash, and birch. Old Hickory Bat Company" but it's not really relevant .  A skull is given with cm scale attached and the blade is required to be 2mm thick.

I am sorry you chose to be so contemptuous at my attempt to help and ignored my request for a better problem description.

It is not about baseball bats, but the leading edge of boomerang type devices.

Why didn't you say this in the first place?

Now I would challenge you understanding of the mechanics of bone, you are completely off beam there.

There is a whole range of bone type with varying mechanical properties, best suited to their function in the body.

Firstly you cannot in general divide the applied force by the cross sectional area of the bone to obtain the compressive stress since bones have significant internal structure.
In particular those in compression are tubes not solid bars.

Secondly you are talking about penetrating trauma (wounds) to the skull.

The skull is a (thin) shell and obeys thin shell mechanics.
So the shell would only respond in compression to widely distributed force loading.
Concentrated applied normal forces put the shell into bending and normal shear.
You need to consider two separate effects: The shell global response to the bending and shear and the local effect of the sharpened penetrating object.
Penetration mechanics is the basis of other mechanical properties - Fracture toughness, resilience and hardness.

So can we start again with a proper problem description, that doesn't lead everyone up the garden path?

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10 hours ago, nameless said:

swansont,

The bat  F of 18,000N is divided between stopping and then repelling the ball., with about equal velocity .

That's not how physics works. The force is not "divided" between those two actions. The ball's exit velocity can be zero (e.g. a bunt) or it can be significantly higher than the pitch speed.

10 hours ago, nameless said:

Then given the similarity in actions , the blade had roughly that F , delivered immediately on accelerating to stop on the bone.

The similarity is superficial. The bone is not moving nearly as fast as the ball is, and bones don't compress the way a ball does. You can't conclude that a blade will exert the same force, because that number was derived under different conditions. It's not like that value of force is something that applies in multiple situations. The physics concepts apply, but each analysis will differ in the details.

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studiot,

The topic can't really be described any more clearly .  The following was quoted above and is not my words:

"An approximate minimum cross-sectional area of the femur is ( .0003 sq m). We divide the compressive force by the cross-sectional area to find the compressive stress on the bone. Our approximate value for the ultimate strength of bone that would be required to support 30x body weight was 80 MPa, which is actually less than the measured value of 205 MPa ( compressive fail) , so the claim that the femur can support 30x body weight seems  reasonable."

I used it as a precedent for my ideas.  Bone is listed in other articles at 205MPa as a bald fact about ultimate fail . I'm claiming that the wooden blade lacks the potential to excise bone due to limits on wood thickness which limit MPa due to limit on weight and human arm-speed.

The image of the skull defines what occurred to the bone.

swansont,

The article listed the two velocities of the ball , around 100mph /40m.sec as in the opening post about the blade. Logically, the stopping force against the ball must equal the next force to repel it.  I can't imagine a closer example to the wooden blade and reasonably the times/ weights and velocities allow a general fail / no fail decision.   Again , the baseball forces are only used to demonstrate my ideas and get to measured MPa and how it is handled. The resulting values seem to match reality of stress.

Thanks for the comment "the physics concepts apply" as this is my purpose in posting here.   Is "average force" the correct way to express the strike?  Is the term  "peak force" not needed in calculating ultimate fail or how does it relate to that?

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Posted (edited)
29 minutes ago, nameless said:

studiot,

The topic can't really be described any more clearly .  The following was quoted above and is not my words:

"An approximate minimum cross-sectional area of the femur is ( .0003 sq m). We divide the compressive force by the cross-sectional area to find the compressive stress on the bone. Our approximate value for the ultimate strength of bone that would be required to support 30x body weight was 80 MPa, which is actually less than the measured value of 205 MPa ( compressive fail) , so the claim that the femur can support 30x body weight seems  reasonable."

I used it as a precedent for my ideas.  Bone is listed in other articles at 205MPa as a bald fact about ultimate fail . I'm claiming that the wooden blade lacks the potential to excise bone due to limits on wood thickness which limit MPa due to limit on weight and human arm-speed.

The image of the skull defines what occurred to the bone.

swansont,

The article listed the two velocities of the ball , around 100mph /40m.sec as in the opening post about the blade. Logically, the stopping force against the ball must equal the next force to repel it.  I can't imagine a closer example to the wooden blade and reasonably the times/ weights and velocities allow a general fail / no fail decision.   Again , the baseball forces are only used to demonstrate my ideas and get to measured MPa and how it is handled. The resulting values seem to match reality of stress.

Thanks for the comment "the physics concepts apply" as this is my purpose in posting here.   Is "average force" the correct way to express the strike?  Is the term  "peak force" not needed in calculating ultimate fail or how does it relate to that?

The femur is not part of the skull, and no impact of the type you describe is going to load it axially in compression.

So which bones are you talking about?

The problem is that you keep mixing matters up and It is difficult to impossible to keep track of what is needed or relevent.

What does yield and ultimate fail in battle refer to?

What battle?

Edited by studiot

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23 minutes ago, nameless said:

swansont,

The article listed the two velocities of the ball , around 100mph /40m.sec as in the opening post about the blade. Logically, the stopping force against the ball must equal the next force to repel it.  I can't imagine a closer example to the wooden blade and reasonably the times/ weights and velocities allow a general fail / no fail decision.   Again , the baseball forces are only used to demonstrate my ideas and get to measured MPa and how it is handled. The resulting values seem to match reality of stress.

The forces do not necessarily mean anything in any other context than that of a wooden bat and a ball made of string would around a core and covered in cowhide.

23 minutes ago, nameless said:

Thanks for the comment "the physics concepts apply" as this is my purpose in posting here.   Is "average force" the correct way to express the strike?  Is the term  "peak force" not needed in calculating ultimate fail or how does it relate to that?

It depends on the situation. The peak force might matter, it might not. It depends on the details of how the system fails. Some systems do not fail immediately; for example, a crack may need to propagate through the material, and it takes time to do that, so a large peak force might not cause failure in that instance. But in other circumstances, it might.

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Posted (edited)

studiot

The skull image shows a low-angle entry which resembles axial parallel compression.  Figures on bone vary widely and frontal skull compression fail ranges from 23-70 MPa. This is greater than the mulga perpendicular fail of 8-14MPa which is exceeded by  the possible 63MPa stress , from the baseball analogy. Femur and frontal are both trabecular diploe bone . This author tested pigskull which didn't yield at all and he says animal leg-bone is closer to human skull ..https://www.cambridge.org/core/journals/antiquity/article/death-of-kaakutja-a-case-of-perimortem-weapon-trauma-in-an-aboriginal-man-from-northwestern-new-south-wales-australia/3E957293B27AB3CD1A30FA7F3DB2BC80

studiot , having part of one's head removed by a blade is a "battle".

swansont,

One source of force was the baseball pitcher who could have chucked a coke can , bag of bitcoins or 145g of fake news at max human effort to display that MPa on impact. Can you suggest another way to find whether the blade did the bone surgery?

Edited by nameless

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40 minutes ago, nameless said:

swansont,

One source of force was the baseball pitcher who could have chucked a coke can , bag of bitcoins or 145g of fake news at max human effort to display that MPa on impact. Can you suggest another way to find whether the blade did the bone surgery?

How the baseball was launched is irrelevant. You could have used a trebuchet for all it matters.

A moving ball capable of deformation hitting a moving bat is not the same as a piece of metal hitting a stationary object that does not deform in nearly the same manner as a ball.

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1 hour ago, nameless said:

The skull image shows a low-angle entry which resembles axial parallel compression.﻿

studiot , having part of one's head removed by a blade is a "battle".  ﻿

Fine puncture wounds in the front of the skull may be fatal but in no way can they be described as

So which is it?

Note head removing sharp blades such as massive swords and guillotines operate on quite different mechanics than piercing implements.

Industrial machines, knowns as 'shears' dont even use sharpness, they rely on close fitting massive blocks passing each other very closely.

Incidentally the skull piercing woulds tell you something about the implement used since there are no exit wounds as bullets or spike might create.

But I still don't know what your actual point or question is.

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Posted (edited)

swansont,

A sports record gives an approximate limit of human energy output for each muscular process.   There is a constant ~40m/sec for cricket bowling , baseball pitching and ice-hockey puck hit.  Would you agree that the recorded .0007sec for ball impact is reasonable for striking a skull with layer of skin ? Maybe the ball deforms half-way when stopped,  then full-deformation on return velocity. A person's head likely has similar movement on impact to the half-way deformed  ball. A spherical ball will not give major resistance until the bat is near final depth in contrast to the flatter bone surface.

Reality is that a .4kg  boomerang didn't cause failure in pig-skull or Synbone forensic skull. Thinner than 2mm the wood-edge buckles out to 2mm at impact.

studiot,

What "fine puncture wounds" ?  Isn't a slice of forehead bone a part of your head?  What sort of pub do you patronise? Do customers normally lose sections of skull or was Richard III's skull modification evidence that Yorkist politics involved pikes and swords?

Can you see that I'm questioning the physics of 2mm wood excising a plate of cortical bone?

Edited by nameless

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10 hours ago, nameless said:

swansont,

A sports record gives an approximate limit of human energy output for each muscular process.   There is a constant ~40m/sec for cricket bowling , baseball pitching and ice-hockey puck hit.

Those are all irrelevant, since nobody is chucking anything at 40 m/s in your scenario. The bone is stationary.

10 hours ago, nameless said:

Would you agree that the recorded .0007sec for ball impact is reasonable for striking a skull with layer of skin ?

No, I would not. I would expect the strike time would become shorter for objects less capable of deformation.

10 hours ago, nameless said:

Maybe the ball deforms half-way when stopped,  then full-deformation on return velocity. A person's head likely has similar movement on impact to the half-way deformed  ball. A spherical ball will not give major resistance until the bat is near final depth in contrast to the flatter bone surface.

So you agree that the situations are different (though not to the same degree as I would), and yet are also arguing that they are identical?

Does bone deform and regain its original shape, or does it shatter, in this exercise?

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Posted (edited)

First post " My inquiry is about the ultimate failure of 2mm thick wood blade at maximum human arm-speed of 40metre/sec ".  Aircraft can be tested by windflow against stationary plane. The MPa force would be the same in a moving blade or stationary blade  as with the bat if the same energy system ( time , distance) was used.

The width of the skull trauma is around 3cm (crudely) and cortical plate excision is comparable with baseball deformation distance. The different spherical ball in fact becomes like hemi-sphere after the low-resistance spherical cap absorbs limited force. The ball may then also have about 3cm of distance like the bone chip.   The bone trabeculae which was sliced is marrow and softer than cortical plates and could also resemble the baseball material.  Both have skin / leather.   The bone is neither plastic nor shatters but by the image is cleanly split.

I believe the remaining force in the blade struck his shoulder and sliced his humeral bone , which may be compared with  deformation energy of ball.

A shorter time is more pressure which tends to perpendicular destruction of 2mm wooden blade.

Edited by nameless

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30 minutes ago, nameless said:

You're not suggesting that I think the victim threw himself at the boomerang held rigid by 4 warriors..

So we have had bats, balls, 30 people, various other objects and now boomerangs. What exactly are you trying to discuss?

31 minutes ago, nameless said:

I believe the remaining force in the blade struck his shoulder and sliced his humeral bone , comparable to deformation energy of ball.

You seem to be talking about a specific injury here. Can you provide a reference?

23 hours ago, nameless said:

studiot , having part of one's head removed by a blade is a "battle".

It is an injury, not a battle. It might occur in a battle, an industrial accident or a homicide attempt.

35 minutes ago, nameless said:

A shorter time is more pressure which tends to perpendicular destruction of 2mm wooden blade.﻿

Depends on the type of wood. As well as the shape, how it has been treated, etc.

I don't see how you can possibly get an answer to this without more specific information.

I imagine that the only way to know the strength of such a wooden blade would be to perform a series of rigorous experiments on multiple samples.

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Strange,

All the data is used to find the MPa force on a mulga-blade boomerang when cutting bone. Here's the humeral trauma:

Sep 15, 2016 - trauma was observed on the posterior of the right humeral head (see Figure 8a & b). A circular segment of cortical bone had been entirely removedexposing the trabeculae, in a wound with clearly defined sharp edges, measuring 14mm in diameter.

Oxford dictionary . battle fight ,..join ~ enter into combat .   v. struggle.

First post :
"Parallel ultimate failure is 40-70MPa and by analogy the perpendicular fail is ~8 - 14MPa for hardwood."
That is specific for mulga , density ~1ton cu.m , used to make boomerangs.   From 4 posts back :" Reality is that a .4kg  boomerang didn't cause failure in pig-skull or Synbone forensic skull. Thinner than 2mm the wood-edge buckles out to 2mm at impact. "

The baseball shows the approximate forces in a similar action , blade on bone at max velocity. It is well above the ultimate fail of 2mm boomerang .  Doubling the thickness will double the force-weight and halve the pressure giving the same MPa. And above 2mm the damage is random , unlike the skull image of smooth entry , floor and exit slice . The scoop-shaped cut is unlikely for hardwood compared with flexible steel.

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1 hour ago, nameless said:

All the data is used to find the MPa force on a mulga-blade boomerang when cutting bone. Here's the humeral trauma:

It probably would have been really helpful to provide that context right at the beginning.

Rather than assume people can guess what you are thinking. (Hint: they can't.)

1 hour ago, nameless said:

As I assume you have read that paper (unfortunately, I don't have access) can you quote the data and calculations they use?

And then show why you think they are incorrect?

1 hour ago, nameless said:

Oxford dictionary . battle fight ,..join ~ enter into combat .   v. struggle.   ﻿

Yep. Note that the definition of "battle" is NOT "having part of one's head removed by a blade".