##### Author(s):

Emad Eldieb*

Forensic medicine and ballistics expert, Egypt

**Received:**03 March, 2015;

**Accepted:**17 March, 2015;

**Published:**19 March, 2015

Dr. Emad Eldieb, Forensic medicine and ballistics expert, Egypt, Email:

Eldieb E (2015) Physics of the Permanent Cavity of the Firearm Wound. Peertechz J Forensic Sci Technol 1(1): 001-005. DOI: 10.17352/pjfst.000001

© 2015 Eldieb E. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

Permanent and temporary cavity; Yaw; Drag; Critical energy of penetration

In this paper we would put a mathematic relation between the displacement(x_{2}) of penetration of the lunched bullet inside the victim and the displacement(x1) in air before impact. The cardinal achievement of the article is that we use the physical definition of the inertial own mass of the matter as defined in the paper published by the author in 2014 under the title; physics of the giant atom [5].

From the definition we can state that if a projectile stroke a defined object with a force more than the binding force between its physical units (like macromolecules of the soft tissue) then the localized stroke zone would lose its physical relative inertia of the mother object and would be related to the inertial mass of the striking projectile in the form of conservation of kinetic energy as;

Ep – Eb = (m + δm) v_{2} Where v is the velocity after strike and the subscripts p and b means projectile and binding respectively and where m and δm means mass of the projectile and the incremental mass of the stroked zone respectively.

Since the tissue has some degree of elasticity and consequently each type of tissue has defined tensile strength and defined ultimate elongation, so the dot product of these physical quantities gives the defined toughness energy density of the tissue. The cardinal idea of our work is that: since toughness energy brings each incremental section area (of an elastic rode) in ultimate elongation meaning that offering each section enough energy to break, simultaneously the bullet throughout its pass brings each cross section of the tissue at break.

This means that the energy lost to do cavitations is equivalent to the toughness energy density times the volume of the permanent cavitations. This means that the bullet throughout its pass succeeds to bring the successive section areas of the tissue at break but without elongation. Since we have two unknowns; the stoppage energy of the bullet inside the victim and the displacement before striking the body, so we would divide our work into two indistinctive sections: where in each one we would put an equation to relate each unknown with the other so the two equations would solve the problem and finally we can define the displacement before impact.

### Introduction

The two major forces decelerating velocity of the bullet are gravity and drag.

Gravity is a vector affecting the trajectory of the motion rather than affecting the scalar speed especially on the short range and if the bullet passes horizontally (perpendicular with gravity vector)

On these two conditions drag is the considerable affecting force. Our work studies the effect of drag on the bullet kinetic energy.

### Discussion

##### 1-Defining the impact velocity

The bullet throughout its pass continuously loses kinetic energy to do: break the binding bonds of molecules of air (or the macromolecules of tissue) and to carry and push these broken molecules (or tissue). Since the volume of the bullet have to equal the poured broken volume of the stroked medium so the pushed mass is

Proportional with; the density of the medium σ, the pushed area a and the displaced length [1]

m = c σ a Δx

Where c is a coefficient measuring the efficiency of the bullet to do the action.

m / δt = c σ a v

m δv / δt = − ½ c σ a v^{2} = F_{d}

m v δv / δx = − ½ c σ a v^{2}

∫δv / v = − k ∫ δx where the integration on the left side from the initial velocity vo to the impact velocity v(x1) and the integration on the right side from x = o to x = xo

lnv – lnv_{o} = − k x

v / vo = e^{- kx} (in air (1)

Let us call k_{1} and x1 for fighting in air and let k_{2} and x_{2} for penetration in the victim tissue.

Let v(X1) and v(x2) refer to velocity of impact and the final velocity (= v_{f} = final velocity inside the tissue) respectively.

Vf ÷ {v_{(x1)} – t} = e^{- (k}_{2}^{ x}_{2}^{)} (inside the victim (2)

From (1) and (2)

V_{f} ÷ (v_{o} – t) = e-(k_{1}x_{1} + k_{2}x_{2} (3)

Ln {(v_{o} – t) ÷ v_{f}} = k_{1}x_{1}+ k_{2}x_{2}

Where t is a quantity of energy lost due to a cause other than the pushing effect, and we will study it under studding toughness of the tissue.

##### 2- Defining k

From above then,

k = ½ c σ a / m

c = c_{g} m ÷ d^{2} B.

Where c_{g} is a dimensionless predetermined factor of G1 model used by the manufacture [2], (-for a rounded head bullet like 9mm ×19mm and weighting 124 grain = 8gram- it equals o.517).

d is the diameter of the projectile (caliper of the bullet), a = the cross section area = π (d / 2)^{2} and B is the dimensionless ballistic coefficient.

From above; k = o.2σ / B.

For the same bullet mentioned above, the empirical studies [3] showed that B = o.13 and consequently;

From the above relation, k = 1.54σ (4)

To be adequate with estimated ballistic coefficient we have to use the measurement units in pound and inch. Taking the mean value of the human body density = 1 ∕ 29.6 Ib per inch^{3}. So from above;

K_{1} = 1 ∕ 1.55×10^{4} (where density of air –in room temperature and pressure- equals 4.13×10^{-5} Ib/inch^{3})

K_{2} = 1 ∕ 19.2.

This means that we can determine- by equation (3) - the distance of firing if we determined the final velocity inside the tissue. This is the next discussion.

##### 3- Bullet energy at stoppage inside the victim body

The bullet throughout its pass loses continuously kinetic energy to do evacuation (anti-toughness) and to push these evacuated tissues (conservation of kinetic energy and of momentum)

##### A- Toughness energy

Suppose we had a rode of a matter has some degree of elasticity. The rode is fixed in an end and pulled from the other one in z- direction where there would be coetaneous elongation proportion with the stress – F / A. The total elongation could be expressed as;

Δz = ∑Δ_{i} = nΔ where Δ is the elongation in each cross section area and n is the number of cross section areas. If we increased the stress till elongation at break then this ultimate stress (which in this case is called tensile) times the ultimate elongation is the toughness energy density.

Although elongation at break means cut in one cross section area of the rode yet this doesn't mean that the gained energy was localized only in this cut area. Each of the gained potential energy and the elongation Δz are distributed equivalently among the cross section areas as; ΔE_{i} = ΔE_{1} = ΔE_{2}. Where ∑E_{i} = ΔE.

The cut occurs at the weakest area defined by the rode anisotropy. We have to notice that at ultimate elongation the gained potential energy is wasted in the cut section area to break the binding bonds in z- direction while the equivalent gained potential energies in all the other cross section areas are converted into kinetic energy where rebound and recoil occurs.

Simultaneously the bullet does the same action but with three different points. When the bullet passes in the tissue (or in the air) it breaks the binding bonds along its pass. So,

No elongation in the matter instead there is displacement of the bullet (elongation in the pass v δt) and this is the first different point. No cut in one area instead there is pass in all cross section areas (penetration) which means break down of the binding bonds in all the cross section areas along the pass of the bullet (z- direction) and this is the second different point.

The third different point although is not significant yet it appears from the following discussion;

Suppose that a projectile was penetrating a muscle in z- direction and each cross section of the stroked muscle has a volume equal x y δz. If the projectile had a constant diameter along its length then the gradient penetration of the projectile would lead to break of the bonds among the macromolecules along z- direction only. The number of the cut areas = n = z ÷ δz where z is the depth of the penetrated tissue and δz is the length of the macromolecule.

But the bullet has a variable diameter along its head. If the length of the curve of the bullet head = H and the length (from the base till the tip) of the head = h, then there would be additional cut pieces within xy level = N.

If the number z of broken macromolecules in a tissue area (having the same area of the bullet head base) = z = xy ÷ δx δy where δx δy is the surface area of the macromolecule. Then, the number of the broken macromolecules in the radius of this area = N = (x ÷ δx) (h÷ H)

The total broken macromolecules = z +(x h ÷ δx H)

Since δy is measured in nanometer [2] and y is measured in millimeter so the additional number approaches zero.

The total number = z which means that the broken macromolecules in the level of xy is so small that it is not effective in the number broken in z direction.

This discussion means that the bullet to do permanent cavity loses energy equals tensile strength of the defined tissue times its ultimate elongation times the volume of the cavity.

During autopsy we can collect the toughness energies T of the different types of the tissues weighted by the depth L of the permanent of the cavity such that;

A (T_{1} L_{1} + T_{2} L_{2} + T_{i} L_{i}) = A∑ T_{i} L_{i} (5)

Where; A is the width of the cavity.

The empirical studies [4,5] showed that the human muscle has tensile S = 15 ±3 N/c^{2} and has ultimate elongation U about 150 percent so, the dot product gives toughness T = 22 j /c^{3}. It also showed that rabbit intestine and soft tissue has S = o.o^{3} N/c^{2} and U = 50 percent [6,7].

So, toughness energy of muscles and of skin is hundreds times that of the other tissues so if the bullet stroke a bulk muscle like that of the arm before penetrating the body cavity we can consider only the toughness of the muscle T_{m} and neglect that of the other tissues. In this case we can rewrite the lost kinetic energy to do cavitations as;

E = A T_{m} L_{m}.

Where; L_{m} is the depth of the bulk penetrated muscle.

##### The critical speed

It is the least speed required for the bullet to pierce the victim skin.

Suppose we have an elastic rode of human skin with cut section area A equal one cubic centimeter and a length z equal one centimeter. The energy need for this rode to elongate till cut (ultimate elongation) = (F ÷ A) (Δz ÷ z).

If a bullet with a caliper equal –as example- 9mm lunched towards a victim then the stroked skin area = π r^{2} = π (9÷2)^{2} = o.63 c^{2}

The empirical studies showed that tensile of skin [8] = F ÷ A = 25 N per unit area and its ultimate elongation equal 25 percent hence its toughness energy equal the tensile times the ultimate elongation = (F ÷ A) (Δz ÷ z) = 6 j per cubic centimeter.

The energy need to cut this rode (with diameter = 9mm) or to penetrate it without elongation.

= T = (F ÷A) (Δz ÷ z) A z = 6.25 ×o.63 = 4j = ½mv^{2}.

The empirical studies [9,10] showed that this critical speed (which is termed in the textbooks as the speed of the stoppage power) for the human skin equals 63 meter per second. This empirical magnitude could be concluded as follows;

The reverse of the cut process is to search for the maximum energy need for this defined area (o.63 c^{2}) not to be pierced or by other meaning to vanish Δz.

Then, this maximum energy is equivalent to the least need kinetic energy for the skin to be penetrated = E_{c} = ½ mv_{c}^{2} = {(F ÷A) (Δz ÷ z) (A z)} ÷ Δz = 4×100 / 25 = 16j which is equivalent to the tensile of this defined surface area (o.63c^{2}).

Putting the mass m = 8 gram then v_{c} = 63 m/s.

We might refer here to that the magnitude of critical energy depends on the stroked section area (caliper of the bullet) but not depend on the depth of the tissue because of z in the above brackets vanish each other so no penetration.

##### B-The lost energy to push the evacuated tissues

Let us call the kinetic energy of the bullet at the inlet of the victim body as E_{x1} and at stoppage inside the body as E _{(x1,x2)}

If the target zone – relative to the bullet- has kinetic energy = 0 (at rest) and if it's mass at the level of action of toughness (molecule of air or macromolecule of tissue) = Δm

½ mv_{b}^{2} + 0 = ½ (m + Δm) v_{a}^{2}

Where the subscripts b, and a, refers to before and after strike respectively.

This result is realized by conservation of momentum as follows;

i) In x- component (i direction);

mv_{b} i = mv_{a} i + Δm v_{a} i (where the vector is denoted by the bold letter)

0 = mv_{a} cosθ + Δmv_{a} cosɸ (6)

Where, θ is the angle between x- coordinate and the bullet motion vector while ɸ is the angle between the same coordinate and direction of the motion of Δm after strike.

Δm ∕ m = − cosθ ∕ cosɸ

Since Δm ∕ m = Δ →0

0 = − cos (90 + ɸ) ∕ cosɸ = −sinɸ ∕ cosɸ = tanɸ

ɸ = 0

θ = π ∕ 2

ii-) y- component (j direction)

mv_{b} j = mv_{a} j + Δmv_{a} j

mv_{b} sin90 = mv_{a} sinθ + Δmv_{a} sinɸ

mv_{b} = mv_{a} cosɸ+ Δmv_{a}sinɸ (7)

From (6) the x- component is in the form;

0 = – mv_{a} sinΔ + Δm cosɸ (8)

Squaring (7) and (8) then adding to each other → m^{2}vb^{2} = va^{2} (m^{2} + Δm^{2})

= v_{a}^{2} (m + Δm) (m – Δm)

Multiplying the left side by 1 = (m ∕ m) = {(m – Δm) ∕ m} gives;

½ mv_{b}^{2} = ½ (m + Δm) v_{a}^{2}

This form realizes conservation of kinetic energy in each cross section with infinitesimal length and infinitesimal mass along the pass of the bullet. Allover the complete pass the conservation is realized on the form of;

½ mv_{b}^{2} = ½ (m + ∑Δm) v_{a}^{2}

Now we will try to estimate the total pushed mass ∑ Δm. It is simply the mean density of the human body tissue times the cross section area of the bullet times the depth of penetration. But we have to introduce the effect of Yaw. For simplification and as approximation let us study the effect of Yaw on the round head, 9mm×1.9mm, bullet (which is characterized by the comparatively small effect of Yaw [9]. The empirical studies [3] showed that Yaw appears effectively in the middle third of the penetration depth, so if we supposed –as example- that the depth equals 51c then we can consider that Yaw appears effectively at the second 17c. The deviation angle of Yaw appears at entrance as 1degree where it increases gradually and dramatically especially at the second third. In the first third the angle begins at 1 degree and increases till it reach to about 15 degree at the end of the first third. In the second third it increases gradually till it reaches at the end of this third to 180 degree. The last third it return the phase of the first one. Now we have to define the pushing face of the bullet. It is simply equals the base of the bullet =A = {π (o.9 ÷ 2)^{2}} = o.63 c^{2}. But Yaw affects greatly the pushing surface as follows;

In the middle third; if sin90 and sin 15 are the maximum and minimum sin respectively then;

A1 = π R l {(sin90 + sin15) ÷ 2} where R = r ÷ 1.o5 = o.45 ÷ 1.o5 = 4.2c, such that the factor 1.o5 is to compensate for the gradient decrease in the radii of the bullet head.

Putting l = 1.9, then;

A_{1} = 2.5× o.625 = 1.56c^{2}. Now in this third we have two pushing faces as;

A_{t} = A_{1} + A = 1.56 + o.63 = 2.2 c^{2}. Then; The pushed volume of the tissue in the middle third of the pass = v_{1} = 2.2 × 17 = 38c^{3}. In the other two thirds;

A_{1} = 2.5 sin7.5 = o.4 c^{2}

A_{t} = o.4 + o.63 = 1c^{2}. Then the pushed volume in these two thirds = v_{2} = 1×34 = 34c^{3}. The total volume = v = v_{1} + v_{2} = 73c^{3}.

Since the mean density of the missile (the led and the copper cover) of the studied bullet = 10 times the mean density of the human body then; m + Δm = b = 7.3 + 1 = 8.3 times the mass of the missile.

The effect of Yaw on the energy lost due to toughness if the depth of the stroke muscle = 8 c and where the angle reaches within this depth to 10 degree is as follows;

A = o.63 + (2.5 sin5) = o.82 c^{2}. Then;

T = 22×o.82×8 = 142 j.

##### Now we can define the required function as

If a projectile stroke an object with a force more than the inter macromolecules binding force then the projectile would loss kinetic energy equal the binding energy of the penetrated zone as discussed under the toughness energy T. The remainder kinetic energy R of the bullet would gradually reduced due to the load of the pushed evacuated mass as discussed under conservation of kinetic energy .So; if the energy lost to destroy the binding energy allover the volume = T and if

E(x_{1}) = E_{1} then; E_{1}– T= R and the final kinetic energy E_{(x1,x2)} = (R ÷ b) where (m + ∑ Δm) ÷ m = b_{(x2)} = b

Let the kinetic energy of the bullet at the inlet = E_{(x1)} = E1and the remainder kinetic energy of the bullet after its stoppage inside the victim body = E_{(x1,x2)} = E_{2}

Then, (E_{1} – T) ÷ b = R÷ b = E_{2} (9)

From equation (1);

E_{1} = E_{o} e–2^{k}_{1}^{x}_{1} (10)

From (9) and (10);

E_{2} = (E_{o}÷ be^{2k}_{1}^{x}_{1}) – (T ÷ b) (11)

From (2);

E_{2} = (E_{1}– T) ÷ e^{2k}_{2}^{x}_{2} (12)

From (12) and (10);

E_{2} = (E_{o} ÷ e^{2k}_{1}^{x}_{1}+ 2^{k}_{2}^{x}_{2}) – (T ÷ e^{2k}_{2}^{x}_{2}) (13)

From (11) and (13);

E_{2} = (E_{o} ÷ e^{2k}_{1}^{x}_{1}+ ^{2k}_{2}^{x}_{2}) – (T ÷ e^{2k}_{2}^{x}_{2}) = (E_{o}÷ be^{2k}_{1}^{x}_{1}) – (T ÷ b) (14)

By dividing on T;

E_{2} = (E_{o} ÷ T e^{2k}_{1}^{x}_{1}+ ^{2k}_{2}^{x}_{2}) – (1 ÷ e^{2k}_{2}^{x}_{2}) = (E_{o}÷ T be^{2k}_{1}^{x}_{1}) – (1 ÷ b) (15)

Where E_{o} ÷ T is a ratio of energies so it is dimensionless and b is ratio of masses so it is also dimensionless. Each of x_{1} and x_{2} is in inches to compensate with k which depends on the ballistic coefficient whose magnitudes calculated using the inch.

Equation (14) gives the following trivial solution;

b = e^{2k}_{2}^{x}_{2} (16)

Actually this result gives the mathematic meaning of the mass (and the volume) of the destroyed tissue and gives –also- the physical meaning of Yaw and its effect on the destroyed tissue. The volume V of the broken tissue of the permanent cavity could be estimated directly as; V = (b – 1) ÷ σ where σ is the mean density of the human body.

### A Case Study

A bullet stoke a victim where the missile entered from the bulk muscle of the arm to outer from the interior side of the same arm and then to enter again to the chest without strike the bone where it passed obliquely after penetrating the lungs to pass though the liver and finally to be impacted behind the liver in front of the muscle of the back with a total penetrating depth = 51 c. We measured the depth of the muscle of the arm and chest where it was about 8 c. After autopsy we examined the missile where it had 9mm caliber, 1.9mm length and mass = 8 gram.

Since the energy lost due to toughness depends mainly on toughness of the muscle so, toughness equals toughness density times the cross section area of the bullet time the penetrated depth of the muscle then;

T = 142 j.

Putting;

b = 8.3 (as estimated in the same studied case), and if the muzzle energy [3] E_{o} = 435 joule, then We can define E_{2} as follows;

Since The critical kinetic energy of the bullet to pass the muscle of the back = E_{c} =

T_{m} A = 17 ×0.82 = 13.9 j (where tensile of the muscle = 15±3 N.)

Since E_{2} have to be equal or less than the critical energy then matching in equation (13) gives x1 equal at least 100m.

We suggested that the shooter aimed the muzzle barrel towards the head of the victim to compensate for the drop trajectory of the bullet. Since the 9mm bullet drop after 100m about half a meter so the bullet stroked the arm of the victim.

Checking the result of equation (15);

Put x_{2} = 51c in the exponential; e^{2k}_{2}^{x}_{2} = 8.3 (17)

Comparing the above magnitude with that of b then,

e^{2k}_{2}^{x}_{2} = b. So we can define b –directly- from x_{2}.

Actually the empirical studies [9] on the bullet of our case showed that it loses about 1.3 of its initial speed after displacing 100 m which is fit with our calculus. Also many empirical(12) studies showed that the same bullet (9mm) at speed of entrance on the range of 255 m/s (like our case) can penetrate the body victim (or the simultaneous prepared gelatin) for about 51c which is –also- fit with our results.

### Conclusion

The paper suggests a new theory to explain the reduced velocity of the bullet versus displacement graphs. The theory depends on dealing the medium (molecules of air or macromolecules of the tissue) in which the bullet passes as a matter with some degree of elasticity having tensile (ultimate stress). Permanent cavitations –on this view- means break down (of the inter-particles binding bonds) but without elongation. The bullet to break and overcome the binding forces loses energy. This provides the first term of the lost energy E1 and gives the physical interpretation of the permanent cavity. The second term E2 arises from that these broken particles are load, carried and pushed by the bullet into it's besides throughout its pass. Since the bullet kinetic energy and its momentum before and after loading have to be conserved so this provides the second term of the lost energy and gives the mathematic interpretation of the temporary cavitations through the effect of the broken pushed tissue into the besides of the bullet pass.

If the initial kinetic energy of the bullet (at x = 0) = E_{o} then, the remainder energy

E_{R} = E_{o} – (E_{1} + E_{2}). Consequently vR is the velocity at stoppage inside the victim body. The paper succeeded to give the physical meaning of the critical speed vc of the bullet to pierce the outer surface of the victim body and its relation with vR as; velocity of stoppage inside the victim body is equal or less than the critical speed. The forensic medicine expert is advised during autopsy to measure the site of the impacted bullet (to determine Ec), the depth of penetration along the bullet pass inside the dead body (to determine from equation 16 the magnitude of b) and -also- is advised to stress on the depth of the penetrated muscles and skin (to define E_{2} and x_{1}). We studied a case and solved it mathematically and compared the results with the empirical studies where we found it fit. The author hopes scientific cooperation in this developing field.

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