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This analysis correctly predicts ranking of bullet stopping power for the eight alloys and one polymer we have tested.

Equilibrium for a membrane-like ring element centered on the impact point

f=2*p*r*r*t*SIN(q)

solving for the elevation angle of the ring

q=ASIN(f/(2*p*r*r*t))

radial strain, assuming zero hoop strain and neglecting bending

e=1/COS(q)-1

solving the strain equation for tilt angle

q=p/2-ASIN(1/(e+1))

q=ASIN(1/(e+1))-p/2

substitute the strain version of the tilt angle into the equilibrium equation

f=2*p*r*r*t*SIN(p/2-ASIN(1/(e+1)))

simplify

f=2*p*o(e)*r*r*t*o(e+2)/(e+1)

solve explicitly for strain

e=(2*p*r*r*t-SQRT(4*p^2*r^2*r^2*t^2-f^2))/SQRT(4*p^2*r^2*sigma^2*t^2-f^2)

e=-(SQRT(4*p^2*r^2*r^2*t^2-f^2)+2*p*r*r*t)/SQRT(4*p^2*r^2*sigma^2*t^2-f^2)

rearrange

e=2*p*r*t*r/SQRT(4*p^2*r^2*t^2*r^2-f^2)-1

substituting rigid/linear strain-hardening function of e for s

e=2*p*r*t*(e/emax*(su-sy)+sy)/SQRT(4*p^2*r^2*t^2*(e/emax*(su-sy)+sy)^2-f^2)-1

substituting same stress relation for s in equilibrium equation

f=2*p*SQRT(e)*r*(e/emax*(su-sy)+sy)*t*SQRT(e+2)/(e+1)

rearranging

f=2*p*SQRT(e)*r*t*SQRT(e+2)*(e*(su-sy)+sy*emax)/(emax*(e+1))

f*emax*(e+1)=2*p*SQRT(e)*r*t*SQRT(e+2)*(e*(su-sy)+sy*emax)

f^2*emax^2*(e+1)^2=4*p^2*e*r^2*t^2*(e+2)*(e*(su-sy)+sy*emax)^2

trying to get explicit solution for strain to integrate for energy

This expression is quartic in e, radicals would be hard to integrate--give up.

f^2*emax^2*(e+1)^2-4*p^2*e*r^2*t^2*(e+2)*(e*(su-sy)+sy*emax)^2=0

instead, solve for radius as a function of strain and try a change of variable

r=f*emax*(e+1)/(2*p*SQRT(e)*t*SQRT(e+2)*(e*(su-sy)+sy*emax))

DIF(r=f*emax*(e+1)/(2*p*SQRT(e)*t*SQRT(e+2)*(e*(su-sy)+sy*emax)),e)

dr/de=-f*emax*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*p*t*e^(3/2)*(e+2)^(3/2)*(e*(su-sy)+sy*emax)^2)

Assembling the integrand:

dr/de*2*p*t*sav*r*de

(-f*emax*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*p*t*e^(3/2)*(e+2)^(3/2)*(e*(su-sy)+sy*emax)^2))*p*t*r*e*(e/emax*(su-sy)+2*sy)

rearranging

-f*r*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*SQRT(e)*(e+2)^(3/2)*(e*(su-sy)+sy*emax)^2)

the maximum force is assumed to be that required to break in tension at twice the original bullet diameter

fmax=2*p*SQRT(emax)*db*su*t*SQRT(emax+2)/(emax+1)

Substitutions for force and radius

-2*p*SQRT(emax)*db*su*t*SQRT(emax+2)/(emax+1)*(f*emax*(e+1)/(2*p*SQRT(e)*t*SQRT(e+2)*(e*(su-sy)+sy*emax)))*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+~

3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*SQRT(e)*(e+2)^(3/2)*(e*(su-sy)+sy*emax)^2)

simplify

-db*f*su*emax^(3/2)*(e+1)*SQRT(emax+2)*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*e*(e+2)^2*~

(emax+1)*(e*(su-sy)+sy*emax)^3)

one last substitution for force

-db*(2*p*SQRT(emax)*db*su*t*SQRT(emax+2)/(emax+1))*su*emax^(3/2)*(e+1)*SQRT(emax+2)*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+~

3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(2*e*(e+2)^2*(emax+1)*(e*(su-sy)+sy*emax)^3)

simplify

-p*db^2*t*su^2*emax^2*(e+1)*(emax+2)*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(e*(e+2)^2*(emax+1)^2*~

(e*(su-sy)+sy*emax)^3)

 

INT(-p*db^2*t*su^2*emax^2*(e+1)*(emax+2)*(e*(su-sy)+2*sy*emax)*(e^3*(su-sy)+3*e^2*(su-sy)+3*e*(su-sy)+sy*emax)/(e*(e+2)^2*~

(emax+1)^2*(e*(su-sy)+sy*emax)^3),e,emax,emin)

 

p*db^2*t*su^2*emax*(su-sy*(emax+1))^2*(emax+2)*LN(su*emax)/(sy*(sy-su)*(emax+~

1)^2*(2*su-sy*(emax+2)))+p*db^2*t*su^2*emax*(su-sy*(emax+1))^2*(emax+2)*LN(su~

*emin+sy*(emax-emin))/(sy*(su-sy)*(emax+1)^2*(2*su-sy*(emax+2)))+p*db^2*t*su^~

2*emax^2*(emax+2)*LN(SQRT(emax+2))/((emax+1)^2*(2*su-sy*(emax+2)))+p*db^2*t*s~

u^2*emax*(emax+2)*LN(SQRT(emax))/(sy*(emax+1)^2)+p*db^2*t*su^2*emax^2*(emax+2~

)*LN(SQRT(emin+2))/((emax+1)^2*(sy*(emax+2)-2*su))-p*db^2*t*su^2*emax*(emax+2~

)*LN(SQRT(emin))/(sy*(emax+1)^2)+p*db^2*t*emax*(su-sy*(emax+1))*(emax-emin)*(~

su^2*emin*(emax+emin+2)+su*sy*emax*(emax+emin+2)*(emin+2)+sy^2*(emax^2+emax*(1~

-emin)-emin)*(emin+2))/(2*(emax+1)^2*(emin+2)*(su*emin+sy*(emax-emin))^2)

 

E(db,t,su,sy,emax,emin):=p*db^2*t*su^2*emax*(su-sy*(emax+1))^2*(emax+2)*LN(su~

*emax)/(sy*(sy-su)*(emax+1)^2*(2*su-sy*(emax+2)))+p*db^2*t*su^2*emax*(su-sy*(~

emax+1))^2*(emax+2)*LN(su*emin+sy*(emax-emin))/(sy*(su-sy)*(emax+1)^2*(2*su-sy~

*(emax+2)))+p*db^2*t*su^2*emax^2*(emax+2)*LN(SQRT(emax+2))/((emax+1)^2*(2*su-~

sy*(emax+2)))+p*db^2*t*su^2*emax*(emax+2)*LN(SQRT(emax))/(sy*(emax+1)^2)+p*d~

b^2*t*su^2*emax^2*(emax+2)*LN(SQRT(emin+2))/((emax+1)^2*(sy*(emax+2)-2*su))-p~

*db^2*t*su^2*emax*(emax+2)*LN(SQRT(emin))/(sy*(emax+1)^2)+p*db^2*t*emax*(su-s~

y*(emax+1))*(emax-emin)*(su^2*emin*(emax+emin+2)+su*sy*emax*(emax+emin+2)*(emi~

n+2)+sy^2*(emax^2+emax*(1-emin)-emin)*(emin+2))/(2*(emax+1)^2*(emin+2)*(su*emi~

n+sy*(emax-emin))^2)

(This expression is evaluated in less than a second on my copy of the computer algebra system, Derive.)

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