Solving Fortran Runtime Error 112 - Mach Number Calculation

[Hejmore]

Hi, I try to write program to calculating mach number, with using bisection method. When I run program, fortran write to me an error in line 40. Can you help? I tried to calculating function by using wolfram sucesfully but I need to run it on fortran, An equation is 1.7795 - (0.334898 (1 + 0.2 x^2)^6.)/x^2. Can you help me? Thx

My error code:

program mach
     real pp,z1,z2,sucin,a,b,x,machx,a1

      pp=1
      a1=0.1
      b=3

      z1=pp**2-(0.334898*(1+0.2*a1**2)**6)/a1**2
      z2=pp**2-(0.334898*(1+0.2*b**2)**6)/b**2

      sucin=z1*z2

      if(sucin.lt.0) go to 1

      if(sucin.gt.0) go to 3

    1 continue

      do
          a=a1
          a1=(b+a)/2

          z1=pp**2-(0.334898*(1+0.2*a1**2)**6)/a1**2
          z2=pp**2-(0.334898*(1+0.2*b**2)**6)/b**2

          sucin=z1*z2
          machx=a
          x=abs(b-a)

          if(x.lt.0.0001) exit

          if(sucin.lt.0) cycle

          if(sucin.gt.0) go to 2
      end do

    2 continue

      do
          b=a1
          a1=(b+a)/2

          z1=pp**2-(0.334898*(1+0.2*a**2)**6)/a**2
          z2=pp**2-(0.334898*(1+0.2*a1**2)**6)/a1**2

          sucin=z1*z2
          machx=b
          x=abs((b-a))

          if(x.lt.0.0001) exit

          if(sucin.lt.0) cycle

          if(sucin.gt.0) go to 1
      end do

    3 continue

      print*,'Mach number is'
      write(5,*) machx

      stop

end program mach

 

[Hejmore]

Problem is solved.

 

[jedishrfu]

That's great.

What was the actual error and how did you solve it?

 

[Hejmore]

You can see, the condition if( sucin.gt.0)go to 3 in line 10 is wrong. If I use only bisection method, the program will not work correctly, because i must to find extreme of the function and then must to choose right interval. I had to use a next numerical method to find extrem and then I could solve this problem. I would be glad if you find some more elegant solution for solving this function. Thx