;
;  This procedure calculates the start time of a CME from coefficients
; 	generated by POLYFITW or some other polynomial fitting routine.
;	coeffs are the returned coefficients from the fitting routine, 
;
;
FUNCTION df_starttimes,coeffs
  
  ; Get coeffs for cubic polynomial
      c1=coeffs(0)  ; constant term
      c2=coeffs(1)  ; linear term
      c3=coeffs(2)  ; quadratic term
      c4=coeffs(3)  ; cubic term
  
  starttimes=fltarr(2)

  ht=[c1-1,c2,c3,c4]   ; Need to set h(t)=1 and then find the roots.
  dhdt=[c2,2*c3,3*c4]  ; first derivative of h(t)
  d2hdt2=[2*c3,6*c4]   ; second derivative of h(t)

  ;
  ; Now, lets get the roots we need.  We need to know where h(t)=1 and
  ;  where dh/dt=0 and where second derivative, d2hdt2=0.  This will tell
  ;  us what we have to do.

  ; Here are the solutions to h(t)=1
  roots=FZ_ROOTS(ht)
  ht_roots=FLOAT(roots)	; Get the solutions from h(t)=1
  tmp=SIZE(ht_roots)
  num_roots_ht=tmp(1)   ; here are the number of solutions to h(t)=1

  ; Here is where dh/dt=0
  roots=FZ_ROOTS(dhdt)
  dhdt_roots=FLOAT(roots) ; Get the roots of dh/dt=0
  tmp=SIZE(dhdt_roots)
  num_roots_dhdt=tmp(1)	; Get the number of real roots from dh/dt

  ; Here is where d2hdt2=0
  roots=FZ_ROOTS(d2hdt2)
  d2hdt2_roots=FLOAT(roots)  ; get the roots of second derivative
  tmp=SIZE(d2hdt2_roots)
  num_roots_d2hdt2=tmp(1)    ; number of roots from d2h/dt2

  ;
  ;--- Start times for a Cubic Polynomial
  ;
  ; Start with the case that c4 > 0, rate of acceleration is increasing
  ;  There are three possibilities:
  ;	1.  dh/dt has no roots.
  ;	2.  dh/dt has one root.
  ;     3.  dh/dt has two roots.
  ;
  ; We also want to look at the possibility that h(t) may have 1 solution
  ;  at r=1 or it may have 3 solutions at r=1.  For 1 solution, take it, for
  ;  3 solutions, take the latest one.
  ;
  IF (c4 GT 0) THEN BEGIN  ;{  Begin case 1
  IF (num_roots_dhdt EQ 0) THEN BEGIN
    starttimes[0]=ht_roots[0]     ; one time is where h(t)=1
    starttimes[1]=d2hdt2_roots[0] ; the other is where d2h/dt2=0
    RETURN,starttimes
  ENDIF

  IF (num_roots_dhdt EQ 1) THEN BEGIN
    starttimes[0]=d2hdt2_roots[0]  ;one time is where second derivative=0
    starttimes[1]=ht_roots[0]  ; The other is where h(t)=1
    RETURN,starttimes
  ENDIF

  IF (num_roots_dhdt EQ 2) THEN BEGIN
    starttimes[0]=(dhdt_roots(0) > dhdt_roots(1))  ;take largest root of dh/dt 
    IF (num_roots_ht EQ 1) THEN $
	starttimes[1]=ht_roots[0]  ;take the only root
    IF (num_roots_ht EQ 3) THEN $
	starttimes[1]=ht_roots[2]  ;take the latest one
    RETURN,starttimes
  ENDIF
  ENDIF  ;}  end case 1

  ;
  ;  Now, if c4 < 0 there is only one physically meaningful situation, that
  ;    of dh/dt having two roots.  Anythings else (no roots or one root), 
  ;    means that the velocity of the CME never became positive.
  ;
  IF (c4 LT 0) THEN BEGIN   ;{ begin case 2
  IF (num_roots_dhdt EQ 2) THEN BEGIN
    ;IF (num_roots_ht EQ 1) THEN starttimes[0]=d2hdt2[0] ;pick earliest time
    ;IF (num_roots_ht EQ 3) THEN starttimes[0]=d2hdt2[1] ;pick middle time
    IF (num_roots_ht EQ 1) THEN starttimes[0]=dhdt_roots[0] ;pick v=0 starttime
    IF (num_roots_ht EQ 3) THEN starttimes[0]=ht_roots[1] ;pick middle time
    starttimes[1]=-1
    RETURN,starttimes
  ENDIF  ELSE BEGIN
    print,'There are not two roots here, not physically meaningful."
    RETURN,starttimes=[-1,-1]
  ENDELSE
  ENDIF  ;}  end case 2

END