Distance on a great circle

Average Accuracy: 62.7% → 80.2%

Time: 2.2min

Precision: binary64

Cost: 164480

Math

\[R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)}}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

↓

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_0 \cdot t_0\right) + t_1}}{\sqrt{\log \left(e^{1 + \left(t_2 \cdot \left(-0.5 + 0.5 \cdot \left(\sin \lambda_1 \cdot \sin \lambda_2 + \cos \lambda_1 \cdot \cos \lambda_2\right)\right) - t_1\right)}\right)}}\right) \end{array} \]

FPCore

(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (*
  R
  (*
   2.0
   (atan2
    (sqrt
     (+
      (pow (sin (/ (- phi1 phi2) 2.0)) 2.0)
      (*
       (* (* (cos phi1) (cos phi2)) (sin (/ (- lambda1 lambda2) 2.0)))
       (sin (/ (- lambda1 lambda2) 2.0)))))
    (sqrt
     (-
      1.0
      (+
       (pow (sin (/ (- phi1 phi2) 2.0)) 2.0)
       (*
        (* (* (cos phi1) (cos phi2)) (sin (/ (- lambda1 lambda2) 2.0)))
        (sin (/ (- lambda1 lambda2) 2.0))))))))))

↓

(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (let* ((t_0 (sin (/ (- lambda1 lambda2) 2.0)))
        (t_1
         (pow
          (-
           (* (cos (* 0.5 phi2)) (sin (* phi1 0.5)))
           (* (sin (* 0.5 phi2)) (cos (* phi1 0.5))))
          2.0))
        (t_2 (* (cos phi1) (cos phi2))))
   (*
    R
    (*
     2.0
     (atan2
      (sqrt (+ (* t_2 (* t_0 t_0)) t_1))
      (sqrt
       (log
        (exp
         (+
          1.0
          (-
           (*
            t_2
            (+
             -0.5
             (*
              0.5
              (+
               (* (sin lambda1) (sin lambda2))
               (* (cos lambda1) (cos lambda2))))))
           t_1))))))))))

C

double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	return R * (2.0 * atan2(sqrt((pow(sin(((phi1 - phi2) / 2.0)), 2.0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0))) * sin(((lambda1 - lambda2) / 2.0))))), sqrt((1.0 - (pow(sin(((phi1 - phi2) / 2.0)), 2.0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0))) * sin(((lambda1 - lambda2) / 2.0))))))));
}

↓

double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = sin(((lambda1 - lambda2) / 2.0));
	double t_1 = pow(((cos((0.5 * phi2)) * sin((phi1 * 0.5))) - (sin((0.5 * phi2)) * cos((phi1 * 0.5)))), 2.0);
	double t_2 = cos(phi1) * cos(phi2);
	return R * (2.0 * atan2(sqrt(((t_2 * (t_0 * t_0)) + t_1)), sqrt(log(exp((1.0 + ((t_2 * (-0.5 + (0.5 * ((sin(lambda1) * sin(lambda2)) + (cos(lambda1) * cos(lambda2)))))) - t_1)))))));
}

Fortran

real(8) function code(r, lambda1, lambda2, phi1, phi2)
    real(8), intent (in) :: r
    real(8), intent (in) :: lambda1
    real(8), intent (in) :: lambda2
    real(8), intent (in) :: phi1
    real(8), intent (in) :: phi2
    code = r * (2.0d0 * atan2(sqrt(((sin(((phi1 - phi2) / 2.0d0)) ** 2.0d0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0d0))) * sin(((lambda1 - lambda2) / 2.0d0))))), sqrt((1.0d0 - ((sin(((phi1 - phi2) / 2.0d0)) ** 2.0d0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0d0))) * sin(((lambda1 - lambda2) / 2.0d0))))))))
end function

↓

real(8) function code(r, lambda1, lambda2, phi1, phi2)
    real(8), intent (in) :: r
    real(8), intent (in) :: lambda1
    real(8), intent (in) :: lambda2
    real(8), intent (in) :: phi1
    real(8), intent (in) :: phi2
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: t_2
    t_0 = sin(((lambda1 - lambda2) / 2.0d0))
    t_1 = ((cos((0.5d0 * phi2)) * sin((phi1 * 0.5d0))) - (sin((0.5d0 * phi2)) * cos((phi1 * 0.5d0)))) ** 2.0d0
    t_2 = cos(phi1) * cos(phi2)
    code = r * (2.0d0 * atan2(sqrt(((t_2 * (t_0 * t_0)) + t_1)), sqrt(log(exp((1.0d0 + ((t_2 * ((-0.5d0) + (0.5d0 * ((sin(lambda1) * sin(lambda2)) + (cos(lambda1) * cos(lambda2)))))) - t_1)))))))
end function

Java

public static double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	return R * (2.0 * Math.atan2(Math.sqrt((Math.pow(Math.sin(((phi1 - phi2) / 2.0)), 2.0) + (((Math.cos(phi1) * Math.cos(phi2)) * Math.sin(((lambda1 - lambda2) / 2.0))) * Math.sin(((lambda1 - lambda2) / 2.0))))), Math.sqrt((1.0 - (Math.pow(Math.sin(((phi1 - phi2) / 2.0)), 2.0) + (((Math.cos(phi1) * Math.cos(phi2)) * Math.sin(((lambda1 - lambda2) / 2.0))) * Math.sin(((lambda1 - lambda2) / 2.0))))))));
}

↓

public static double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = Math.sin(((lambda1 - lambda2) / 2.0));
	double t_1 = Math.pow(((Math.cos((0.5 * phi2)) * Math.sin((phi1 * 0.5))) - (Math.sin((0.5 * phi2)) * Math.cos((phi1 * 0.5)))), 2.0);
	double t_2 = Math.cos(phi1) * Math.cos(phi2);
	return R * (2.0 * Math.atan2(Math.sqrt(((t_2 * (t_0 * t_0)) + t_1)), Math.sqrt(Math.log(Math.exp((1.0 + ((t_2 * (-0.5 + (0.5 * ((Math.sin(lambda1) * Math.sin(lambda2)) + (Math.cos(lambda1) * Math.cos(lambda2)))))) - t_1)))))));
}

Python

def code(R, lambda1, lambda2, phi1, phi2):
	return R * (2.0 * math.atan2(math.sqrt((math.pow(math.sin(((phi1 - phi2) / 2.0)), 2.0) + (((math.cos(phi1) * math.cos(phi2)) * math.sin(((lambda1 - lambda2) / 2.0))) * math.sin(((lambda1 - lambda2) / 2.0))))), math.sqrt((1.0 - (math.pow(math.sin(((phi1 - phi2) / 2.0)), 2.0) + (((math.cos(phi1) * math.cos(phi2)) * math.sin(((lambda1 - lambda2) / 2.0))) * math.sin(((lambda1 - lambda2) / 2.0))))))))

↓

def code(R, lambda1, lambda2, phi1, phi2):
	t_0 = math.sin(((lambda1 - lambda2) / 2.0))
	t_1 = math.pow(((math.cos((0.5 * phi2)) * math.sin((phi1 * 0.5))) - (math.sin((0.5 * phi2)) * math.cos((phi1 * 0.5)))), 2.0)
	t_2 = math.cos(phi1) * math.cos(phi2)
	return R * (2.0 * math.atan2(math.sqrt(((t_2 * (t_0 * t_0)) + t_1)), math.sqrt(math.log(math.exp((1.0 + ((t_2 * (-0.5 + (0.5 * ((math.sin(lambda1) * math.sin(lambda2)) + (math.cos(lambda1) * math.cos(lambda2)))))) - t_1)))))))

Julia

function code(R, lambda1, lambda2, phi1, phi2)
	return Float64(R * Float64(2.0 * atan(sqrt(Float64((sin(Float64(Float64(phi1 - phi2) / 2.0)) ^ 2.0) + Float64(Float64(Float64(cos(phi1) * cos(phi2)) * sin(Float64(Float64(lambda1 - lambda2) / 2.0))) * sin(Float64(Float64(lambda1 - lambda2) / 2.0))))), sqrt(Float64(1.0 - Float64((sin(Float64(Float64(phi1 - phi2) / 2.0)) ^ 2.0) + Float64(Float64(Float64(cos(phi1) * cos(phi2)) * sin(Float64(Float64(lambda1 - lambda2) / 2.0))) * sin(Float64(Float64(lambda1 - lambda2) / 2.0)))))))))
end

↓

function code(R, lambda1, lambda2, phi1, phi2)
	t_0 = sin(Float64(Float64(lambda1 - lambda2) / 2.0))
	t_1 = Float64(Float64(cos(Float64(0.5 * phi2)) * sin(Float64(phi1 * 0.5))) - Float64(sin(Float64(0.5 * phi2)) * cos(Float64(phi1 * 0.5)))) ^ 2.0
	t_2 = Float64(cos(phi1) * cos(phi2))
	return Float64(R * Float64(2.0 * atan(sqrt(Float64(Float64(t_2 * Float64(t_0 * t_0)) + t_1)), sqrt(log(exp(Float64(1.0 + Float64(Float64(t_2 * Float64(-0.5 + Float64(0.5 * Float64(Float64(sin(lambda1) * sin(lambda2)) + Float64(cos(lambda1) * cos(lambda2)))))) - t_1))))))))
end

MATLAB

function tmp = code(R, lambda1, lambda2, phi1, phi2)
	tmp = R * (2.0 * atan2(sqrt(((sin(((phi1 - phi2) / 2.0)) ^ 2.0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0))) * sin(((lambda1 - lambda2) / 2.0))))), sqrt((1.0 - ((sin(((phi1 - phi2) / 2.0)) ^ 2.0) + (((cos(phi1) * cos(phi2)) * sin(((lambda1 - lambda2) / 2.0))) * sin(((lambda1 - lambda2) / 2.0))))))));
end

↓

function tmp = code(R, lambda1, lambda2, phi1, phi2)
	t_0 = sin(((lambda1 - lambda2) / 2.0));
	t_1 = ((cos((0.5 * phi2)) * sin((phi1 * 0.5))) - (sin((0.5 * phi2)) * cos((phi1 * 0.5)))) ^ 2.0;
	t_2 = cos(phi1) * cos(phi2);
	tmp = R * (2.0 * atan2(sqrt(((t_2 * (t_0 * t_0)) + t_1)), sqrt(log(exp((1.0 + ((t_2 * (-0.5 + (0.5 * ((sin(lambda1) * sin(lambda2)) + (cos(lambda1) * cos(lambda2)))))) - t_1)))))));
end

Wolfram

code[R_, lambda1_, lambda2_, phi1_, phi2_] := N[(R * N[(2.0 * N[ArcTan[N[Sqrt[N[(N[Power[N[Sin[N[(N[(phi1 - phi2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision] + N[(N[(N[(N[Cos[phi1], $MachinePrecision] * N[Cos[phi2], $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(lambda1 - lambda2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(lambda1 - lambda2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[(1.0 - N[(N[Power[N[Sin[N[(N[(phi1 - phi2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision] + N[(N[(N[(N[Cos[phi1], $MachinePrecision] * N[Cos[phi2], $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(lambda1 - lambda2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(lambda1 - lambda2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[R_, lambda1_, lambda2_, phi1_, phi2_] := Block[{t$95$0 = N[Sin[N[(N[(lambda1 - lambda2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[Power[N[(N[(N[Cos[N[(0.5 * phi2), $MachinePrecision]], $MachinePrecision] * N[Sin[N[(phi1 * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] - N[(N[Sin[N[(0.5 * phi2), $MachinePrecision]], $MachinePrecision] * N[Cos[N[(phi1 * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$2 = N[(N[Cos[phi1], $MachinePrecision] * N[Cos[phi2], $MachinePrecision]), $MachinePrecision]}, N[(R * N[(2.0 * N[ArcTan[N[Sqrt[N[(N[(t$95$2 * N[(t$95$0 * t$95$0), $MachinePrecision]), $MachinePrecision] + t$95$1), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[Log[N[Exp[N[(1.0 + N[(N[(t$95$2 * N[(-0.5 + N[(0.5 * N[(N[(N[Sin[lambda1], $MachinePrecision] * N[Sin[lambda2], $MachinePrecision]), $MachinePrecision] + N[(N[Cos[lambda1], $MachinePrecision] * N[Cos[lambda2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Initial program 62.7%

   \[R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)}}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

2. Simplified 62.7%

   \[\leadsto \color{blue}{R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}{\sqrt{\left(1 - {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\right) - \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right)} \]
   Proof

   [Start] 62.7

   \[ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)}}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

   associate-*l* [=>] 62.7

   \[ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \color{blue}{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

3. Applied egg-rr 63.7%

   \[\leadsto R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\color{blue}{\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(\phi_2 \cdot 0.5\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(\phi_2 \cdot 0.5\right)\right)}}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}{\sqrt{\left(1 - {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\right) - \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

4. Applied egg-rr 79.6%

   \[\leadsto R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(\phi_2 \cdot 0.5\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(\phi_2 \cdot 0.5\right)\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}{\sqrt{\left(1 - {\color{blue}{\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(\phi_2 \cdot 0.5\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(\phi_2 \cdot 0.5\right)\right)}}^{2}\right) - \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}\right) \]

5. Applied egg-rr 79.6%

   \[\leadsto R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(\phi_2 \cdot 0.5\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(\phi_2 \cdot 0.5\right)\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}{\sqrt{\color{blue}{\log \left(e^{1 - \left({\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(0.5 \cdot \phi_2\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(0.5 \cdot \phi_2\right)\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(0.5 + -0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right)}\right)}}}\right) \]

6. Applied egg-rr 80.2%

   \[\leadsto R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(\phi_2 \cdot 0.5\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(\phi_2 \cdot 0.5\right)\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right)}}{\sqrt{\log \left(e^{1 - \left({\left(\sin \left(\phi_1 \cdot 0.5\right) \cdot \cos \left(0.5 \cdot \phi_2\right) - \cos \left(\phi_1 \cdot 0.5\right) \cdot \sin \left(0.5 \cdot \phi_2\right)\right)}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(0.5 + -0.5 \cdot \color{blue}{\left(\sin \lambda_1 \cdot \sin \lambda_2 + \cos \lambda_1 \cdot \cos \lambda_2\right)}\right)\right)}\right)}}\right) \]

7. Final simplification 80.2%

   \[\leadsto R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(\sin \left(\frac{\lambda_1 - \lambda_2}{2}\right) \cdot \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\right) + {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}}}{\sqrt{\log \left(e^{1 + \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(-0.5 + 0.5 \cdot \left(\sin \lambda_1 \cdot \sin \lambda_2 + \cos \lambda_1 \cdot \cos \lambda_2\right)\right) - {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\right)}\right)}}\right) \]

Alternatives

Alternative 1
Accuracy	79.1%
Cost	132425

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\\ t_2 := \sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + t_1}\\ \mathbf{if}\;\lambda_1 \leq -510000 \lor \neg \left(\lambda_1 \leq 4.1 \cdot 10^{-13}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_2}{\sqrt{1 - \left(t_1 + \cos \phi_2 \cdot \left(\cos \phi_1 \cdot \left(0.5 + -0.5 \cdot \cos \lambda_1\right)\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_2}{\sqrt{1 + \left(\cos \phi_2 \cdot \left(\cos \phi_1 \cdot \left(-0.5 + 0.5 \cdot \cos \left(-\lambda_2\right)\right)\right) - t_1\right)}}\right)\\ \end{array} \]

Alternative 2
Accuracy	73.6%
Cost	132360

\[\begin{array}{l} t_0 := \cos \phi_1 \cdot \cos \phi_2\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := \sin \left(\lambda_2 \cdot -0.5\right)\\ t_3 := {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\\ t_4 := \sqrt{t_0 \cdot \left(t_1 \cdot t_1\right) + t_3}\\ \mathbf{if}\;\lambda_2 \leq -102000:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_4}{\sqrt{\log \left(e^{1 + \left(\cos \phi_1 \cdot \left(\cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right) - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}\right)}\right)}}\right)\\ \mathbf{elif}\;\lambda_2 \leq 6.5 \cdot 10^{-5}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_4}{\sqrt{1 - \left(t_3 + \cos \phi_2 \cdot \left(\cos \phi_1 \cdot \left(0.5 + -0.5 \cdot \cos \lambda_1\right)\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_3 + t_2 \cdot \left(t_0 \cdot t_2\right)}}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + t_1 \cdot \left(t_0 \cdot t_1\right)\right)}}\right)\\ \end{array} \]

Alternative 3
Accuracy	79.6%
Cost	132224

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_0 \cdot t_0\right) + t_1}}{\sqrt{1 + \left(t_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right) - t_1\right)}}\right) \end{array} \]

Alternative 4
Accuracy	79.6%
Cost	132224

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_0 \cdot t_0\right) + t_1}}{\sqrt{\left(1 + t_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right) - t_1}}\right) \end{array} \]

Alternative 5
Accuracy	63.7%
Cost	112384

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}}}{\sqrt{1 + \left(\cos \phi_1 \cdot \left(\cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right) - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}\right)}}\right) \end{array} \]

Alternative 6
Accuracy	63.7%
Cost	112384

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\left(\cos \left(0.5 \cdot \phi_2\right) \cdot \sin \left(\phi_1 \cdot 0.5\right) - \sin \left(0.5 \cdot \phi_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)}^{2}}}{\sqrt{\left(1 + \cos \phi_1 \cdot \left(\cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right)\right) - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}\right) \end{array} \]

Alternative 7
Accuracy	63.2%
Cost	111680

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}}{\sqrt{\sqrt{{\left(\mathsf{fma}\left(\cos \phi_1, \cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right), 1 - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}\right)\right)}^{2}}}}\right) \end{array} \]

Alternative 8
Accuracy	63.2%
Cost	105216

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}}{\sqrt{\left|\mathsf{fma}\left(\cos \phi_1, \cos \phi_2 \cdot \left(0.5 + -0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right), {\sin \left(-0.5 \cdot \left(\phi_2 - \phi_1\right)\right)}^{2}\right) + -1\right|}}\right) \end{array} \]

Alternative 9
Accuracy	41.2%
Cost	92936

\[\begin{array}{l} t_0 := \cos \phi_1 \cdot \cos \phi_2\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := t_0 \cdot \left(t_1 \cdot t_1\right)\\ t_3 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ \mathbf{if}\;t_1 \leq -0.12:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 + t_3}}{\sqrt{\log \left(1 + \mathsf{expm1}\left(0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right)}}\right)\\ \mathbf{elif}\;t_1 \leq 0.05:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_3 + t_1 \cdot \left(t_0 \cdot t_1\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 + {\left(\sin \left(\phi_1 \cdot 0.5\right) + -0.5 \cdot \left(\phi_2 \cdot \cos \left(\phi_1 \cdot 0.5\right)\right)\right)}^{2}}}{\sqrt{1 - {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}}}\right)\\ \end{array} \]

Alternative 10
Accuracy	62.8%
Cost	92864

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := t_0 \cdot \left(\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot t_0\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + t_1}}{\sqrt{1 + \left(\left(-0.5 + \frac{\cos \left(\phi_1 - \phi_2\right)}{2}\right) - t_1\right)}}\right) \end{array} \]

Alternative 11
Accuracy	63.1%
Cost	92617

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}\\ t_2 := {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -0.0037 \lor \neg \left(\phi_1 \leq 6.2 \cdot 10^{-6}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_1}{\sqrt{{\cos \left(\phi_1 \cdot 0.5\right)}^{2} - \cos \phi_1 \cdot t_2}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_1}{\sqrt{\frac{1}{\frac{1}{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot t_2}}}}\right)\\ \end{array} \]

Alternative 12
Accuracy	62.7%
Cost	92544

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}}{\sqrt{1 + \left(\cos \phi_1 \cdot \left(\cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right) - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}\right)}}\right) \end{array} \]

Alternative 13
Accuracy	62.7%
Cost	92544

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}}{\sqrt{\left(1 + \cos \phi_1 \cdot \left(\cos \phi_2 \cdot \left(-0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right)\right) - {\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}\right) \end{array} \]

Alternative 14
Accuracy	58.0%
Cost	92361

\[\begin{array}{l} t_0 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ \mathbf{if}\;\phi_1 \leq -0.0145 \lor \neg \left(\phi_1 \leq 6.2 \cdot 10^{-6}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\cos \phi_1 \cdot {\sin \left(\lambda_2 \cdot -0.5\right)}^{2} + {\sin \left(\phi_1 \cdot 0.5\right)}^{2}}}{\sqrt{1 - \left(t_0 + t_1 \cdot \left(t_2 \cdot t_1\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_1 \cdot t_1\right) + t_0}}{\sqrt{\left(0.5 - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\right) + 0.5 \cdot \cos \left(2 \cdot \left(\phi_2 \cdot -0.5\right)\right)}}\right)\\ \end{array} \]

Alternative 15
Accuracy	58.0%
Cost	92361

\[\begin{array}{l} t_0 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ \mathbf{if}\;\phi_1 \leq -0.45 \lor \neg \left(\phi_1 \leq 6.2 \cdot 10^{-6}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\cos \phi_1 \cdot {\sin \left(\lambda_2 \cdot -0.5\right)}^{2} + {\sin \left(\phi_1 \cdot 0.5\right)}^{2}}}{\sqrt{1 - \left(t_0 + t_1 \cdot \left(t_2 \cdot t_1\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_1 \cdot t_1\right) + t_0}}{\sqrt{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}}}\right)\\ \end{array} \]

Alternative 16
Accuracy	63.1%
Cost	92361

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_0 \cdot t_0\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}\\ t_2 := {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -0.0037 \lor \neg \left(\phi_1 \leq 5.6 \cdot 10^{-6}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_1}{\sqrt{{\cos \left(\phi_1 \cdot 0.5\right)}^{2} - \cos \phi_1 \cdot t_2}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_1}{\sqrt{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot t_2}}\right)\\ \end{array} \]

Alternative 17
Accuracy	57.7%
Cost	92105

\[\begin{array}{l} t_0 := {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_3 := \sin \left(\lambda_2 \cdot -0.5\right)\\ t_4 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ \mathbf{if}\;\phi_1 \leq -5 \cdot 10^{-23} \lor \neg \left(\phi_1 \leq 6.2 \cdot 10^{-6}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_3 \cdot \left(t_1 \cdot t_3\right) + t_2}}{\sqrt{{\cos \left(\phi_1 \cdot 0.5\right)}^{2} - \cos \phi_1 \cdot t_0}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_1 \cdot \left(t_4 \cdot t_4\right) + t_2}}{\sqrt{\left(0.5 - \cos \phi_2 \cdot t_0\right) + 0.5 \cdot \cos \left(2 \cdot \left(\phi_2 \cdot -0.5\right)\right)}}\right)\\ \end{array} \]

Alternative 18
Accuracy	53.4%
Cost	86281

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -0.008 \lor \neg \left(\phi_1 \leq 0.00185\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_2 + t_0 \cdot \left(t_1 \cdot t_0\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_1 \cdot \left(t_0 \cdot t_0\right) + t_2}}{\sqrt{\left(0.5 - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\right) + 0.5 \cdot \cos \left(2 \cdot \left(\phi_2 \cdot -0.5\right)\right)}}\right)\\ \end{array} \]

Alternative 19
Accuracy	41.4%
Cost	86153

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ \mathbf{if}\;t_0 \leq -0.12 \lor \neg \left(t_0 \leq 0.072\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_1 \cdot \left(t_0 \cdot t_0\right) + t_2}}{\sqrt{0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_2 + t_0 \cdot \left(t_1 \cdot t_0\right)\right)}}\right)\\ \end{array} \]

Alternative 20
Accuracy	41.4%
Cost	86152

\[\begin{array}{l} t_0 := \cos \phi_1 \cdot \cos \phi_2\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_3 := \sqrt{t_0 \cdot \left(t_1 \cdot t_1\right) + t_2}\\ t_4 := 0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\\ \mathbf{if}\;t_1 \leq -0.12:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_3}{\sqrt{\log \left(e^{t_4}\right)}}\right)\\ \mathbf{elif}\;t_1 \leq 0.072:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_2 + t_1 \cdot \left(t_0 \cdot t_1\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_3}{\sqrt{t_4}}\right)\\ \end{array} \]

Alternative 21
Accuracy	41.4%
Cost	86152

\[\begin{array}{l} t_0 := 0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_3 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_4 := \sqrt{t_1 \cdot \left(t_2 \cdot t_2\right) + t_3}\\ \mathbf{if}\;t_2 \leq -0.12:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_4}{\sqrt{\log \left(1 + \mathsf{expm1}\left(t_0\right)\right)}}\right)\\ \mathbf{elif}\;t_2 \leq 0.072:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_3 + t_2 \cdot \left(t_1 \cdot t_2\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_4}{\sqrt{t_0}}\right)\\ \end{array} \]

Alternative 22
Accuracy	53.4%
Cost	86089

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -0.032 \lor \neg \left(\phi_1 \leq 0.00075\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_2 + t_0 \cdot \left(t_1 \cdot t_0\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_1 \cdot \left(t_0 \cdot t_0\right) + t_2}}{\sqrt{0.5 + \left(0.5 \cdot \cos \left(-\phi_2\right) - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\right)}}\right)\\ \end{array} \]

Alternative 23
Accuracy	40.3%
Cost	85252

\[\begin{array}{l} t_0 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := \cos \phi_1 \cdot \cos \phi_2\\ \mathbf{if}\;\lambda_1 - \lambda_2 \leq -500000000000:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\cos \phi_2 \cdot {\sin \left(\lambda_2 \cdot -0.5\right)}^{2} + {\sin \left(\phi_2 \cdot -0.5\right)}^{2}}}{\sqrt{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}}}\right)\\ \mathbf{elif}\;\lambda_1 - \lambda_2 \leq 2000000000:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}^{2}}}{\sqrt{1 - \left(t_0 + t_1 \cdot \left(t_2 \cdot t_1\right)\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{t_2 \cdot \left(t_1 \cdot t_1\right) + t_0}}{\sqrt{\log \left(1 + \mathsf{expm1}\left(0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)\right)\right)}}\right)\\ \end{array} \]

Alternative 24
Accuracy	34.8%
Cost	66633

\[\begin{array}{l} t_0 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ \mathbf{if}\;\lambda_1 - \lambda_2 \leq -2 \cdot 10^{-11} \lor \neg \left(\lambda_1 - \lambda_2 \leq 4 \cdot 10^{-156}\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sqrt{\left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot \left(t_1 \cdot t_1\right) + {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}}}{\sqrt{0.5 + 0.5 \cdot \cos \left(\lambda_1 - \lambda_2\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{1 - \left({t_0}^{2} + \cos \phi_2 \cdot \left(\cos \phi_1 \cdot {\sin \left(0.5 \cdot \lambda_1\right)}^{2}\right)\right)}}\right)\\ \end{array} \]

Alternative 25
Accuracy	20.0%
Cost	66372

\[\begin{array}{l} t_0 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_1 := \cos \phi_1 \cdot \cos \phi_2\\ t_2 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ t_3 := \sin \left(\lambda_2 \cdot -0.5\right)\\ \mathbf{if}\;t_0 \leq 4 \cdot 10^{-13}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_2}{\sqrt{1 - \left({t_2}^{2} + t_1 \cdot {t_3}^{2}\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_3}{\sqrt{1 - \left({\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2} + t_0 \cdot \left(t_1 \cdot t_0\right)\right)}}\right)\\ \end{array} \]

Alternative 26
Accuracy	20.0%
Cost	66372

\[\begin{array}{l} t_0 := \sin \left(\lambda_2 \cdot -0.5\right)\\ t_1 := \sin \left(\frac{\lambda_1 - \lambda_2}{2}\right)\\ t_2 := \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot t_1\\ t_3 := {\sin \left(\frac{\phi_1 - \phi_2}{2}\right)}^{2}\\ \mathbf{if}\;t_1 \leq 4 \cdot 10^{-13}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}{\sqrt{1 - \left(t_3 + t_0 \cdot t_2\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{1 - \left(t_3 + t_1 \cdot t_2\right)}}\right)\\ \end{array} \]

Alternative 27
Accuracy	16.3%
Cost	59264

\[\begin{array}{l} t_0 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{1 - \left({t_0}^{2} + \cos \phi_2 \cdot \left(\cos \phi_1 \cdot {\sin \left(0.5 \cdot \lambda_1\right)}^{2}\right)\right)}}\right) \end{array} \]

Alternative 28
Accuracy	16.3%
Cost	59264

\[\begin{array}{l} t_0 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{1 - \left({t_0}^{2} + \left(\cos \phi_1 \cdot \cos \phi_2\right) \cdot {\sin \left(\lambda_2 \cdot -0.5\right)}^{2}\right)}}\right) \end{array} \]

Alternative 29
Accuracy	16.5%
Cost	53001

\[\begin{array}{l} t_0 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ t_1 := {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -2.2 \cdot 10^{+15} \lor \neg \left(\phi_1 \leq 110000000\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{{\cos \left(\phi_1 \cdot 0.5\right)}^{2} - \cos \phi_1 \cdot t_1}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{1 - \left(\cos \phi_2 \cdot t_1 + {\sin \left(\phi_2 \cdot -0.5\right)}^{2}\right)}}\right)\\ \end{array} \]

Alternative 30
Accuracy	16.5%
Cost	52873

\[\begin{array}{l} t_0 := \sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)\\ t_1 := {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}\\ \mathbf{if}\;\phi_1 \leq -1.2 \cdot 10^{+15} \lor \neg \left(\phi_1 \leq 82000000\right):\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{{\cos \left(\phi_1 \cdot 0.5\right)}^{2} - \cos \phi_1 \cdot t_1}}\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(2 \cdot \tan^{-1}_* \frac{t_0}{\sqrt{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot t_1}}\right)\\ \end{array} \]

Alternative 31
Accuracy	14.1%
Cost	52608

\[R \cdot \left(2 \cdot \tan^{-1}_* \frac{\sin \left(0.5 \cdot \left(\phi_1 - \phi_2\right)\right)}{\sqrt{{\cos \left(\phi_2 \cdot -0.5\right)}^{2} - \cos \phi_2 \cdot {\sin \left(0.5 \cdot \left(\lambda_1 - \lambda_2\right)\right)}^{2}}}\right) \]

Reproduce

herbie shell --seed 2023129 
(FPCore (R lambda1 lambda2 phi1 phi2)
  :name "Distance on a great circle"
  :precision binary64
  (* R (* 2.0 (atan2 (sqrt (+ (pow (sin (/ (- phi1 phi2) 2.0)) 2.0) (* (* (* (cos phi1) (cos phi2)) (sin (/ (- lambda1 lambda2) 2.0))) (sin (/ (- lambda1 lambda2) 2.0))))) (sqrt (- 1.0 (+ (pow (sin (/ (- phi1 phi2) 2.0)) 2.0) (* (* (* (cos phi1) (cos phi2)) (sin (/ (- lambda1 lambda2) 2.0))) (sin (/ (- lambda1 lambda2) 2.0))))))))))