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(HP-67) Barkers's Equation - SlideRule - 12-06-2019 01:27 PM
An extract from An Efficient Method for Solving Barkers's Equation, British Astronomical Association, R. Meire, Journal of the British Astronomical Association, Vol. 95, NO.3/APR, P.113, 1985
"In a parabolic orbit, the true anomaly v (as a function of the time) can be obtained by solving a cubic equation for tan y, the so-called Barker equation. A modified method of solution is described and it is shown that this new form is very efficient from a computational point of view: it needs less program statements, is faster, and it has greater accuracy than the normally used trigonometric solution.
…
In programming this trigonometric method on a computer or on a calculator, a problem occurs with the cubic root in equation (6) if W < 0. We must include a conditional test in the program to solve this problem. As an example, Appendix A shows the program for an HP-67 calculator, and it can be seen that the cubic root problem takes four additional steps.
…
Appendix B gives the program for the HP-67 calculator. Although the trigonometric solution has a certain 'beauty', it cannot compete with this new solution (equation (12)) from a practical point of view."
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there are THREE pages to the article. Use the navigation links at the bottom to 'see' ALL three. The program listings are on page 115.
RE: (HP-67) Barkers's Equation - Albert Chan - 12-06-2019 06:39 PM
Solving cubic with Cardano's formula, x³ + 3x - 2W = 0
y = ³√(W + √(W²+1))
x = y - 1/y
Note: discriminant = W²+1 > 0, we have only 1 real root for x
If W<0, y may be hit with subtraction cancellation.
We can avoid catastrophic cancellation by solving x'³ + 3x' - 2|W| = 0
x = sign(W) x'
Or, we can go for the big |y|: // assumed acot(W) = atan(1/W)
y = ³√(W + sign(W) √(W²+1)) = ³√(cot(acot(W)/2))
We still have the cancellation error issue when y ≈ 1
A better non-iterative formula is to use hyperbolics.
x = 2 sinh(sinh-1(W)/3)
RE: (HP-67) Barkers's Equation - Albert Chan - 12-07-2019 09:39 PM
(12-06-2019 06:39 PM)Albert Chan Wrote: y = ³√(W + sign(W) √(W²+1)) = ³√(cot(acot(W)/2))
Prove: let θ = acot(W), c = cot(θ/2)
Half angle formula: cot(θ) = W = (c²-1) / (2c)
c² - 2W c - 1 = 0 → c = W ± √(W²+1)
Since |θ| < pi/2, c has same sign of θ, which has same sign of W (assumed sign(0)=1)
However, 2 roots of c have opposite sign. Matching c and W signs, we have:
c = cot(acot(W)/2) = W + sign(W) √(W²+1)
RE: (HP-67) Barkers's Equation - Albert Chan - 01-31-2020 03:38 PM
To complete the symmetry (again, assume sign(0)=1):
c² - 2W c + 1 = 0 → c = W ± √(W²-1)
c = cot(csc-1(W)/2) = W + sign(W) √(W²-1)
Note: domain of csc-1(W) = sin-1(1/W) is |W| ≥ 1, otherwise c is a complex root.
RE: (HP-67) Barkers's Equation - Mathias Zechmeister - 04-10-2020 10:34 PM
Hi,
I arrived here by googling for "Barker's equation" and "sinh".
Because I derived the same equation as noted here:
(12-06-2019 06:39 PM)Albert Chan Wrote: x = 2 sinh(sinh-1(W)/3)
So it seems this equation is already known.
But is there any reference to this equation?
I can't find any!?
RE: (HP-67) Barkers's Equation - Albert Chan - 04-11-2020 03:29 AM
Welcome to the forum, Mathias Zechmeister
Hyperbolic solutions to the cubics is simply matching hyperbolic triple angle formula.
see https://mathworld.wolfram.com/CubicFormula.html, eqn 78,79,80
(12-06-2019 06:39 PM)Albert Chan Wrote: Solving cubic with Cardano's formula, x³ + 3x - 2W = 0
y = ³√(W + √(W²+1))
x = y - 1/y
Another way is with identity: sinh-1(z) = ln(z + √(z²+1))
→ y = e^(sinh-1(W)/3)
→ x = 2 sinh(sinh-1(W)/3)
RE: (HP-67) Barkers's Equation - Mathias Zechmeister - 04-11-2020 10:18 PM
Thanks for the reference. So the relation is mathematically known, but it was likely not applied to Barker's equation.
With the relation:
(12-06-2019 06:39 PM)Albert Chan Wrote: x = 2 sinh(sinh-1(W)/3)it is even simplier to compute Barker's equation with a pocket calculator than in R. Meire (1985).
Well, it seems the HP-67 had no hyperbolic functions.
I'm working on a follow-up paper of Zechmeister (2018) and I'm going to mention this relation.
(04-11-2020 03:29 AM)Albert Chan Wrote: Another way is with identity: sinh-1(z) = ln(z + √(z²+1))Indeed, that is, how I found it. I noted the term z + √(z²+1) in Barker's equation, and Fukushima (1997, Eq. 73) as well as Raposo-Pulido+ (2018, Eq. 43) reminded of this identity.
→ y = e^(sinh-1(W)/3)
→ x = 2 sinh(sinh-1(W)/3)
RE: (HP-67) Barkers's Equation - Mathias Zechmeister - 08-10-2020 08:10 AM
For your information, my paper is accepted. A preprint is available here:
https://arxiv.org/abs/2008.02894
In Sect. 7.2., I briefly discuss Barker's equation, more as a side note. I put you, Albert Chan, in the acknowledgements.
Please let me know, if something should be adjusted (e.g. pseudonym?).