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(HP-67) Barkers's Equation
12-06-2019, 01:27 PM
Post: #1
(HP-67) Barkers's Equation
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.
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12-06-2019, 06:39 PM (This post was last modified: 12-07-2019 09:21 PM by Albert Chan.)
Post: #2
RE: (HP-67) Barkers's Equation
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', where sign(W) = 1 if W=|W|, else -1

Or, we can go for the big |y|.

y = ³√(W + sign(W) √(W²+1)) = ³√(cot(cot-1(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)
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12-07-2019, 09:39 PM (This post was last modified: 01-13-2020 02:57 AM by Albert Chan.)
Post: #3
RE: (HP-67) Barkers's Equation
(12-06-2019 06:39 PM)Albert Chan Wrote:  y = ³√(W + sign(W) √(W²+1)) = ³√(cot(cot-1(W)/2))

Prove: let θ = cot-1(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(cot-1(W)/2) = W + sign(W) √(W²+1)
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01-31-2020, 03:38 PM (This post was last modified: 01-31-2020 03:41 PM by Albert Chan.)
Post: #4
RE: (HP-67) Barkers's Equation
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.
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