Now we can calculate the quadratic regression coordinates. Of course, Shackleton couldn’t have had Microsoft Excel with him, so he could only use the following equations 21 :
= 2 + +
(∑ ) 2
324 2 10
= ∑ 2 −
𝑆
= 14694 −
= 4196.4
∑ ∗ ∑
324 ∗ 500.2 10
𝑆
= ∑ −
= 16177.6 −
= −28.88
∑ ∗ ∑ 2
324 ∗ 14694 10
= ∑ 3 −
𝑆 2
= 723204 −
= 247118.4
∑ 2 ∗ ∑
14694 ∗ 500.2 10
= ∑ 2 −
𝑆 2
= 732723.4 −
= −2270.48
(∑ 2 ) 2
14694 2 10
= ∑ 4 −
𝑆 2 2
= 37129398 −
= 15538034.4
(𝑆 2
) − (𝑆
∗ 𝑆
∗ 𝑆 2
)
= (−2270.0.48 ∗ 4196.4) − (−28.88 ∗ 247118.4) 4196.4 ∗ 15538034.4 − 247118.4 2
=
) 2
(𝑆
) − (𝑆 2
∗ 𝑆 2 2
= −0.005780675…
(𝑆
) − (𝑆 2
∗ 𝑆 2 2
∗ 𝑆 2
)
= (−28.88 ∗ 15538034.4) − (−2270.88 ∗ 247118.4) 4196.4 ∗ 15538034.4 − 247118.4 2
=
) 2
(𝑆
) − (𝑆 2
∗ 𝑆 2 2
= 0.02718315685…
∑ 2
∑
∑
=
− ∗
− ∗
500.2 10
324 10
14694 10
=
− 0.02718315685 ∗
− (−0.005780675) ∗
= 49.988678…
Hence the top of the parabola has the coordinates (p,q):
= 23.5120957… = − 2 −4 4
= − 2
= 50.308244597….
So the local noon occurred at 11:30+23.5120957 mins=11:53:31
And the altitude of the sun then was: 50.308244597°
The declination of the sun for noon 22 nd April was: 12°11’=12.8333333
21 “Declination of Sun,” [Online]. Available: http://www.susdesign.com/popups/sunangle/diagram- graphics/declination1.gif. [Accessed 15 April 2015].
31
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