Question 7.7: Where in the World Are You? With your watch set for Pacific ......

Where in the World Are You? With your watch set for Pacific Standard Time (PST), you travel somewhere and when you arrive you note that the upper limb sunrise is at 1:11 A.M. (by your watch) and sunset is at 4:25 P.M. It is July 1^{st} (δ = 23.1º, E = −3.5 min). Where are you?

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Between 1:11 am and 4:25 P.M. there are 15 h and 14 min of daylight (15.233 h). With solar noon at the midpoint of that time—that is, 7 h and 37 min after sunrise—we have

\text{Solar noon} \ = \ 1:11 \ \text{A.M.} \ + \ 7:37 \ = \ 8:48 \ \text{A.M.} \ \text{PST}

Longitude can be determined from (7.14):

\begin{matrix} \text{Solar Time} \ \left(ST\right) & = \ \text{Clock Time} \ \left(CT\right) \ + \ \frac{4 \ \text{min}}{\text{degree}} \ ( \text{Local Time Meridian} \\ & – \ \text{Local longitude} )º \ + \ E \left(\text{min}\right) \quad \quad \quad \quad \quad \quad \quad \end{matrix}     (7.14)
\begin{matrix} \text{Solar Time} & = \ \text{Clock Time} + \ \frac{4 \ \text{min}}{\text{degree}} \ ( \text{Local Time Meridian} \\ & – \ \text{Local longitude} )º \ + \ E \left(\text{min}\right) \quad \quad \quad \quad \quad \quad \quad \end{matrix}

Using the 120º Local Time Meridian for Pacific Time gives

\begin{matrix} 12:00 \ − \ 8:48 & = \ 192 \ \text{min} \ = \ 4\left(120 \ − \ \text{Longitude}\right) \ + \ \left(−3.5\right) \ \text{min} \\ \quad \quad \text{Longitude} & = \ \frac{480 \ − \ 3.5 \ − \ 192}{4} \ = \ 71.1º \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \end{matrix}

To find latitude, it helps to first ignore the correction factor Q. Doing so, the daylength is 15.233 h, which makes the hour angle at sunrise equal to

H_{SR} \ = \ \frac{15.233 \ \text{hour}}{2} \ \cdot \ {15º}/{\text{hour}} \ = \ 114.25º

A first estimate of latitude can now be found from (7.16)

\cos \ H \ = \ – \frac{\sin \ L \ \sin \ \delta}{\cos \ L \ \cos \ \delta} \ = \ – \tan \ L \ \tan \ \delta (7.16)

L \ = \ \tan^{-1} \ \left(-\frac{\cos \ H_{SR}}{\tan \ \delta}\right) \ = \ \tan^{-1} \ \left(-\frac{\cos \ 114.25º}{\tan \ 23.1º}\right) \ = \ 43.9º

Now we can find Q, from which we can correct our estimate of latitude

Q \ = \ \frac{3.467}{\cos \ L \ \cos \ \delta \ \sin \ H_{SR}} \ = \ \frac{3.467}{\cos \ 43.9º \ \cos \ 23.1º \ \sin \ 114.25º} \ = \ 5.74 \ \text{min}

Geometric daylength is therefore 2 × 5.74 min = 11.48 min = 0.191 h shorter than daylength based on the upper limb crossing the horizon. The geometric hour angle at sunrise is therefore

H_{SR} \ = \ \frac{\left(15.233 \ − \ 0.191\right)h}{2} \ \cdot \ {15º}/{h} \ = \ 112.81º

Our final estimate of latitude is therefore

L \ = \ \tan^{-1} \ \left(-\frac{\cos \ H_{SR}}{\tan \ \delta}\right) \ = \ \tan^{-1} \ \left(-\frac{\cos \ 112.81º}{\tan \ 23.1º}\right) \ = \ 42.3º

Notice we are back in Boston, latitude 42.3º, longitude 71.1º. Notice too, our watch worked fine even though it was set for a different time zone.

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