Daylight Model, Maintenance Window, and a Broken Promise
A solar-panel installer in a city at 52°N uses a sinusoidal model for hours of daylight, (L(d) = 12 + 3.6\sin!\left(\dfrac{2\pi(d-80)}{365}\right)), where d is the day of the year and d = 1 is 1 January. The model was fitted to two measured values: 15.6 h on day 172 and 8.4 h on day 355. Maintenance crews need at least 10 hours of daylight, and the company currently schedules work between day 60 and day 300. Last season a crew was caught out on day 288, when the model predicted 10.4 h but the measured value was 9.3 h. A rival firm uses a straight line joining the same two measured points. The operations manager wants to know whether the sinusoidal model can be trusted for scheduling, and whether the 'at least 10 hours' promise should stand for the whole of the current window.
1. State the amplitude, mean and period of the model, and find the days on which it gives exactly 12 hours of daylight.
Read the worked answer
Writing the model as (L(d) = 12 + 3.6\sin\theta) with (\theta = \frac{2\pi(d-80)}{365}):
- mean = 12 h (the annual average)
- amplitude = 3.6 h, so the model ranges from 8.4 h to 15.6 h
- period = 365 days
12 hours means (3.6\sin\theta = 0), so (\theta = 0) or (\theta = \pi):
(d = 80) and (d = 80 + \frac{365}{2} = 262.5).
That is about 21 March and 19–20 September, the equinoxes, when day and night are equal. The maximum occurs at (\theta = \pi/2), i.e. d = 171.25 (≈ 21 June), giving 15.6 h; the minimum at d = 353.75 (≈ 20 December), giving 8.4 h. Those two extrema reproduce exactly the two measured values used to fit the model, which is a consistency check but not independent evidence that the shape is right.
2. Solve \(L(d) \ge 10\) and interpret the answer against the company's scheduling window of day 60 to day 300.
Read the worked answer
(12 + 3.6\sin\theta \ge 10 \Rightarrow \sin\theta \ge -0.5556).
Solving (\sin\theta = -0.5556) gives (\theta = 3.7301) (or (\theta = -0.5890), equivalently 5.6942).
Since (d = 80 + \frac{365}{2\pi}\theta = 80 + 58.086\theta):
- (d = 80 + 58.086(-0.5890) = 45.8)
- (d = 80 + 58.086(3.7301) = 296.7)
(The solution (\theta = 5.6942) gives d = 410.8, outside the year - it is the same date in the next cycle.)
So the model gives at least 10 h of daylight for 45.8 ≤ d ≤ 296.7, roughly 15 February to 23 October.
Interpretation: the scheduling window day 60–300 is not fully inside this. Days 297–300 fall below 10 h by the model's own numbers, so the promise is already false for the last few days of the window. Two defensible fixes: shorten the window to day 290 to leave a margin, or keep day 300 but treat the promise as a target rather than a guarantee.
3. Evaluate whether the sinusoidal model should be trusted for scheduling, comparing it with the rival's linear model.
Read the worked answer
Check the day that actually failed. Day 288: (\theta = \frac{2\pi(208)}{365} = 3.5806), (\sin\theta = -0.4250), so the model gives (12 - 1.53 = 10.47) h against a measured 9.3 h - a residual of about −1.2 h, and in the direction that matters (the crew gets less light than promised).
The linear model through (172, 15.6) and (355, 8.4) has gradient (-7.2/183 = -0.03934) h per day, giving (15.6 - 0.03934(116) = 11.0) h on day 288 - a residual of −1.7 h, worse still.
The sinusoid is the better of the two, but neither is safe for a hard guarantee. Real daylight is not symmetric about the equinoxes: Earth moves faster in its orbit near perihelion, so autumn light falls more quickly than spring light rises, and the sinusoid smooths that asymmetry away. The largest errors appear near the ends of the season, which is exactly where the day-300 promise sits.
Judgement: keep the sinusoid for planning, but do not advertise 'at least 10 hours' past about day 270 without measured data; a better model adds a second harmonic or uses a table of measured values.