Gnomon Measurements at 39.7° N
At a school in Akita (39.7° N, 140.1° E) students set a vertical 1.000 m gnomon on levelled ground and record the length of its shortest shadow: 20 March 0.831 m; 21 June 0.291 m; 22 December 1.966 m. Each shadow is measured to the nearest millimetre with a steel tape. The students use solar declinations of 0° on 20 March, +23.44° on 21 June and −23.44° on 22 December. On 20 March the Akita group's shortest shadow was recorded at 11:47 JST, while a partner school on the 135° E meridian recorded its shortest shadow at 12:07 JST the same day. Japan Standard Time is the local mean time at 135° E. On 21 June a second group, using a rod they checked as vertical with a spirit level, record a shadow of 0.340 m when the clock reads 12:00 JST.
1. Calculate the Sun's noon altitude and the latitude implied by each of the three readings. Comment on the agreement and on what the three values together test.
Read the worked answer
For a vertical gnomon, tan(altitude) = height / shadow, and latitude = 90° − altitude + declination.
20 March: 1.000 / 0.831 = 1.203, altitude = 50.28°, declination 0°, latitude = 39.72° N. 21 June: 1.000 / 0.291 = 3.436, altitude = 73.76°, latitude = 90 − 73.76 + 23.44 = 39.68° N. 22 December: 1.000 / 1.966 = 0.5087, altitude = 26.96°, latitude = 90 − 26.96 − 23.44 = 39.60° N.
All three cluster at 39.6–39.7° N, matching the school's true latitude within about 0.1° (around 11 km). The mm-level tape readings do not justify claiming better than 0.05–0.1°, so this is consistent with a sound setup.
The three values together do more than one value alone: the equinox altitude minus the December altitude is 23.3°, and the June minus equinox altitude is 23.5°, both close to the obliquity 23.44° that was assumed. Internal consistency of the seasonal differences is a check that the declination values and dates are right, and the close agreement of three independent shadow readings makes a single gross blunder - wrong rod height, tilted rod, wrong date - unlikely.
2. Explain why the shortest shadow at Akita fell at 11:47 JST rather than at 12:00, and why the partner school on 135° E recorded 12:07.
Read the worked answer
Noon by the clock is not noon by the Sun. Japan Standard Time is defined by the mean Sun at 135° E, so at that meridian mean solar noon is 12:00 JST - but the partner school's shortest shadow came at 12:07, seven minutes later. That offset is not a longitude effect: it is the difference between apparent solar time (the real Sun) and mean solar time, which on 20 March runs about seven minutes behind. Both schools share it because it depends only on the date.
Akita is at 140.1° E, which is 5.1° east of the standard meridian. The Earth turns 1° in 4 minutes, so the Sun crosses Akita's meridian 5.1 × 4 ≈ 20 minutes earlier than it crosses 135° E. Subtracting that from the partner school's 12:07 gives about 11:47 JST, exactly what the students recorded.
The reasoning to use is comparative: the two sites differ only in longitude, so the difference between their times (20 minutes) must be the longitude term, and the shared offset from 12:00 by the 135° E school (7 minutes) must be the date-dependent term. Neither number needs to be memorised; each is read off the other's data.
3. On 21 June the expected shortest shadow is 0.291 m at solar noon. Evaluate the second group's 12:00 JST reading of 0.340 m: first correct for the time of day, then judge three proposed explanations - the Sun's angular size, refraction, and a tilt of the rod.
Read the worked answer
At 12:00 JST the Sun is not yet on the meridian: apparent solar noon in Akita on 21 June is near 11:42, so the hour angle is about 4.6°. Then sin(altitude) = sin39.7 sin23.44 + cos39.7 cos23.44 cos4.6° = 0.254 + 0.704 = 0.958, so altitude ≈ 73.3° and the shadow should be 1.000/tan73.3 ≈ 0.30 m. The timing correction explains only about 0.01 m of the 0.05 m excess.
- Sun's angular size (0.53°): a finite disc produces a penumbra whose width is roughly 0.0093 × shadow length ≈ 3 mm, and a slightly enlarged blur of the edge cannot make the shadow 40 mm longer. Not sufficient.
- Refraction: at 73° altitude the lift is about half an arcminute, changing the shadow by well under a millimetre. Not sufficient.
- Tilt: a rod leaning towards the Sun by angle θ displaces its top horizontally by 1.000 × sinθ. For θ = 2.5°, that is 0.044 m - almost exactly the discrepancy. A 2.5° lean is easy to miss even when checked against a bubble that is itself out of adjustment or read against a non-level base.
So the tilt explanation is the only one of the three with the right magnitude. The general lesson is to test proposed errors quantitatively against the size of the anomaly before accepting them.