One slope creates all three results. In the familiar 440 by 280 royal-cubit model of Khufu's pyramid, the half-base is 220 cubits. That gives a 5½-palm seked, a perimeter-to-height ratio near 2π, and a slant-to-half-base ratio near the golden ratio φ. Both comparisons follow from the same slope. To see how close they are, hold the base fixed and calculate the height each exact match would require.
01 · BasisPetrie’s survey and the cubit model
Flinders Petrie reconstructed a mean original casing side of 9,068.8 inches from surviving casing and socket evidence. He gave about 0.5 inch as the probable error of that mean. He used that mean and his estimate of the casing angle to reconstruct an original height of 5,776 ± 7 inches.1
Petrie discussed a 440-cubit side and 280-cubit height as a theory. He drew attention to the approximation 22/7 for π in the ratio of half the perimeter to the height.1 For a fuller account of the survey and its uncertainty, see the Great Pyramid study. The calculations below start with Petrie’s two proposed cubit counts.
02 · MethodCalculate the 5½-palm seked
Take a vertical section through the apex and the midpoint of one base side. Its horizontal run is half the base, 440 ÷ 2 = 220 cubits. Its rise is 280 cubits. An Egyptian seked expresses horizontal run in palms for a rise of one cubit. With seven palms to a cubit, the model gives 7 × 220 ÷ 280 = 5½ palms. A larger seked means a flatter face.
Problem 56 in the later Rhind Mathematical Papyrus uses the same operation. In Chace’s translation, the scribe halves a 360-cubit base and converts the run per cubit of rise into palms. With a height of 250 cubits, the exercise gives 7 × 180 ÷ 250 = 5.04 palms. It describes an unnamed pyramid in a mathematical exercise.2
03 · CalculationDerive both near-matches from the same triangle
The square base has perimeter 4 × 440 = 1,760 cubits. Dividing by the height gives 1,760 ÷ 280 = 44/7 ≈ 6.285714, about 0.040% above 2π. The face’s slant height is the diagonal of the right triangle, √(280² + 220²) cubits. Divide that by its 220-cubit horizontal leg to get √317/11 ≈ 1.618590, about 0.034% above φ.
Scroll the table sideways to read every column.
| Seked | Perimeter ÷ height | Slant ÷ half-base | Basis |
|---|---|---|---|
| 5½ palmsKhufu model | 6.285714 near 2π | 1.618590 near φ | 440-cubit side, 280-cubit height |
| 6 palmsIllustrative change | 6.857143 | 1.536591 | Same square-pyramid formula; no monument attribution |
The six-palm row changes the slope as a mathematical test. For any seked s in palms, the perimeter divided by height is 8s/7. The slant divided by half-base is √(1 + (7/s)²). A larger seked raises the first result and lowers the second.
04 · Counter-testExact matches require different heights
Hold the side B at Petrie’s reconstructed 9,068.8 inches. The 440-by-280 proportions give a height of B × 280/440. To make perimeter divided by height exactly 2π, divide the perimeter 4B by 2π. To make slant divided by half-base exactly φ, set the slant to φ × B/2. Pythagoras then gives the height as (B/2) × √(φ² − 1).
Scroll the table sideways to read every column.
| Condition | Height (inches) |
|---|---|
| 440-by-280 proportions | 5,771.1 |
| Exact 2π ratio | 5,773.4 |
| Exact φ ratio | 5,767.8 |
Subtract the two exact-target heights before rounding and the gap is about 5.5 inches. That gap is smaller than the 7 inches Petrie gave on either side of his reconstructed height. The comparison shows the scale of the difference. Petrie’s estimate shares the base used here and comes with its own assumptions about the casing angle.
Use the pyramid calculator with the Khufu preset to compare this model with other pyramid slopes. The False Precision study explains how rounding can change the apparent fit.
Sources and Further Reading
- W. M. Flinders Petrie, The Pyramids and Temples of Gizeh (London: Field & Tuer; New York: Scribner & Welford, 1883), pp. 39–43 for the casing-base and height reconstruction; p. 183 for the 440-by-280-cubit theory and Petrie's π reading. Digital Giza catalogue record.
- Arnold Buffum Chace et al., The Rhind Mathematical Papyrus, vol. I (Oberlin: Mathematical Association of America, 1927), p. 1 for the manuscript context and pp. 96–97 for Problem 56. Digitised volume.