A361601 Decimal expansion of the maximum possible disorientation angle between two identical cubes (in radians).

1, 0, 9, 6, 0, 5, 6, 8, 1, 5, 2, 4, 0, 6, 2, 5, 4, 8, 9, 0, 6, 1, 7, 2, 6, 5, 6, 5, 6, 4, 1, 2, 5, 7, 3, 5, 6, 9, 5, 9, 4, 2, 4, 7, 2, 7, 3, 1, 8, 4, 0, 8, 6, 3, 3, 9, 9, 1, 0, 9, 6, 8, 7, 7, 7, 2, 0, 6, 7, 8, 8, 7, 1, 0, 9, 2, 9, 7, 0, 9, 1, 0, 7, 7, 9, 8, 7, 0, 6, 3, 1, 4, 8, 8, 8, 2, 5, 7, 5, 7, 5, 7, 6, 9, 1

Offset: 1

Author

Amiram Eldar, Mar 17 2023

Keywords

• nonn
• cons

Comments

Mackenzie and Thomson (1957) attributed the idea of finding this angle to the British theoretical physicist Frederick Charles Frank (1911-1988), who proposed this problem in 1949.

The disorientation angle between two identical bodies is the least angle of rotation about an axis through the center of mass of one of the bodies that is needed to bring it into the same orientation as the other body. For two cubes with indistinguishable faces, there are 24 rotations angles that will bring the first cube into coincidence with the second, and the disorientation angle is the least of them.

The rotation which achieves this maximum disorientation can be described as a rotation by 90 degrees about any axis parallel to a face diagonal of the cube.

The angle in degrees is 62.7994296198...

The solution to the analogous two-dimensional problem with two squares is the trivial value Pi/4 (A003881).

Examples

			1.09605681524062548906172656564125735695942472731840...
		

Links

• D. C. Handscomb, On the random disorientation of two cubes, Canadian Journal of Mathematics, Vol. 10 (1958), pp. 85-88.
• J. K. Mackenzie, Second Paper on Statistics Associated with the Random Disorientation of Cubes, Biometrika, Vol. 45, No. 1-2 (1958), pp. 229-240.
• J. K. Mackenzie and M. J. Thomson, Some Statistics Associated with the Random Disorientation of Cubes, Biometrika, Vol. 44, No. 1-2 (1957), pp. 205-210.
• Wikipedia, Misorientation.
• Index entries for transcendental numbers.

Crossrefs

Cf. A003881, A201488, A361602, A361603, A361604.

Programs

Mathematica

RealDigits[ArcCos[Sqrt[2]/2 - 1/4], 10, 100][[1]]

PARI

acos(sqrt(2)/2 - 1/4)

Formula

Equals arccos(sqrt(2)/2 - 1/4).

Equals 2 * arccos(1/2 + sqrt(2)/4).

Equals 2 * arctan((sqrt(2)-1) * sqrt(5-2*sqrt(2))).