SAME DISTANCES
Same distances. Different handedness.
Every edge agrees. Every label has its partner. Try turning this tetrahedron onto its mirror, then ask for the best rotation possible.
Can a turn close the gap?
Centroids coincide · fixed orthographic view
Drag to turn the moving body. With the field focused, arrow keys turn x / y by 5°; [ and ] turn z. The sliders offer the same controls.
This is a 2D view of a 3D object. Warm connectors join matching labels; every error below includes depth.
units RMS
Rotate the moving tetrahedron.
Reflection is excluded. Moving and target have opposite signed volumes.
Six agreements. One missing sign.
The labels stay attached. A proper rotation preserves the sign of the volume formed by A, B, C, D. Reflection reverses it. These nonzero signs cannot agree through turning alone.
- Target signed volume
- −2.667
- Moving signed volume
- +2.667
Cubic units · determinant of (B − A, C − A, D − A), divided by 6. The order of the labels defines the sign.
Six edge lengths agree.
| Pair | Target | Moving |
|---|
Values rounded here; the comparison uses full floating-point coordinates.
Read the study
What stays the same
Four labels name the same corresponding vertices throughout. All six pairwise Euclidean distances are fixed. The target and its mirror are noncoplanar, and the six edge lengths are distinct. The moving body is never stretched or sheared.
Only this labeled pair is being compared. Permuting labels is a different matching problem. This study does not claim that every unlabeled shape, or every distance dataset, has two different hands.
The error is three-dimensional
RMS = √[Σᵢ ‖Rpᵢ − qᵢ‖² / 4]
Corresponding points are compared in all three coordinates, in fixed units. The orthographic projection is only a viewing device: screen overlap is not the test. Both centroids sit at zero, so zero translation minimizes the error for every rotation. Moving either centroid apart can only increase it.
The viewing angle and scale never change when you request a fit. Dragging rotates the moving object, not the camera.
Why the best rotation is certified
The centered target has covariance C = diag(4, 1, ¼). Its mirror reverses z. For any proper rotation R, put B = R diag(1, 1, −1). B is orthogonal with determinant −1, so tr B ≤ 1 and each diagonal entry is at most 1.
RMS² = 2 Σⱼ Cⱼⱼ(1 − Bⱼⱼ)
≥ ½(3 − tr B) ≥ 1
The identity rotation attains RMS = 1 exactly. “Best rotational fit” uses that analytic solution; it does not stop an iterative search or infer impossibility from your attempt. The minimum includes every proper rotation and every translation.
The operation that changes the answer
“Allow reflection” reflects the original moving body through z = 0 and resets its rotation. This changes its orientation sign and aligns all four corresponding points exactly. The operation has determinant −1; it cannot be performed by a rigid turn in three dimensions.
With reflection allowed, the sliders still rotate that reflected body. “Best rotational fit” returns to the original mirror pair and its rotation-only optimum. Reset restores the opening attempt. There is no automatic motion.