 Floating-pinion ellipsoidal uncertainty bound

A floating-pinion ellipsoidal uncertainty bound gives the smallest two-contact compression margin when the uncertain parameter vector lies in a specified ellipsoid. It retains shared parameter effects, as does the floating-pinion affine uncertainty bound, but uses a different admissible-set shape.

Let approach and imbalance be affine functionsand describe the set by $u=u_0+Lz$ with $\Vert z{\Vert }_2\le 1$. Put $\bar{\delta }={\delta }_0+a^{\mathsf{T}}{u}_0$ and $\bar{b}=b_0+c^{\mathsf{T}}{u}_0$. The floating-pinion robust contact margin then satisfies the exact formulaTo prove it, write $\delta -|b|=\min (\delta -b,\delta +b)$. For either linear coefficient vector $q$, the Cauchy-Schwarz inequality givesIf $L^{\mathsf{T}}q\ne 0$, equality is attained at $z=-L^{\mathsf{T}}q/\Vert L^{\mathsf{T}}q{\Vert }_2$. Thus the result supplies both a margin and an attaining direction. The image representation also handles rank-deficient sets without inverting a shape matrix.

For an original example in micrometres, take $u_0=0$, $L$ the identity andOver $u_1^2+u_2^2\le 1$, the two branch minima are $4.5-\sqrt{0.1}$ and $5.5-\sqrt{0.1}$, so the margin is about $4.1838\mu \mathrm{m}$. The first branch attains it at $u=(-1,-3)/\sqrt{10}$. The encompassing box $|u_1|,|u_2|\le 1$ instead gives $4.1\mu \mathrm{m}$ because it admits combinations excluded by the disk.

A covariance-derived ellipse does not automatically certify an absolute engineering bound. Its probability content requires distributional assumptions and an appropriate radius; a hard guarantee requires a defensible admissible set. NASA's face-gear mounting and slow-roll load-sharing observations motivate the physical contact question, not the particular coefficients or set used here.^[ref] The two-contact force law must also remain valid; contact opening requires the floating-pinion contact-loss check.