In FASTSIM, there is a way to combine the flexibility parameters L1, L2, L3 into a single L. The formula is given in Kalker's book and papers, as well as in papers by other authors who extend the applicability or performance of FASTSIM, or just use it in their work:
L = (|sx|*L1 + |sy|*L2 + |phi|*sqrt(a*b)*L3)/sqrt(sx2+sy2+a*b*phi2),
with sx, sy, phi longitudinal, lateral and spin creepages, respectively. The formula is described to represent a "weighted mean" of the flexibilities. However, I can see that the result doesn't always look like a mean. For instance, with sx = 1, sy = 1, phi = 0 (i.e. equal weights x and y, no spin), it is
L = (L1 + L2)/sqrt(2)
– rather than the expected (L1 + L2)/2 – and that may not be between L1 and L2. Is this a correct behaviour, or am I making a mistake in the calculation?
You are right, this is incorrect, and this has been overlooked for a long time.
Kalker wrote that "there is a way to make the agreement perfect, and that is to substitute for L a weighted mean [..]". He showed this "perfect agreement" for pure longitudinal creepage (L=L1), pure lateral creepage (L=L2) and pure spin creepage (L=L3). Yet this weighted mean breaks down in other situations.
Consider for example an elliptical contact with a/b = 3.333, nu = 0.25. This has L1 = 0.421 \approx L2 = 0.416. With phi = 0, sx = sy, we then want to have L = L1 = L2, yet the formula gives L = 1.4 * L1.
A linear weighted mean would be
L = ( |sx| * L1 + |sy| * L2 + c * |phi| * L3 ) / ( |sx| + |sy| + c * |phi| )
Proper quadratic weighting is possible as follows:
L = ( sx2 * L1 + sy2 * L2 + a * b * phi2 * L3 ) / ( sx2 + sy2 + a * b * phi2 )
Both could work well, I don't expect big differences between them.
We studied possible ways to correct and improve FASTSIM. Our solution is published in a new paper:Article Improved accuracy for FASTSIM using one or three flexibility values
(Open Access).- Badges
- Science topic
- Similar topics
- Organizational Psychology
- Training