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My method:
percent-interval :=> PI old-interval :=> OI
PI(a, p1) -> OI(a-a*p1, a+a*p1) PI(b, p2) -> OI(b-b*p2, b+b*p2)
PI(c, p) = PI(a, p1)*PI(b, p2) :=> OI(a-a*p1, a+a*p1) * OI(b-b*p2, b+b*p2) = OI((a-a*p1)*(b-b*p2), (a+a*p1)*(b+b*p2)) = OI(cL, cU) c = (cL + cU)/2 = ((a-a*p1)*(b-b*p2) + (a+a*p1)*(b+b*p2))/2 = a*b*(1+p1*p2) p = (c-cL)/c = (a*b*(1+p1*p2) - (a-a*p1)*(b-b*p2))/a*b*(1+p1*p2) = a*b*(p1 + p2)/a*b*(1+p1*p2) = (p1+p2)/(1+p1*p2)
c = a*b*(1+p1*p2); p = (p1+p2)/(1+p1*p2);
Example:
PI(10, 0.005)*PI(10, 0.004) = PI(10*10*(1+0.004*0.005), (0.004+0.005)/(1+0.004*0.005)) = PI(100.002, 0.00899982);
No scheme, just math. If Ca is center of a, and Ta is tolerance of a, a is the interval
and b is
If the endpoints are positive, a*b has the endpoints (after simplifying):
Ta*Tb will be a wee number, so it can be ignored. So, it appears that for small tolerances, the tolerance of the product will be approximately the sum of the component tolerances.
A quick check:
LisScheSic
Small correction: By definition of tolerance, 0.5 is not needed.