Close turn counts amplify distance errors in a two-point slope

Give every total an explicit error allowance

Construct a response of forty centimeters per turn and a fixed two-centimeter record offset. The selected exact totals at one, two, three and five turns are forty-two, eighty-two, 122 and 202 centimeters. Now assign each recorded total an error between minus 0.2 and plus 0.2 centimeters. That allowance is chosen for arithmetic; it is not a claim about a real ruler, person or mouse.

Compare counts one and two. The most positive difference error occurs when the second reading is 0.2 high and the first is 0.2 low. The distance difference grows by 0.4 centimeters, giving 40.4 instead of forty. Reversing those two signs gives 39.6. Opposite errors are possible under the stated bounds, even though no probability of either combination has been specified.

Separate the counts while retaining the same allowance

Compare one and three turns instead. The same worst-case 0.4-centimeter difference error is divided by two, producing a slope allowance of 0.2 centimeters per turn. Using one and five divides it by four, producing 0.1. The corresponding constructed ranges are 39.8 to 40.2 and 39.9 to 40.1. Every result refers to the same selected per-record bound and an unchanged response.

Write recorded totals as L1 plus delta1 and L2 plus delta2. Subtracting the exact slope leaves (delta2 minus delta1)/(n2 minus n1). The triangle inequality bounds its magnitude by (|delta2| + |delta1|)/|n2 minus n1|. If both are at most e, the result is 2e divided by the absolute count separation. Unequal bounds e1 and e2 instead give (e1 + e2)/|n2 minus n1|.

A longer record can have a larger allowance

The improvement above held the per-record allowance constant. Change that assumption: keep the one-turn allowance at 0.2 centimeters, but select a five-turn allowance of two centimeters. The difference-based slope bound becomes (0.2 + 2)/4 = 0.55 centimeters per turn. This is larger than the earlier one-versus-two bound of 0.4. More separated counts did not compensate for the newly selected larger error allowance.

This counterexample does not assert that longer mouse travel necessarily has that error. It shows why the denominator alone cannot rank two procedures when the numerator assumptions change. Lifting, repositioning or losing track of motion requires an explicit method assessment; it cannot be represented by a smaller bound merely because the record contains more revolutions.

Shared errors and opposite errors behave differently

For any fixed count pair, equal signed reading errors cancel from the difference. Opposite signed errors add in magnitude. The worst-case bound deliberately covers both without asserting a distribution. Repeating the same number in a table does not establish which pattern real observations follow, nor does it establish independence.

If a separate statistical model supplies equal standard uncertainties u and zero covariance for the two total estimates, its slope standard uncertainty is sqrt(2)u/|n2 minus n1|. Selecting u = 0.2 and separation two gives approximately 0.141421 centimeters per turn. That is a different input interpretation from a hard bound of 0.2, so it is not a substitute for the earlier 0.2 worst-case slope allowance. With covariance c retained, the numerator becomes sqrt(u1 squared + u2 squared − 2c). Unknown covariance must not silently be set to zero.