PYTHON · NUMPY

Tolerances: the band a part must live in

Assign tolerances to ≥2 critical dimensions and predict how tightening or loosening each changes cost and fit, reaching a functioning fit at the lowest cost.

01Challenge

Make the bracket's bore accept a Ø11.98–12.00 dowel pin as cheaply as you can. Type a tolerance band on the bore (D1), a ± value or asymmetric +x/−y, then Run Fit Check across 200 simulated parts. Enter something absurdly tight and cost explodes; loose and pins jam or rattle. Both are the intended productive failure.

02Model

Your CAD model says 12.000, but to how many zeros? No machine makes exactly twelve. A real dimension isn't a number, it's a band: the tolerance. And the band isn't arbitrary, it comes from function. This bore's job is to accept this pin: too small and it won't go, too big and it rattles.

Now the trap: tighter feels safer, but tightness costs money (slower cuts, finer tools, more inspection, more scrap), and cost climbs steeply as the band shrinks. So the right answer is the loosest band that still functions. Tighter than that, you're burning money for precision the part never needed.

A part must land inside its tolerance band.

03Guided practice

Step A (worked): bore 12.00 +0.05/+0.02 → band 0.03, PASS but cost 3.33× (over-tight).
Step B (fade): widen to keep PASS and drop cost: bore 12.010 +0.080/−0.000 → band 0.080, cost 1.25×; the 0.080 band is both cheapest and loosest (push wider and it fails loose; the band has a ceiling).
Step C (independent): tolerance the slot D3 (transition fit on a 7.95–7.99 rib) from scratch and predict that halving the bore band doesn't change the slot's cost.

04Feedback

PASS WHEN Bore D1 and slot D3 both PASS (every sampled part functions) AND total relative cost ≤ 1.40× (mean of per-dimension multipliers), on the locked seed.

If your run fails, check:

FAIL_TIGHT on D1: smallest bore is only 0.005 mm over the largest pin; the pin can jam. Raise the bore's lower limit by ≥0.005 mm.
FAIL_LOOSE on D1: largest bore gives 0.130 mm clearance, past the 0.110 mm limit. Lower the upper limit (loosest passing band is 0.080 mm).
Fit works, but total cost is 2.17×: your 0.03 mm bore band (3.33×) is tighter than function needs. Each band can open to 0.080 mm; widen toward there.
D3 has no band; an untoleranced critical dimension can't be quoted or inspected. Assign a band yielding −0.010 to +0.110 mm clearance on the rib.

05Retrieve & space

From M1 (DFM): which earlier choice (wall, draft, or fillet) most affects whether a tight bore is even machinable? → fillet/edge access & rigidity.
From M2: to hold a tighter bore band, which way do feed and depth move, and what does it cost? → lower feed / lighter finishing pass → more time.
Within 3.1: 'Loosest band that still ____ = lowest cost' → functions.

06Mastery & project

On a fresh seed with different pin measurements, tolerance both bore and slot so all 200 sampled parts function AND total relative cost ≤ 1.40×, with a correct prediction of how halving one band moves cost. Unlocks 3.2.

Feeds the M5 capstone: the bore and slot tolerances here become the LSL/USL spec limits the M5 line must hit at yield. This lesson defines the target band everything downstream must repeatably hit.

← Sequencing a process plan Where variation comes from →

PYTHON · NUMPY

Tolerances: the band a part must live in

The twin's fit/cost model in numpy: worst-case clearance for the fit verdict, cost ∝ 1/band, total = mean of per-dimension multipliers. Edit your bands and run.

You read

the arrays and values already in scope

You change

the code you write in each cell

Fixed

the dataset and the checks that grade you