Final Assessment
The course ends with a practical assessment: knowledge questions, record reviews, and two real selection/diagnosis projects. It is designed like the rest of the course — applied, not rote. Aim to justify every answer, not just recall it.
Concept
The assessment has three parts, scored out of 100: basic questions (40) testing the core concepts, record reviews (20) checking you can read a worn tool / selection record, and two projects (40) requiring a written tool selection and a wear diagnosis. Pass when you reach the threshold and every key technical point is correct.
Why It Matters
Recalling that "G00 is rapid" proves little. Applying the locked numbers, choosing a grade for a material, and diagnosing a worn edge prove you can do the job. The projects mirror the way tooling decisions actually happen on the shop floor.
How It Works
Scoring
| Part | Items | Points |
|---|---|---|
| Basic questions | 20 short | 40 |
| Record reviews | 2 | 20 |
| Practical projects | 2 (selection + diagnosis) | 40 |
| Total | 100 |
Pass threshold: reach 80/100 and correct every key technical point (e.g. the locked numbers and the wear-to-cause mapping).
How the locked numbers are derived
The basic-question arithmetic is the same on every locked example; learn it once and it repeats. Spindle speed from cutting speed: n = 1000·vc / (π·D). With vc = 120 m/min and D = 10 mm that is 1000·120 / (3.1416·10) = 3819.72 RPM — you would run it down to about 3800 RPM on the machine. Feed rate then is vf = n·zc·fz = 3800·4·0.04 = 608 mm/min. Surface finish from feed f and nose radius rε: Ra ≈ f²/(32·rε) = 0.10²/(32·0.4) = 0.781 µm. And the drill point for a 118° drill: h = (D/2)/tan(59°) = 5·tan(31°) = 3.004 mm of extra depth beyond the measured hole. If you can reproduce these four by hand, the numeric basic questions are secure.
How to prepare
Work through the locked examples by hand before the assessment: compute 3819.72 RPM from vc and diameter, compute 608 mm/min feed, convert f and nose radius to 0.781 µm, and recall the 3.004 mm drill point. Be ready to explain why a higher-price insert can win on cost per part (page 29), and to name the wear mode from a description (page 25/27). The two projects use the printable worksheets — fill them as if for a real job, justifying each field rather than guessing.
Worked Example (sample questions)
- Locked baseline: Ø10 tool, vc = 120 m/min. Compute spindle RPM. (Answer: n = 1000·vc /(π·D) = 3819.72 RPM.)
- n = 3800, z = 4, fz = 0.04 — feed rate? (608 mm/min.)
- Material welded to the insert nose — name it and give the first corrective direction. (Built-up edge; first check edge geometry, chip load and coolant, then test speed within the grade range — not an automatic raise.)
- Ø10 drill at 118° through a plate — point height to add? (3.004 mm.)
Record reviews and the two projects use the printable worksheets: fill a selection card and a wear-trial log, then justify each choice.
Worked scoring example
How the threshold actually decides. Suppose a candidate scores 34/40 on the basic questions, 16/20 on the record reviews, and 32/40 on the two projects — a total of 82. The raw score is above 80, so it would pass on points alone. But the pass rule has a second condition: every key technical point must be correct. If that same candidate computed the locked RPM as 380 (a factor-of-ten unit slip) or called built-up edge a flank-wear problem, the key-point condition fails despite 82 points. Conversely, a candidate on 84 points who reproduces all four locked numbers and names the wear modes correctly passes. The two conditions are both required: the points confirm breadth, the key points confirm that the essentials are not guessed. This is why memorizing a list is not enough — the arithmetic and the cause-mapping have to be right.
What the two projects actually ask for
The 40 project points are not essays — they are structured deliverables on the worksheets. The selection project gives you a short brief (a material, a feature, a tolerance, a quantity) and asks you to fill the selection card: read the drawing and material, choose operation and geometry, choose substrate and coating, set starting vc/fz from the catalog, and check holder/overhang/coolant — each field justified, exactly the page-13 workflow. The diagnosis project gives you a description of a failing edge (a welded nose, a checked rake, a chipped corner) and asks you to name the wear mode, separate wear from overload, state the single first change, and describe how you would verify it on the wear-trial log — exactly the page-27 loop. Grading is against the same rubric used through the course: a correct tool choice with no justification earns partial credit, and a correct number with the wrong unit earns none. Treat the projects as mini versions of cases 32 and 33, not as new material.
Why a right number with the wrong unit still fails
The rule "a correct number with the wrong unit earns no points" is not pedantic — on the shop floor the unit error is the accident. Compute the locked feed as 608 and enter 608 inches per minute instead of mm/min, and the tool moves about 25 times too far and fast, crashing the cut. Drop the ×1000 in the RPM formula and get 3.8 instead of 3800, which looks like a decimal slip but is a thousand-fold speed error. These are exactly the failures the course drills against (page 03): the arithmetic is the easy part; the unit bookkeeping is where a real program goes wrong. That is why the assessment weights units and asks you to reproduce the locked conversions by hand, and why the practice questions deliberately mix mm/min, RPM and µm. A candidate who can derive 3819.72 RPM but cannot say which diameter or speed produced it has not actually learned the system — they have memorised an answer. The key-point condition exists to catch that gap, not to be difficult.
How to use a failed assessment as a study map
The point of the two-condition pass rule is not to gate-keep; it is to tell you exactly what to re-learn. If you score 82 but fail a key point, look at which one: a slip on the RPM formula means revisit page 03 and re-derive n, vf and Ra by hand until it is automatic; naming built-up edge as flank wear means re-read pages 25–27 until the morphology-to-cause mapping is instant; a selection project that picked a brittle grade for an interrupted cut means revisit pages 09–13 and re-run case 32. Each failed key point points at one chapter, not at "the whole course." This is why the assessment is best treated as a diagnostic you can take more than once: the first attempt exposes the weak link, you study that chapter specifically, and a second attempt confirms it closed. A candidate who skips the practical half because recall feels easy, or who memorises the four locked numbers without understanding their units, learns the wrong lesson — the assessment rewards the same applied judgement the shop requires, and retaking it after targeted review is how the module actually turns into skill.
Why the record reviews are weighted at 20 points
The 20 record-review points test something the multiple-choice questions cannot: reading an actual worn insert or a filled selection card and deciding what it means. On the shop floor you rarely start from a clean problem; you start from a used tool, a half-filled worksheet, or a strange finish, and you have to infer what went wrong. The record review gives you that artefact — a description of a worn edge, a proposed speed, a recorded life — and asks you to judge whether the record is internally consistent, whether the unit is right, and what the next action should be. This is why it sits between recall and the full projects: it is the step that proves you can read evidence rather than only generate it. A common trap is treating the record review as a free recall question and skimming past the numbers; the points live in noticing that the logged feed lacks a unit, that the proposed grade contradicts the interruption, or that the recorded life does not match the claimed speed. Reading a record critically is itself the skill page 27 trains, and the 20 points confirm it.
Why 60 of 100 points are practical, not recall
The weighting is deliberate: recall questions carry only 40 points, while record reviews and the two projects carry the other 60. That distribution mirrors how tooling decisions are actually judged on the shop floor, where nobody is asked to recite the RPM formula from memory but everyone is expected to pick a sensible grade for a described job and explain why. A candidate who can memorise all four locked numbers but cannot justify a selection, or who misreads a worn insert, would score high on paper and fail in practice — so the assessment puts the weight where the job is. This also tells you how to study: doing the worksheets and walking cases 32–34 is worth more than re-reading the recall chapters, because the practical half is where the points are lost. The two-condition pass rule reinforces the same point — breadth across the basic questions plus correctness on the applied tasks, never one alone. Treat the assessment as a simulation of a tooling review, not a vocabulary quiz.
Why justification carries the project marks
In the two projects, a correct final choice earns only part of the available points; the rest comes from the reasoning shown on the worksheet. This mirrors how a real tooling decision is reviewed: a supervisor does not just want to know that you picked a coated carbide insert, they want to see that you read the material condition, recognised the interruption, matched the chipbreaker feed window, and checked the overhang and coolant. A lucky guess that lands on the right grade for the wrong reason cannot be trusted to repeat on the next, slightly different job, and the grading deliberately does not reward it. This is also why the assessment is so forgiving about an imperfect number but strict about the logic: if your RPM is 3800 instead of 3819.72 because you rounded to the nearest machine step, that is fine and expected; if you cannot say which diameter and speed produced it, the answer is meaningless. When you write up a project, treat each field as a sentence of evidence — "material is 280 HB QT, so a coated steel grade; cut is interrupted, so a tougher edge" — rather than a bare selection.
How the locked numbers reappear across modules
The assessment deliberately reuses the same four locked numbers from page 03 across later pages because real tooling thinking is built on a few anchors rather than a new calculation for every cut. The 3819.72 RPM feeds the 608 mm/min feed, which feeds the 0.781 µm finish example used on the turning and boring pages, and the 3.004 mm drill point reappears in the hole-quality case. Seeing the same numbers again in a selection or diagnosis project is therefore not repetition for its own sake: it tests whether you can carry a baseline through a multi-step decision instead of treating each page as isolated facts. A practical way to prepare is to start from vc = 120, D = 10 and rebuild the whole chain on one blank sheet — RPM, feed, expected finish, drill depth, and which wear mode each setting would produce — until it runs without notes. If you can walk the chain forward and backward, the assessment's numeric and applied halves stop being separate questions and become variations on the same system.
Common Mistakes
- Answering numbers without units. Fix: always state mm/min, RPM, µm.
- Describing a tool choice without justification. Fix: link it to material and operation.
- Skipping the record review. Fix: the projects and record reviews carry 60 of 100 points — a perfect recall score still fails the practical half.
Practice
Part A — Basic Questions (Q01–Q20, 2 pts each = 40)
Answer all 20. Each is worth 2 points; a right number with the wrong unit earns 0. Answers are given in the collapsibles for self-check.
- Q01. Locked baseline: Ø10 tool, vc = 120 m/min. Compute spindle RPM.
Answer
n = 1000·vc /(π·D) = 1000×120 /(3.1416×10) = 3819.72 RPM (run ~3800).
- Q02. n = 3800, z = 4, fz = 0.04 mm/tooth — feed rate vf?
Answer
vf = n·zc·fz = 3800×4×0.04 = 608 mm/min.
- Q03. f = 0.10 mm/rev, rε = 0.4 mm — ideal Ra?
Answer
Ra ≈ f²/(32·rε) = 0.01/(32×0.4) = 0.781 µm.
- Q04. Ø10 drill at 118° through a plate — point height to add?
Answer
h = (D/2) / tan(α/2) = 5 / tan(59°) = 5·tan(31°) ≈ 3.004 mm. This is the ideal axial point height for a Ø10 drill with a 118° included point angle; through-hole travel also requires verified exit clearance.
- Q05. CTT-M1: Ø8 end mill, vc = 200 m/min. n?
Answer
n = 1000×200 /(3.1416×8) = 7958 RPM.
- Q06. CTT-M1: n = 7958, z = 3, fz = 0.04. vf?
Answer
vf = 7958×3×0.04 ≈ 955 mm/min.
- Q07. CTT-M1: ap = 2, ae = 1.6, vf = 955. MRR Q?
Answer
Q = ap·ae·vf/1000 = 2×1.6×955/1000 ≈ 3.06 cm³/min.
- Q08. CTT-T1: radial ap = 2 mm on a lathe — how much does the diameter drop per pass?
Answer
4 mm per pass (radial stock 2 mm on each side).
- Q09. CTT-T1: measured Ø40.028 vs upper limit 40.020. Accept?
Answer
No — over by 0.008 mm; the mean being inside does not release it.
- Q10. CTT-H1: pre-drill Ø19.80 to ream Ø20 — radial stock?
Answer
(20−19.80)/2 = 0.10 mm radial.
- Q11. CTT-H1: third section 20.023 vs upper limit 20.021. Over by?
Answer
0.002 mm.
- Q12. A twist drill cuts mainly with which features?
Answer
The two main cutting lips; the margins support/guide, the chisel edge extrudes.
- Q13. Doubling boring-bar overhang L from 1:1 to 2:1 — deflection grows by about?
Answer
δ∝L³, so about 8×.
- Q14. Doubling a solid round bar diameter — stiffness grows by?
Answer
I ∝ D4, so stiffness 16× (deflection ~1/16).
- Q15. ISO group P / M / K refer to which families?
Answer
P = steel, M = stainless, K = cast iron (N non-ferrous, S superalloys+titanium, H hardened).
- Q16. Cost: insert A 120/4 edges/50 parts + 0.04 downtime. Per part?
Answer
120/4/50 = 0.60 + 0.04 = 0.64.
- Q17. Cost: insert B 160/4 edges/80 parts + 0.025 downtime. Per part?
Answer
160/4/80 = 0.50 + 0.025 = 0.525.
- Q18. Welded material on the edge (BUE) — first move?
Answer
It is adhesion; check edge/geometry/chip-load/coolant first, and test speed within the manufacturer range — not an automatic "raise speed".
- Q19. Why ream last, never to correct a wandering hole?
Answer
A reamer follows the existing hole; it sizes/finishes but cannot move position. Bore for position, ream to size.
- Q20. A tool wins on cost per part. The locked case says B is cheaper — why?
Answer
More good parts per edge spreads its price lower (0.50 vs 0.60) and less indexing (0.025 vs 0.04) — not because B cuts a shorter cycle.
Part B — Record Reviews (CR1–CR2, 10 pts each = 20)
You are given a record. State whether it is internally consistent, find the error, and give the corrected action. 10 pts per record: 4 for identifying the defect, 3 for the corrected value/action, 3 for the justification.
CR1 (10 pts)
A selection card reads: "Ø8 3-flute end mill on 6061-T6, vc = 80 m/min, n = 7958 rpm, vf = 955 mm/min." Retain fz = 0.04 mm/tooth and z = 3. Find the inconsistency and correct it.
Model answer (10)
The n = 7958 corresponds to vc = 200 on Ø8, not vc = 80. At vc = 80, n = 1000×80/(π×8) = 3183 rpm, and vf = 3183×3×0.04 = 382 mm/min. Either the vc should be 200 (keeping n/vf = 7958/955) or the n/vf must be recomputed for 80. (4 points: vc–n mismatch; 3: correct n/vf; 3: vf follows n·zc·fz.)
CR2 (10 pts)
A turning log records three consecutive parts: "diameters 40.006 / 40.014 / 40.028, upper limit 40.020. Mean = 40.016 — OK, release." Find the error and correct the decision.
Model answer (10)
The mean is inside, but part 3 at 40.028 exceeds the upper limit by 0.008 mm. Acceptance is by the worst of the three consecutive parts, not the average. Decision: reject part 3, plan a wear limit/offset before the diameter can exceed 40.020, and rule out setup/thermal first. (4: average-vs-per-part error; 3: reject part 3 + plan wear limit; 3: worst of the three parts controls.)
Part C — Projects (P1–P2, 20 pts each = 40)
Use the printable selection card and wear-trial log. Each project: correct choice with justification earns full marks; a bare answer earns partial.
P1 (20 pts) — Tool selection
Brief: 6061-T6 plate, open pocket 40×30×8, inside corners R5, finished wall/floor, moderate quantity on a VMC. Fill the selection card: material → operation → geometry (diameter/flutes) → substrate/coating → starting vc/fz/n/vf → holder/overhang/coolant. Justify each field.
Rubric (20)
Material/BUE reasoning (4); Ø8 3-flute chosen because radius 4 < R5 and 3 flutes clear chips (5); polished-edge carbide with an aluminum-appropriate coating supported by product data (3); vc=200/fz=0.04 stated as classroom assumptions, n≈7958, vf≈955 (5); short ER holder, flood coolant (3). No justification → half marks.
P2 (20 pts) — Wear diagnosis
Brief: OD turning 42CrMo4 QT, diameters creep 40.006→40.014→40.028 against upper limit 40.020; finish degrades together; edge shows a smooth growing flank land. Name the wear mode, separate wear from overload, state the single first change, and describe how you verify on the trial log.
Rubric (20)
Gradual flank wear, not sudden fracture/chipping (5); rule out seating/offset/thermal before compensating (5); first verify wear and size trend, then within tolerance compensate or set a wear limit / index as evidence supports (5); verify by VB-vs-time and re-measure diameter, record on the log (5). Diameter compensation cannot repair a damaged edge or poor surface finish. Calling it BUE or "raise speed" fails the key-point condition.
Pass: total ≥ 80/100 and every key technical point (the four locked numbers, the wear-to-cause mapping, and the worst-cross-section rule) correct.
Q1. What score is needed to pass?
80/100, with every key technical point correct.
Q2. What do the two projects ask you to produce?
A written tool selection and a wear diagnosis, each justified.
Sources
- Reference (general principles, not a recommendation for a specific insert): Sandvik Milling Guide and Sandvik Turning Handbook (C-1020-18). All worked numbers on this page are classroom assumptions unless stated as measured.
- Locked numeric baselines from page 03; case methods from M6 and M8.