§1Coordinates and the straight line
Analytical geometry replaces drawing with arithmetic: a point is a pair of numbers, a line an equation, and every intersection a solvable system.
Two lines are parallel when their slopes match and perpendicular when the slopes multiply to −1. The angle between slopes m₁ and m₂ is tan β = (m₂ − m₁)/(1 + m₁m₂).
A slot runs from (20, 15) to (80, 55). Slope m = 40/60 = 0.6667, so the slot climbs at arctan 0.6667 = 33.69°, and its equation is y = 0.6667x + 1.667. A pin centre at (60, 10) sits |0.6667 × 60 − 10 + 1.667|/√(1.4444) = 26.35 mm off the slot axis — clearance settled without a drawing.
§2Polar coordinates
Some parts are born polar — anything with holes on a pitch circle, cams, anything a rotary table touches. Converting between systems is two lines of trigonometry.
The point (86.603, 50.000) has r = √(7500 + 2500) = 100.000 and θ = arctan(50/86.603) = 30° — and rotating back, (100, 30°) returns (86.603, 50.000). The bolt-circle ordinate table on the Solution of Triangles page is this conversion run seven times.
§3The circle
The most manufactured shape in existence. Its equation, its parts and its tangent behaviour cover most layout questions.
x² + y² − 12x + 8y − 48 = 0: complete the squares — (x − 6)² + (y + 4)² = 48 + 36 + 16 = 100. Centre (6, −4), radius 10.
Tangent facts that settle layout arguments: a tangent is perpendicular to the radius at its point of contact; the two tangent lengths from an external point are equal; and an angle inscribed in a semicircle is a right angle — the geometry behind checking a bore with a square.
§4Ellipse, parabola and hyperbola
Slice a cone at different angles and the three conics appear. In the workshop the ellipse turns up wherever a cylinder is cut on a slant; the parabola wherever something is focused or thrown.
No exact closed form exists for the perimeter — the approximation above is astonishingly good. Parabola: y² = 4ax, focus at (a, 0); every ray parallel to the axis reflects through the focus. Hyperbola: x²/a² − y²/b² = 1, asymptotes y = ±(b/a)x.
a = 50, b = 30: h = (20/80)² = 0.0625, so C ≈ π × 80 × [1 + 0.1875/11.953] = 255.27 mm. Numerical integration of the exact perimeter integral gives 255.270 — Ramanujan’s formula is right to the last figure shown.
§5Areas of plane figures
The mensuration table — every flat figure a drawing office meets, one line each.
| Figure | Area | Notes |
|---|---|---|
| Triangle | ½ × base × height | or Heron’s rule from three sides |
| Rectangle | l × w | diagonal = √(l² + w²) |
| Parallelogram | base × height | height ⊥ base, not the slant side |
| Trapezium | ½(a + b) × h | a, b the parallel sides |
| Regular polygon, n sides s | ¼ n s² cot(180°/n) | coefficients tabulated in §6 |
| Circle | πr² = πd²/4 | C = πd |
| Annulus (ring) | π(R² − r²) | = π(R−r)(R+r) |
| Sector | ½ r² θ | θ in radians |
| Segment | ½ r² (θ − sin θ) | tabulated in §7 |
| Ellipse | π a b | perimeter: §4 |
| Parabolic segment | ⅔ × chord × height | chord ⊥ axis |
| Cycloid (one arch) | 3πr² | arch length = 8r |
| Fillet (square corner, radius r) | r²(1 − π/4) ≈ 0.2146 r² | corner area outside the arc |
Irregular flat areas yield to the trapezoidal rule — divide into strips of equal width w and sum A ≈ w(½y₀ + y₁ + ··· + yₙ₋₁ + ½yₙ) — or, with an odd trick more accuracy, Simpson’s rule. Both are the Geometry page’s handshake with the Integrals section of Algebra and Equations.
Contents§6Regular polygons
For a regular polygon everything scales from the side: circumscribed radius, inscribed radius and area follow the coefficients below, computed for unit side.
| n — name | Interior ∠ | Circumradius R | Inradius r | Area |
|---|---|---|---|---|
| 3 — triangle | 60.0° | 0.57735 | 0.28868 | 0.43301 |
| 4 — square | 90.0° | 0.70711 | 0.50000 | 1.00000 |
| 5 — pentagon | 108.0° | 0.85065 | 0.68819 | 1.72048 |
| 6 — hexagon | 120.0° | 1.00000 | 0.86603 | 2.59808 |
| 7 — heptagon | 128.6° | 1.15238 | 1.03826 | 3.63391 |
| 8 — octagon | 135.0° | 1.30656 | 1.20711 | 4.82843 |
| 9 — nonagon | 140.0° | 1.46190 | 1.37374 | 6.18182 |
| 10 — decagon | 144.0° | 1.61803 | 1.53884 | 7.69421 |
| 11 — hendecagon | 147.3° | 1.77473 | 1.70284 | 9.36564 |
| 12 — dodecagon | 150.0° | 1.93185 | 1.86603 | 11.19615 |
| For side s: multiply R and r by s, area by s². R is centre-to-corner (across corners = 2R), r is centre-to-flat (across flats = 2r). | ||||
Hex bar of 24 mm across flats: across corners = AF × 2/√3 = 27.71 mm — the minimum round it can be milled from. A 50 mm square’s diagonal is 50√2 = 70.71 mm.
§7Circular segments and tank contents
The segment — the sliver between chord and arc — prices every partial fill, every flat on a bar, every sight-glass reading. Its table below is computed for unit radius; scale areas by r².
| θ | Arc s | Chord c | Height h | Area |
|---|---|---|---|---|
| 10° | 0.17453 | 0.17431 | 0.00381 | 0.00044 |
| 20° | 0.34907 | 0.34730 | 0.01519 | 0.00352 |
| 30° | 0.52360 | 0.51764 | 0.03407 | 0.01180 |
| 40° | 0.69813 | 0.68404 | 0.06031 | 0.02767 |
| 50° | 0.87266 | 0.84524 | 0.09369 | 0.05331 |
| 60° | 1.04720 | 1.00000 | 0.13397 | 0.09059 |
| 70° | 1.22173 | 1.14715 | 0.18085 | 0.14102 |
| 80° | 1.39626 | 1.28558 | 0.23396 | 0.20573 |
| 90° | 1.57080 | 1.41421 | 0.29289 | 0.28540 |
| 100° | 1.74533 | 1.53209 | 0.35721 | 0.38026 |
| 110° | 1.91986 | 1.63830 | 0.42642 | 0.49008 |
| 120° | 2.09440 | 1.73205 | 0.50000 | 0.61418 |
| 130° | 2.26893 | 1.81262 | 0.57738 | 0.75144 |
| 140° | 2.44346 | 1.87939 | 0.65798 | 0.90034 |
| 150° | 2.61799 | 1.93185 | 0.74118 | 1.05900 |
| 160° | 2.79253 | 1.96962 | 0.82635 | 1.22525 |
| 170° | 2.96706 | 1.99239 | 0.91284 | 1.39671 |
| 180° | 3.14159 | 2.00000 | 1.00000 | 1.57080 |
| Scale s, c, h by r and area by r². Given chord and height only: r = (c²/4 + h²)/2h. | ||||
A horizontal tank Ø1.2 m × 3.0 m long holds fluid 400 mm deep. r = 0.6, h = 0.4: cos ½θ = 0.2/0.6, so θ = 141.06° = 2.4619 rad. Segment area = ½ × 0.36 × (2.4619 − sin 141.06°) = 0.330 0 m²; volume = 0.3300 × 3.0 = 0.990 m³ = 990 litres.
The same segment arithmetic gives the depth of a milled flat (worked on the Solution of Triangles page, §9) and the area lost to a keyway in a shaft section.
§8Volumes of solids
One table for the standard solids, then two rules — prismoidal and Pappus — that dispatch the shapes the table forgot.
| Solid | Volume | Surface (curved / total) |
|---|---|---|
| Prism / cylinder | A_base × h | cyl: 2πrh (curved) |
| Pyramid / cone | ⅓ A_base × h | cone: πr × slant |
| Frustum of cone | ⅓πh(R² + Rr + r²) | π(R + r) × slant |
| Sphere | 4⁄3 πr³ | 4πr² |
| Spherical cap, height h | ⅓πh²(3r − h) | zone: 2πrh |
| Torus (mean R, section r) | 2π²Rr² | 4π²Rr |
Sphere r = 60: V = 4⁄3π × 216 000 = 904 779 mm³, A = 4π × 3600 = 45 239 mm². Grinding a cap 15 mm high off it removes ⅓π × 225 × 165 = 38 877 mm³ and exposes a zone of 2π × 60 × 15 = 5 655 mm².
A hopper tapers from a 400 mm square to a 200 mm square over 300 mm. Aₘ (side 300) = 90 000: V = 50 × (160 000 + 360 000 + 40 000) = 28.0 litres — and the exact frustum formula returns the identical figure, because the prismoidal rule is exact for any solid whose section varies quadratically.
A full torus of section radius 6 mm on mean radius 40 mm: V = 2π × 40 × π × 36 = 28 425 mm³, surface 4π² × 40 × 6 = 9 475 mm². Pappus turns every ring, rim and flange into one multiplication.
§9Circles in circles; rollers on a shaft
A ring of equal circles — rollers round a shaft, wires in a cable, pins in a bore — obeys one relation between count, circle size and ring size.
| n | Centre circle E/d | Enclosing bore D/d | Central shaft D/d |
|---|---|---|---|
| 3 | 1.15470 | 2.15470 | 0.15470 |
| 4 | 1.41421 | 2.41421 | 0.41421 |
| 5 | 1.70130 | 2.70130 | 0.70130 |
| 6 | 2.00000 | 3.00000 | 1.00000 |
| 7 | 2.30476 | 3.30476 | 1.30476 |
| 8 | 2.61313 | 3.61313 | 1.61313 |
| 9 | 2.92380 | 3.92380 | 1.92380 |
| 10 | 3.23607 | 4.23607 | 2.23607 |
| 11 | 3.54947 | 4.54947 | 2.54947 |
| 12 | 3.86370 | 4.86370 | 2.86370 |
| E = d / sin(180°/n). Bore = E + d encloses the ring; shaft = E − d fills its centre. | |||
Eight 12 mm rollers touching each other and a central shaft: E = 12/sin 22.5° = 31.358 mm, so the shaft is Ø19.358 mm and the measurement over the rollers is E + d = 43.358 mm — a needle-bearing layout settled in two lines.
Packing circles into a circle without the ring constraint is a genuinely irregular problem — optimal counts jump unpredictably as the ratio grows, and layout work should rely on published packing data or a drawing trial rather than a formula. The single-ring case above, and rings nested with a centre circle (add one: the centre takes D_bore ≥ 3d at n = 6), cover most machine-element situations.
Contents§10Quick reference
The working core of the page on one card rack.
Line
y = mx + b
⊥ slopes: m₁m₂ = −1
Circle parts
s = rθ A_seg = ½r²(θ − sinθ)
c = 2r sin ½θ
Ellipse
A = πab
C ≈ π(a+b)(1 + 3h/(10+√(4−3h)))
Polygon
A = ¼ns² cot(180°/n)
hex AC = AF × 2/√3
Solids
sphere V = 4⁄3πr³
cap V = ⅓πh²(3r−h)
Two big rules
V = (h/6)(A₁+4Aₘ+A₂)
Pappus: V = 2πȳA
