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ArticlePublished 11 Jul 2026Updated 12 Jul 20269 min readBy Kevin Jogin
KEVOS® Knowledge Library · Engineering → Mathematics

Engineering / Mathematics

Geometry

Lines, circles, polygons and solids — the shapes parts are actually made of. This reference carries the equations of analytical geometry, the mensuration formulas for area and volume, and the computed tables that turn both into shop-floor numbers.

  • Reading time · 9 min
  • 10 sections
  • Polygon & segment tables computed
  • Tank contents worked in litres
O r θ chord tangent sector A = ½ r² θ · c = 2r sin ½θ · tangent ⊥ radius
Doc №KL-ENG-MATH-006
SectionEngineering → Mathematics
Sheet1 of 1
DrawnKEVOS®
Date2026-07-11

§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.

The line y = mx + b  m = y₂ − y₁x₂ − x₁ = tan (slope angle)  y − y₁ = m(x − x₁)
Distances between points: (x₂−x₁)² + (y₂−y₁)²  point to line mx − y + b = 0: |mx₀ − y₀ + b|m² + 1

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).

Example 1 — a slotted link as an equation

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.

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§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.

x = r cos θ y = r sin θ   r = x² + y² θ = arctan(y/x) — quadrant by the signs of x, y
Example 2 — both directions

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.

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§3The circle

The most manufactured shape in existence. Its equation, its parts and its tangent behaviour cover most layout questions.

Equation (x − h)² + (y − k)² = r²  centre (h, k), radius r
Example 3 — centre and radius from the general form

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.

Parts (angle θ at centre, radians) circumference = 2πr arc s = rθ chord c = 2r sin ½θ sector A = ½r²θ segment A = ½r²(θ − sin θ)
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§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.

a b F₁ F₂
Fig. 1. Ellipse: the sum of distances to the two foci is constant (= 2a); the foci sit at ±√(a² − b²) from centre.
Ellipse + = 1  A = πab
Perimeter (Ramanujan) C ≈ π(a+b)[1 + 3h/(10 + 4 − 3h)], h = (a−b)²(a+b)²

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.

Example 4 — perimeter of a 100 × 60 elliptical cover

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.

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§5Areas of plane figures

The mensuration table — every flat figure a drawing office meets, one line each.

Areas of plane figures
FigureAreaNotes
Triangle½ × base × heightor Heron’s rule from three sides
Rectanglel × wdiagonal = √(l² + w²)
Parallelogrambase × heightheight ⊥ base, not the slant side
Trapezium½(a + b) × ha, b the parallel sides
Regular polygon, n sides s¼ n s² cot(180°/n)coefficients tabulated in §6
Circleπr² = πd²/4C = πd
Annulus (ring)π(R² − r²)= π(R−r)(R+r)
Sector½ r² θθ in radians
Segment½ r² (θ − sin θ)tabulated in §7
Ellipseπ a bperimeter: §4
Parabolic segment⅔ × chord × heightchord ⊥ axis
Cycloid (one arch)3πr²arch length = 8r
Fillet (square corner, radius r)r²(1 − π/4) ≈ 0.2146 r²corner 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.

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§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.

Regular polygons, side s = 1 (computed)
n — nameInterior ∠Circumradius RInradius rArea
3 — triangle60.0°0.577350.288680.43301
4 — square90.0°0.707110.500001.00000
5 — pentagon108.0°0.850650.688191.72048
6 — hexagon120.0°1.000000.866032.59808
7 — heptagon128.6°1.152381.038263.63391
8 — octagon135.0°1.306561.207114.82843
9 — nonagon140.0°1.461901.373746.18182
10 — decagon144.0°1.618031.538847.69421
11 — hendecagon147.3°1.774731.702849.36564
12 — dodecagon150.0°1.931851.8660311.19615
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).
Example 5 — hexagon across corners, square across diagonal

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.

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§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.

Circular segments for radius r = 1 (computed)
θArc sChord cHeight hArea
10°0.174530.174310.003810.00044
20°0.349070.347300.015190.00352
30°0.523600.517640.034070.01180
40°0.698130.684040.060310.02767
50°0.872660.845240.093690.05331
60°1.047201.000000.133970.09059
70°1.221731.147150.180850.14102
80°1.396261.285580.233960.20573
90°1.570801.414210.292890.28540
100°1.745331.532090.357210.38026
110°1.919861.638300.426420.49008
120°2.094401.732050.500000.61418
130°2.268931.812620.577380.75144
140°2.443461.879390.657980.90034
150°2.617991.931850.741181.05900
160°2.792531.969620.826351.22525
170°2.967061.992390.912841.39671
180°3.141592.000001.000001.57080
Scale s, c, h by r and area by r². Given chord and height only: r = (c²/4 + h²)/2h.
O θ h liquid
Fig. 2. Part-filled horizontal tank: depth h fixes the half-angle through cos ½θ = (r − h)/r; the wetted cross-section is the segment ½r²(θ − sin θ).
Example 6 — litres in a part-filled tank

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.

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§8Volumes of solids

One table for the standard solids, then two rules — prismoidal and Pappus — that dispatch the shapes the table forgot.

Volumes and surfaces of solids
SolidVolumeSurface (curved / total)
Prism / cylinderA_base × hcyl: 2πrh (curved)
Pyramid / cone⅓ A_base × hcone: πr × slant
Frustum of cone⅓πh(R² + Rr + r²)π(R + r) × slant
Sphere4⁄3 πr³4πr²
Spherical cap, height h⅓πh²(3r − h)zone: 2πrh
Torus (mean R, section r)2π²Rr²4π²Rr
Example 7 — a 60 mm ball and its cap

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.

Prismoidal formula V = h6(A₁ + 4Aₘ + A₂)  A₁, A₂ the end sections, Aₘ the true mid-height section
Example 8 — a square hopper transition

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.

Pappus (Guldinus) rules revolve a plane area: V = 2πȳ · A  revolve a curve: S = 2πȳ · L  (ȳ = centroid radius from the axis)
Example 9 — an O-ring groove’s ring of metal

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.

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§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.

Ring of n circles, diameter d, centres on circle E E = dsin(180°/n)  around a shaft: D_shaft = E − d · inside a bore: D_bore = E + d
A ring of n touching circles of diameter d (computed)
nCentre circle E/dEnclosing bore D/dCentral shaft D/d
31.154702.154700.15470
41.414212.414210.41421
51.701302.701300.70130
62.000003.000001.00000
72.304763.304761.30476
82.613133.613131.61313
92.923803.923801.92380
103.236074.236072.23607
113.549474.549472.54947
123.863704.863702.86370
E = d / sin(180°/n). Bore = E + d encloses the ring; shaft = E − d fills its centre.
Example 10 — eight rollers round a shaft

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.

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§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

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