Chapter 2: Carry out mensuration

Carry out mensuration is a fundamental skill in Biomedical Engineering, especially when dealing with the design and analysis of medical devices and equipment. Accurate measurement of lengths, areas, and volumes is crucial for ensuring that biomedical instruments fit anatomical requirements or operate within specified parameters. In Kenya, where biomedical engineers often work in hospitals and medical research institutions, proficiency in mensuration supports effective problem-solving and innovation.

2.1 Basic Concepts of mensuration

Mensuration is the branch of mathematics that measures geometric figures. From a few measured dimensions it gives rules for finding lengths, areas and volumes. A technician who knows the diameter and length of a tube can work out how much fluid it holds, and a technician who knows the sides of a sensor pad can work out how much conductive gel covers it.

Three quantities are used throughout mensuration. Length measures distance along a line or around a boundary, area measures the amount of flat or curved surface, and volume measures the space a solid occupies.

Every mensuration problem follows the same path: measure the dimensions, choose the formula for the shape, substitute the values with their units, calculate, and check that the answer has a sensible size and the right unit. The formulas for plane figures and solids are developed in Plane Geometry and Solid Geometry.

2.1.1 Importance of mensuration in engineering and real-world applications

Many engineering decisions depend on a measured length, area or volume. In biomedical engineering these numbers decide whether a device fits the body, holds the right amount of fluid, or is made from enough material. A wrong measurement is not only a wrong answer: an undersized housing will not hold its circuit board, and a wrongly calculated reservoir volume gives a wrong refill time.

Where mensuration is used:

• Fitting devices to the body: a cuff, brace or strap is sized from circumference and length measured on the person, and the same measurements decide how much material is cut.
• Capacity of containers: the volume of a syringe barrel, a reservoir or a collection container is found from its cross-sectional area multiplied by its length, so each millilitre mark on a barrel sits a fixed length along it, set by the bore area.
• Tubing and flow: the bore area and length of infusion tubing give the volume of fluid needed to fill it before use, and the bore area also decides how fast the fluid moves for a given flow rate.
• Contact area of sensors: the area of an electrode or sensor pad decides how much of the skin it covers, so pads are specified by their area as well as their shape.
• Material estimates: the surface area of a housing, panel or tube tells the workshop how much paint, coating, foam or fabric to order.
• Storage and sterilising capacity: the internal volume of a sterilising chamber or storage cabinet, compared with the volume of the trays, shows how many items fit.
• Building and general work: floor area for tiling, volume of concrete for a base, and capacity of a water tank all use the same ideas.

Example: A workshop cuts rectangular silicone pads 5 cm by 4 cm from a sheet 40 cm by 25 cm. Find the greatest number of pads from one sheet.

Area of the sheet = 40 × 25 = 1 000 cm²
Area of one pad = 5 × 4 = 20 cm²
Upper limit from area = 1 000 ÷ 20 = 50 pads

Layout A (5 cm along the 40 cm side):
  along 40 cm: 40 ÷ 5 = 8 pads
  along 25 cm: 25 ÷ 4 = 6.25, so 6 whole pads
  total = 8 × 6 = 48 pads

Layout B (4 cm along the 40 cm side):
  along 40 cm: 40 ÷ 4 = 10 pads
  along 25 cm: 25 ÷ 5 = 5 pads
  total = 10 × 5 = 50 pads

Answer: 50 pads, with no waste. The area calculation gives the upper limit, and the layout decides whether the limit is reached: layout A leaves a strip of 40 cm² (1 000 − 48 × 20), which is the area of two pads.

Accuracy and error in measurement

A small error in a length becomes a larger error in an area or a volume, because the length is squared or cubed in the formula. The area of a circle is π × d² ÷ 4, so a diameter measured 1 mm too large on a 10 mm part changes the area by about 21%.

Diameter 10 mm: area = 3.1416 × 10² ÷ 4 = 78.54 mm²
Diameter 11 mm: area = 3.1416 × 11² ÷ 4 = 95.03 mm²
Change = (95.03 − 78.54) ÷ 78.54 = 0.21, which is 21%
Check: (11 ÷ 10)² = 1.21, so the area grows by 21%

Measuring the dimension twice, using an instrument suited to the size (a vernier calliper for small parts, a tape for long ones) and carrying every working figure through to the end before rounding keeps the final answer reliable.

2.1.2 Units of measurement

A measurement is a number together with a unit: 12 mm, 0.45 L and 68 kg are measurements, while 12 alone is not. The unit tells the reader what standard the number is counted in, so two people measuring the same part get the same result.

Two systems of units are met in engineering work. The SI (metric) system is the one used for engineering measurement in Kenya. The Imperial system survives in older drawings, in some imported catalogues and in fitting and tubing lists that quote inches. A technician must read both and convert correctly.

Rules that keep calculations correct:

• Write the unit with every number: a number without a unit cannot be checked and is often misread.
• Use one system in one calculation: convert all lengths to the same unit before multiplying, never multiply millimetres by metres.
• Square the unit for area: a length in cm multiplied by a length in cm gives cm².
• Cube the unit for volume: three lengths in cm multiplied together give cm³.
• Convert before you calculate: converting the dimensions first is safer than converting the answer, because squared and cubed factors are easy to forget at the end.
• Give the answer in the unit the user needs: a reservoir is quoted in litres or millilitres, not cubic millimetres.

2.1.2.1 SI units

The SI is the International System of Units, the modern form of the metric system. Every quantity has a base or derived unit, and larger or smaller units are made by adding a prefix that multiplies the unit by a power of ten. This makes SI conversions simple: they are moves of the decimal point.

The prefixes widely used with length, area and volume are:

The litre (L) is not an SI unit but is accepted for fluid volume: 1 L = 1 000 cm³ = 0.001 m³, and 1 mL = 1 cm³.

Writing SI symbols correctly:

• Leave a space between the number and the symbol: 25 mm, not 25mm.
• Never add an s to a symbol: 5 kg, not 5 kgs.
• Use the correct capitals: kg and mm are lower case, L for litre is a capital, and µm uses the Greek letter mu.
• Choose a prefix that keeps the number easy to read: 0.00025 m is clearer as 0.25 mm or 250 µm.
• Write area and volume with powers: cm² and cm³, not sq cm or cc.

Example: the width of a microfluidic channel measured as 0.00025 m is 0.25 mm (× 1 000) or 250 µm (× 1 000 000).

2.1.2.2 Imperial units

The Imperial system uses units such as the inch, foot, pound and gallon, which are related to each other by whole numbers that are not powers of ten. Each Imperial unit has an exact SI equivalent fixed by international agreement, so conversion is exact when the factors below are used.

The square and cube factors follow from the length factor: 1 in² = 25.4² = 645.16 mm², and 1 in³ = 25.4³ = 16 387.064 mm³ = 16.387064 cm³.

Two cautions apply when reading Imperial data. The US gallon (3.785411784 L) is smaller than the Imperial gallon (4.54609 L), and the US fluid ounce (29.5735 mL) is larger than the Imperial fluid ounce (28.4131 mL), so a data sheet must be checked for which system it uses. Tubing and fittings are often listed by fractions of an inch, such as 1/4 in, which is 6.35 mm.

2.1.2.3 Conversion between different units

Converting a unit means expressing the same quantity in a different unit without changing its size. The method is to multiply by a conversion factor written as a fraction equal to one, arranged so that the unit to be removed cancels.

Conversion factors for length, area and volume:

Procedure for converting a unit:

1. Write the quantity with its unit, for example 2.5 m.
2. Find the relationship between the two units from the tables, for example 1 m = 1 000 mm.
3. Decide the direction: changing to a smaller unit (m to mm) gives a larger number, and changing to a larger unit (mm to m) gives a smaller number.
4. For area, square the length factor (1 m = 100 cm, so 1 m² = 100² = 10 000 cm²), and for volume, cube it (1 m³ = 100³ = 1 000 000 cm³).
5. Multiply or divide so that the unit to be removed cancels.
6. Round only the final answer, and write the new unit beside it.
7. Check the size: if the number moved in the opposite direction to step 3, the factor was used the wrong way round.

Example 1: A length of infusion tubing is 2.5 m long. Express its length in millimetres.

1 m = 1 000 mm
2.5 m = 2.5 × 1 000 mm = 2 500 mm

Answer: 2 500 mm

Example 2: A sensor has an area of 12 mm². Express the area in cm².

1 cm² = 100 mm²
12 mm² = 12 ÷ 100 cm² = 0.12 cm²
Check: 12 mm² is about 3.5 mm by 3.5 mm, and 0.35 cm by 0.35 cm gives 0.12 cm²

Answer: 0.12 cm²

Example 3: A steriliser tray has a surface area of 0.35 m². Express the area in cm².

1 m² = 10 000 cm²
0.35 m² = 0.35 × 10 000 cm² = 3 500 cm²

Answer: 3 500 cm²

Example 4: A pump reservoir holds 450 cm³. Express its capacity in litres and in millilitres.

1 L = 1 000 cm³, so 450 cm³ = 450 ÷ 1 000 L = 0.45 L
1 cm³ = 1 mL, so 450 cm³ = 450 mL

Answer: 0.45 L, which is 450 mL

Example 5: A data sheet lists the outside diameter of a connector as 2.25 in. Express the diameter in millimetres.

1 in = 25.4 mm
2.25 in = 2.25 × 25.4 mm = 57.15 mm

Answer: 57.15 mm

Example 6: A mounting plate is 2 in wide and 3 in long. Find its area in cm², by two methods.

Method 1 (convert the area):
  area = 2 × 3 = 6 in²
  1 in² = 6.4516 cm²
  6 × 6.4516 = 38.7096 cm²

Method 2 (convert the sides first):
  2 in = 2 × 2.54 = 5.08 cm
  3 in = 3 × 2.54 = 7.62 cm
  area = 5.08 × 7.62 = 38.7096 cm²

Answer: 38.71 cm²

Example 7: A scale reads 150 lb. Express the mass in kilograms.

1 lb = 0.45359237 kg
150 lb = 150 × 0.45359237 = 68.0388555 kg

Answer: 68.04 kg

Example 8: A tube is 127 mm long. Express its length in inches (converting from metric to Imperial divides by the factor).

1 in = 25.4 mm
127 mm = 127 ÷ 25.4 in = 5 in

Answer: 5 in

Example 9: A pump delivers 125 mL each minute. Express the flow rate in litres per hour.

Per hour: 125 mL/min × 60 min/h = 7 500 mL/h
Litres: 7 500 ÷ 1 000 = 7.5 L/h

Answer: 7.5 L/h

Common mistakes in conversion:

• Using the length factor for area: 1 m² is 10 000 cm², not 100 cm².
• Using the length factor for volume: 1 m³ is 1 000 000 cm³, not 100 cm³.
• Rounding too early: 2.54 is exact, but a rounded intermediate result carried forward spreads the error.
• Confusing mL and cm³ with L and m³: 1 mL equals 1 cm³, and 1 L equals 1 000 cm³, not 1 m³.
• Mixing systems: a calculation with some lengths in inches and some in millimetres must be converted first.

Chapter Summary

This chapter introduced mensuration as the measurement of length, area and volume, and showed why every measured dimension must carry its unit through the whole calculation. It covered SI and Imperial units and the conversion of lengths, areas and volumes, with the length factor squared for areas and cubed for volumes. Plane geometry gave the area and perimeter of squares, rectangles, parallelograms, triangles (including Heron's formula) and trapezoids, the circumference and area of circles and rings, the area and approximate perimeter of ellipses, and the addition, subtraction and decomposition methods for composite figures. Solid geometry gave the volume and surface area of prisms, cylinders, pyramids, cones, spheres and spheroids, and showed how composite solids are handled by adding volumes, subtracting voids and leaving out hidden joints, with mass found from density. Frustums of cones and pyramids were treated with the volume rule built on the two face areas, the slant height for the sloping area, and a check by subtracting the cut-off solid. Sectors and segments of a circle were solved with the radian formulas for arc length, sector area and segment area. The chapter closed with engineering applications: the capacity and lining of a tapered reservoir, the priming volume and transit time of tubing, coating estimates, and a procedure for applied problems.