AV Math Fundamentals

Introduction

Nobody goes into AV to do maths, and almost everybody in AV does maths every day. How many amps the rack will draw. How many panels make a six metre wall, and what they weigh. How far back the projector goes. How much delay the second loudspeaker tower needs. Whether a two terabyte drive will hold a three-day conference.

None of it is hard, and almost all of it is the same handful of ideas used in different places. What makes it matter is that the answers are acted on. A wrong load trips a circuit in the middle of a keynote. A wrong weight goes over somebody's head. A wrong screen size is bought and shipped. The technicians people trust with numbers are not the ones who are best at arithmetic. They are the ones who set the problem out clearly, keep the units straight, and check whether the answer could possibly be right.

What this book covers

Units, conversions, ratios, rounding and estimating; the tools — powers, roots, Pythagoras, basic trigonometry and logarithms; electrical arithmetic from Ohm's law to three-phase power, voltage drop and heat; weights, load sharing, bridles and design factors; decibels, from the formulas to loudspeaker level at the back row and safe exposure times; the speed of sound, delay and wavelength; screen sizing, projection throw and brightness, contrast and LED wall arithmetic; light levels and beam sizes; and bits, bytes, bit rates and file sizes.

Every calculation is worked. It was measured against AVIXA's AV maths training, the published standards for display image size and projected contrast, the standard texts on sound system engineering, entertainment rigging and automated lighting, so that nothing a technician is taught elsewhere about the maths of AV is missing here.

What this book does not do

It does the arithmetic. It does not authorise electrical work or rigging decisions. When a calculation in this book is about power or weight overhead, the result belongs to the qualified electrician or rigger responsible, and the point of doing it yourself is to understand their numbers and to notice when something does not add up.

How to Use This Book

Each chapter follows the same shape, and the shape matters more than it might look. You read the ideas, then you see them play out in a real situation, then you get the procedure, then you are shown what goes wrong and how to recognise it early. Reading only the parts that look new is the fastest way to arrive on a job holding half of something.

1. Read the chapter body first, with a calculator or your phone beside you. Work each example yourself as you read it.
2. Read On the job. The same material as a real situation on a real show, which is where it stops being abstract.
3. Work through How to do it. These are procedures you can follow on a job this week.
4. Study What goes wrong. Most wrong answers in AV maths are the same few mistakes, and they are easy to recognise once you have seen them.
5. Do the exercise before reading the answer key. Reading a worked answer feels like being able to do it and is not.
6. Take the checklist with you. Photograph it on your phone. Use it until you no longer need it.

Keep a notebook

One habit is worth starting today, before the first chapter: keep a small notebook and write the formulas and the numbers in it as you meet them — amps equals watts divided by volts, 20 log for voltage, 343 metres per second, 2.9 milliseconds per metre, the width of a 16:9 screen as 0.87 of its diagonal, eight bits in a byte — and, more importantly, work at least one example of each in your own hand.

Write it by hand rather than photographing this page. Copying a formula puts it on paper; working an example yourself is what puts it in your head, and setting it out is most of the understanding.

Then use it twice. Read a few pages on the way to a job, which takes five minutes and turns formulas you half-know into ones you know. And carry it on the job, because the formula you cannot remember is always the one somebody is waiting for.

A note on the words

The same quantity is written differently in different places, and it is worth expecting that. Amps are A, amperes or current; watts are W or power. A metre is a meter in American spelling, and a foot is written ft or with a single quote mark. Decibels are written dB, but dB alone means nothing without knowing what it is relative to, which is what the letters after it are for: dBu, dBV, dBFS, dB SPL. Throw ratio is written 1.5:1 or simply 1.5. Megabits and megabytes differ only by a capital letter, Mb against MB, and by a factor of eight.

None of that is worth arguing about. What matters is that you always write the unit next to the number, because a number without its unit is the start of most mistakes in this book.

A note on safety

Some calculations in this book decide whether a circuit is overloaded, a load is too heavy, or a room is too loud to stay in. Those numbers are safety numbers. They are rounded in the safe direction, checked by a second method or a second person, and handed to the qualified person responsible, who has the final word.

Nothing in this book authorises electrical work, rigging, or the design of anything that holds weight over people. Doing the arithmetic yourself is how you understand and check those decisions, not how you make them.

Where this book sits

This book assumes you have read AV Fundamentals 101, or that you already know what is in it.

Experience counts instead. If you have time on crews where you have already met watts, decibels, screen sizes and weights, and want the arithmetic underneath them, start here.

If none of that is true yet, read AV Fundamentals 101 first.

What pairs with this

• The Physics of Sound — the why behind the decibel formulas

Where it leads

These are not steps in an order. Take whichever one fits.

• Electrical Fundamentals for AV — the power arithmetic, put to work on a real show
• Rigging Fundamentals — the load arithmetic, where it matters most
• Acoustics Fundamentals — decibels and wavelengths, applied to rooms

Before You Begin

Why Math Matters in AV

The everyday calculations behind AV work.

Practical AV math

AV work involves practical math constantly — calculating power and electrical loads, weights for rigging, signal levels, distances and coverage, screen sizes, and more. This isn't abstract mathematics; it's everyday, applied calculation that AV professionals use to do the work safely and correctly. Getting the math right matters — for safety (power and rigging must be calculated correctly), quality (correct levels, coverage, and sizing), and professionalism (accurate numbers). AV math is the everyday, applied calculation behind the work.

This matters because so much of AV depends on getting numbers right — an electrical load miscalculated can trip power or be unsafe, a weight miscalculated in rigging can be dangerous, a screen sized wrong won't work for the room. Practical AV math is a real, foundational skill (and central to the CTS certification). Practical math is foundational to safe, correct AV work.

Where AV uses math

AV uses math across several areas. Power and electrical — calculating power (watts), current (amps), and electrical loads (essential and safety-critical). Weight and loads — calculating weights for rigging and staging (also safety-critical). Signal and audio — levels, gain, and decibels (how audio levels work). And geometry — distances, angles, coverage (where sound/light reaches), and sizing (screens, projection, throw). These are the main areas where AV math is used. AV math spans power, weight, signal, and geometry.

These areas (power, weight, signal, geometry) cover most everyday AV calculations, and this book introduces the practical math in each. AV math is used across power, weight, signal, and geometry — the practical calculations of the job.

Units and prefixes

Every number in AV has a unit, and most calculation errors on a show are unit errors rather than arithmetic errors: milliseconds treated as seconds, megabits treated as megabytes, millimetres treated as metres.

The metric system builds large and small units from one base unit with prefixes, each a power of ten. A kilo is a thousand: a kilowatt is 1,000 watts, a kilogram 1,000 grams. A mega is a million: a megabit is 1,000,000 bits, a megahertz a million hertz. A giga is a billion. Going smaller, a milli is a thousandth: a millisecond is 0.001 seconds, a milliamp 0.001 amps, a millimetre 0.001 metres. A micro is a millionth, written with the Greek letter µ, as in µV for microvolts.

Converting between prefixes is moving the decimal point three places for each step. 2,400 watts is 2.4 kilowatts. 0.016 seconds is 16 milliseconds. 5 milliamps is 0.005 amps.

The habit that prevents most mistakes is simple: write the unit next to every number, every time, including in the middle of a calculation. When you divide watts by volts, the answer is in amps; when you divide metres by metres per second, it is in seconds. If the units in your working do not produce the unit you wanted, the calculation is set up wrong, and you have found the error before it reached the answer.

Metric and imperial

North American venues, riggers and many spec sheets work in feet, inches and pounds. Most manufacturers and most of the world work in metres, millimetres and kilograms. A technician converts between them several times a day.

The conversions worth knowing by heart are few. One inch is exactly 25.4 millimetres. One foot is 0.3048 metres, so one metre is about 3.28 feet. One kilogram is about 2.205 pounds, so one pound is about 0.454 kilograms. One square metre is about 10.76 square feet.

To convert, multiply by the factor that makes the unit you do not want cancel out. A 500 pound load is 500 × 0.454 = 227 kilograms. A 12 metre throw is 12 × 3.28 = 39.4 feet. A 1,200 millimetre panel is 1,200 ÷ 25.4 = 47.2 inches.

Temperatures are the odd one out, because the scales start in different places. Degrees Fahrenheit are degrees Celsius × 9 ÷ 5 + 32. A 30 °C outdoor stage is 86 °F.

Two warnings. A ton is not a tonne: a US ton is 2,000 pounds, about 907 kilograms, and a metric tonne is 1,000 kilograms, about 2,205 pounds, so a "one ton" chain hoist needs to be read carefully. And never mix units within one calculation. Convert everything to one system first, calculate, and convert the answer if you need to.

Order of operations

The same numbers give different answers if the steps are done in a different order, and calculators do not always do what you meant.

The convention is: brackets first, then powers and roots, then multiplication and division, then addition and subtraction. Within the same level, work left to right.

Here is a real example. A rack holds four processors at 150 watts and two media servers at 750 watts, on 120 volts. The total current is the total watts divided by the volts: (4 × 150 + 2 × 750) ÷ 120. Inside the brackets, multiply first: 600 + 1,500 = 2,100 watts. Then divide: 2,100 ÷ 120 = 17.5 amps.

Typed into a phone calculator without the brackets, 4 × 150 + 2 × 750 ÷ 120 gives 600 + 12.5 = 612.5, which is not a current anybody should believe. The fix is to use brackets, or to do the calculation in written steps with the result of each step and its unit written down.

Ratios, proportions and percentages

A ratio compares two quantities. An enormous amount of AV arithmetic is ratios in different clothes: a 16:9 aspect ratio, a 1.5:1 throw ratio, a 5:1 design factor, a signal-to-noise ratio, a contrast ratio of 50:1.

A ratio can be written as a fraction or a decimal. A 16:9 screen is 16 ÷ 9 = 1.78 times as wide as it is tall. A throw ratio of 1.5:1 means the throw distance is 1.5 times the image width.

A proportion says two ratios are equal, and it is how you scale. If 16:9 means 16 units wide for every 9 high, a screen 4 metres wide is 4 × 9 ÷ 16 = 2.25 metres high.

A percentage is a ratio out of a hundred. To find a percentage of a number, multiply by the percentage and divide by 100: 80 per cent of a 20 amp circuit is 20 × 80 ÷ 100 = 16 amps. To find what percentage one number is of another, divide and multiply by 100: a drop of 8.3 volts on 120 volts is 8.3 ÷ 120 × 100 = 6.9 per cent.

Percentages of a change work the same way. A wall drawing 15 amps that increases by 20 per cent draws 15 × 1.2 = 18 amps.

Rounding safely

Most answers need rounding, and on a show the right direction to round depends on what the number is for, not on whether the digit after the decimal point is above or below five.

Round in the safe direction. When the answer is how many devices fit on a circuit, round down: if 8.7 panels fit, the answer is 8, because the ninth would overload it. When the answer is how many circuits are needed, round up: if 7.9 circuits are needed, the answer is 8. When the answer is a weight or a power draw for planning, round up. When the answer is a capacity or a limit, round down.

Keep precision until the end. Rounding in the middle of a long calculation carries the rounding error into every later step, and several small errors can add up. Carry the full number through and round the final answer.

And do not report more precision than the inputs had. If a weight was estimated to the nearest ten kilograms, a total of 437.25 kilograms is not more accurate than 440; it only looks it. The number of digits you write is a claim about how well you know the answer.

Estimate first, then calculate

The most useful habit in AV maths takes five seconds: before you calculate, estimate roughly what the answer should be. Then, when the calculated answer arrives, compare them.

Estimates use easy numbers. On 120 volts, amps are roughly watts divided by 100, which slightly overestimates and is safe. On 230 volts, amps are roughly watts divided by 200. Sound travels about 3 milliseconds per metre. A 16:9 screen is a little under nine tenths as wide as its diagonal. A byte is eight bits, and a megabit per second for an hour is about 450 megabytes.

Using the rack from the order of operations example, 2,100 watts at 120 volts is roughly 2,100 ÷ 100 = 21 amps. The calculated 17.5 amps is in the right range. The typed mistake, 612.5 amps, is thirty times the estimate, and the estimate catches it instantly.

A sanity check asks whether the answer makes physical sense. Could a rack on an ordinary circuit draw six hundred amps? Could a delay be two seconds for a tower fifty metres away? Could a screen be taller than it is wide at 16:9? When the answer fails the sanity check, stop and find the mistake. It is almost always a unit, a misplaced decimal point or the wrong order of operations.

Metric prefixes

Conversions worth knowing

Round in the safe direction

• How many fit — round down.
• How many are needed — round up.
• A weight or a draw for planning — round up.
• A capacity or a limit — round down.

On the job

A conference in Toronto. The rigging plot from the European production company lists every point load in kilograms, and the venue's house rigger wants them in pounds for the venue's own records. A production assistant has been asked to convert them and has emailed a spreadsheet.

The rigger reads the first line: a 450 kilogram LED wall point, listed as 204 pounds. She frowns, because the wall she is looking at is not a two hundred pound object.

You check it with an estimate. A kilogram is about two pounds, so 450 kilograms is about 900 pounds. The spreadsheet is less than a quarter of that. The assistant divided by 2.205 instead of multiplying, which converts pounds to kilograms, the wrong way round.

You correct the formula to multiply, and 450 kilograms becomes 992 pounds. Every other point in the spreadsheet moves by the same factor. The rigger checks three of them against her own estimate and signs the plot. The only tool that caught it was knowing roughly what the answer should be before looking at the calculated one.

How to do it

1. Write down what you are trying to find, and its unit.
2. Write down every number you know, each with its unit.
3. Convert everything into one system of units.
4. Estimate the answer with easy numbers.
5. Calculate in written steps, with brackets where order matters, keeping units on every line.
6. Round the final answer in the safe direction.
7. Compare with the estimate; if they disagree, find out why before using the answer.

What goes wrong

Practice it

Answer before checking. Estimate each one first.

Powers, Roots, Triangles and Logarithms

Five mathematical tools do almost all the work in this book, and none of them is harder than the arithmetic most people already know. Squares and square roots. Powers of ten. Pythagoras, for the length of a diagonal. A little trigonometry, for angles. And logarithms, which sound intimidating and are simply the idea behind decibels.

This chapter introduces each with an AV example and nothing more. Every later chapter uses them, and a phone calculator does the button-pressing.

Squares, square roots and powers of ten

A number squared is the number multiplied by itself: 4 squared, written 4², is 16. The square root is the reverse: the square root of 16, written √16, is 4. Squares appear whenever something spreads over an area, which is why sound and light fall off with the square of distance, and roots appear whenever you work backwards from an area to a length.

Powers of ten are a compact way of writing very large and very small numbers. 10² is 100, 10³ is 1,000, 10⁶ is 1,000,000. Negative powers are fractions: 10⁻³ is 0.001. A 1 gigabit link is 10⁹ bits per second, and a 5 milliamp trip current is 5 × 10⁻³ amps.

Multiplying powers of ten adds the powers: 10³ × 10² = 10⁵, that is 1,000 × 100 = 100,000. That single fact is the whole secret of logarithms, later in this chapter.

On a phone calculator, turn it sideways for the scientific keys: x² for squares, √ for roots, and 10ˣ for powers of ten.

Pythagoras

In any right-angled triangle, the square of the longest side, the hypotenuse, equals the sum of the squares of the other two sides. Written out, a² + b² = c², so c = √(a² + b²).

It turns up everywhere a diagonal hides in AV.

A screen's size is sold as its diagonal. A screen 3 metres wide and 1.69 metres high has a diagonal of √(3² + 1.69²) = √(9 + 2.86) = √11.86 = 3.44 metres.

The distance from a loudspeaker to a seat is a diagonal. A loudspeaker hung 4 metres higher than a listener's ears and 12 metres in front of them is √(12² + 4²) = √(144 + 16) = √160 = 12.65 metres away.

A bridle leg is a diagonal. With beams 8 metres apart and the apex 3 metres below them, each leg spans half the gap, 4 metres across, and 3 metres down, so it is √(4² + 3²) = √25 = 5 metres long.

Working backwards is the same formula rearranged: if you know the diagonal and one side, the other side is √(c² − a²).

One triangle is worth memorising: sides of 3, 4 and 5 make a perfect right angle. Riggers and stage crews use it with a tape measure to mark a square corner on a floor: 3 metres along one line, 4 metres along the other, and when the diagonal between the two marks measures exactly 5 metres, the corner is square.

Angles, and the tangent

Trigonometry connects the angles of a right-angled triangle to the ratios of its sides. Of its three functions, one does most of the work in AV, and the other two appear occasionally.

Name the sides from the angle you care about: the opposite side is across from it, the adjacent side is next to it, and the hypotenuse is the longest side. Then:

The tangent of the angle is opposite ÷ adjacent, written tan.

The sine is opposite ÷ hypotenuse, written sin.

The cosine is adjacent ÷ hypotenuse, written cos.

The tangent is the useful one, because AV so often knows an angle and a distance and wants a width. A lighting fixture with a 20 degree beam, 10 metres from the stage, makes a triangle with half the beam, 10 degrees, at the fixture. The adjacent side is the 10 metres; the opposite side is half the beam's width. So half the width is 10 × tan(10°) = 10 × 0.176 = 1.76 metres, and the full beam is 3.53 metres across.

Working the other way, when you know two sides and want the angle, use the inverse function, written tan⁻¹ or atan on a calculator. A projector 1.5 metres above the centre of a screen and 8 metres back is tilted down by tan⁻¹(1.5 ÷ 8) = tan⁻¹(0.1875) = 10.6 degrees.

Make sure the calculator is set to degrees, not radians. A tangent of 10 that comes out as 0.648 instead of 0.176 is a calculator in radians.

Logarithms, as much as decibels need

A logarithm answers one question: what power of ten gives this number? The logarithm of 100 is 2, because 10² is 100. The logarithm of 1,000 is 3. The logarithm of 10 is 1, and of 1 is 0. Numbers between these have logarithms between them: the logarithm of 2 is about 0.301, and of 4 about 0.602.

Logarithms turn multiplying into adding. Because 10³ × 10² = 10⁵, log(1,000 × 100) = log(1,000) + log(100) = 3 + 2 = 5. That is the entire reason they are used for sound. Hearing responds to ratios: the difference between 1 and 2 watts sounds about the same as between 100 and 200 watts. On a logarithmic scale, equal ratios become equal steps, so a doubling is always the same number of decibels whatever the starting level.

That is what a decibel is: ten times the logarithm of a power ratio. Doubling the power is 10 × log(2) = 10 × 0.301, about 3 decibels. Ten times the power is 10 × log(10) = 10 decibels. A hundred times the power is 20 decibels.

The decibels chapter builds everything from there. For now, the idea to hold is that decibels add when the underlying quantities multiply, which is why sound engineers can add and subtract gains and losses along a chain instead of multiplying them.

The tools on one page

Values worth recognising

On the job

A lighting crew is marking out a stage floor in a warehouse venue so the set can be placed square to the audience. There is no drawing of the floor, only the centre line snapped in chalk, and the stage manager wants a line exactly at right angles to it for the downstage edge.

Nobody has a large square. You take a tape measure. From the point where the edge meets the centre line, you measure 3 metres along the centre line and make a mark. Your colleague holds the tape at the point and swings 4 metres towards where the edge should go. You hold the tape from your 3 metre mark and swing 5 metres. Where the two arcs meet is the second point of a perfect right angle.

The chalk line snapped through that point is square to the centre line. The set goes down against it, and the scenic carpenter checks the corner with his own square and nods.

Pythagoras is two thousand years old and still the fastest way to mark a square corner on an empty floor.

How to do it

1. Mark the corner point on the line you already have.
2. Measure 3 units along that line and mark it.
3. From the corner, swing an arc of 4 units towards the new line.
4. From the 3 unit mark, swing an arc of 5 units to cross it.
5. Mark where the arcs cross; the line from the corner through it is square.
6. For more accuracy on large floors, use multiples — 6, 8, 10 or 9, 12, 15.

What goes wrong

Practice it

Answer before checking. Use a calculator.

Power & Electrical Math

Watts, amps, and electrical loads.

Power, current, and the basics

A core area is power and electrical math — calculating electrical power and loads. The key relationship is power (watts) = voltage (volts) × current (amps) — the basic formula linking watts, volts, and amps. From it, you can calculate the current (amps) that equipment draws (amps = watts ÷ volts), which is essential for knowing electrical loads. Understanding this basic power formula and how to find power and current is foundational AV electrical math. Power (W) = volts (V) × amps (A) is the key relationship.

This lets you work out, for example, how much current a piece of gear draws (from its wattage and the voltage) — the basis for understanding electrical loads. It's simple but essential and safety-relevant math. The basic power/current relationship is foundational AV math. (Electrical work itself must follow codes and qualified guidance.)

Electrical loads and limits

The reason power math matters is electrical loads and limits — knowing how much current your equipment draws (the load) and keeping it within the safe limits of the circuits and distribution supplying it (every circuit has a current limit, e.g., a 20-amp circuit). You calculate the total load (adding up the current the gear draws) and ensure it stays within the circuit's safe capacity (not overloading it). This load calculation is essential and safety-critical — overloading circuits trips power or causes hazards. Calculate loads and keep them within safe circuit limits.

This is why AV pros calculate power: to distribute equipment across circuits safely (within limits), avoiding overloads. It's practical, safety-relevant math done on real jobs. AV power math keeps electrical loads safely within limits. (Always follow electrical codes and qualified guidance — overloading is dangerous.)

Ohm's law, worked

Ohm's law connects voltage, current and resistance: volts equal amps multiplied by ohms, V = I × R. Rearranged, I = V ÷ R and R = V ÷ I. Combined with the power formula, it gives two more useful forms: power equals current squared times resistance, P = I² × R, and power equals voltage squared divided by resistance, P = V² ÷ R.

Here is where they appear. An amplifier driving an 8 ohm loudspeaker with 40 volts has a current of 40 ÷ 8 = 5 amps, and delivers 40² ÷ 8 = 1,600 ÷ 8 = 200 watts. Into a 4 ohm load, the same 40 volts drives 10 amps and 400 watts, which is why an amplifier works much harder into a lower impedance.

A cable with a total resistance of 0.5 ohms carrying 16 amps loses 16 × 0.5 = 8 volts along its length, and turns 16² × 0.5 = 128 watts into heat in the cable.

Strictly, loudspeakers have impedance rather than simple resistance, and it varies with frequency. For the arithmetic of a patch, the nominal impedance printed on the cabinet is used as if it were a resistance.

Series and parallel loudspeakers

Loudspeakers can be connected to one amplifier channel in series, one after another, or in parallel, side by side. The load the amplifier sees depends entirely on which.

In series, the impedances add. Two 8 ohm cabinets in series are 16 ohms. Series connections are rare in live sound, because if one cabinet fails open, all of them go silent.

In parallel, the load falls. For identical cabinets, divide one cabinet's impedance by the number of cabinets: two 8 ohm cabinets are 4 ohms, three are 2.67 ohms, four are 2 ohms. For different impedances, use the reciprocal formula: 1 ÷ Z = 1 ÷ Z1 + 1 ÷ Z2 + … An 8 ohm and a 4 ohm cabinet in parallel are 1 ÷ (1/8 + 1/4) = 1 ÷ 0.375 = 2.67 ohms. A quick check: a parallel total is always lower than the smallest single impedance.

The result is compared with the amplifier's minimum rated load. An amplifier rated down to 4 ohms can drive two 8 ohm cabinets on one channel, but not three.

Circuits, 80 percent and three-phase power

A circuit's capacity in watts is its voltage times its breaker rating. For continuous loads, planned to 80 per cent, it is volts × amps × 0.8. A 20 amp circuit at 120 volts is 120 × 20 × 0.8 = 1,920 watts. A 16 amp circuit at 230 volts is 230 × 16 × 0.8 = 2,944 watts.

The number of identical devices that fit is the capacity divided by one device's draw, rounded down. LED panels drawing 180 watts at maximum fit on that 120 volt circuit 1,920 ÷ 180 = 10.67 times, so ten panels.

The number of circuits needed for a large quantity is the total divided by the capacity per circuit, rounded up. Eighty-four panels at ten per circuit need 84 ÷ 10 = 8.4, so nine circuits.

Three-phase supplies deliver power on three lines. For a balanced load on a wye supply, the simplest way to find the total power is to work one phase and multiply by three: a supply with 60 amps per phase at 120 volts to neutral carries 120 × 60 × 3 = 21,600 watts, 21.6 kilowatts. Electricians often write the same thing with the line-to-line voltage and the square root of three, √3 × 208 × 60, which comes to the same 21.6 kilowatts, because 208 is 120 × √3.

Working backwards, a 21.6 kilowatt balanced load on a 120/208 volt supply draws 21,600 ÷ (120 × 3) = 60 amps per phase.

Voltage drop, calculated

Voltage drop is Ohm's law applied to the cable. The drop equals the current times the resistance of the whole circuit path, which is the conductor out and the conductor back: V drop = I × R per metre × length × 2.

Copper conductor of 12 AWG has a resistance of about 5.2 milliohms per metre, 0.0052 ohms. A 50 metre run carrying 16 amps drops 16 × 0.0052 × 50 × 2 = 8.3 volts. On 120 volts that is 8.3 ÷ 120 × 100 = 6.9 per cent, and the gear at the end receives about 112 volts.

The same run in 10 AWG, at about 3.3 milliohms per metre, drops 16 × 0.0033 × 50 × 2 = 5.3 volts, 4.4 per cent. On a 230 volt supply the same current and cable would drop the same 8.3 volts, but that is only 3.6 per cent of the supply, one reason higher voltages travel further.

Planning guidance commonly aims to keep voltage drop to a few per cent. The exact limit, and the design of any long feeder, belongs to the electrician; the calculation shows a technician why a long thin run misbehaves.

Heat from equipment

Almost every watt a piece of AV equipment draws ends up as heat in the room, the rack or the road case. An amplifier's output that reaches the loudspeakers is the main exception, and even most of that becomes heat eventually.

Heating and cooling are often specified in BTU per hour, British thermal units, especially in North America. One watt is about 3.412 BTU per hour. A rack drawing 2,000 watts puts about 2,000 × 3.412 = 6,824 BTU per hour of heat into its surroundings.

The number matters for control rooms, equipment closets and enclosed racks, whose cooling must remove it, and for the venue's air conditioning in a small room packed with gear and people. An equipment rack in a cupboard with no ventilation is a heater with no thermostat.

Electrical formulas

Copper conductor resistance

Approximate values at room temperature, for voltage drop estimates.

On the job

A regional sales meeting. The audio technician is asked whether one amplifier channel rated to 4 ohms can drive the four 8 ohm ceiling loudspeakers along one side of the ballroom, because the second channel is already feeding the other side.

You work it out before anyone connects them. Four identical 8 ohm loudspeakers in parallel are 8 ÷ 4 = 2 ohms, half the amplifier's minimum. Wired that way, the amplifier would run hot and protect itself, probably in the middle of the CEO's speech.

There are two ways to make the arithmetic work. Two cabinets in series make 16 ohms, and two such pairs in parallel make 16 ÷ 2 = 8 ohms, comfortably within the rating. The installer's wall plates allow it, so you wire the four as two series pairs in parallel, and check the load on the amplifier's own display, which reads close to 8 ohms.

The ballroom's left side sounds the same as the right, and the amplifier stays cool all day.

How to do it

1. Read the nominal impedance of each cabinet.
2. Draw the wiring — which cabinets are in series and which in parallel.
3. Work out series groups first by adding.
4. Then combine parallel groups, identical ones by dividing and different ones by the reciprocal formula.
5. Compare the total with the amplifier's minimum rated load.
6. If it is below, rewire or add an amplifier channel; never connect it and hope.

What goes wrong

Practice it

Answer before checking.

Weight, Signal & Decibels

Loads, levels, and the decibel.

Weight and load math

Weight and load math is calculating weights — for rigging and staging especially (also safety-critical). This means knowing the weights of gear, adding them up (the total load being hung or supported), and ensuring the rigging points, hardware, and structure can safely bear it (with safety factors). It connects directly to rigging (where load calculation is life-safety-critical and done by qualified riggers). Knowing and summing weights is essential for safe rigging and staging. Weight calculation is essential and safety-critical for rigging and staging.

This practical weight math (knowing and totalling weights, checking against limits) underpins safe rigging and staging. It's simple arithmetic with serious safety stakes. (Real rigging load calculations must be done by qualified riggers — Rigging Fundamentals is the next step.) Weight math underpins safe rigging — done carefully and, for rigging, by the qualified.

Signal levels and decibels

Audio (and signal) math involves levels and the decibel (dB) — the unit used for audio levels and gain. Decibels measure relative levels (how much louder/quieter, how much gain), and they work logarithmically (e.g., +6 dB is double the voltage or sound pressure and +3 dB is double the power, every doubling of distance drops sound level by about 6 dB) — which is why a basic understanding of dB helps with audio. You don't need deep math, but understanding that audio levels are in dB and roughly how dB works is useful AV math. The decibel (dB) is the key unit for audio levels and gain.

This basic dB understanding (levels are logarithmic, dB measures relative level/gain) helps make sense of audio levels, gain structure, and how sound behaves — practical knowledge for audio work. A basic grasp of decibels is useful, practical AV math.

Totalling a load, and weight per metre

A load is everything that hangs, not only the thing people think of. Totalling one is addition, done completely.

List every item and its weight, from the manufacturer's data, the label or a scale, and add them. For a lighting position: the fixtures, their clamps, their safety bonds, the truss, the cable including the length that hangs down to the floor, and the hoist and its chain if it hangs with the load.

Things with length are given as weight per metre, or per foot. The total is weight per metre × length. A truss that weighs 7 kilograms per metre and is 12 metres long weighs 7 × 12 = 84 kilograms. A multicore cable at 0.9 kilograms per metre with 25 metres hanging from the truss adds 0.9 × 25 = 22.5 kilograms. Chain hanging from a hoist adds its weight per metre times the length hanging.

Then round up, because the purpose of the number is to be sure the load is not heavier than planned.

Sharing a point load between two supports

A load on a beam or truss held at two points is shared between them, and the nearer support carries more. The arithmetic is a proportion.

Each support carries the load multiplied by the distance to the other support, divided by the span. For a load W, a span L, and the load a distance a from the left support: the left support carries W × (L − a) ÷ L, and the right carries W × a ÷ L.

A 60 kilogram fixture 2.5 metres from the left end of a 10 metre span puts 60 × 7.5 ÷ 10 = 45 kilograms on the left support and 60 × 2.5 ÷ 10 = 15 kilograms on the right. The check is that the two shares add up to the whole load: 45 + 15 = 60.

With several point loads, work each one separately and add the shares at each support.

Uniformly distributed loads

A load spread evenly along a span is a uniformly distributed load, a UDL. The truss's own weight is one; so are a row of identical fixtures at equal spacing, cable run along a truss, and an LED wall hung along its length.

For a UDL centred between two supports, each support carries half the total. A 10 metre truss at 7 kilograms per metre weighs 70 kilograms, and each end carries 35 kilograms.

Real loads combine both. The same truss with the 60 kilogram fixture from the last section puts 35 + 45 = 80 kilograms on the left support and 35 + 15 = 50 kilograms on the right.

Everything here assumes a simple span on two supports. Truss with three or more points, cantilevers past a support and loads that move do not follow these simple shares, and those calculations belong to a qualified rigger.

Bridle legs — length and tension

A bridle hangs a load from two points using two legs that meet at an apex. Two pieces of arithmetic describe it, and both use the same right-angled triangle.

The leg length is Pythagoras. The horizontal side is the distance from the beam to the point directly above the apex, and the vertical side is the drop from the beams to the apex. For beams 8 metres apart with the apex midway and 3 metres below, each leg is √(4² + 3²) = 5 metres.

The tension in each leg is the share of the load that leg carries, multiplied by the leg length and divided by the vertical height. For a symmetrical bridle each leg carries half the load vertically, so with a 500 kilogram load, each leg's tension is 250 × 5 ÷ 3 = 417 kilograms.

The legs also pull sideways on each beam: the share × horizontal ÷ vertical, 250 × 4 ÷ 3 = 333 kilograms.

The surprising result is that each leg carries more than half the load, and the flatter the legs, the more it carries. Legs at 45 degrees carry about 71 per cent of the whole load each; at 60 degrees from vertical, 120 degrees between them, each leg carries the entire load. This is why bridle angles are never changed on site by anyone except the rigger who calculated them.

Design factor and working load limit

Every piece of rigging hardware has a working load limit, WLL, the most it is rated to carry in normal use, and a minimum breaking strength, the load at which it is expected to fail. The ratio between them is the design factor, sometimes called the safety factor.

Design factor = breaking strength ÷ working load limit. A wire rope sling with a minimum breaking strength of 5,000 kilograms and a design factor of 5:1 has a working load limit of 5,000 ÷ 5 = 1,000 kilograms.

The design factor exists to cover everything the arithmetic cannot see: wear, small shocks, manufacturing variation and slight errors in estimated weights. It is not spare capacity to be used. A load is compared with the working load limit, never with the breaking strength, and a sling rated at 1,000 kilograms is not used for 1,200 because it breaks at 5,000.

Where the decibel comes in

This chapter has introduced the decibel as the unit of audio level, and noted that it works logarithmically. The next chapter works it through properly, from its two formulas to loudspeaker level at the back row, because it is used in almost every audio calculation a technician makes.

Load arithmetic

Bridle leg tension by angle

Tension in each leg of a symmetrical two-leg bridle, as a share of the whole load. This shows the shape of the relationship; real bridles are calculated by the rigger.

On the job

A conference general session. The rigging plot shows a 12 metre truss hung from two hoists at its ends, carrying six moving lights evenly spaced, plus a projection screen frame of 90 kilograms hung 3 metres from the left end. The lighting programmer has asked to add two more fixtures near the left end, and the production manager wants to know, roughly, what that does before asking the head rigger.

You work it the way the plot does. The truss at 7 kilograms per metre is 84 kilograms, 42 on each hoist. The six lights at 35 kilograms each, with their clamps, spread evenly, are a 210 kilogram UDL, 105 on each hoist. The screen frame puts 90 × 9 ÷ 12 = 67.5 kilograms on the left hoist and 22.5 on the right. The left hoist carries 42 + 105 + 67.5 = 214.5 kilograms; the right carries 42 + 105 + 22.5 = 169.5. Together, 384 kilograms: the truss, the lights and the frame add up.

Two more 35 kilogram lights, 1.5 metres from the left end, add 70 × 10.5 ÷ 12 = 61.3 kilograms on the left hoist and 8.8 on the right, taking the left to about 276 kilograms.

You take the numbers to the head rigger rather than to the programmer, because it is the head rigger's plot. He checks them against the hoists, the points and the truss's load table, agrees the addition, and updates the plot.

How to do it

1. List every load, and whether it is a point load or spread along the span.
2. Convert all weights to one unit and round up.
3. For each spread load, give each support half, if it is centred.
4. For each point load, give each support its share by the distance to the other support.
5. Add the shares at each support.
6. Check that the support totals add up to the total load.
7. Give the numbers to the qualified rigger, who compares them with the plot and the ratings.

What goes wrong

Practice it

Answer before checking.

Decibels, Worked

The decibel is the most useful and most misunderstood unit in AV. It is not a quantity like a volt or a metre. It is a way of describing a ratio between two quantities on a logarithmic scale, which is why a decibel on its own means nothing until you know what it is compared to, and why decibels are added where ordinary quantities would be multiplied.

This chapter takes the idea from the logarithms chapter and works it through: the two formulas and why there are two, the reference levels, adding levels together, the inverse square law, how loud a loudspeaker will be at a distance, and how long a person can safely stay in a loud room.

Two formulas — 10 log for power, 20 log for voltage

The decibel was defined for power. The difference in decibels between two powers is 10 log of their ratio: dB = 10 × log(P2 ÷ P1).

Doubling the power is 10 × log(2) = 3 dB. Ten times the power is 10 dB. Half the power is −3 dB. An amplifier turned up from 100 watts to 400 watts is 10 × log(4) = 6 dB louder.

Voltages, currents and sound pressures are not powers. Power is proportional to the square of voltage, and squaring inside a logarithm doubles it, so for voltage the formula becomes 20 log: dB = 20 × log(V2 ÷ V1).

Doubling the voltage is 20 × log(2) = 6 dB, which is four times the power. Ten times the voltage is 20 dB, a hundred times the power.

So both +3 dB and +6 dB can be called "double", and they are doubles of different things: +3 dB doubles the power, +6 dB doubles the voltage or sound pressure. Using the wrong formula is the most common decibel mistake. The rule is to use 10 log for anything measured in watts, and 20 log for volts and pascals.

How much louder something sounds is a third question. As a rough guide, around 10 dB more is heard as about twice as loud. A 3 dB change is clearly noticeable, and a 1 dB change is hard to hear in normal listening.

Reference levels — dBu, dBV, dBFS and dB SPL

A decibel figure with a reference attached becomes an actual level, because the reference says what 0 dB is.

dBu is voltage relative to 0.775 volts. Volts = 0.775 × 10^(dBu ÷ 20). Professional line level, +4 dBu, is 0.775 × 10^(0.2) = 1.23 volts. A microphone signal at −50 dBu is 0.775 × 10^(−2.5) = 0.00245 volts, 2.45 millivolts. Going the other way, dBu = 20 × log(volts ÷ 0.775).

dBV is voltage relative to 1 volt. Consumer line level, −10 dBV, is 10^(−0.5) = 0.316 volts. In dBu that is 20 × log(0.316 ÷ 0.775) = −7.8 dBu. So professional line level is about 12 dB hotter than consumer line level, which is exactly why a consumer device into a professional input sounds quiet.

dBFS is relative to the full scale of a digital system; 0 dBFS is the maximum, and every real level is negative.

dB SPL is sound pressure relative to 20 micropascals, roughly the quietest sound a young ear can detect. A quiet room might be 30 dB SPL, conversation about 60, a loud concert 100 to 110. Sound level meters often apply a weighting that follows the ear's sensitivity, shown as dBA or dBC.

A gain is simply a difference in decibels. An amplifier with an input of 1.23 volts and an output of 40 volts has a voltage gain of 20 × log(40 ÷ 1.23) = 30.2 dB.

Adding decibel levels

Two loudspeakers each producing 90 dB at a seat do not produce 180 dB. Levels in decibels cannot be added directly, because they are logarithms.

For two unrelated sources — two different instruments, two crowd noises, a generator and an air handler — the powers add. Two equal levels give double the power, which is +3 dB: 90 dB and 90 dB make 93 dB. Ten equal sources make 10 dB more.

When the two levels differ, the louder one dominates, and the addition to it shrinks quickly. The general formula is total = 10 × log(10^(L1 ÷ 10) + 10^(L2 ÷ 10)), and a small table covers most situations: a difference of 0 dB adds 3 dB to the louder, 3 dB adds about 1.8, 6 dB adds about 1, and 10 dB adds only about 0.4. A noise 10 dB below a sound barely changes the total.

Two copies of the same signal are different. Two identical loudspeakers playing the same audio, arriving at exactly the same time and in phase, add their pressures, which is up to +6 dB. Arriving at different times, they partly cancel at some frequencies and add at others, which is why two loudspeakers side by side can sound worse than one.

The inverse square law

Sound from a small source spreads out in all directions, and the same energy covers a larger and larger area as it travels. The area grows with the square of the distance, so the intensity falls with the square of the distance. That is the inverse square law.

In decibels it is simple: the level falls by 20 × log(d2 ÷ d1). Every time the distance doubles, the level drops by 20 × log(2) = 6 dB. A source measuring 110 dB SPL at 4 metres is 104 dB at 8 metres, 98 dB at 16 metres and 92 dB at 32 metres.

For distances that are not doublings, use the formula. From 4 metres to 30 metres, the drop is 20 × log(30 ÷ 4) = 20 × log(7.5) = 17.5 dB, so the level at 30 metres is 92.5 dB.

Real loudspeakers follow the law closely only far enough away and in the open air. Close to a large cabinet, or with line arrays, which lose less per doubling over part of their range, and in rooms, where reflections add level, the numbers differ. The law is still the right first estimate, and it explains why the front row is so much louder than the back.

How loud at the back row — sensitivity, power and distance

A loudspeaker's sensitivity is the sound level it produces with 1 watt of input, measured at 1 metre, written as dB SPL at 1 W / 1 m. Combine it with the amplifier power and the distance, and you have a first estimate of how loud the system can be at any seat.

SPL at the listener = sensitivity + 10 × log(power in watts) − 20 × log(distance in metres).

A loudspeaker with a sensitivity of 97 dB, driven with 400 watts, reaches 97 + 10 × log(400) = 97 + 26 = 123 dB at 1 metre. At a listener 30 metres away, it has dropped by 20 × log(30) = 29.5 dB, to 93.5 dB SPL.

That is the maximum, on paper, before the headroom a show needs. If the programme needs 95 dB at the back row with 10 dB of headroom for peaks, the system needs 105 dB of capability there, and this loudspeaker is about 11.5 dB short. Doubling the amplifier power adds only 3 dB. The remedy is more loudspeakers, more sensitive ones, or putting a delay loudspeaker closer to the back.

Real results are lower than the formula, because loudspeakers lose some output when driven hard, called power compression, and higher in rooms with reflections. The estimate is where planning starts, not where it ends.

Noise exposure — level and time

Hearing damage depends on both how loud and for how long. Occupational noise limits express this as a level that is acceptable for a working day and an exchange rate that says how much less time is allowed for each step louder.

The widely used recommendation from NIOSH in the United States is 85 dBA for 8 hours with a 3 dB exchange rate: every 3 dB louder halves the safe time. So 88 dBA is 4 hours, 91 is 2 hours, 94 is 1 hour, 97 is 30 minutes, 100 is 15 minutes, and 103 is under 8 minutes. That follows directly from the decibel: +3 dB is double the sound energy, so the same dose arrives in half the time.

The US OSHA regulation uses 90 dBA for 8 hours with a 5 dB exchange rate, which allows considerably more exposure; many other countries use 85 dBA and 3 dB, as NIOSH recommends. A technician at front of house for a long day at 100 dBA reaches a full recommended daily dose in about a quarter of an hour on the stricter rule, which is why crews wear hearing protection and why many venues and festivals monitor audience levels.

Decibel formulas

Adding unrelated levels

Noise exposure, 85 dBA and a 3 dB exchange rate

On the job

A corporate client is holding their annual party in a long hall, with the band at one end. The production manager asks whether the band's two main loudspeakers can deliver 100 dB at the back wall, 40 metres away, because the client wants the back of the room dancing too.

The loudspeakers have a sensitivity of 99 dB at 1 W / 1 m and each is driven by a 1,000 watt amplifier channel. One cabinet reaches 99 + 10 × log(1,000) = 99 + 30 = 129 dB at 1 metre. At 40 metres it has fallen by 20 × log(40) = 32 dB, to 97 dB. Two cabinets playing the same programme add a little more, at best 3 to 6 dB depending on how their sound combines at that seat, so somewhere around 100 to 103 dB on paper.

On paper, just. But that is flat out, with no headroom for a snare hit, and in practice loudspeakers compress before they reach their paper maximum. It would sound strained at the front long before it was loud enough at the back, and the front rows would be at well over 110 dB.

You suggest a pair of delay loudspeakers halfway down the hall, 20 metres from the back wall. They need far less level to reach the back, the main system can run with headroom, and the front of the room is no longer the loudest place in the building. The production manager takes the numbers to the client, who agrees.

How to do it

1. Find the loudspeaker's sensitivity at 1 W / 1 m.
2. Find the amplifier power it will be driven with.
3. Measure the distance to the listener in metres.
4. Calculate sensitivity + 10 log(watts) − 20 log(metres).
5. Subtract the headroom the programme needs for peaks.
6. Compare with the level wanted; if short, consider more or closer loudspeakers, not just more power.
7. Treat the result as a first estimate, and measure in the room.

What goes wrong

Practice it

Answer before checking.

Sound, Time and Distance

Sound is slow. In the time light crosses a stadium, sound has barely left the stage. That slowness is behind delay loudspeakers, lip sync, microphones that sound hollow when two of them pick up one voice, and the way low frequencies seem to ignore walls while high frequencies are blocked by a person's head.

This chapter is the arithmetic of sound in time and space: the speed of sound and how temperature changes it, delay in milliseconds per metre, aligning a delay loudspeaker, wavelength and what it explains, and converting between video frames and milliseconds.

The speed of sound, and delay per metre

Sound travels through air at about 343 metres per second at 20 °C, or about 1,125 feet per second. Turned round, that is about 2.9 milliseconds for every metre, or about 0.89 milliseconds for every foot. The easy estimate is 3 milliseconds per metre.

The time for sound to travel a distance is the distance divided by the speed. A loudspeaker 20 metres away is heard 20 ÷ 343 = 0.058 seconds, 58 milliseconds, after it plays. A microphone 3 metres from a source picks it up about 9 milliseconds late.

The speed changes with temperature: roughly 331.3 + 0.606 × the temperature in °C metres per second. At 30 °C on a summer stage it is about 349.5 metres per second; at 5 °C in a cold tent, about 334. Over short distances the difference is negligible. Over the length of a large field, it is a few milliseconds, enough to matter to a carefully aligned delay system as the evening cools.

Aligning a delay loudspeaker

A delay loudspeaker, placed partway back in a long room or field, reinforces the sound for the listeners behind it. Its signal arrives from the console electrically, in a fraction of a millisecond, while the sound from the main loudspeakers takes time to travel through the air. Without delay, listeners behind the delay loudspeaker hear its sound first and the main system's sound as a late echo.

So its signal is delayed by the time the main system's sound takes to reach it. The distance is measured from the main loudspeakers to the delay loudspeakers, and the delay time is that distance divided by the speed of sound. Delay loudspeakers 45 metres from the main system need 45 ÷ 343 = 131 milliseconds.

Many engineers then add a small extra delay, often around 10 milliseconds, so the main system's sound arrives at the listener very slightly first. Because hearing locates a sound by what arrives first, the audience then hears the sound as coming from the stage, with the delay loudspeaker adding level without drawing attention to itself.

The measured delay is confirmed by listening or with a measurement system at a seat behind the delay loudspeaker, and it is adjusted if temperature changes the speed of sound noticeably through a long outdoor event.

Wavelength

The wavelength of a sound is the distance from one peak of the wave to the next. It is the speed of sound divided by the frequency: wavelength = 343 ÷ frequency.

At 50 hertz, a deep bass note, the wavelength is 343 ÷ 50 = 6.9 metres. At 100 hertz, 3.4 metres. At 1,000 hertz, 34 centimetres. At 10,000 hertz, 3.4 centimetres.

That range, from the length of a truck to the width of a thumb, explains much of how sound behaves in a room. Objects much smaller than a wavelength barely affect it: a 7 metre bass wave flows round a person, a column or a loudspeaker cabinet as if they were not there, which is why bass is heard behind loudspeakers and through walls. Objects larger than a wavelength block and reflect it: a 3 centimetre treble wave is stopped by a head, a lectern or a banner, which is why a loudspeaker hidden behind scenery sounds dull.

It also explains why two identical sounds arriving at slightly different times cancel at some frequencies. When one copy is delayed by half a wavelength, its peaks line up with the other's troughs. A delay of 1 millisecond is half the period of a 500 hertz wave, so the two cancel there, and the pattern repeats up the spectrum — the hollow, phasey sound of two microphones on one voice at different distances.

Frames and milliseconds

Video delay is measured in frames, and audio delay in milliseconds. Matching one to the other means converting.

The duration of one frame is 1,000 milliseconds divided by the frame rate. At 25 frames per second one frame is 40 milliseconds; at 30, 33.3 milliseconds; at 50, 20 milliseconds; at 60, 16.7 milliseconds; at 24, 41.7 milliseconds. The fractional rates used in North America, 29.97 and 59.94, give 33.4 and 16.7 milliseconds.

A video chain that adds three frames at 50 frames per second delays the picture by 3 × 20 = 60 milliseconds. If the audio chain adds 5 milliseconds, the audio needs 55 more to line up.

Timecode counts frames too. A timecode of 00:01:30:12 at 25 frames per second is 1 minute, 30 seconds and 12 frames: 90 seconds plus 12 × 40 milliseconds, 90.48 seconds.

Sound in time and space

Wavelengths across the audio range

Frame durations

On the job

An outdoor festival with a delay tower 60 metres from the main stage. The system engineer set the delay at 5 pm, in the heat of the afternoon, when the air was 32 °C, and set it to the bare travel time, without the usual extra margin. By the headline set at 10 pm the air has dropped to 16 °C, and the audience near the delay tower says the sound has become smeared, with the kick drum doubled.

At 32 °C sound travels at about 331.3 + 0.606 × 32 = 350.7 metres per second, so 60 metres took 171.1 milliseconds, and that is what went into the processor. At 16 °C it travels at about 341.0 metres per second, so the same 60 metres takes 176.0 milliseconds. The main system's sound now arrives about 5 milliseconds later than it did when the delay was set. The delay tower is now firing before the stage sound arrives, instead of just after.

The engineer resets the delay to 176 milliseconds plus the usual 10 milliseconds so the stage sound leads, 186 milliseconds in all, checks with the measurement microphone behind the tower, and the kick drum pulls back into one hit. The next night, the delay is checked again after dark as a matter of routine.

How to do it

1. Measure the distance from the main loudspeakers to the delay loudspeakers.
2. Note the air temperature and calculate the speed of sound.
3. Divide the distance by the speed to get the delay in seconds, and convert to milliseconds.
4. Set the delay on the delay loudspeakers' processor.
5. Add a small extra delay, often around 10 ms, so the stage sound leads.
6. Check at a listening position behind the delay loudspeakers, by ear or measurement.
7. Recheck when the temperature changes significantly.

What goes wrong

Practice it

Answer before checking.

Screens, Projection and LED Walls

Video in a room is geometry and light. How big an image must be for the farthest person to read it. How wide a screen of a given diagonal is. How far back a projector goes for that width. Whether it is bright enough against the room's light, and whether the contrast survives. How many LED panels make the wall, what its resolution is, what it weighs and what it draws.

Each of those is a short calculation, and each is done before anything is ordered, because the answer decides what arrives on the truck.

Aspect ratio, and width and height from a diagonal

Aspect ratio is width to height. 16:9 is the standard for video and most modern displays; 16:10 is common on laptops and some projectors; 4:3 appears on older displays; and ultra-wide formats such as 21:9 and custom LED walls can be anything.

Displays are sold by their diagonal, and rooms need their width and height. Pythagoras connects them. For an aspect ratio of A:B, the diagonal of a screen A units wide and B high is √(A² + B²). For 16:9 that is √(256 + 81) = √337 = 18.36. So width = diagonal × 16 ÷ 18.36 = diagonal × 0.872, and height = diagonal × 9 ÷ 18.36 = diagonal × 0.490.

A 100 inch 16:9 display is 87.2 inches wide and 49.0 inches high. A 65 inch display is 56.7 by 31.9 inches, which is about 1.44 by 0.81 metres.

For 4:3 the factors are exactly 0.8 and 0.6, from the 3-4-5 triangle. For 16:10 they are 0.848 and 0.530.

When content and screen do not share an aspect ratio, something gives: 4:3 content on a 16:9 screen has bars at the sides, or is stretched, or is cropped. The arithmetic of a mismatch is the ratio of the two ratios.

Sizing an image for the farthest viewer

The size of an image is not set by the size of the wall. It is set by the person at the back, and by the smallest thing on the screen they need to read.

The published standard for this is ANSI/AVIXA V202.01, Display Image Size for 2D Content in Audiovisual Systems, known as DISCAS. It works from two measurements. The farthest viewer distance is how far the most distant viewer who needs to see the content sits from the screen. The element height is the height of the smallest element the viewer must be able to read — typically a lowercase letter in the smallest text on a slide — as a percentage of the whole image height.

It then distinguishes the viewer's task. Basic decision making is when the viewer needs to understand the content, for instance reading a presentation, but not scrutinise every detail. Analytical decision making is when they must examine fine detail, such as a spreadsheet, an engineering drawing or a medical image, and it requires a considerably larger image for the same distance. From the farthest viewer distance, the element height and the task, the standard gives the minimum image height. It also addresses the closest viewer, who should not be so close that their eyes cannot take in the image comfortably.

AVIXA publishes calculators for the standard, and the calculation should be done with it. The practical results are always the same shape: the farther the viewer or the smaller the text, the bigger the image; and detailed work needs much more image than a keynote.

Older rules of thumb, still quoted, set the farthest viewer at a multiple of the image height — around six times for reading typical content, closer for detailed work, and further for simple video. They are a quick check, not a design method.

Throw ratio and throw distance

A projector's throw ratio says how far back it must be for a given image width: throw ratio = throw distance ÷ image width. It is a property of the lens, and a zoom lens has a range.

So throw distance = throw ratio × image width. A lens with a throw ratio of 1.5:1 filling a 4 metre wide image needs to be 1.5 × 4 = 6 metres from the screen. The same image with a 0.8:1 short-throw lens needs 3.2 metres; with a 2.5:1 long lens, 10 metres.

Working the other way, a projector that must sit 12 metres back to clear the audience, filling a 5 metre wide screen, needs a throw ratio of 12 ÷ 5 = 2.4:1, and the lens chosen must include 2.4 in its range.

Throw distance is measured from the lens, not the back of the projector, and manufacturers' calculators include lens offset — how far above or below the lens centre the image sits. Lens shift moves the image without moving the projector, within limits; keystone correction distorts the image digitally to fix a tilted projector and loses resolution, so it is a last resort.

Screen brightness from lumens

A projector's light output is rated in lumens. What the audience sees is the screen's luminance: how much light comes off each area of it towards them, measured in nits (candela per square metre) or foot-lamberts.

On a screen with a gain of 1, a projector's lumens spread over the screen's area give: nits = lumens × gain ÷ (π × area in square metres). In imperial units, foot-lamberts = lumens × gain ÷ area in square feet. One foot-lambert is about 3.43 nits.

A 10,000 lumen projector filling a 4 by 2.25 metre screen, 9 square metres at a gain of 1, gives 10,000 ÷ (3.14 × 9) = 354 nits, or about 103 foot-lamberts.

Gain is how much the screen concentrates light towards the audience compared with a perfect matte white surface. A gain of 1.3 gives 30 per cent more luminance straight on and less at wide angles.

Real results are lower: lumens decline as a lamp or laser ages, and zoom lenses and settings reduce output. And whether 354 nits is enough depends on the room. A dark cinema is designed around roughly 48 nits. A conference room with lights on, windows or a lit stage needs far more, because the room's light washes into the image — which is the subject of contrast.

Contrast ratio in the room

A bright image in a bright room can still be unreadable, because what matters to the eye is the difference between the image's whites and blacks, and the room's light lands on both.

Contrast ratio is the luminance of white divided by the luminance of black, with the room's light included. A projector might be rated at thousands to one in a black test room. In a real room, the ambient light reflected off the screen adds to black and white alike.

Take the 354 nit screen from the last section, with a projector black level of 5 nits, in a room where ambient light reflects 20 nits off the screen. White is 354 + 20 = 374 nits. Black is 5 + 20 = 25 nits. The contrast ratio in the room is 374 ÷ 25 = 15:1, however good the projector's rated contrast.

ANSI/INFOCOMM 3M, Projected Image System Contrast Ratio, sets out how to measure contrast in the actual room and minimum ratios by viewing requirement: at least 15:1 for passive viewing, 50:1 for basic decision making, 80:1 for analytical decision making, and 240:1 for full-motion video. On those figures, the example room is only good enough for passive viewing. The fixes follow directly from the arithmetic: more light on the screen, less room light reaching it, or a screen material that rejects ambient light.

LED wall arithmetic

An LED wall is built from panels, and nearly every number about it follows from the panel's size, its pixel pitch and how many there are.

Pixel pitch is the distance between the centres of neighbouring pixels, in millimetres. The pixels across one panel are the panel width divided by the pitch. A 500 millimetre panel at 2.6 millimetre pitch has 500 ÷ 2.6 = 192 pixels across, rounded to the manufacturer's actual figure.

The panel count is the wall size divided by the panel size, in each direction. A wall 6 metres wide and 3.5 metres high, in 500 millimetre panels, is 12 panels by 7, which is 84 panels.

The wall's resolution is panels × pixels per panel: 12 × 192 = 2,304 pixels wide and 7 × 192 = 1,344 high. That is not a standard video format. Its aspect ratio is 2,304 ÷ 1,344 = 1.71, a little narrower than 16:9 at 1.78, so a 16:9 source either leaves small bars, is cropped slightly or is scaled to fit, and the content team needs the exact pixel size.

Weight and power are panels × per-panel figures. At 7.5 kilograms each, 84 panels weigh 630 kilograms before frames, hardware and cable. At a maximum of 180 watts each they draw 15,120 watts at full white — 126 amps at 120 volts — and at a typical 60 watts, 5,040 watts.

Viewing distance is the last number. A common rule of thumb puts the minimum comfortable viewing distance in metres at about the pixel pitch in millimetres, so about 2.6 metres for this wall, with further back looking smoother; closer, and the individual pixels are visible. Some manufacturers suggest two to three times that figure for the best image.

Display and projection formulas

Aspect ratio factors

Minimum contrast ratio by viewing requirement

From ANSI/INFOCOMM 3M, measured in the room with its real lighting.

On the job

A hotel asks for a projector and screen for a board meeting in its ballroom. The farthest board member will sit 14 metres from the screen, and the content is financial spreadsheets. The hotel's in-house screen is 3 metres wide.

You check the screen before anything else. A 3 metre wide 16:9 screen is 1.69 metres high. For spreadsheets — analytical decision making, with small figures — at 14 metres, the image-size calculator returns a minimum image height considerably larger than that. The farthest viewers would be squinting at numbers they cannot read.

Then brightness and contrast. The ballroom has chandeliers the client wants on. You measure the ambient light on the screen with the chandeliers on at their meeting level. With the hotel's 6,000 lumen projector, the arithmetic gives a contrast ratio in the room below 50:1, never mind the 80:1 analytical content needs.

You take the numbers to the client with two options: a larger screen and a much brighter projector with the chandeliers dimmed over the screen end, or a larger screen and pairs of large displays along the table for the spreadsheets, with the projection used for headlines. They choose the displays. Nobody spends the meeting leaning forward and reading numbers aloud to each other.

How to do it

1. Measure the farthest viewer distance.
2. Identify the smallest element the viewers must read, and whether their task is basic or analytical.
3. Use the image size standard's calculator to find the minimum image height.
4. Convert the proposed screen's size into image height and compare.
5. Choose the lens by throw ratio from where the projector can go.
6. Calculate screen luminance from lumens, gain and area.
7. Measure ambient light on the screen and calculate the contrast ratio in the room.
8. Compare with the minimum contrast for the viewing requirement.

What goes wrong

Practice it

Answer before checking.

Geometry & Applying Math

Distances, coverage, sizing — and using math well.

AV geometry

AV uses practical geometry — distances, angles, coverage, and sizing. Examples: throw distance and image size for projection (how far the projector is and how big the image), screen sizing for a room (how big a display needs to be for the viewing distance), coverage (where speakers' sound or lights' beams reach), and angles and positions. This practical geometry helps you place and size AV correctly for the space. You don't need advanced math, but practical geometry (distances, sizes, coverage) is everyday AV math. Practical geometry sizes and places AV correctly in a space.

This geometry (sizing screens, calculating throw, planning coverage) is everyday practical math for getting AV right in a room — the display big enough, the projector positioned right, the coverage adequate. Practical geometry is everyday AV math for sizing and placement.

Applying and checking calculations

Beyond the formulas, using AV math well means applying it correctly and checking it — knowing what to calculate, getting the right numbers, applying the right formula correctly, and checking the result (does it make sense? is it safe? did I make an error?), especially for safety-critical calculations (power, weight). Good practice is to calculate carefully and double-check, because errors can be costly or dangerous. Applying and checking calculations correctly is essential to using AV math. Calculate carefully and always check — especially for safety-critical math.

This careful, checked approach (right calculation, sensible result, double-checked) is what makes AV math reliable and safe in practice — the math is only useful if done correctly. Combining why math matters, power, weight, signal, geometry, and applying it carefully is AV math. AV professionals calculate correctly and check their work. This is AV math in action. Careful, checked calculation is how AV math is used well. (For safety-critical calculations, follow qualified guidance and codes.)

Light levels — candela, lux and foot-candles

Three units describe light, and the arithmetic between them is the same inverse square law that describes sound.

Luminous intensity, in candela, is how bright a source is in a particular direction. A lighting fixture's photometric data gives its intensity at the centre of its beam, often tens of thousands of candela.

Illuminance is how much light arrives on a surface, measured in lux in metric units, which is lumens per square metre, or foot-candles in imperial units, which is lumens per square foot. It is what a light meter reads at a presenter's face, and what a camera operator cares about. One foot-candle is about 10.76 lux.

For a source aimed straight at a surface: illuminance in lux = candela ÷ distance in metres squared. In foot-candles, divide by the distance in feet squared instead.

A fixture with a centre intensity of 20,000 candela, 10 metres from a presenter, gives 20,000 ÷ 10² = 200 lux, about 18.6 foot-candles. Moved to 20 metres, it gives 20,000 ÷ 400 = 50 lux: double the distance, a quarter of the light. That is the inverse square law for light, and it is why a follow spot at the back of an arena needs to be vastly more powerful than a fixture on a truss above the stage.

Light arriving at an angle spreads across more surface and lights it less, by the cosine of the angle from straight on. Photometric tools and lighting designers account for it; for a quick estimate, the straight-on figure is an upper limit.

Beam angle, field angle and beam diameter

A fixture's beam angle is the cone within which its intensity is at least half of the centre value. Its field angle is the wider cone within which it is at least a tenth. Inside the beam angle the light is reasonably even; between the beam and field angles it falls away to the soft edge.

The diameter of the beam at a distance comes from the tangent: diameter = 2 × distance × tan(angle ÷ 2).

A fixture with a 20 degree beam angle and a 30 degree field angle, 10 metres from the stage, makes a beam 2 × 10 × tan(10°) = 3.53 metres across, and a field 2 × 10 × tan(15°) = 5.36 metres across. Two such fixtures need to be placed close enough that their beams overlap if the stage is to be evenly lit, rather than showing dark gaps between pools of light.

A zoom fixture has a range of angles, and the same arithmetic tells a lighting technician which zoom setting covers a given area from a given hanging position.

Coverage width and projector tilt

The same tangent calculation works for anything with a coverage angle.

A loudspeaker's horizontal coverage angle, commonly the angle within which its level is no more than 6 dB below the on-axis level, covers a width of 2 × distance × tan(angle ÷ 2). A 90 degree loudspeaker covers 2 × 20 × tan(45°) = 40 metres of width at 20 metres. A 60 degree loudspeaker covers 2 × 20 × tan(30°) = 23.1 metres. Rows wider than that at that distance are outside the loudspeaker's coverage and hear a duller, quieter sound.

A camera lens's horizontal field of view does the same: a 40 degree field of view shows 2 × 10 × tan(20°) = 7.3 metres of stage width at 10 metres.

Angles also come back out of distances. A projector 1.5 metres above the centre of a screen and 8 metres back must tilt down by tan⁻¹(1.5 ÷ 8) = 10.6 degrees. A loudspeaker hung 5 metres above ear height, aimed at a listener 25 metres away, tilts down by tan⁻¹(5 ÷ 25) = 11.3 degrees.

Light and angle formulas

Beam and field

• Beam angle — where intensity falls to half the centre value.
• Field angle — where it falls to a tenth.
• Double the distance, a quarter of the light.

On the job

A corporate video shoot in a hotel conference room. The camera operator says the presenter needs at least 500 lux on her face for the look they want, and the only fixtures are two LED profile spots on a truss 8 metres away, each with a centre intensity of 15,000 candela and a 19 degree beam angle.

One fixture at 8 metres gives 15,000 ÷ 64 = 234 lux. Two, both aimed at her face from similar angles, give about 469 lux — just under, before allowing for any angle.

And the beam at 8 metres is 2 × 8 × tan(9.5°) = 2.68 metres across, so if she walks more than about 1.3 metres either side of her mark she leaves the brightest part of the light.

You propose two changes: lower the truss to bring the fixtures to 6 metres, which gives each 15,000 ÷ 36 = 417 lux and the pair comfortably over 500, and add a third fixture from the other side for coverage when she moves. The gaffer agrees, and the camera operator's meter confirms it at the rehearsal.

How to do it

1. Find the fixture's centre intensity in candela and its beam and field angles.
2. Measure the distance from the fixture to the subject.
3. Calculate illuminance as candela divided by distance squared.
4. Add the contributions of each fixture on the subject.
5. Calculate the beam diameter at the subject's distance.
6. Compare level and coverage with what is needed, and adjust distance or add fixtures.
7. Confirm with a light meter at the subject's position.

What goes wrong

Practice it

Answer before checking.

Data, Bit Rates and File Sizes

Recording, streaming and networked AV all run on data, and the arithmetic of data is where a single capital letter can mean a factor of eight. A drive fills up on the second day of a three-day conference. A stream stutters because the venue's upload was exactly the stream's bit rate. Twelve video feeds are sent down a network link that can carry nine.

This chapter is bits and bytes, bit rates, file sizes from bit rate and time, why uncompressed video needs the connections it does, and how to budget bandwidth for a stream or a network.

Bits, bytes and bit rates

A bit is a single binary digit, a 0 or a 1. A byte is eight bits. Data rates, how fast data moves, are nearly always given in bits per second: megabits per second, Mbps, and gigabits per second, Gbps. Storage, how much data is kept, is nearly always given in bytes: megabytes, MB, gigabytes, GB, and terabytes, TB.

The abbreviations differ by case: a lower-case b is bits and an upper-case B is bytes. So 100 Mbps is 100 megabits per second, which is 12.5 megabytes per second.

Two more wrinkles are worth knowing. In data, kilo, mega and giga usually mean powers of a thousand, but some operating systems count storage in powers of 1,024, which is why a drive sold as 1 terabyte shows as about 931 gigabytes on a computer. Nothing is missing; the units differ. And a network link's rated speed is the maximum under ideal conditions; real throughput is lower, and a link shared with other traffic has less still.

File size from bit rate and time

A recording's size is its bit rate multiplied by its duration, divided by eight to turn bits into bytes.

File size in megabytes = bit rate in megabits per second × seconds ÷ 8.

A stream recorded at 6 megabits per second for one hour is 6 × 3,600 ÷ 8 = 2,700 megabytes, about 2.7 gigabytes. A camera recording at 220 megabits per second — a common bit rate for a high-quality HD editing codec at 30 frames per second — for three hours is 220 × 10,800 ÷ 8 = 297,000 megabytes, 297 gigabytes.

For a whole event, multiply by every device recording. Four cameras recording 297 gigabytes each for each day of a three-day conference need about 3.6 terabytes, before the program recording, the backups and a margin for sessions that run long. That is why media is planned from the bit rate, not bought by guessing.

The useful rule of thumb from this: one megabit per second for an hour is 450 megabytes. Any bit rate multiplied by 450 gives megabytes per hour.

Uncompressed video, and why SDI speeds are what they are

Uncompressed video carries every pixel of every frame, and its bit rate can be worked out from the picture: pixels per frame × bits per pixel × frames per second.

A 1920 by 1080 frame has 2,073,600 pixels. Professional video usually stores colour as 4:2:2 at 10 bits, which averages 20 bits per pixel. At 60 frames per second that is 2,073,600 × 20 × 60 = 2,488,320,000 bits per second, about 2.5 gigabits per second, for the picture alone.

That is why 1080p60 needs 3G-SDI, a link of about 2.97 gigabits per second: the rest carries the blanking between lines and frames, audio and data. At 30 or 25 frames per second, or interlaced, the picture needs about half, which fits the older HD-SDI rate of 1.485 gigabits per second. A UHD frame has four times the pixels, so 2160p60 needs about 10 gigabits per second for the picture, which is why 12G-SDI exists.

The same arithmetic explains compression. A 6 megabit per second stream of 1080p60 is about 400 times smaller than the uncompressed picture. The encoder throws away information viewers are unlikely to notice, and the more aggressively it compresses, the more it can go wrong on fast motion, fine detail or confetti.

Budgeting bandwidth

Bandwidth is capacity, and a network or an internet connection used at its full capacity performs badly. Every budget includes headroom.

For a live stream, the upload capacity of the connection should be comfortably more than the stream's bit rate. A common working rule is at least double: a 6 megabit per second stream wants 12 megabits per second of dedicated, reliable upload, more if a backup stream runs at the same time. The upload is tested at the venue, on the actual connection, at the time of day of the event, because a hotel's connection at 9 am with five hundred guests online is not the connection a salesperson tested on a quiet Sunday.

On a production network, add up the streams. Full-bandwidth NDI at 1080p60 uses around 125 megabits per second per stream, so eight camera feeds are around 1 gigabit per second — the whole capacity of a 1 gigabit link, which is therefore too small. Networked audio uses far less: dozens of channels fit in a few tens of megabits. Video over IP at higher quality uses much more. The network design matches the switches and links to the total, with room to spare.

Data formulas

Links and what they carry

On the job

A three-day medical conference is recording every session in four breakout rooms plus the main hall, one program recording per room, at 50 megabits per second. Each room runs 9 hours a day. The client has sent five 2 terabyte drives, one per room, and wants to know whether they are enough.

Each room records 50 × 450 = 22,500 megabytes per hour. Over 9 hours that is 202,500 megabytes a day, about 202.5 gigabytes, and over three days about 608 gigabytes per room. A 2 terabyte drive, about 1.8 terabytes as the computer counts it, holds that three times over.

The main hall is different: the client also wants every camera isolated, four cameras at 50 megabits per second plus the program. Five recordings at 608 gigabytes each over three days is 3 terabytes, which does not fit on one 2 terabyte drive.

You tell the client on the first morning rather than the third afternoon, and ask for two more drives for the main hall. They arrive at lunch. The backup copies, which nobody had budgeted, go on a further drive the production company supplies.

How to do it

1. List every recording device and its bit rate in megabits per second.
2. Multiply each bit rate by 450 to get megabytes per hour.
3. Multiply by the hours recorded per day and by the number of days.
4. Add up all devices recording to the same media.
5. Add backups and a margin of at least 20 per cent for overruns.
6. Convert to the drive's size as the computer reports it, allowing for the 1,000 against 1,024 difference.
7. Compare with the media available, and ask for more before the event, not during it.

What goes wrong

Practice it

Answer before checking.

AV Math as a Foundation

The practical math behind professional AV.

AV math mastered

Pulling it together, AV math is the practical, everyday calculation behind AV work — power and electrical, weight and loads, signal levels and decibels, and geometry — applied carefully and checked, underpinning safe, correct, professional work. Understanding and using AV math is a foundational skill that supports many AV competencies (and the CTS certification, which includes this math). It's not about being a mathematician — it's about confidently doing the practical calculations the job requires. AV math mastered is confidently doing the everyday calculations behind safe, correct AV work. AV math is the practical numeric foundation of the work.

This foundational math underpins much of AV (rigging, power, audio, video, design) and is part of being a competent, professional AV practitioner. Understanding AV math — power, weight, signal, geometry, applied carefully — is a valuable, foundational capability. AV math is a practical foundation that supports safe, correct, professional AV work and certification.

Building your AV math

Developing your AV math means practising the practical calculations (power/loads, weights, levels, geometry) until they're comfortable, applying them on real jobs (where the math becomes second nature), and deepening it as needed for your role and for certification (the CTS exam includes AV math). It connects to rigging, power, audio, video, and design. The more you use AV math, the more confident and capable you become — and the safer and more professional your work. Deliberately practising AV math is the path. Practise the practical calculations until they're second nature.

From here, growing means using AV math confidently across your work, deepening it for your specialty and for certification, and always calculating carefully (especially for safety). It's a foundational skill that makes you more capable and professional. The practical math this book covers — power, weight, signal, geometry, applied carefully — is the foundation; mastery comes from using it regularly. Keep practising and applying AV math, and you build a confident, professional numeric foundation for your AV work. (For safety-critical calculations, always follow codes and qualified guidance.)

Practical sign-off

AV math is learned by doing calculations. The learner does practical power, weight, signal, and geometry calculations correctly and checks them. A mentor signs off when the learner can do everyday AV math confidently and safely.

• Did basic power and electrical (load) calculations
• Did weight and load math for rigging/staging
• Understood signal levels/decibels and did AV geometry
• Applied and checked AV calculations correctly and safely

A method for any calculation

Every calculation in this book, from a load to a delay to a drive size, can be done with the same five steps, and the steps matter more than any formula.

First, say what you are finding, in words and with its unit: "the current on circuit 3, in amps". Second, write what you know, each number with its unit, converted into one system. Third, estimate the answer with easy numbers. Fourth, calculate in written steps, choosing the formula from the units — if you have watts and volts and want amps, the only arrangement that produces amps is watts divided by volts. Fifth, check the result against the estimate and against common sense, and round it in the safe direction.

Written down, the working is also a record. When somebody questions a number, or when the gear changes and it must be recalculated, the working shows exactly what was assumed. A number without its working can only be trusted or distrusted; a number with its working can be checked.

Checking by a second route

The strongest check on an important number is to reach it a second way.

Work it backwards. If 2,100 watts at 120 volts is 17.5 amps, then 17.5 × 120 should be 2,100. If a delay of 131 milliseconds covers 45 metres, then 0.131 × 343 should be about 45.

Work it in different units. A load calculated in kilograms, converted to pounds, should match the same load calculated in pounds from the start.

Check a part against the whole. Support shares must add to the total load. Phase currents must add to the total draw. The pixels across an LED wall must equal the panels times the pixels per panel.

And for safety numbers, have a second person do the calculation independently, from the same information, without seeing your answer. Two people making the same mistake independently is far less likely than one person making it once.

Show maths on one page

The formulas a technician reaches for most often fit on one page, and it is worth copying them into the notebook in your own hand.

Power: amps = watts ÷ volts. Continuous circuit capacity: volts × amps × 0.8. Parallel identical loudspeakers: impedance ÷ number. Heat: watts × 3.412 = BTU per hour.

Weight: supports share a point load by the distance to the other support. Bridle leg tension: share × leg length ÷ vertical drop. Working load limit: breaking strength ÷ design factor.

Decibels: 10 log for power, 20 log for voltage and pressure. −6 dB per doubling of distance. Two equal unrelated sources: +3 dB.

Time: 2.9 milliseconds per metre of sound. Wavelength: 343 ÷ frequency. One frame: 1,000 ÷ frame rate milliseconds.

Screens: 16:9 width = diagonal × 0.872. Throw distance = throw ratio × width. Nits = lumens × gain ÷ (π × square metres). LED pixels = panel width ÷ pitch.

Light: lux = candela ÷ distance². Beam width = 2 × distance × tan(half the angle).

Data: megabytes = megabits per second × seconds ÷ 8, or megabits per second × 450 per hour.

The method, and second-route checks

The method for any calculation, and the checks that catch a wrong answer.

Every calculation

• Say what you are finding, with its unit.
• Write what you know, with units, in one system.
• Estimate with easy numbers.
• Calculate in written steps, letting the units choose the formula.
• Check against the estimate and common sense, and round safely.

Second-route checks

• Work it backwards.
• Work it in different units.
• Check that the parts add to the whole.
• Have a second person calculate safety numbers independently.

On the job

A production manager is writing the technical rider response for a touring show that has never played this venue. The venue's rigging data says each point is rated to 1,000 pounds. The show's rigging plot, from a European production company, lists its heaviest point at 420 kilograms, and the lighting designer has already written "fits" in the margin.

You set it out. Finding: the heaviest point load, in pounds, against a 1,000 pound rating. Known: 420 kilograms. Estimate: a kilogram is about two pounds, so about 840 pounds, which fits. Calculate: 420 × 2.205 = 926 pounds. Check backwards: 926 × 0.4536 = 420. It fits — with a margin of only 74 pounds.

Then the part that is not arithmetic. The plot's 420 kilograms is the load on the hoist hook. The venue's 1,000 pound rating is for the point, and the hoist itself, its chain and the rigging hardware hang from that point too. You ask the production's rigger for the hoist and hardware weights. They add another 65 kilograms, 143 pounds, taking the point to 1,069 pounds, over the rating.

The "fits" comes out of the margin. The production's head rigger and the venue's rigger agree a bridle to share that load across two points, and the numbers go into the advance with the working shown.

How to do it

1. Confirm exactly what the number describes and what is included in it.
2. Set out the calculation with the five steps.
3. Check it by a second route.
4. Have a second person calculate it independently.
5. Round in the safe direction.
6. Give it, with the working, to the qualified person responsible.

What goes wrong

Practice it

Answer before checking. Set each one out with the five steps.

Glossary

What the gear actually looks like

Every device named in this book, in one place. Recognition is the whole point of Chapter 4 — use this page to attach a picture to each name before you meet it on a dock at six in the morning.

Audio

Dynamic microphone

Condenser microphone

Wireless handheld

Radio rack

Mixing console

Stage box / snake

Power amplifier

Main loudspeaker

Subwoofer

Stage monitor wedge

In-ear monitors

DI box

Video

Camera

Video switcher

Scaler / processor

Projector

LED wall panel

Confidence monitor

Lighting & control

Conventional fixture

Moving-head fixture

Lighting console

Dimmer

Show controller

Media server

Networking & infrastructure

Network switch

Patch bay

Equipment rack

Power distribution

Cable tester

A properly coiled cable

Connectors, close up

XLR

TRS / TS ¼-inch

SDI on BNC

HDMI

Cat5e / Cat6 (RJ45)

Fiber optic

DMX (5-pin XLR)

AC power — not signal