Tracy Gunn
- Aug 20, 2022
- 12 min read

IDC Dive Theory - Complete Physics "At a Glance"

Updated: Aug 20

If you already have a firm grasp of IDC physics, you may not want to read through all the other posts and only need quick reminders. This post will present you all the dive theory in one brief "at a glance" presentation. If you need more detailed information, click each heading to be redirected to that blog.

Understanding About Water, Heat, Light, Sound, and Gases.

The bending of light is called REFRACTION, making objects appear 33% LARGER and 25% CLOSER underwater (Ratio 4:3).

Water ABSORBS COLOR. Red is the first colour to go.

VISUAL REVERSAL is when objects appear further away, caused by TURBIDITY creating suspended particles in the water.

Conduction refers to heat transmission via direct contact. This type of heat transfer affects divers most, so it is important to wear a well-fitted wetsuit.

Water conducts heat 20 times better than air.

Convection: involves heat transmission via fluids.

Radiation: heat transmission via electromagnetic waves also affects divers.

Underwater sound travels 4 times faster because of its elasticity and density, which makes sounds appear to be coming from overhead.

Above water, we can determine a sound’s direction by the minute difference between when a sound arrives at one ear and the other.

Underwater, sounds arrive at our ears almost simultaneously, making it difficult to determine the direction from which a sound is coming.

This can be a problem when a boat is overhead, as it may be difficult to determine how close the boat is or the direction it travels. When you hear a boat, the best thing to do is to remain close to the bottom until the sound disappears and then surface by your dive float or your own dive vessel.

Water, Heat, Light, Sound, and Gases Exam

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Archimedes' Principle

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

CONSTANT 1 Litre of Salt Water = 1.03 kg

1 litre of Fresh Water = 1.00 kg

The definition of a kilogram is exactly 1 litre of fresh water.

Part 1

1. An object that weighs less than the water it displaces floats and is positively buoyant. The buoyancy is expressed as a positive number, such as “two kilograms positive.”

2. An object that weighs the same as the water it displaces neither floats nor sinks. It is called neutrally buoyant; adding or removing weight will make it sink or float.

3. An object that weighs more than the water it displaces will sink and is called negatively buoyant. Its buoyancy is expressed as a negative number, such as “two kilograms negative.”

If an object has neutral buoyancy in Seawater and you take it to Freshwater, the object will sink.

If an object has neutral buoyancy in Fresh Water and you take it to Sea water, it will float.

Part 2

To determine the buoyancy of an object, you need to determine 3 things.

Think Open Water Diver

1 Open (O) objects weight =(kg)

2 Water (W) Water = (Seawater or Freshwater)

3 Diver (D) Displacement = (litres)

From there, you can determine the weight of the water displaced.

EQUATIONS

Archimedes states that we need to determine the weight of the displacement, so this is where we always start the equation.

Think Master Scuba Diver

1 Master (M)ultipy (lt to kg)

Displacement (lt) × salt water or fresh water = weight of water displaced (kg)

2 Scuba (S)ubtract

Weight of displaced water – weight of object = Buoyancy (+ve. -ve or neutral)

3 Diver (D)ivide (kg to lt)

Buoyancy in kg ÷ seawater or freshwater = amount of displacement (lt)

Depth (in either salt water or fresh water) has no effect on determining buoyancy.

Ex 1 You Plan to recover a 150kg outboard motor in seawater that displaces 60 litres and lies at 30m. How much air must you put in a lifting device to make the motor neutrally buoyant?

Open (O) Object weigh = 150 kg

Water (W) Water = Sea (1.03kg)

Diver (D) Displacement = 60 Litres

Master (M) Multiply 60lt X 1.03kg = 61.8kg

Scuba (S) Subtract 61.8kg-150kg = -88.20kg (sink) -ve sign is only to show if it floats or sinks.

Diver (D) Divide 88.2kg ÷ 1.03kg = 85.6lt

Ex2 You must sink an object that weighs 50 kg and displaces 300 Litres into fresh water; how much lead do you need to affix to the object to make it 10kg negative on the bottom?

Open (O) Object weight = 50kg

Water (W) Water = Fresh (1.00kg)

Diver (D) Displacement = 300 Litres

Master (M) Multiply 300lt X 1kg =300kg

Scuba (S) Subtract 300kg – 50kg =250kg +ve sign means it floats; we do not need to work out displacement.

Diver (D) Divide N/A

Therefore, 250kg weight to make it neutral + 10kg to make 10kg negatively buoyant

250kg + 10kg = 260kg

Archimedes' Exam part 1

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Archimedes' Exam Part 2

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UNDERSTANDING PRESSURE

Bars in Saltwater

Salt Water 10.0 meters = 1 bar

Fresh Water 10.3 meters = 1 bar

Freshwater needs more depth for the same ATM/Bar as it is less dense than saltwater (weighs less).

Gauge Pressure

= Water pressure, EXCLUDING atmospheric

* Only used when asking for Gauge

Saltwater = Depth÷10

Fresh Water = Depth÷10.3

Ambient / Absolute Pressure

= Gauge AND Atmospheric

* Always use this for physics equations

Saltwater = Depth÷ 10+1

Fresh Water = Depth÷ 10.3 +1

Important Equations to find atmospheric pressure

Always check

Type of pressure requested (gauge, ambient or absolute)
The type of water (sea or fresh)
The depth before starting the equation.

Examples

What is the Gauge pressure at 28mt of Fresh Water?

28/10.3= 2.7ata

What is the Absolute/Ambient pressure at 28mt Fresh Water?

28/10.3+1=3.7ata

What is the Gauge pressure at 53mt Salt Water?

53/10=5.3ata

What is the Absolute/Ambient pressure at 53mt of Salt Water?

53/10+1= 6.3ata

To convert to another pressure measure, multiply by:

Metric Imperial

10 msw (meters of seawater) 33 for fsw (feet of seawater)

10.3 mfw (meters of fresh water) 34 for ffw (feet of fresh water)

1.03 for kg/cm² (kg per cm squared) 14.7 for psi (pounds per square inch)

Under Pressure Exam

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BOYLE'S LAW

If temperature remains constant both volume and density of gas are affected in proportion to the atmospheric pressure and inversely to each other.

DO NOT CONFUSE BOYLES LAW WITH ARCHIMEDES PRINCIPLE

Boyle's law talks about pressure/ volume/ density relationships. For the purposes of explanation and calculation, we assume no change in temperature.

This law affects you in every dive. It is why you need to equalize your ears, why your wetsuit compresses upon descent, why your BCD expands upon ascent and why bubbles get bigger going to the surface. It is the reason why you use more air at depth and have less bottom time.

How to examine the questions for Boyles Law

Part 1

STEP 1 Find the unit of measurement

Look at the answers. The unit of measurement will be stated there (always multiple choice in final exams)
What is the question asking for?
Minutes
PSI or Bar
Litres
Find the number from inside the question that is in that unit and write it down.

STEP 2 Think about the question

TIME

The shallower you are, the slower you consume air (you have more time)

The deeper you go, the faster you consume air. (you have less time).

More minutes X, less minutes ÷

PSI or Bar

The shallower you are, the less air you breathe with each breath, and the amount of air consumed in PSI/Bar will decrease.

The deeper you go, the more air you breathe with each breath, and the amount of air consumed in PSI/Bar will increase.

More PSI/Bar X, Less PSI/Bar ÷

Litres

If you take a balloon up from depth, its size increases (more volume)

If you take a balloon down to depth, its size decreases (less volume)

Bigger volume X, less volume ÷

This is THE most important step and where the most errors occur. Think it through first.

STOP, BREATHE, THINK, ACT

STEP 3 Find the surrounding pressure

READ THE QUESTION and find the depth in meters (or feet)
Determine the pressure at that depth

Think Saltwater (10) vs Freshwater (10.3).

Always +1 (you are determining ambient/absolute pressure)

Write that pressure down

STEP 4 Find the answer

Step 1 Measurement
Step 2 ×/÷
Step 3 Surrounding pressure
= Step 4

EXAMPLES OF SINGLE SURFACE/DEPTH CHANGES

Air consumption in minutes

Q It takes 60 minutes to breathe through a tank of air at the surface. If all factors remain unchanged, how long will it take to breathe through the same tank at 20 mt?

A 180 minutes

B 60 minutes

C 30 minutes

D 20 minutes

STEP 1 Answer is asking for minutes. So, we look for minutes in the question and write it down - 60 minutes.

STEP 2 Think! Do you use air faster at 20 meters than at the surface? We know we use air much faster at depth; therefore, the answer will be LESS TIME than 60 minutes. (÷)

STEP 3 The depth is 20 meters. 20÷10+1=3ATA

STEP 4 60÷3=20 minutes (therefore, the answer is D)

Volume (flexible container)

Q If we have a balloon with a volume of 8 litres at the surface, what would its volume be at 30mt?

A 32 litres

B 24 litres

C 2 litres

D 2.6 litres

STEP 1 Answer is asking for litres. So, we look for litres in the question. 8 litres

STEP 2 Think! Does the volume increase or decrease as we descend? The air becomes denser, so the volume becomes smaller; therefore, the answer will be LESS VOLUME than 8 litres (÷)

STEP 3 The depth is 30 meters. 30÷10+1=4ATA

STEP 4 8÷4=2litres (therefore, the answer is C)

Air consumption in PSI/Bar

Q A diver breathes through 60 Bar of air in 60 minutes at the surface. If all factors remain unchanged, how many BAR will the diver breathe in 40 minutes at 30 meters?

A 160 Bar

B 120 Bar

C 10 Bar

D `45 Bar

STEP 1 Answer is looking for BAR. So, we look for BAR in the question and write it down

60 bar in 60 minutes (here is the trick. We need to know how many bar per minute)

60÷60=1Bar per min

STEP 2 Think! Do we consume more air at depth or less? The air is denser, so we use more air with each breath. Therefore, the answer will be MORE PSI than 1 bar per min (X)

STEP 3 The depth is 30 meters. 30÷10+1= 4ATA

STEP 4 The answer is to multiply bar per min by ATA and then again for 40 min!!!

1X4X40= 160Bar (therefore, the answer is A)

Part 2

LET'S DO MULTIPLE DEPTH CHANGES

VOLUME

1. You Multiply the volume to go UP

2. You Divide the volume to go DOWN

3. If you have 2 different depths underwater, FIRST take the volume to the surface and then back down again!!!

Depth	Atmospheres	Volume
0 (Surface)	1	Vol (1) × ata	= Vol 2
Depth	ata	Litres	Vol (1)
Depth	ata	Vol (2) ÷ ata	= Vol (3)

ANSWER IS ??? LITRES

FILL IN ALL KNOWN VALUES FIRST (IN BOLD)

First, take Volume (1) to the surface (2) (times to go up)

Then take down to Volume (3) (divide to go down, remember volume gets less at depth)

Example 1: A flexible container with a volume of 35 litres is at 28m of freshwater and is taken up to 24m of seawater; what would the new volume be?

Depth	Atmospheres	Volume
0 (Surface)	1 ata	35 × 3.7	= 129.5 litres (2)
24 meters	24÷10+1=3.4 ata	129.5÷ 3.4	= 38 litres (3)
28 meters	28÷10.3+1=3.7 ata	35 litres	(1)

The answer is 38 litres at 24 meters.

V1P1=V2P2

V2=V1P1÷P2

V2=35×3.7÷3.4

V2=38 litres

DENSITY (= atmospheres)

1. You Divide (÷) the volume of air (BAR per min) the diver uses when the diver goes UP (Shallower)

2. You Multiply (×) the volume of air (Bar per min) the diver uses when the diver does Down (Deeper)

3. If you have two different depths underwater, FIRST take the density to the surface and then go back down again!!

4. Volume of air used at the surface is referred to as Surface Air Consumption or SAC

Air becomes denser (more pressure) as you descend underwater. A diver will use more air due to the increased (more) pressure (He’s going deeper…more pressure). A diver will use less air due to the decreased (less) pressure as he ascends. (He’s going up… less pressure)

Depth	Atmospheres	Surface Air Consumption (SAC)
0 (Surface)	1 ata	AC (1) ÷ ata	= SAC (2)
Depth	ata	Bar (Air Consumption)	(1)
Depth	ata	AC(2) × ata	AC (3)

ANSWER IS ??? BAR

FILL IN ALL KNOWN VALUES FIRST

First, take Air consumption (Bar) (1) to the surface; Surface Air Consumption (SAC) (2) (Divide to go up)

Then take down to new Air consumption (Bar) (3) (Multiply to go down, remember you use more air at depth)

Example: A diver consumes 4bar/60psi per minute at 10m/33ft in seawater with a given cylinder. What is the diver’s consumption rate at 33m/108ft using the same cylinder?

Depth	Atmospheres	(SAC)
0 (Surface)	1 ata	4÷2=2 Bar SAC	=SAC (2)
10 meters	10÷10+1=2ata	4 Bar (AC)	(1)
33 meters	33÷10+1=4.3ata	2×4.3=8.6 Bar	=AC (3)

The answer is 8.6 Bar

D1/P1-D2/P2

D2=D1/P1×P2

D2=4÷2×4.3

D2=8.6 Bar

Suppose the question states Bar consumption in X minutes. Divide until you determine Air Consumption per Minute and then begin the equation, i.e. 120 bar per hour is 120÷60= 2 bar per min.

Boyle's Law Exam Part 1

Single-Level Depth Changes

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Boyle's Law Exam Part 2

Multi-Level Depth Changes

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CHARLE'S LAW

As the temperature increases (Rises) the volume of a flexible container will increase (get bigger) and the pressure inside the container (only a nonflexible) will increase

The opposite happens when the temperature decreases (Drops). As the temperature decreases (becomes less), the volume of a flexible container will decrease (get smaller), and the pressure inside a nonflexible container will decreases (becomes less)

FOR EVERY 1 DEGREE CELCIUS CHANGE (UP OR DOWN) IN TEMPERATURE, THERE IS A 0.6 BAR CHANGE IN PRESSURE

For a flexible container (balloon). If you increase the temperature, the volume will increase but not the pressure.

If you leave a balloon outside in the sun, the volume will increase, not the pressure.

For a nonflexible container (tank). If you increase the temperature, the pressure will increase, not the volume.

If you leave a full tank outside in the sun, the pressure inside will increase, not the volume.

Also, if you decrease the pressure, you will decrease the temperature (but not the volume)

Example: 220 Bar filled at 28 degrees. Will use the tank in 3-degree water; what is the pressure change inside the tank?

28-3 degrees = 25 degree change 25 x 0.6 bar = 15 Bar

220 Bar – 15 Bar = 205 Bar

Charles' Law Exam

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HENRY'S LAW

At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

Application to Diving: The gas that affects us while diving is nitrogen because it is inert, and the liquid that Henry describes is related to our body (not the ocean). This law helps us explain and avoid decompression sickness.

SATURATION: When the pressure (an amount) of gas that has been forced into a liquid is the same as the surrounding pressure. This happens every time we ASCEND or DESCEND to a different depth. As we descend or increase the pressure of a gas in contact with a liquid, we force more gas to dissolve into that liquid until saturation is reached (Same when we ascend, except gas will be released until saturation is again reached). The pressure is the same outside the liquid as the pressure inside the liquid.

The longer time a diver spends underwater at a certain depth (under pressure), the more Nitrogen the body absorbs until saturation is reached; no more nitrogen can be absorbed.

SUPERSATURATION: When the pressure becomes less, like when you open a bottle of soft drink, the gas that was dissolved into the liquid starts to come out of the liquid. If it is controlled, the bubbles come out slowly. If the pressure is released too fast, the bubbles will also come out too fast. This happens if we ASCEND too quickly. The pressure inside the liquid is greater than the gas in contact with the liquid, and bubbles may occur.

If a diver comes up too fast from a dive, pressure is released too fast, the nitrogen will come out too fast, and bubbles will form, causing DECOMPRESSION SICKNESS.

This is Supersaturation

Example: If you shake a coca cola bottle, you build up pressure inside, forcing the gas into the cola. When you quickly open the bottle, you decrease the pressure; all the gas comes out of the cola! BUBBLES!!!

Henry's Law Exam

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DALTON'S LAW

The sum of all the gases will always equal a whole (100%)

As the pressure Increases (Get more), the Partial Pressure (NOT THE PERCENTAGE, THE PERCENTAGE NEVER CHANGES) will increase (Get more).

Dalton's law (also called Dalton's law of partial pressures) states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

Air = 21% Oxygen 79% Nitrogen (ALWAYS)

Dalton's law can always have the same equation viewed 3 ways

1 The percentage of the gas (never changes)

2 The partial pressure (expressed in decimal. Think partial…decimal)

3 The equivalent of breathing what percentage at the surface (expressed in %)

EXAMPLES

1. 21% Oxygen + 79% Nitrogen =100% Air

Remember, the effect of a gas changes as does the surface equivalent, but the percentage remains the same

2. Surface = 1 Bar

1Bar x 0.21 oxygen = .21 oxygen

1Bar x 0.79 nitrogen = .79 nitrogen

1 Bar

10m = 2 Bar (10/10+1=2 Bar)

2Bar x 0.21 oxygen = .42 oxygen

2Bar x 0.79 nitrogen = 1.58 nitrogen

2 Bar

30m = 4Bar (30/10+1=4 Bar)

4Bar x 0.21 oxygen = 0.84 oxygen

4Bar x 0.79 nitrogen = 3.16 nitrogen

4 Bar

This is important to be able to calculate the partial pressure of gasses to any depth. Doing this allows you to see when oxygen reaches high levels that may cause oxygen toxicity.

1.4 Bar (Max)

1.6 Bar (contingency)

This can be used with any gas mix, such as Helium, Nitrogen, Carbon Dioxide etc…

Nitrogen levels become narcotic when it reaches +/- 3 Bar.

3 Bar/ (divided by) 0.79 (If using air)

= 3.7 Bar = 27m (Ambient)

Example: Enriched Air 36% Oxygen at 40m (5bar)

1 The percentage at 40mt will still be the same

Oxygen 36%, Nitrogen 64%

Remember, the % does not change; only the surface air equivalent

2 PO2= 0.36 x 5 bar =1.8 Bar * oxygen toxicity in this range

PN2= 0.64 x 5 bar =3.2 Bar