• Tracy Gunn

IDC Dive Theory - Complete Physics "At a Glance"

Updated: Aug 22

If you already have a firm grasp of IDC physics, maybe you do not want to read through all the other posts and just need quick reminders. This post is to present to you all the dive theory in one quick "at a glance", presentation. If you need more detailed information, click on each heading to be redirected to that specific 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 color to go.

VISUAL REVERSAL is when objects appear to be further away; this is 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, which is why 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 when it arrives at 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 there is a boat overhead, as it may be difficult to determine how close the boat is, or the direction it is traveling. The best thing to do when you hear a boat 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 being “two kilograms positive.”

2. An object that weighs the same as the water it displaces neither floats nor sinks. It is called neutrally buoyant and 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.


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 the 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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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

  1. Type of pressure requested (gauge, ambient or absolute)

  2. The type of water (sea or fresh)

  3. The depth before starting the equation.


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?


What is the Gauge pressure at 53mt Salt Water?


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


Boyle's law is talking 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 the reason 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.


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


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 that 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



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!!!




0 (Surface)


Vol (1) × ata

= Vol 2




Vol (1)



Vol (2) ÷ ata

= Vol (3)



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 fresh water, is taken up to 24m of Sea Water, what would the new volume be?




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


The answer is 38 litres at 24 meters





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)



Surface Air Consumption (SAC)

0 (Surface)

1 ata

AC (1) ÷ ata

= SAC (2)



Bar (Air Consumption)




AC(2) × ata

AC (3)



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. Using the same cylinder, what is the diver’s consumption rate at 33m/108ft?




0 (Surface)

1 ata

4÷2=2 Bar SAC

=SAC (2)

10 meters


4 Bar (AC)


33 meters


2×4.3=8.6 Bar

=AC (3)

The answer is 8.6 Bar





D2=8.6 Bar

If 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

Boyles' Law Exam Part 1

Single Level Depth Changes

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