- 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
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.
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 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
Archimedes' Exam Part 2
UNDERSTANDING PRESSURE
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
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 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.
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 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
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 fresh water, is taken up to 24m of Sea Water, 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
OR
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. Using the same cylinder, what is the diver’s consumption rate at 33m/108ft?
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
OR
D1/P1-D2/P2
D2=D1/P1×P2
D2=4÷2×4.3
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