# Buoyancy – PADI Instructor Course and Divemaster Exams

So far we have basically ignored the type of water.

I will return to some of the topics addressed in parts 1,2 and 3 to discuss how we would answer them if the question is about FRESH WATER. Everything we have done to date assumed salt water.

With Buoyancy, the type of water is critical. We have to know what type of water the object is in to answer buoyancy questions.

The difference in the weight of salt water compared to fresh water is what matters here.

You probably already know that you are MORE buoyant in salt water than fresh. You need to wear more weight in salt water than when you are diving in fresh water.

There is one number that we need to calculate the difference. That number is 1.03

1 Litre of Salt Water weighs 1.03kg – That’s crucial knowledge. Put another way, 1 Litre of Salt Water weighs a little bit more than 1kg.

There are 4 factors that we need to consider when looking at the buoyancy of an object.

1. The weight of the object,

2. The volume (or displacement) of the object,

3. The type of water it is in

4. The current state of the object (floating, sinking or neutral).

In general, if we know three of the four things, we should be able to provide the missing fourth piece of information. These questions seem to cause more confusion than many others. I mark buoyancy questions incorrect more often than any other in the PADI instructor development course than any other type of question.

## Part 1- Describing what would happen to an object

If an object is positively buoyant in salt water, we simply cannot say what would happen to the object if it was moved into fresh water. However, if it is negatively buoyant in salt water it will certainly be negatively buoyant in fresh water.

If an object is negatively buoyant in fresh water we cannot say what would happen to the object if it was moved into salt water. However, if it is positively buoyant in fresh water it will certainly be positively buoyant in salt water.

If we were told how many kilograms of positive or negative buoyancy it had, we would be able to calculate the change, more on that later.

If an object is NEUTRAL, we can describe what would happen to it.

## Steps to find an object’s buoyancy:

An object’s buoyancy is the difference between its weight, and the weight of the water it displaces. If it is heavier than the weight of the water it displaces, it will sink. If it is lighter than the weight of the water it displaces, it will float. If they are equal, it will be neutrally buoyant.

**Step 1** Calculate the weight of the water it displaces. You need to find the volume in Litres and multiply that by the constant, 1 for fresh water or 1.03 if it is in salt water.

**Step 2** Once you have this information you subtract the weight of the object.

### Examples

1. An object with a weight of 210 kg is in salt water. Its volume is 200 ltr. Will the object’s buoyancy be

a. Positive

b. Negative

c. Neutral

d. The answer cannot be determined

**Step 1** Calculate the weight of the water it displaces. The volume is 200, the water is salt.

200 x 1.03 = 206.

**Step 2** Once you have this information you subtract the weight of the object. The weight is 210.

206 – 210 = -4 kg It will be negatively buoyant Answer B

2. An object with a weight of 51 kg is in salt water. Its volume is 50 ltr. Will the object’s buoyancy be

a. Positive

b. Negative

c. Neutral

d. The answer cannot be determined

**Step 1** Calculate the weight of the water it displaces. The volume is 50, the water is salt.

50 x 1.03 = 51.5

**Step 2** Once you have this information you subtract the weight of the object. The weight is 210.

51.5 – 51 = 0.5 kg It will be positively buoyant Answer A

3. An object with a weight of 309 kg is in salt water. Its volume is 300 ltr. Will the object’s buoyancy be

a. Positive

b. Negative

c. Neutral

d. The answer cannot be determined

**Step 1** Calculate the weight of the water it displaces. The volume is 50, the water is salt.

300 x 1.03 = 309

**Step 2** Once you have this information you subtract the weight of the object. The weight is 210.

309 – 309 = 0 kg It will be neutrally buoyant Answer C

## Part 2 - Finding an object’s weight or volume

In the PADI instructor course, one of the questions I see more people get wrong than any other is calculating an objects weight or volume.

We learned in video 1 of this section that we can ONLY calculate what would happen to an object if it moved from fresh water to salt water or vice-versa if it is NEUTRALLY buoyant.

The same can be said for calculating an object’s volume or weight. If an object I neutral in one type of water and we know either its volume or its weight, we can calculate the missing piece of information.

With FRESH water this is extremely easy. An object that neutral in FRESH water has a volume equal to its weight. NO CALCULATIONS NEEDED!

**In salt water:**

If we know a neutrally buoyant object’s volume; we can MULTIPLY the volume by the water constant (1 for fresh or 1.03 for salt) and that will give us its weight.

If we know a neutrally buoyant object’s weight; we can DIVIDE the weight by the water constant (1 for fresh or 1.03 for salt) and that will give us its volume.

**THE RULE TO LEARN IS:**

**VOLUME TO WEIGHT MEANS MULTIPLY**

**WEIGHT TO VOLUME DIVIDE**

I don’t like learning rules or formulas and just applying them. To feel confident in the PADI instructor exam, you need to understand what you are doing and why. That is why I recommend using the diagram I introduced in my video Buoyancy Calculations Part 1.

Here is how you would use it to calculate an object’s volume of weight:

As we learned in Video 2, buoyancy questions are all about calculating the upward force. Once we know that we can figure out the rest.

### Examples

1) An object with a volume of 150 Litres is neutrally buoyant in fresh water. What is its weight?

a. 146 Kg

b. 150 Kg

c. 155 KG

d. The answer cannot be determined

**Step 1 **Write down the volume.

**Step 2 **That’s it, you just wrote down the answer because its in FRESH water.

2) An object with a volume of 150 Litres is neutrally buoyant in SALT water. What is its weight?

a. 146 Kg

b. 150 Kg

c. 155 KG

d. The answer cannot be determined

**Step 1 **Write down the volume 150 Ltrs

**Step 2 **Multiply by 1.03 150x1.03= 154.5 Kg

150 x 1.03 = 154.5 We have calculated the upward force. Since the object is Neutral, the upward force and the downward force are the same. So the weight must be 154.5 Answer C

3) An object with a weight of 150 Kg is neutrally buoyant in SALT water. What is its volume?

a. 146 Kg

b. 150 Kg

c. 155 KG

d. The answer cannot be determined

**Step 1 **Write down the weight. 150 Kg

**Step 2 **Divide by 1.03 150/1.03=145.6 Ltrs

We can now see that when an object is neutral it is a simple process to calculates its weight OR volume. There is another more complicated way questions such as this could be asked. We could be told that an object is negative or positive by a known number of kilograms. If we know how negative or positive it is, we CAN calculate its weight or volume.

Step 1 We need to work out what the upward force is

4) An object has a negative buoyancy of 20kg in salt water. Its volume is 75 Litres. What is its weight?

a. 77 Kg

b. 73 Kg

c. 97 Kg

d. 57 Kg

**Step 1 **Write down the volume 75

**Step 2 **Multiply by 1.03 75x1.03=77.25

**Step 3 **add the negative buoyancy 77.25+20=97.25

5) An object has a positive buoyancy of 20kg in salt water. Its volume is 75 Litres. What is its weight?

a. 53 Kg

b. 93 Kg

c. 57 Kg

d. 97 Kg

**Step 1 **Write down the volume 75

**Step 2 **Multiply by 1.03 75x1.03=77.25

**Step 3 **Subtract positive buoyancy or add the negative buoyancy the object has 77.25-20=57.25

6) An object has a negative buoyancy of 20kg in salt water. Its weight is 75 kg. What is its volume?

a. 55 Ltrs

b. 53 Ltrs

c. 73 Ltrs

d. 58 Ltrs

**Step 1 **Write down the weight 75

**Step 2 **Subtract negative buoyancy 75-20=55

**Step 3 **Divide by the salt water constant 1.03. – 55/1.03=53.39

7) An object has a positive buoyancy of 20kg in salt water. Its weight is 75 kg. What is its volume?

a. 98 Ltrs

b. 92 Ltrs

c. 73 Ltrs

d. 53 Ltrs

**Step 1 **Write down the weight 75

**Step 2 **add positive buoyancy. 20 + 75 = 95

**Step 3 **Divide by the salt water constant 1.03 – 95/1.03 = 92.23

## Part 3 - Lift bag Questions

To find the amount of air that is needed to lift and object you must know what type of water the object is in, its weight and its volume.

In fresh water, it is simple.

· You subtract the volume from the weight, and you have your answer!

Salt water is a little bit trickier.

· Divide the weight by 1.03. This tells you the number of litres of water that must be displaced in total.

· Subtract the volume/displacement of the object.

· Now you have the remaining amount of water that must be displaced after taking into the account the objects volume. This is the amount of water you need to displace using the lift bag.

1) An object with a weight of 175 Kg is in salt water. It has a volume of 45 Litres. How much water would have to be displaced using a lifting device to bring this object to the surface?

a. 130 Litres

b. 170 Litres

c. 125 Litres

d. 129 Litres

2) An object with a weight of 325 Kg is in salt water. It has a displacement of 160 Litres. How much air would have to be added to a lifting device to bring this object to the surface?

a. 156 Litres

b. 160 Litres

c. 165 Litres

d. 170 Litres

3) An object with a weight of 175 Kg is in fresh water. It has a volume of 45 Litres. How much water would have to be displaced using a lifting device to bring this object to the surface?

a. 130 Litres

b. 170 Litres

c. 125 Litres

d. 129 Litres

Here is a question a You Tube subscriber sent me:

True or False If an object weighs 82kg in a freshwater lake and it displaces 93 liters of water, you would need to add 15 kgs of weight to make it 4kg negatively buoyant.