OBJECTIVES:
1.  Describe the types of calculations necessary to determine absolute atmospheres, water weight of an object, number of lift bags necessary, repiratory minute volume and the length of time that a tank will last underwater.
2.  Explain four patterns used for underwater searches and when they are used.
3.  Describe the necessary steps to follow in order to safely lift and object.

Consider the following questions as you read.

1.  Being as specific as possible, list or explain how water is different from air.  (For example:  air is a gas, water is a liquid.)
2.  In what ways are the effects of water on your senses different from the effect of air on the same senses.  (For example:  at 76º F air is warm enough to wear shorts and a tank top; however, diving in the same water temperature will cause a person to get cold very quickly.)

 Chances are most of the answers that you have listed in the above two questions can be explained by one simple concept:  density.  Density is figured as mass per unit volume:  What is the weight of something in a given space.  Take for example the comparison of water to air.  How much more would one cubic foot of water weigh than one cubic foot of air?  The difference is due to the densities of the substances.  Because water is denser than air it would weigh more when confined to the same size space.  In fact, fresh water has a density of 62.4 pounds per cubic foot (lbs/cf).  Salt water has a density of 64 lbs/cf.  And, air has a density of .08 lbs/cf.  This means that salt water is 800 times more dense than air, and being such effects a divers movement, vision, hearing, and of course, heat loss.  It also explains why it is much easier to float in the ocean than in a lake, and why it is impossible (in a normal condition, without modification) to float in the air.  But, why is it that a mega ton, steel aircraft carrier can float on top of the water?  Isn’t steel more dense than water?
 Archimede’s principle states that an object is buoyed up by a force that is equal to the weight of the water that is displaced.  For example:  an object with a volume of 4 cubic feet will displace 249.6 lbs of fresh water (4 X 62.4 lbs/cf = 249.6) or 256 lbs of salt water (4 X 64 lbs/cf = 256).  If the object weights 253 lbs it will sink in the fresh water, because it weighs more than the water that it displaces; however, in the salt water it will float because it weighs less than the water that it displaces.  More specifically the water weight of the object in fresh water is -3.4 lbs (249.6 - 253 = -3.4 lbs, sink), and the water weight of the object in salt water is +3 lbs (256 - 253 = +3 lbs, float).  Therefore, the water weight of an object is determined by the difference between the weight of the object and the weight of the water displaced.  

3.  What is the water weight of an object in salt water that has a volume of 2 cubic feet and weighs 150 lbs?
4.  One atmosphere of pressure at the surface can be defined as 14.7 psi.  How many feet underwater makes up an additional 14.7 psi (or one atmosphere)?

 Experience has shown us that gases and liquids have different characteristics, and, therefore behave differently under similar circumstances.  For instance, air is compressible and water is, for all practical purposes, not compressible.  Water, in fact, can be stacked upon itself with relatively little effect on the underlying water molecules(the same cannot be said for air, where the underlying molecules become more dense).  It is for this reason that there is a linear relationship between the depth of the water and the amount of pressure that is being exerted by the water.  From basic scuba class you might remember that every 33 feet of salt water (or 34 feet of fresh water) exerts one atmosphere of pressure.  Sixty-six feet of salt water exerts a total pressure of three atmospheres (one atmosphere of air plus two atmospheres of water).  This total pressure is referred to as absolute pressure or absolute atmospheres.  It is important when completing calculations at depth that a person uses absolute values and not strictly water pressure values.  Absolute pressure at any depth is determined by adding the depth to 33 feet of salt water (fsw) or 34 feet of fresh water (ffw) and then dividing by 33 fsw (or 34 ffw).  The answer is recorded in atmospheres absolute (ata).  For example:  at 60 fsw the pressure is 2.82 ata  (60 + 33 ÷ 33 = 2.8181 ata).  

5.  What is the absolute pressure at 70 fsw?  at 70 ffw?
6.  What happens to a flexible container (i.e. a balloon) as it is taken deeper and deeper under water?

 As you dive deeper into the water column the pressure increases linearly.  At 33 fsw there is twice as much pressure as there was at the surface, and at 66 fsw there is three times as much pressure as there was at the surface, etc.  Because of this pressure increase, the volume in a flexible container also changes.  In fact it changes inversely proportional to the absolute pressure.  More specifically, Boyle’s Law tells us that as the pressure increases the volume decreases.  You can think of volume and pressure as being on opposite ends of a see-saw:  as one goes up the other must go down.  For example:  at 33 fsw there is twice as much pressure, so the volume inside the flexible container must be 1/2 the original volume.  At 66 fsw (three times the pressure), the  volume is 1/3 the original volume, and at 99 fsw (four times the pressure), the volume is 1/4 the original volume.
 An important concept can be reached by asking yourself what is happening to the air inside this flexible container?  Is it leaking out?  Are there fewer particles of air than when we originally started at the surface?  You should realize that the answer is “no” and that air has not gone anywhere.  It is simply getting more dense as the air molecules are squeezed more tightly together.  One of the reason that we breathe more air at depth is because we need more air to fill the same volume of our lungs due to the air becoming more dense.  This is also why it is important not to hold your breath while ascending to a shallower depth.  As you ascend the air becomes less dense as it expands (due to less pressure).  Expanding air can easily rupture your lungs, because your lungs are not as flexible as most people seem to think (they are not balloons).  

7.  At the surface a balloon has a volume of 12 cubic inches.  What is the balloon’s volume at 33 fsw?  at 66 fsw?  at 80 fsw?
8.  In order to lift an object underwater using a lift bag (not muscle), what must have to happen?

 Archimede’s principle (buoyancy of an object in a fluid) is still in operation when considering to lift an object underwater.  Even though the object may be negatively buoyant and sitting on the bottom, all that needs to be accomplished in order to  raise the object is to either decrease its weight or increase its volume.  If we are talking about lifting a treasure chest full of gold, decreasing its weight maybe an option.  However, this is not always a feasible task.  A better option is to increase the volume of the object.  This can be accomplished by attaching lift bags and filling them with air.  The proper number of lift bags can be determined by calculating the water weight of the object using the same procedure as at the beginning of this session.  Suppose that an object with a volume of 35 cubic feet  weights 2300 lbs in salt water.  Its water weight is 60 lbs (35 X 64 = 2240, 2300 - 2240 = -60 lbs).  It will take two 50 lb lift bags each with a volume of .78 cubic feet to lift the object.  (The thought here is that you need to be able to lift more than the water weight of the object, so that physical exertion is not necessary.  You should never use your BCD to assist in lifting the object.  The closer the lifting capacity of the bags are to the actual water weight of the object the safer the lift will be.)  

9.  How many 20 lb lift bags should be used in the preceding problem?
10.  How much air is needed to lift the object in the preceding problem?

 Calculating how much air is needed to lift an object must take into consideration the depth of the object in absolute atmospheres.  For example, a lift bag will need 3 times more air at 66 fsw (3 times more pressure) than it would at the surface due to the increase in pressure.  In the previous problem .937 of a cubic foot would be needed to lift the object (60 lbs ÷ 64 lbs/cf = .937 cf).  A standard size 80 cf  scuba tank at 3000 psi has 37.5 psi of pressure for every cubic foot of volume (3000 psi ÷ 80 cf = 37.5 psi/cf).  Therefore, at the surface 35.13 psi are needed to fill the lift bag (37.5 psi/cf X .937 cf = 35.13 psi).  If the object is at 66 feet, which is three times more pressure than at the surface, three times more air will be needed to fill the bag, or 105.41 psi.  It is important to always take more air than is needed for contingency purposes.

11.  How many pounds per square inch (psi) are in every cubic foot of a steel 72 at 2250 psi?
12.  How many cubic feet are left in a steel 72 if the gauge pressure is at 1000 psi?

 To calculate the above problem, divide the amount of air remaining in the tank by the  psi/cf value (1000 ÷ 31.25 psi/cf = 32 cf  of air in the tank).  This can be a helpful calculation for planning a dive and determining the volume of air remaining in the tank for a given pressure.  Important gear needed for light salvage include:  lift bags, rope, spare tank, regulator with submersible pressure gauge (spg) and air fill nozzle, along with standard scuba equipment for personal use.  

13.  Summarize the steps necessary for successfully lifting an object to the surface.
14.  What types of things must be considered when planning and executing a search pattern?

 Divers who have received training in search and recovery have an interesting attitude regarding items missing underwater, “With proper training and planning, no object is really lost.  It is merely wet!”  This posture reflects the very essence of search and recovery and shows that given enough information on the object and proper planning, any object missing within the boundaries of recreational scuba diving can be found.  The three steps necessary for developing a successful search include:  1) Selecting a search pattern that not only takes into consideration the training and experience of the divers involved, but also utilizes witnesses accounts and the current conditions of the water,  2) establishing depth and time limits for the dive based on prior dives made that day and the general health and condition of the divers involved, and  3)  discussing emergency procedures and a contingency plan for the dive that includes any special forms of underwater communications.  
 Searchers must choose between four general types of patterns based on weather or not there is a current underwater and weather or not the last known point of the object is a specific place or a general location.  Two general categories of search patterns are either tethered patterns or free swimming patterns.  Tethered patterns are necessary when the current is so strong that the diver’s search pattern will be effected.  Free swimming patterns allow more flexibility in searching, but are more navigation dependent (divers need to be proficient with navigation to increase the chance of finding the object lost).  After current conditions have been determined the searchers need to determine if there is a pin-point location or a general area that is being searched.  The proper pattern can then be determined based on these two criteria.
 Pin-Pointed Search Area:  In a pin-pointed area the searchers will start their dive from a specific spot (where the object was seen going into the water).  If there is no current, divers will dive an “expanding square” pattern.  One diver will navigate the pattern while the other diver looks for the object.  In an expanding square pattern, as with all patterns, the divers must take into consideration the size of the object, the bottom composition, and visibility.  For small objects the overlap of the pattern needs to be greater, than if the object is large.  Additionally, if the bottom composition is flat and smooth, the overlap of the pattern can be smaller than if the bottom is filled with large rocks and crevices.  For the purpose of explanation this example will use exact directions for searching.   This may not always be practical or possible.  Begin the search by heading due north.  Swim exactly one visibility from the starting point (i.e. 5 feet).  Stop, make a right hand turn due east, and swim exactly one visibility from the turning point (again 5 feet).  Stop, make a right hand turn due south, and swim exactly two risibilities from the turning point (10 feet).  Stop, make a right hand turn due west, and swim exactly two risibilities from the turning point (10 feet).  Stop, make a right hand turn due north, and swim exactly three risibilities from the turning point (15 feet).  Divers should continue this pattern, adding one visibility every two legs until they have reached their time or air limit, or until they find the object.  If necessary several of these patterns can be run overlapping each other in order to cover a large area.
 If there is a current the divers will need a tether line to run between them.  One diver will act as the anchor and remain stationary, while the other divers swims circles around him/her.  The first pass of the pattern (circle) should be one visibility away from the anchor diver.  The second pass will be two risibilities from the anchor diver.  And, the third pass will be three risibilities from the anchor diver.  This will continue until they have reached their time or air limit, or until they have found the object.  There are a couple of keys to running a good tethered pattern.  First, the tether must remain tight so that the pattern is complete.  Second, the anchor must signal to the diver each time they come to the starting point, so that the diver can stop and extend the tether to the next distance.  Finally, the divers must practice and use simple, yet relevant signals.  (i.e. one tug can mean stop/ go, two tugs ??????????).
 General Search Area:  If only the general area is know the divers may need to search a larger area to find the missing object.  Again there are two options for the searcher to choose.  If there is no current, the divers will dive a “U” shaped pattern (often called a parallel pattern).   Begin the search by heading a specific direction starting at the edge of the search area (i.e. due north).  Travel this direction for a set distance (this is the long side of the pattern).  At the end of the distance, move to the west (or east) two risibilities.  Travel due south for the same length as with the previous leg of the pattern.  Move to the west two risibilities.  Travel due north for the same length as with the previous two legs of the pattern.  Continue this pattern until one of the preset limits is reached (time, depth, or air supply), or until the object is found.  Divers can overlap the “U” pattern with additional patterns until the object has been found.
 If there is a current the divers will need to set up a “jack stay” to allow for a systematic search of the area and control drift.  A jack stay is set up by setting four corners using anchors and buoys.  The corners should mark the area that is being searched.  two lines should be set to run between each of two buoys.  The two lines should run parallel.  The jack stay runs between the two lines and is perpendicular to each.  (This line should be movable.)  Divers start at opposite ends of the jack stay line and swim towards each other looking for the object on the same side of the line as their travel.  When the divers reach the end of the line, they each unhook the line and move it down the stationary line two visibility distances.  They then simultaneously swim to the opposite end of the line, unhook it, and move it two visibility distances.  This pattern is continued until the length of the set lines is traveled.  The anchor lines can be moved any direction to expand the search to another adjacent area.  Divers should observe time, air and depth limits during their dive.

15.  What major obstacles must be overcome when searching for an object?
16.  What three pieces of information are necessary to determine a divers surface air consumption (SAC) or respiratory minute volume (RMV)?

 Knowing how long you can stay down on a tank of air can be an important factor in predive planning.  Knowing your RMV is critical when converting between tanks of different sizes.  Standard SAC can be calculated by finding out how much air (in psi) you breathe in a specific amount of time, at a specified depth.  For example, a diver who breathes 1500 psi of air in 20 minutes, at 60 feet has an SAC of 26.7 psi/min (1500/20 = 75 psi/min at depth, 75/2.81 ata = 26.69 psi/min).  This can be converted to an RMV by calculating the psi per cubic foot of the tank you used and dividing it into your psi per minute value.  For example, if you figured your SAC using an aluminum 80 tank (at 3000 psi), then the psi per cubic foot value is 37.5 (3000 psi/80 cf =  37.5 psi/cf).  To figure your RMV divide your SAC of 26.7 psi/min by the 37.5 psi/cf tank ratio to get .71 cf/ min (26.7 psi/min / 37.5 psi/cf = .71 cf/min).  Unlike using SAC, the RMV can be used on any size tank to determine how long the tank will last.  For example, at the surface an 80 cubic foot tank will last this diver 112.7 minutes (80 cf/.71 cf/min = 112.7 minutes).  A steel 72 will last this diver 101.4 minutes (72 cf/.71 cf/min = 101.4 minutes) at the surface.

17.  How long can the above diver dive if they leave 500 psi in their 80 cubic foot tank and dive to a depth of 50 feet?

 There are two important factors that effect how long a tank will last:  depth and amount of work.   Consider, first,  that at work your air consumption will increase.  With heavy work air consumption will double.  This means that your tank will last half as long at the surface.  Additionally, depth has a profound effect on air consumption.  For example, at 66 feet (3 ata) your air only lasts one third the length of time as at the surface.  If you are working hard at 66 feet, your air will be used six times more quickly than at the surface (3 for depth X 2 for work = 6 faster).  To calculate how long your tank will last first remove the amount of air being left in the tank (500 psi).  This equates to about 13 cubic feet (500 psi/37.5 psi/cf).  That leaves 67 cubic feet to use during the dive.  A dive to 50 feet (2.5 ata)  means that you will use your air 2.5 times faster than at the surface, so instead of .71 cf/min, you will use 1.78 cf/min (.71 X 2.5 = 1.78 cf/min).  If you then divide the amount of air in your tank by your RMV you will see that your tank will last 37.6 minutes (67/1.78 = 37.6 minutes). 
 Several other factor may also effect a divers RMV.  For instance, cold conditions will increase air consumption.  Heat is transferred underwater by conduction (direct contact) as opposed to convection (circulation and slow continuous removal) at a rate that is more than 20 times faster than air.  This means that a diver can become very cold, very quickly.  When using dive tables it is important to always use the next greater time anytime the dive is cold or strenuous.  Additionally, cold water also decreases the volume of a gas required to fill a lift bag.
 Being over weighted can also increase your RMV.  Proper weighting requirements includes having a minimum amount of weight for neutral buoyancy at 15 fsw at the end of the dive.  To determine the proper weight, a diver should float at eye level in full scuba with a full breath of air (no air in BCD).  The diver should begin sinking upon exhaling.  Therefore, to decrease air consumption stay warm and be properly weighted.

18.  What four things contribute to a reduced RMV?