Question on SDI Leadership Manual

[ScubaSarus]

Has anyone read over the physics chapter and noted the errors in conversions, units, how units are presented, traspositions of numbers, deriving of formulas etc.

I have a good physics and chemistry background and was told it is hard. Ive caught all the errors so it isnt that hard but to someone without a physics or chemistry background, this physics chapter would be SUPER HARD!!!!

Im using the 2003 edition. I have emailed SDI to touch base with them on the physics chapter.

I am going to make a nice formula sheet soon especially for this chapter with all the conversion needed and may post it for others to use.

I can get into specifics but there is quite a few funny ways this manual presents the math.

 

[sharkattack]

I'm an SDI instructor, I do not have a physics or chemistry background. I agree, it was hard. I might of had that same version doing my DM course. We found a lot of errors too.

 

[captndale]

I think part of learning is verifying everything you are taught. Not just in scuba, but in any educational endeavor, you should question every piece of information that is presented to you, It is quite common for there to be errors in formulae and worked-out problems, including those on tests. It is also common for there to be errors and incorrect reasoning in the basic tennants of a class or philosophy. The purpose of education is to learn, not to follow. Remember that getting a good grade is a secondary, but not unimportant, goal.

Finding the incorrect information in your text can be more educational than blindly following whatever is presented.

When the NOAA dive manual was first published I found several mistakes among the formulae in the first edition.

All that said, I wish that they would fix the errors.

 

[ScubaSarus]

I agree Captn. The book is expensive, its been around for a while (all black and white), and I feel they should have so much pride in their training that they would fix those errors. Im going to make my own addendum and present it to the dive shop in a few weeks or so and maybe email it to SDI.

My symapthys Sharky, I can only imaggine how much harder it is when you cant work problems backwards or derive formulas (convesion factors) with errors.

 

[Mr. Bubble]

I am not a physics fan, and I have been studying this book for a month now. After cramming all that info together, trying to make sense of it...I'm glad to learn that its incorrect

 

[ScubaSarus]

Just some help non-metric Sea Water in consideration kg or lbs for net weight

Air needed to lift object =(object weight - weight water displacement)/(64 lbs or 1.025 kg)

or simply (net downward weight or force ) / (64 lbs) to get cf of air or /(1.025 kg) to get Liters of Air

1cf = .028 meter cubed = 28.316 Liters

1 meter cubed = 1000 Liters

1 kilogram = 2.205 lbs

1 Liter SW = 1.025 kg

1 Meter cubed SW = 1025 kg

Page 31 bottom example. They have transposed 45kg with 34 Liters

should be 1.025 x 34 = 34.85 not (1.025 x 45 =46.125 kg)
45kg - 34.85 kg = 10.15 kg net downward weight

Air needed = (10.15/1.025kg/L) = 9.902 Liters of air needed not 12.125

Page 36

should be Trace elements = .001 x 2ATA/BAR = .002 not (.01)

Page 40 Question #2

Should be 56.63 Liters not 56 cm3

Were ever you see an underscore "_" the printer has left off a unit or some sort of needed text.

Final Volume Vf at depth = Intial Volume Vi at surface diviede by # of ATA's.

Vf= Vi/# ATA's

***Example more clearly

What volume of air at 1ATA will provide .363 cf at 2.52 ATA

Vi x Pi = Vf x Pf

Initial Volume at surface = unknown
Initial Pressure = 1 ATA at surface
Final Volume = .363 cf
Final Pressure = 2.52 ATA

Simply (.363 x 2.52) / (1) = Vi = .907 cf of 1ATA air needed

Just some clarifications for volume of air need problems.

1) Find Weight of Displaced of water in kg or lbs (convet Volume in cf or L to lbs or kg)
2) Find net weight of downward force in kg or lbs. (Weigh of object - Weight of displaced water)
3)Find volume of air need to lift weight. (Net Weight/ 64 lbs per cf or Net Weightr /1.025 kg per L)
4) Calculate ATA/BAR tahts easy [Depth/(33 or 10)] + 1
5) Find surface volume of air needed at the depth Vi x Pi=Vf x Pf so Vi = [(Vf x Pf)/(Pi)] note Pi usually = 1

Question 9 on page 40 seems to be answered on page 48 in the next chapter.

 

[Blox]

1 Meter cubed SW = 10025 kg

that's gotta be a typo - should be 1025 kg

 

[ScubaSarus]

Good Job Blox

Fixed

My glass house has a crack in it.

***Now who can put into laymans terms how they derived the pressure ratings for the innertube?

 

[sharkattack]

Just some help non-metric Sea Water in consideration kg or lbs for net weight

Air needed to lift object =(object weight - weight water displacement)/(64 lbs or 1.025 kg)

or simply (net downward weight or force ) / (64 lbs) to get cf of air or /(1.025 kg) to get Liters of Air

1cf = .028 meter cubed = 28.316 Liters

1 meter cubed = 1000 Liters

1 kilogram = 2.205 lbs

1 Liter SW = 1.025 kg

1 Meter cubed SW = 1025 kg

Page 31 bottom example. They have transposed 45kg with 34 Liters

should be 1.025 x 34 = 34.85 not (1.025 x 45 =46.125 kg)
45kg - 34.85 kg = 10.15 kg net downward weight

Air needed = (10.15/1.025kg/L) = 9.902 Liters of air needed not 12.125

Page 36

should be Trace elements = .001 x 2ATA/BAR = .002 not (.01)

Page 40 Question #2

Should be 56.63 Liters not 56 cm3

Were ever you see an underscore "_" the printer has left off a unit or some sort of needed text.

Final Volume Vf at depth = Intial Volume Vi at surface diviede by # of ATA's.

Vf= Vi/# ATA's

***Example more clearly

What volume of air at 1ATA will provide .363 cf at 2.52 ATA

Vi x Pi = Vf x Pf

Initial Volume at surface = unknown
Initial Pressure = 1 ATA at surface
Final Volume = .363 cf
Final Pressure = 2.52 ATA

Simply (.363 x 2.52) / (1) = Vi = .907 cf of 1ATA air needed

Just some clarifications for volume of air need problems.

1) Find Weight of Displaced of water in kg or lbs (convet Volume in cf or L to lbs or kg)
2) Find net weight of downward force in kg or lbs. (Weigh of object - Weight of displaced water)
3)Find volume of air need to lift weight. (Net Weight/ 64 lbs per cf or Net Weightr /1.025 kg per L)
4) Calculate ATA/BAR tahts easy [Depth/(33 or 10)] + 1
5) Find surface volume of air needed at the depth Vi x Pi=Vf x Pf so Vi = [(Vf x Pf)/(Pi)] note Pi usually = 1

Question 9 on page 40 seems to be answered on page 48 in the next chapter.

What was the question or paragraph regarding this? " Page 36

should be Trace elements = .001 x 2ATA/BAR = .002 not (.01)"

I do not have my book with me. It wasn't talking about fractions of gas was it?

 

[ScubaSarus]

I was going over Daltons Laws of partial pressures list the percentages of gases in the air we breath and converting them to partial pressures by dividing them (the percentages) by 100.

Trace elements we breath is .1% / 100 = .001 x 2 = .002.

The book had .01 x 2 = .002. No biggie as the answer is right but can be confusing to one whose trying to understand the stuff for the first time and requires reasoning and working the problems backwards.