Physics

[richarddean]

I had to learn it all myself, no help from him whatsoever

 

[Guba]

Is this about to become another "PADI basher" thread? Please let me know now so I can block it. Enough is enough.

 

[Lucy's Diver]

This math makes my head hurt.
Just send me another lift bag.

 

[rawls]

Richard...

I cannot believe I am saying this but I have to agree with Sparticle on this one...You need to know the concepts behind this stuff in order to make any sense of it. It is more than just being able to work the problem.

The concept deals with buoyancy issues...negative, positive and neutral buoyancy. You want to make an object 40kg positively buoyant. You have a 237 kg object that displaces 123 liters of salt water. This means the upward or buoyant force pushing against the object is 123 liters. Salt water weighs 1.03 kg/ liter. You are dealing with kg and liters so you have to convert the liters to kg in order to work through the problem (get apples with apples). 123 lsw weighs approx 126.7 kg. So your 237 kg object, less the buoyant force of 126.7 kg, has a remaining negative buoyancy of 110.3 kg, ie, the sucker is still sunk. So you have to displace an additional 110.3 kg of sw to make the object neutral. To make it 40kg positive you have to add the additional 40 kg totaling 150.3 kg. Now you want to figure out how much water that is in liters. 150.3kg/1.03 = approx 146 liters. These questions are kind of silly because there are other unknowns you have to determine...You are given the information but in the real world you have to determine this information...What's the weight of this anchor I found sitting on the bottom...How much water does it displace...How much air do I need to add to get it to the surface...This is where asking Sparticle to come along on a dive with you would be advantageous

Can I at least make the assumption you paid a 'pro' to teach you this DM course?

Padi by any chance?

Did you come up with this highly intelligent statement by yourself or did you need some help?

 

[Thalassamania]

Hi Guys can anyone help as I seem to have conflicting answers for this

An Object weighs 237kg and displaces 123 litres of water. How much additional seawater do you need to displace to give the object 40 kg positive buoyance

I came up with the answer of 146 litres, but a freind says it is 68.3 litres

Who is right

My math was 237 + 40 = 277

277/1.03=268.93ltrs
268.93-123lt =145.93

first you need to tell me the salinity, temperature and depth of the displaced sea water.

 

[rawls]

first you need to tell me the salinity, temperature and depth of the displaced sea water

:dancinglock: :031:

 

[neil]

http://www.metricsucks.org/

 

[lucybuykx]

http://www.metricsucks.org/

ha ha. Yes it is so difficult to calculate everything using base 10.

I would much rather calculate some things in base 20, some in base 12, some in base 16 and some in base 8. Oh hang on, just for variety theres some base 14 as well. And then remind me, how many yards in a mile??

And don't get me started on cups of ingredients

 

[h90]

first you need to tell me the salinity, temperature and depth of the displaced sea water.

yeah in 12.000 meter deep and minus 5 degree you get complete different figures
I had somewhere a table on how water get compressed under pressure, but can't find it anymore. Didn't use it for at least 10 years

 

[h90]

Richard...

I cannot believe I am saying this but I have to agree with Sparticle on this one...You need to know the concepts behind this stuff in order to make any sense of it. It is more than just being able to work the problem.

The concept deals with buoyancy issues...negative, positive and neutral buoyancy. You want to make an object 40kg positively buoyant. You have a 237 kg object that displaces 123 liters of salt water. This means the upward or buoyant force pushing against the object is 123 liters. Salt water weighs 1.03 kg/ liter. You are dealing with kg and liters so you have to convert the liters to kg in order to work through the problem (get apples with apples). 123 lsw weighs approx 126.7 kg. So your 237 kg object, less the buoyant force of 126.7 kg, has a remaining negative buoyancy of 110.3 kg, ie, the sucker is still sunk. So you have to displace an additional 110.3 kg of sw to make the object neutral. To make it 40kg positive you have to add the additional 40 kg totaling 150.3 kg. Now you want to figure out how much water that is in liters. 150.3kg/1.03 = approx 146 liters. These questions are kind of silly because there are other unknowns you have to determine...You are given the information but in the real world you have to determine this information...What's the weight of this anchor I found sitting on the bottom...How much water does it displace...How much air do I need to add to get it to the surface...This is where asking Sparticle to come along on a dive with you would be advantageous



Did you come up with this highly intelligent statement by yourself or did you need some help?

to be 100 % correct you should calculate the kg into Newton. kg x 9.80665 (if I remember right). So technically 40 kg positiv buoyant does not exist, it must be 392,4 Newton. There would be hugh mistakes if you dive on other planets than earth.
Just a remark for the "Dive on other planets" Speciality course.