Good guess james! I calculated that the DS 125 weighs around 5 ounces in salt water.
Fill a straight sided round bucket almost full of water.
Place a ruler standing vertically inside.
Measure the height of the water.
Add a DS 125
Note new height of the water.
We can now determine the volume or displacement of a DS 125 (or any other negatively buoyant object)
Volume(displacement in this case) =Length times Pi times radius squared
The bucket I used measured 13 inches across. It's radius therefore is 6.5. I put 7.5 inches of water in it. I gently placed a DS 125 in the bucket and the water level raised one half (0.5) inch on the ruler.
The volume of the DS 125 then would be calculated:
0.5(the increase in height of the water) times 3.14 (pi) times 6.5 squared (the radius of 13 inches)
0.5(L) times 3.14(PI) times 42.25(6.5 squared)=volume
0.5x3.14x42.25=66.33
The answer, 66.33 is presented to us in cubic inches.
A cubic inch of salt water weighs .59 ounces
For every cubic inch of water displaced, .59 ounces of lift (buoyancy)is created.
the DS 125 displaces (volume) about 66.33 cubic inches of water. Each cubic inch of salt water weighs about .59 ounces. 66.3 times .59 ounces equals around 39 ounces of weight displaced. The DS weighs around 44 ounces on land. Subtract the 39 ounces of water displaced and the net result is it's negative weight in water of around 5 ounces.
the above calculations are based upon the Archimedes Principle of water displacement......