Derivation of Formulas for the Volume of A Solution of Height "H"
in a Wilmad-Labglass 5-mm NMR Tube, style 535-PP.

Note: 1000-mm^3 = 1-cc = 1-ml


The 5-mm 535-PP Tube is 5.0-mm OD and has a 0.38mm wall thickness.

ID = OD - 2 x wall thickness :: 5.0 - (2 x .38) = 4.24-mm ID

4.24-mm ID = 2 x Inside Radius :: R = 2.12-mm

Let H = the height of the solution in millimeters (mm)

Pi = 3.1416

An approximation of the volume of the 535-PP NMR tube would be:

Volume of Cylinder of height H (mm) and a Diameter of 4.24-mm:

    = Pi x R^2 x H = Pi x 2.12^2 x H = 14.12 x H mm^3

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To allow for the fact that the bottom end of the NMR tube is really a
half-sphere, you must find the difference in volume between a cylinder
of R radius x R height and a half-sphere of R radius and then subtract
this difference from the tube volume as calculated above, thus:

Volume of Tip as a Cylinder = Pi x R^2 x R = Pi x R^3

Volume of Tip as a Half-Sphere = 2/3 x Pi x R^3

Volume Difference = Pi x (R^3 - 2/3 R^3) = Pi x 1/3 R^3 = 9.978 mm^3

A true volume calculation for the 535-PP NMR tube would be:

Volume of Cylinder of height H (mm), Diameter of 4.24-mm, with a
spherically radiused nose

    = ((14.12 x H) - 9.978) mm^3 where H is greater than 2.12-mm