The following diagram shows the required corrections of mass values ∆m for
a few selected density values ρ
W
. The incorrect values result from the effects of air
buoyancy. The corrections depend on the weights displayed on the balance.
Example: aluminum: conventional mass of 2 g (density 2.7)
Readout
Without air buoyancy correction→ + 2000.000 mg
With air buoyancy correction→ + 2000.589 mg
3–6
Add the mass error to the weight
displayed on the balance
Weight displayed
on the balance (grams)
Correction
of the mass ∆m
(milligrams)
Subtract the mass error from
the weight displayed on the balance
g
ρ
w
= 0.8 —–
cm
3
g
ρ
w
= 1.0 —–
cm
3
g
ρ
w
= 1.4 —–
cm
3
g
ρ
w
= 2.0 —–
cm
3
g
ρ
w
= 4.0 —–
cm
3
g
ρ
w
= 22.5 —–
cm
3
g
ρ
w
= 8.0 —–
cm
3
g
ρ
w
= 0.0012 —– (air)
cm
3
Density of the object
to be weighed
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