Spreadsheet Emulator- Calculator for the Solution of a Differential Equation for Bodies Falling
Falling Rain Drops
In June 1904, German physicist and Nobel Prize Winner to be, Philipp Lenard, was
the first to report that rain drops were not of a tear drop shape but rather fell
as spheres when they were less than ~1.00 mm in radius and that larger drops
up to ~ 2.5 mm in radius took on a hamburger like shape, a flat bottom with a rounded
top. He found that drops larger in radius than ~ 2.75 mm were unstable and
broke apart within seconds.
He employed a vertical wind tunnel and adjusted flow rate until drops were suspended
to determine terminal velocities. He found that fall speed increased with
drop radius up to a radius of about 2.25 after which it increased more slowly and
did not increase beyond 8 metres per second. He deduced that the changes in
shape increased the air drag thus slowing the fall rate.
For mention of Lenard's raindrop results, see here.
Clouds of Fog
The foregoing reference describes a typical cloud droplet as having a radius of
about 10 microns with a terminal velocity of about 0.01 metres per second
A comprehensive study of hailstones
can be found in:
Measurement and Interpretation of Hailstone Density and Terminal
Journal of the Atmospheric Sciences,
Volume 40, Issue 6 (June 1983) pp. 1510–1516.
That author uses the term equivalent spherical diameter to describe
the stones. The average stone had an equivalent radius of 0.0053 metres,
a density of about 440 kg per cubic metre, and a Cd of ~ 0.7. A range of terminal velocities is reported that suggests a vt of about 8.5 metres
per second for the average stone.
Stones in the sample ranged in radius from 0.00315 to 0.0077 metres.
This writer has not located an authoritative study
of the size and falling characteristics of snowflakes.
Calculations and Calculator Results
Falling Rain Drops
Take the density of 20o
C rainwater as 998.2 kg/m3
. The volume of a 1 mm sphere is:
4/3 * 3.14159265 * 10-9
~= 4.189 * 10-9
Its mass is:
998.2 * 4.189 * 10-9
~= 4.181 * 10-6
Take the drag Cd of a sphere as ~ 0.4
When these values are plugged into the calculator it provides a vt of ~7.376 metres/s.
As steady state fall is our interest, we use this as the chosen velocity to get R
as ~ 492. This Reynolds number is marginally low for the Rayleigh approximation
to serve and much too high for Stoke's law to apply.
Try a radius of 2.25 mm!
Then the volume is ~4.771 * 10-8
and the mass is ~4.763 * 10-5
Flattening was said to begin when the radius exceeded 1.0 mm. The Aerodynamic
Database suggests a drag coefficient of 1.42 for extreme flattening, a hollow semi
Choose a drag of 0.85.
The calculator returns a vt of ~ 7.59 metres/s. This leads to an R of ~ 1138 which
is nicely within the Rayleigh range.
Try 2.75 mm!
The volume is ~ 8.711 * 10-8
and the mass is ~ 8.696 * 10-5
. Because the
shape should now be quite flattened choose Cd = 1.0.
The calculator returns a vt of ~ 7.736 metres/s. This leads to an R of ~ 1418.
The three estimations foregoing are reasonably in agreement with the observations
of Philipp Lenard made over a century ago.
Clouds of Fog
Although one is quite sure that the Rayleigh approximation will not apply to 10-micron
fog droplets, curiosity drives us to view the calculator's results for that
The volume is ~ 4.189 * 10-15
The mass is ~ 4.181 * 10-12
The calculator returns a vt of ~ .738 metres/s leading to an R of ~ 0.49.
R is much too small for the Rayleigh approximation to apply and the calculated vt is
much larger than the observed value of ~ 0.01 metres per second.
The reported radius of 0.0053 metres for an average
hailstone leads to a volume of ~ 6.236 * 10-7
. Its density of ~ 440 kg per cubic metre
leads to a mass of ~ 2.744 * 10-4
With the reported Cd
of ~ 0.7, the calculator provides a vt of ~ 8.522
metres/s which leads to an R of 3011.
The agreement of our calculator vt with the reported value of ~ 8.5 metres/s is
The largest hailstone had a radius of .0077 metres. Its volume and
mass are ~ 1.912 * 10-6
and ~ 8.414 * 10-4
For this case the calculator provides a vt of ~ 10.272 m/s and an R of ~ 5273.
The terminal velocity converts to ~ 37 km/hr. Larger stones have occurred and in
some case caused considerable property damage.
The Tower of Pisa Revisited
The tower is ~ 55 metres high. Consider a drop distance of 50 metres.
An air filled volleyball has a mass of about 0.0275 and a radius of about 0.108
With a Cd of 0.4,according to the calculator, the ball would drop 50 metres in
~ 9.4 seconds.
Now put the ball on a scale and inject enough water to double the mass and presume
that it is dropped again. In this case the calculator reports that the ball
would fall 50 metres in ~ 6.9 seconds.
How long would it take for an iron ball to fall the same distance? Answer:
~ 3.2 seconds. Check it out!
Note that buoyancy is not inherently taken into account by this calculator. The
second topic of this chapter, under Examples
, shows how the value
employed for g
may be chosen so that the calculator results will
include the effect of the object's buoyancy.
to access Calculator. This calculator takes into account only vertical motion
in atmosphere. As a result, it is termed the "1D Calculator". One may
also access it from the top line of tabs.