Balloon Ascent Speeds


In my book I discuss technical issues like the lift of helium,
descent speeds of parachutes, cosmic rays, etc. The next topic
I'm tackling is the ascent speed of a balloon. Here's what I have for
my notes so far. I'm looking for input, because I don't have the
complete picture yet. Please spend a little time looking these
notes over and making suggestions. Thanks.

A. Intro

We know the balloon will expand as it rises due to decreasing air
pressure. As a gas expands, as in a balloon, it cools. This
cooling decreases the volume of a gas. We can see that during a
flight that the balloon expands a lot more due to reduced air
pressure than it contracts due to cooling air temperatures (both
inside and outside the balloon). So cooling has a minor effect
compared to reducing air pressure. The forces acting on a balloon
are three, payload weight (W) pulling down, lift (L) pulling up, and
drag (D) pulling in the direction opposite of the balloon's motion (so
drag pulls down, just like the weight). Looking at flight data, we
see that the balloon ascends at a constant rate for the entire flight
[1]. As long as there is no net force acting on a balloon (or any
object for that manner), it moves with zero acceleration, or
constant velocity. Since the balloon climbs at a constant
veleocity, the force of lift of the balloon is balanced by drag and
weight, D + W = L. Lift and weight are constant on a balloon flight
[2], therefore the force of drag must be a constant throughout the
flight also.

B. Drag

The equation for drag is given as,
D = Cd * Ra * sq(V)/2 * rho

Cd is a coefficient of drag
Ra is the reference area
sq(V)/2 is the square of the velocity, divided by 2
rho is the density of the air.

Cd is a dimensionaless constant. This means it's simply a
number and does not have any units (like pounds or feet) attached
to it. For a sphere, Cd is about 0.5

Ra is an area you choose to use. If you select a different area as
your Ra, then the Cd for that Ra is different, but Cd is still just a

I only what to determine how ascent speed changes as the
variables change, so I'm going to treat this equation as a
relationship of proportionality and drop the constants for the rest of
this discussion.

When you look at the US Standard Atmosphere (a model of how
air pressure, density, and temperature change as a function of
altitude), you see that the change in air density closely tracks the
change in air pressure. So I'm going to replace air density with air
pressure. The reason I do this is that we can see the change in air
pressure changes the balloon's volume and therefore it's reference
area. As a Ra, I'm going to select the frontal area occupied by the
balloon. This Ra is proportional to the two-thirds power of the
volume. The balloon's volume, ideally, is proportional to the inverse
of the air pressure. Observations show that the balloon's vertical
velocity, its ascent rate, is constant throughout a flight. So sq(V)/2
is also a constant. When I combine like terms and move the
constants out, I end up with the following,

D is proportional to (volume)^2/3 * pressure

Which simplifies to

D pro to 1/(pressure)^2/3 * pressure

Which simplified to

D pro to (pressure)^1/3

However, D must be a constant, whereas pressure is not a
constant! So what's wrong here?

The change in temperature can't be responsible for adjusting the
change in volume (that is due to changing air pressure) because
we see the temperature change in opposite direction in the tropo
and stratosphere. And yet, the ascent rates in both the tropo and
stratosphere are both constant.

So it appears to me that Cd is not a constant throughout a flight.

C. Cd

Cd models drag's depenedencies on things like shape, air
viscosity, and air compressibility. The balloon's shape does not
change significantly during a flight. The air compressiblity is not
siginificant belows speeds of 200 mph. This leaves only air
viscosity. Is the change in the air's viscosity proportional to the
cube root of the air's pressure? If so, the change in Cd would
oppose the chnage in drag caused by the change in air pressure
(or more exactly, air density). Does the Cd of an aircraft change
as the aircraft climbs?

Note 1
There is a knee in the ascent speed of a balloon as you approach
the stratosphere. In the startosphere, the air temperature
increases as the balloon climbs. Perhaps the slight extra increase
in the balloon's diameter due to increasing stratospheric
temperatures increases the balloon's drag slightly, causing it to
slow down a bit. As a future experiment, does a black balloon rise
faster than an identical white balloon or slower?

Note 2
The lifting ability of helium does decrease with decreasing
pressure. But this amounts to only something like a percent or
two. This is not enough to significantly effect our numbers.