The basic equations - Severe Convective Storms and Tornadoes: Observations and Dynamics - page 56

We can now calculate the local accelerations at t ¼ 0 by substituting (2.128) and
(2.129) into (2.99) and (2.100), using (2.101). Then
Dv =

2

Dt ¼

3 B a ð cos

j þ sin k Þ for r < a ð 2 : 130 Þ

3

2 1

¼

3 B a ð a

= r Þ ð

2 cos

j þ sin k Þ for r > a ð 2 : 131 Þ

Since

k z ¼ j cos þ k sin ð 2 : 132 Þ

it follows that

2

Dv = Dt ¼

3 B a k z

for r < a ð 2 : 133 Þ

In other words, within the buoyant sphere the acceleration is uniform and upward;
about two thirds is due to that from buoyancy alone. It follows that there must be
a downward-directed pressure gradient force to counteract buoyancy and cancel
the effect of one third of it. Outside the buoyant sphere

2 k z

2

Dv = Dt ¼

3 B a ð a

= r Þ for r > a,at = /2 ð 2 : 134 Þ

Thus, at r ¼ a, the vertical acceleration is up along the z-axis and decays above
and below the sphere. At ¼ 0, for r > a

3 k z

1

Dv = Dt ¼

3 B a ð a

= r Þ ð 2 : 135 Þ

which is a downward-directed acceleration in the horizontal plane at the equator.
The acceleration field is depicted in Figure 2.10. Away from the sphere, pressure
gradient forces must be driving air motion because there is no buoyancy to do so.
The accelerations inside and up to the edge of the buoyant bubble (at r ¼ a) are
independent of the radius of the bubble (cf. (2.130) and (2.131)). Thus, the accelera-
tions there are a function of the buoyancy—not the size of the bubble. In other

Figure 2.10. Qualitative depiction of the acceleration field (vectors) induced by a buoyant,
spherical bubble (solid circle).