Last Revision March, 2011
Influences on Motion in Earth's Atmosphere
An excerpt from a 2006 textbook, "Classical Mechanics" by R. Douglas Gregory, Cambridge University Press, sets up the challenge for this and the previous topic. It is:
Although it is not known how to obtain analytical solutions for the differential equations that apply to a body encountering resistance that is proportional to the square of its velocity, it is quite possible to obtain practical computer assisted numerical solutions to such equations.
The object is now to consider elements to be taken into account to model both horizontal and vertical motion of bodies in Earth's Atmosphere, 2-dimensional motion.
Does the Atmosphere have much Effect?
Imagine that a cannon projects a spherical maple wood ball vertically upward with a muzzle velocity v. Presume that maple wood has a density of ~755 kg/cubic metre. See
for wood densities.
The ball will encounter a drag force and a slight buoyant force due to the atmosphere as well as the force due to gravity just as did the falling body of the previous topic. The major difference between the two cases is that there is an initial velocity v and effect of buoyancy is now included.
The downward force equation is:
m * a = m * g - m' * g - Drag
where m' is the mass of the displaced fluid
The effect of the buoyancy is
taken into account as a modification to g in the differential equation:
a = g * (1-m'/m) - Drag/m or
a = g' - Drag/m
. Assume a radius of 4 centimetres, g =9.82022, and the mass of a ball as ~ 0.2024. Provide an upward velocity of 100 metres per second. Without air drag or buoyancy how high will the ball rise? This is a mechanics question to which the analytic answer is most often expressed as:
v ^ 2 / (2 * g) ~= 509.15 metres.
This analytic expression presumes that the force of gravity is uniform and does not include buoyancy, which is reasonable as in this case these effects are quite small.
The 2D calculator, described and made available for use by the viewer in the last topic of Chapter 6 or on the upper row of tabs, provides an answer of ~ 509.95 metres without drag but with buoyancy included.
With small air drags
of 0.1 and 0.4 in this same case the calculator provides corresponding heights of ~ 309 metres and 163 metres, buoyancy included. Each height a far cry from the other and both well distanced from the case of no resistance.
Drag Versus Reynolds Number
Some values of the drag coefficient versus the Reynolds number are given in a document by H. Edward Donley
That source provides numerical values for R and for its logarithm, log(R, 10), and then uses the logarithms for the plot. This writer noted an instance where R and its logarithm were not in agreement. The error appears to clearly be a typing error that has been corrected by this author to produce the plot following:
Two additional points have been added to the chart, the diamond shape points. These are said to be average values where the lower value, 0.1, is said to apply to smooth spheres and the higher value, 0.4, is said to be suitable for rough spheres. The position of the lower diamond suggests that the Donley values apply to smooth spheres.
For many purposes the drag coefficient for a sphere is taken to be sufficiently constant for Reynolds numbers in the range 1000 < R < 100,000.
For a given radius of sphere the Reynolds number R is directly proportional to the velocity of the body and inversely proportional to the kinematic viscosity of the medium. The latter is quite dependent on the temperature of the medium. See
Aside from the drag coefficient do we need to consider changes in air density, gravity, wind, spin and the like to get it right?
Following is a quote from "Guns of World War II", available on the Web at the time of writing but since vanished.
Even if, for centuries, ballistics was considered as a science, in (sic) facts, it is all but impossible, even today, to forecast the end-result of a fired shell.
A projectile fired at an angle to the direction of gravitational force can be described as having two orthogonal components of velocity, one in the direction of the gravitational force and one at 90
In the absence of other forces such as wind, these components are presumed to remain in a plane.
Only the first component will be influenced by gravity.
Both components will encounter air resistance. Too determine the amount of resistance affecting each; the resistance due to their combined velocity must be distributed between them in accord with the ratio of their individual velocities to the combined velocity.
Call the vertical component the y component and the horizontal component the x component.
At the beginning of a numerical calculation step an object may have velocity components v
with a combined velocity:
V = (v
The atmospheric resistance factor is presumed to have the form K * V^2 where K is determined by the shape of the object and the density of the atmosphere.
In constructing a spreadsheet, the increments to v
for an increment of time Dt become:
= -Dt * (v
/ V) *K * V^2 and
= -Dt * (v
/ V) *K * V^2 - Dt * g'
Elevation and Maximum Range
At the beginning of a projectile flight, the initial velocity v
is resolved into its x, y components, employing radian measure:
*sin(θ*π/180) and v
The elevation angle θ is with respect to the horizontal.
As time progresses these components will change in value as gravity and the resistance of the medium take their toll on the motion.
Use the 2D calculator link given in the last topic of Chapter 6, to choose the elevation that will provide maximum range for a 4.0 cm. wooden ball projected at 100 metres per second. Do this for drag coefficients of 1E-10, 0.1, and 0.4. Use a step size of 0.02
, ~1020 m); (42.5
, ~530 m); (42
Once again, air resistance matters. The smoothness of the ball plays a very significant role!
In the next topic further reality is added to model motion through Earth's atmosphere by introducing the Standard Atmosphere and the variation in gravitational attraction with altitude.