Free Fall without and with Air Resistance

Theory

The equations without air resistance are quite simple. We assume g=const, s(t=0)=0, and v(t=0)=0:

Taking into account the air resistance force (drag force):

which depends on the drag coefficient c_D, the velocity v, the density of air ρ, and the cross section A of the body,

the equation of motion is

The air resistance force will increase until it equals the gravitational force m^.g, and the body will then move at a constant velocity v_limit as the net force is m^.a=0

The differential equation of motion is solved by

The drag coefficient c_D is a function of the non-dimensional Reynolds number Re, defined by

L is a characteristic linear dimension (e.g. diameter of the body/sphere), ρ the density, and μ the dynamic viscosity (for air 17.9 µPa^.s at 10°C).

>>In theory, the flow is laminar when Reynolds number is below 4,000. However, in practice, turbulence is not effective when Reynolds number is below 200,000; so when the Reynolds number is less than 200,000, you may assume laminar flow. In addition, when the Reynolds number is higher than 2,000,000, you may assume turbulent flow.<< (*).

Using a "Height" input the delay due to air resistance is computed (∆t absolute, and ∆t/T in relation to the time T of free fall).

Ferdinand Reich's experiments were performed at a height of 158.5 m using spheres of 4,034 cm diameter and 270,45 g mass.
The speed limit 86.3 m/s is not reached. The effect of air resistance on time is ∆T = 5.8676s - 5.6845s = 0.1831s, ∆T/T = 3.22%.

Select "∆t - H":

The diagram below is showing the time difference as a function of diameter due to air resistance for iron balls (ρ=7874 kg/m^3) falling from 150 m:

∆t/T ~ 1/diameter (R²=0.999996)

The diagram below is showing the time difference as a function of mass due to air resistance for iron balls (ρ=7874 kg/m^3) falling from 150 m:

∆t/T ~ 1/mass (R²=0.999992)

The time when 95 % of the terminal velocity (v) is reached as a function of mass (iron balls) is shown in the following diagram:

t_95 ~ mass^1/6 (R²=0.999997)

Parachute

The diagram is showing the rate of descent (vLimit, terminal velocity) as a function of mass, and the equivalent height at free fall.

	Enter mass (kg), and diameter (m) of the body,
	or enter mass (kg) and density (kg/m^3), e.g. 7874 kg/m^3 for iron, diameter will be computed
	or enter density (kg/m^3), and diameter (m), e.g. 7874 kg/m^3 for iron, mass will be computed
	Enter acceleration (m/s^2), density of the air (kg/m^3), drag coefficient (dimensionless), for a sphere the dag coefficient is c_D = 0.4 .. 0.5 (0.47)
	enter maximum time (s), or Height (m), and press button "Apply Input"
You may also use the Parachute mode:
	Enter time (s), or Height (m), and press button "Apply Input".
	Using mass = 150 kg, diameter = 9 m, drag coefficient c_D = 1.3, dome-shaped chute c_D = 1.3 .. 1.5, parasheet (flat sheet) c_D = 0.75 the rate of descent will be about 5 m/s.
Using a "Height" input the delay due to air resistance is computed (∆t absolute, and ∆t/T in relation to the time T of free fall).