Computes the position of a falling body:
without drag (as in a vacuum); and with linear drag,
i.e., proportional to velocity, b v,
with b the drag coefficient. (The units are assumed SI.)
The equations, with t time, y position (above ground), and
v (upward) velocity are, with τ = m⁄b
(s, [T]):
| No drag | dv⁄dt = -g |
v = -g t + v0 |
| |
dy⁄dt = v =
-g t + v0 |
y = -(½)g t² + v0t +
y0 |
| Drag, linear | dv⁄dt = -g -
v ⁄ τ |
v = -g τ +
(v0 + g τ)
exp(-t ⁄ τ) |
| |
dy⁄dt = v = … |
y = -g τ t
+ τ Q [1 - exp(-t ⁄ τ)]
+ y0 |
Q = v0 + g τ
[LT−1].
(g = 9.8 m⁄s², thus all values must be SI.)
The drag coefficient, b, can be given by
b = 6 π η R, with η the (dynamic, usual)
viscosity and R the radius of a falling sphere (laminar flow only),
i.e., Stokes law. Viscosity for some fluids, in Pa.s:
air, 1.9E-5; water, 0.001; olive oil, 0.08; and glycerine, 1.3.
A "Newtonian drag"
(proportional to v²) might also be used, but the analytical solution
is complex, so in that case numerical methods (such as Runge-Kutta) are used
(as for other exponents of v).
A plot is shown for the position, y, and
the velocity (its derivative), v, for both cases
(no drag, and linear drag).
Other suggested data:
(a) small drag, r ⪆ 0 (such as 0.01 for the basis data),
giving almost coincident curves;
(b) for a rain drop of diameter 3 mm, mass 1.413E-5 kg (14 mg),
with b = 1.E-5 (if linear drag), the expected terminal velocity
is -13.8 m/s (e.g., raindrop). This case is, however, typically
considered Newtonian. |
