Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-l] Final velocity of bullets



Regarding Ken Caviness' analysis & his claim concerning it:

Roger is correct on both counts:

1. We generally model the high speed case of fluid resistance
(typical example: an object moving in air) by the quadratic
approximation: f = c v^2, where the constant (here: c) depends on
the geometry of the object and the density of the fluid.

2. We can't simply assume that the y-component of the velocity on
impact is the terminal velocity, and the x-component is the same as
the original horizontal velocity. Friction acts in the direction
opposite to the instantaneous velocity, so this is a vector
differential equation. With the common choice of "up" being the +y
direction, and choosing the +x direction to be perpendicular to it
such that the object's initial velocity has a component in the +x
direction, we get:

-mg j_hat - c v^2 v_hat = m a_vector,

where v_hat is a unit vector in the direction of the velocity:
v_hat = v_vector/v, and j_hat is a unit vector in the y-direction
(i_hat in the x-direction).

Writing a_vector = a_x i_hat + a_y j_hat, v_vector = v_x i_hat +
v_y j_hat, this gives

-mg j_hat - c v^2 (v_x i_hat + v_y j_hat)/v = m (a_x i_hat +
a_y j_hat),

or

-mg j_hat - c v (v_x i_hat + v_y j_hat) =m (a_x i_hat + a_y j_hat).

Separating the x- & y- components:

- c v v_x = m a_x and -mg - c v v_y = m a_y

If that second equation had (v_y)^2 instead of v v_y, we'd have our
normal 1-dimensional case (at least as far as the y-direction goes)
with v_y approaching sqrt(m/(cg)) asymptotically. But we don't.
The length of v_vector, v = sqrt(v_x^2+v_y^2), so the equations
really are

- c v_x sqrt(v_x^2+v_y^2) = m a_x
and -mg - c v_y sqrt(v_x^2+v_y^2) = m a_y

The bad news is that both of these involve both x and y
derivatives, there's no way to separate variables and integrate.

The news is not quite that bad. Actually, the problem *is*
separable. In fact, the generalized problem of the air resistance
being proportional to any fixed arbitrary power of the speed is also
a separable problem. It's just that the separation isn't effected
directly in Cartesian coordinates from the start.

Numerical approximations can be done quite easily -- for instance
with an Excel spreadsheet -- and the basic way I think of it is
that we approach terminal velocity in _some_ direction of motion,
while the angle is also approaching vertical downward.

Give it enough time and the object will be heading basically
straight downward, basically at the terminal velocity v_t =
sqrt(m/(cg)). (Notice that the formula for v_t depends on which
approximation we're using to model the fluid resistance! They're
all approximations, anyway.)

True.

Fun to think about this stuff!

Ken Caviness
Physics
Southern Adventist University

As I mentioned above, the problem is fully separable, and its
Cartesian trajectory can be expressed as a set of parametric
equations involving only quadratures containing just the common
parameteric variable which happens to be the local inclination
angle (or equivalently, the local slope) of the trajectory at each
of its points. In the special case's of the exponent in the air
resistance power law being zero (air resistance being independent of
speed) or one (air resistance being proportional to the speed
itself) all the relevant integrals are easily analyticaly doable and
the entire Cartesian trajectory can be explictly solved for by
eliminating the common parametric parameter. But in the (usual)
case of the air resistance being proportional to the 2nd power (&
other powers as well) of the speed not all of the integrals are
doable analytically and the common parameter in the parameteric
equations cannot be eliminated from the system by algebraic
manipulations. But even in this case the problem is still formally
separable and the trajectory is solvable by numerical quadratures in
the explicit formulae for the coordinates, x, y, & t in terms of the
common parametric variable.

The way the problem is solved is to write Newton's 2nd law in terms
of the vector component that is tangential to the local motion and
in terms of the perpendicular centripetal component that is normal
to that local motion. The resulting relevant DE that has the speed
v as the dependent variable and the inclination angle A of the path
above horizontal as the independent variable. This DE is fully
solvable and the solution gives us a first integral for the
problem. Once we have the speed v as an explicit function of the
angle A, i.e. v = v(A), we can put it into appropriate integral
expressions that give the functions t = t(A), x = x(A), and y = y(A).
If we are lucky enough to be able to do the integrals we have an
explicit closed form formulae for the trajectory. In the usual case
when the exponent n in the air resistance power law is not 1 (or 0)
these last integrals become too difficult to get a nice closed form
solution. But the formal solution is still there in terms of the
undone integrals.

In my outline of the problem's solution below I write the air
resistance in terms of the terminal velocity (er, speed) v_t rather
than in terms of the coefficient 'c' that Ken uses above. We can
write the air resistance contribution to the total tangential force
as -m*g*(v/v_t)^n . Note that in the special case of n = 2 for
quadratic velocity damping that the parameter 'c' above is given by
c = m*g/(v_t)^2.

Using this notation the component of Newton's 2nd law that is
tangential to the local motion becomes:

dv/dt = -g*(sin(A) + (v/v_t)^n) eq. 1

and the normal/centripetal/perpendicular component becomes:

-v*dA/dt = g*cos(A) eq. 2

Once we have the function v = v(A) we can use eq. 2 to find the
elapsed time t as a function of the parameteric angle A, i.e.
t = t(A), by integrating the differential

dt = -v(A)*dA/(g*cos(A)) eq. 3

We also know that if ds is the differential of the path length
we have dx/ds = cos(A) where x measures the horizontal Cartesian
coordinate. But since ds = v*dt this means that

dx = cos(A)*ds = cos(A)*v*dt = cos(A)*v*(-v*ds/(g*cos(A))) =

dx = -(v(A)^2)*dA/g eq. 4

So once we find the speed v as a function of angle, i.e. v(A), we can
integrate eq. 4 to get the function x = x(A).

Likewise, we know that for the vertical Cartesian coordinate y we
have dy/ds = sin(A). This means that we can write the differential
dt as

dy = sin(A)*ds = sin(A)*v*dt = sin(A)*v*(-v*ds/(g*cos(A))) =

dy = -(v(A)^2)*tan(A)*dA/g eq. 5

Again, once we have the speed function v = v(A) we can integrate
eq. 5 to get the function y = y(A).

Thus, it is now clear that all we need to get parametric expressions
for the trajectory functions: t(A), x(A), and y(A) is to find the
speed function v = v(A) first. To do this we use eq. 2 and the
chain rule on eq. 1 to eliminate t as the independent variable in
favor of the variable A.

d(...)/dt = (dA/dt)*d(...)/dA = -(g*cos(A)/v)*d(...)/dA

so

-(g*cos(A)/v)*dv/dA = -g*(sin(A) + (v/v_t)^n) or simplifying

d(ln(v))/dA = tan(A) + sec(A)*(v/v_t)^n eq. 6

Eq. 6 is a 1st order DE that can be explicitly solved by making some
substitutions. The formal solution is given by

v(A) = v_t*sec(A)/((v_t*sec(A_0)/v_0)^n + n*I(A))^(1/n) eq. 7

Where the function I(A) is defined as the definite integral

I(A) == INT{A'=A, to A'=A_0 |dA'*(sec(A'))^(n+1)} eq. 8

The quantity A_0 in eq. 7-8 is the initial launch angle of the
projectile above horizontal, and the quantity v_0 in eq. 7 is the
initial launch speed of the projectile.

We now are in a position to see the procedure for formally finding the
trajectory. For any fixed power n in the air resistance we first do
the integral in eq. 8 and get the I(A) function which we then
substitute into eq. 7 which we then substitute the resulting v(A)
function into eq. 3 and integrate the RHS starting from A_0 up to the
current value of A. That gives us the currently elapsed time
t = t(A). Next we substitute the v(A) function into eq. 4 and
integrate the RHS from the initial A_0 angle up to the current value
A. The LHS is just the difference x(A) - x_0. That gives us the
x(A) function. Next, we substitute the v(A) function into eq. 5 and
integrate the RHS from the initial A_0 angle up to the current value
A. The LHS is just the difference y(A) - y_0. That gives us the
y(A) function. At this point we have the 3 functions t(A), x(A), and
y(A). Next, we invert the t(A) fucntion to give the angle A as a
function of time A = A(t). This is then substituted into the
functions x(A) and y(A) giving the overall trajectory equations
x = x(t), and y = y(t).

In the special n = 1 case of linear air resistance in the velocity
all the necessary integrals, functional inversions, and substitutions
are fully doable and the procedure gives us nice explicit formulas
for the trajectory x(t, y(t) and for the path followed y(x). But I
won't give those formulas here. That is an exercise for the reader.
In the special n = 2 case of quadratic air resistance in the velocity
the integral in eq. 8 is fully doable and we can get an explicit
functional form for the speed as a function of angle, v = v(A). This
gives us a first integral for the motion. But when this v(A) function
is substituted into eq. 3-5 none of the resulting integrals for t(A),
x(A), and y(A) are easy enough to do (at least they're not easy enough
for *me* to be able to do) so as to obtain explicit integrated
formulas for the functions t(A), x(A), and y(A) functions. But these
integrals can still be done by numerical integration and numerically
evaluated function values for t(A), x(A), and y(A) can be so obtained.
By way of completeness the exact formula for the v(A) function in the
n = 2 case is given by:

v(A) = v_t/sqrt(cos^2(A)*B - sin(A))

where the quantity B is defined as

B == (v_t*sec(A_0)/v_0)^2 + tan(A_0)*sec(A_0) + arctanh(sin(A_0)) -
- arctanh(sin(A))

Having the sin(A) and the arctanh(sin(A)) terms in the denominator's
square root in the v(A) function makes the integrals in eq. 3 - 5
quite hard to do.

But regardless of our ability to analytically do the integrals the
formal solution procedure outlined above shows us that the problem is
actually separable after all.

David Bowman