Conical Pendulum Launcher

A conical-pendulum, with a vertical axis, is employed to launch a projectile, which we call a ball. Here is a drawing, in which I drew the flight trajectory as a Bezier-curve, rather than the parabola that it actually is. We consider five cases.

the variables

given variables

length of rope = u
height of vertex of the conical pendulum = h
angle between vertical and the rope = theta
acceleration due to gravity = g (It will cancel-out)
mass of the ball = m (It will cancel-out)

derived variables

radius of conical pendulum = r
height of the ball (at the end of the conical pendulum) = alt
time for the ball to drop to the ground = t
the range; that is, the horizontal distance that the ball travels while in flight = range
the distance of the landing-spot of the ball from the axis of the conical pendulum = R
the weight of the ball = w
the tension in the rope = f
the centripetal force = cent
energy: kinetic = ke, potential = pe, total = te
(with precognition) effective height, above launch-point = b
(with precognition) effective height, above impact-point = effh.

calculations

range equations

In general, the range depends upon the altitude alt from which the projectile is launched, its initial speed v, and the launch angle. Here, we consider only a horizontal launch; because the projectile naturally launches horizontally in a tangential direction.

alt = g t^2 / 2
range = v t

Square the second and substitute t^2 from the first, to obtain

range^2 = v^2 (2 alt / g) = 2 v^2 alt / g = 4 b alt

Since the launch is tangential, it is at right-angles to the radius r. Hence, the resultant distance R is the square-root of the sum of the squares of the range and the radius.

R^2 = range^2 + r^2 = 2 v^2 alt / g + r^2 = 4 b alt + r^2.

energy

At the point of launch, the energy is

ke = m v^2 / 2
pe = m g alt
te = ke + pe = m (v^2 / 2 + g alt) = m g (b + alt)
effh = te / (m g) = b + alt.

conical pendulum

geometry:

r = u sin(theta)
alt = h - u cos(theta)

force:

w = m g = f cos(theta)
cent = m v^2 / r = w tan(theta)

combined:

2 b = v^2 / g = u sin(theta) tan(theta).

together

R^2 = 2 (u g sin(theta) tan(theta)) (h - u cos(theta)) / g + (u sin(theta))^2 = (u sin(theta))^2 (1 + 2 h / (u cos(theta)) - 2) = (u sin(theta))^2 (- 1 + 2 h / (u cos(theta)))
R = u sin(theta) sqrt(- 1 + 2 h / (u cos(theta)))
effh = (u sin(theta) tan(theta) / 2) + (h - u cos(theta)) = h - u (2 (cos(theta))^2 - (sin(theta))^2) / (2 cos(theta)) = h - u (1 + 3 cos(2 theta)) / (4 cos(theta)).
impact-energy = m g effh = (f cos(theta)) (h - u (1 + 3 cos(2 theta)) / (4 cos(theta)).) = f (h cos(theta) - u (1 + 3 cos(2 theta)) / 4).

maximize (with respect to u)

R^2 = (sin(theta))^2 (- u^2 + 2 h u / cos(theta))
2 R dR/du = (sin(theta))^2 ( - 2 u + 2 h / cos(theta))
0 = dR/du ==> u = h / cos(theta)
R^2 = (h tan(theta))^2 (- 1 + 2) = (h tan(theta))^2
R = h tan(theta)
effh = h - (h / cos(theta)) (1 + 3 cos(2 theta)) / (4 cos(theta)) = h (4 (cos(theta))^2 - 1 - 3 (2 (cos(theta))^2 - 1)) / (2 cos(theta))^2 = h (2 - 2 (cos(theta))^2) / (2 cos(theta))^2 = h (tan(theta))^2 / 2
impact-energy = m g effh = (f cos(theta)) (h (tan(theta))^2 / 2) = f h sin(theta) tan(theta) / 2.

Hence, for a constant tension f, one would want to increase theta (within its domain [0, pi/2)) to increase the impact-energy. Actually, usability and aerodynamic considerations would preclude any theta much beyond the given value of (pi / 3).

numerical values

the given common values:

h = 2 metres
theta = pi / 3.
Hence, sin(theta) = sqrt(3) / 2, cos(theta) = 1 / 2, tan(theta) = sqrt(3), sec(theta) = 2, cos(2 theta) = - 1 / 2.

case 0) the conical-pendulum by itself, with the given u = 2 metres

u = 2 metres
R = r = (2 metres) (sqrt(3) / 2) = sqrt(3) metres = 1.73 metres
effh = v^2 / (2 g) = (2 metres) (sqrt(3) / 2) (sqrt(3) / 2 = 3 / 2 metres = 1.5 metres.

case 1) projectile launched, with the given u = 2 metres

u = 2 metres
alpha = 0
R = (2 metres) (sqrt(3) / 2) sqrt(-1 + 2 (2 metres) / ((2 metres) (1 / 2))) = 3 metres
effh = (2 metres) - (2 metes) (1 + 3 (- 1 / 2)) / (4 (1 / 2)) = 2.5 metres.

case 2) projectile launched, with the optimum u

u = (2 metres) / (1 / 2) = 4 metres
alpha = 0
R = (2 metres) (sqrt(3)) = 2 sqrt(3) metres = 3.46 metres,
effh = (2 metres) - (4 metres) (1 + 3 (- 1 / 2)) / (4 (1 / 2)) = 3.0 metres.

which is one-and-a-half times the previous value. Likewise, the range is 0.46 metres more than its previous value.

discussion

Is this the best that can be accomplished? The Peruvians employ a pair of small rocks and a very long rope. Then, they launch the rocks in an upward (inclined) direction. They sail farther and then the trailing rope snares upon the prey. The rocks wind around and strike.
The calculation is similar; but, has to take into account the launch-angle. For the greatest range, the launch angle has to be optimized.

range equation, revisited

effective height b

Set the kinetic energy at the low-point equal to the potential energy at the high-point -- b -- (where, by assumption, the kinetic energy is equal to zero), then solve:

(1 / 2) m v^2 = ke = pe = m g b.
b = v^2 / (2 g)
v^2 = 2 g b.

Now that you have the b, go back to the foregoing and evaluate in terms of the b, as appropriate. Actually, since we already had performed these evaluations by precognition, all that you have to do is to verify.

the range from the differential equation

Let R be the displacement vector. In Cartesian coordinates, its second derivative and the two integrals are:

R" = (0, - g)
R' = (v cos(alpha), - g t + v sin(alpha))
R = (v t cos(alpha), - g t^2 / 2 + v t sin(alpha) + alt) = (x, 0),

where alpha is the elevation-angle of the initial velocity, whose magnitude is v, and alt is the initial vertical displacement. We want to maximize x, subject to the constraint of the second component of R. Thus, let us define the function f as

x = v t cos(alpha)
0 = - g t^2 / 2 + v t sin(alpha) + alt
f = f(t, alpha, y) = v t cos(alpha) + y (- g t^2 / 2 + v t sin(alpha) + alt),

where y is an arbitrary variable. Compute the three partial derivatives of f and set each equal to zero:

0 = df/dt = v cos(alpha) - y (g t - v sin(alpha))
0 = df/d-alpha = - v t sin(alpha) + y v t cos(alpha) = vt (- sin(alpha) + y cos(alpha))
0 = df/dy = - g t^2 / 2 + v t sin(alpha) + alt.

This is a set of three simultaneous equations. Solve the second equation for y and substitute into the first equation:

y = tan(alpha)
0 = v cos(alpha) - tan(alpha) (g t - v sin(alpha)) = (v - g t sin(alpha)) / cos(alpha).

Solve for t and substitute into the third of the equations, to obtain

t = v / (g sin(alpha))
0 = - g (v / (g sin(alpha)))^2 / 2 + v sin(alpha) (v / (g sin(alpha))) + alt = - b / (sin(alpha))^2 + 2 b + alt.

Solve for launch-angle alpha, to obtain

(1 - cos(2 alpha)) / 2 = 1 - (cos(alpha))^2 = (sin(alpha))^2 = b / (2 b + alt)
(cos(alpha))^2 = 1 - (sin(alpha))^2 = 1 - b / (2 b + alt) = (b + alt) / (2 b + alt)
(cot(alpha))^2 = (b + alt) / b = 1 + alt / b
cos(2 alpha) = 1 - 2 b / (2 b + alt) = alt / (2 b + alt).

Substitute t into the equation for x, to obtain

range = x = v (v / (g sin(alpha))) cos(alpha) = 2 b cot(alpha) = 2 b sqrt(1 + alt / b)
R^2 = range^2 + r^2 = 4 b^2 (1 + alt / b) + r^2 = 4 b^2 + 4 b alt + r^2.

The range, with the cosine of twice the launch-angle alpha, constitutes the generalized range-equation -- taking into account a non-zero initial altitude alt.
This R^2, with the optimum launch-angle, has the term (4 b^2) in addition to that for a horizontal launch. Substitute the b, r, and alt from the conical-pendulum, to obtain

R^2 = (u sin(theta) tan(theta))^2 + 2 (u sin(theta) tan(theta)) (h - u cos(theta)) + (u sin(theta))^2 = 2 h u sin(theta) tan(theta) + u^2 (sin(theta))^2 ((tan(theta))^2 - 1)
alt = h - u cos(theta)
b = u sin(theta) tan(theta) / 2
cos(2 alpha) = alt / (2 b + alt) = (h - u cos(theta)) / (2 (u sin(theta) tan(theta) / 2) + (h - u cos(theta))) = (h - u cos(theta)) cos(theta) / (h cos(theta) - u ((cos(theta))^2 - (sin(theta))^2)) = (h - u cos(theta)) cos(theta) / (h cos(theta) - u cos(2 theta)).

optimum u

Since each of the terms in the expression for R^2 is positive, the optimum u is at the upper end-point of the domain -- u = h / cos(theta). (This is the same as that of case 2.) Substitute this value of u, to obtain

R^2 = 2 h (h / cos(theta)) sin(theta) tan(theta) + (h / cos(theta))^2 (sin(theta))^2 ((tan(theta))^2 - 1) = h^2 (2 (sin(theta))^2 (cos(theta))^2 + (sin(theta))^4 - (sin(theta))^2 (cos(theta))^2) / (cos(theta))^4 = h^2 (sin(theta))^2 / (cos(theta))^4
R = h tan(theta) sec(theta)
cos(2 alpha) = (h - (h / cos(theta)) cos(theta)) cos(theta) / (h cos(theta) - (h / cos(theta) cos(2 theta))) = 0
alpha = pi / 4 (identically -- regardless of the values of h or theta.)

numerical values

With skill, the launch-angle alpha can be controlled. We will employ the alpha to extend the range -- and, thence, the R. However, since the energy is conserved during the free-flight, we just copy the effh and impact-energy from the preceding respective cases.

case 3) projectile launched at the optimum alpha, with the given u

u = 2 metres (from case 1)
cos(2 alpha) = ((2 metres) - (2 metres) (1 / 2)) (1 / 2) / ((2 metres) (1 / 2) - (2 metres) (- 1 / 2)) = (1 / 2) / 2 = 1 / 4
alpha = 37.8 deg
R^2 = 2 (2 metres) (2 metres) (sqrt(3) / 2) (sqrt(3)) + (2 metres)^2 ((sqrt(3) / 2)^2 (sqrt(3))^2 - 1) = 12 metres^2 + 6 metres^2 = 18 metres^2
R = 3 sqrt(2) metres = 4.24 metres
effh = 2.5 metres (from case 1)

case 4) projectile launched at the optimum alpha, with the optimum u

u = 4 metres (from case 2)
alpha = pi / 4 = 45.0 deg
R = (2 metres) (sqrt(3) (2) = 4 sqrt(3) metres = 6.93 metres
effh = 3.0 metres (from case 2).

Summary

The height h of the vertical-axis conical-pendulum is 2 metres and its angle theta is (pi / 3). As we gain skill, the effective-height effh at impact doubles and the R quadruples!

#	u	cos(2 alpha)	alpha	R^2	R	effh	comments
0	2	n/a	n/a	3	1.73	1.5	conical-pendulum by itself
1	2	1	0	9	3.00	2.5	given u, alpha = 0
2	4	1	0	12	3.46	3.0	optimum u, alpha = 0
3	2	1 / 4	37.8	18	4.24	2.5	given u, optimum alpha
4	4	0	45.0	48	6.93	3.0	optimum u & alpha