In the late 18th and throughout the 19th century, lead shot for muskets was prepared by use of a shot tower. Molten lead was poured from the top of a tower and, during its fall, the drops became spherical under the action of surface tension. In this article, we ask and answer the question: How does the size of the lead shot depend on the height of the tower? In the process, we explain the basic technology underlying an important historical invention (the shot tower) and use simple physics (Newtonian mechanics and the thermodynamic laws of cooling) to model its operation.

Topics
Newtonian mechanics, Thermodynamic functions, Thermodynamic states and processes, Thermodynamic properties, Nuclear reactors, Transition metals, Educational aids, Viscosity, Patents, Public policy and governance
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G. V.
 
Sanders
, “
Gravity molds shot in a modern tower
,”
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An exact calculation proceeds as follows. First compute the fall distance X required for a molten drop to solidify (during which time the temperature of the lead remains constant at its melting point, Ti⁠) by solving
0.185k2η0.64πR1.6(Ti−Ta)∫0Xv−0.4dy=4π3ρR3L
(with the positive y-axis pointing downward). Given a formula for v(y) —e.g., (2gy)1/2 in free fall or vT at terminal speed— this integral can be computed to determine X(R). Next find the distance Y over which the solid sphere cools. The convectional cooling power equals the rate of decrease of the internal energy U of a pellet, so that
Pconvection=−dUdt=−dydtdUdy
using the chain rule. Note that dy/dt equals the speed v of the shot, while dU=mcdT⁠. Hence the right-hand side of Eq. (4) can be equated to −vmc dT/dy. Again given a formula for v (y), a separable differential equation is obtained that can be solved for Y(R). Finally, add X+Y=H to relate the height of the tower to the maximum radius R of the shot that can be produced. For example, if one uses v=vT for both parts of the motion, then one obtains Eq. (6) with α=1.1×10−4m3/8⁠, essentially the same result that was found above by approximating the averages. (Incidentally, in this case Y/X ≈ 2 independent of R, indicating that it takes about one-third of the height of the tower for the sphere to solidify and the remaining two-thirds for it to cool down.)
4.
F.
 
Kreith
 and
M. S.
 
Bohn
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5.

Usingthe numbers cited after Eq. (10), the average Reynolds number is r¯=vTR/η=2500⁠, which is large enough to justify modeling the air resistance as being quadratic rather than linear in the speed of a drop.

© 2012 American Association of Physics Teachers.
2012
American Association of Physics Teachers
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