Archimedes' Determination of the Volumes of Cones, Cylinders & Spheres

  Archimedes used the method of exhaustion to determine the volumes of a number of geometrical solids. It's considered a precursor to calculus which is used to sum infinitesimally small elements of volume, area and length. The method of exhaustion uses a series of inscribed and circumscribed geometrical solids to fill and surround a volume and focuses on the difference between the two. As the difference becomes smaller and smaller the remainder is reduced to zero. Archimedes determined that the volume of a sphere is four time that of a cone whose base is equal to the area cut by a plane through the center of the sphere and whose height is the radius of the sphere. The volume of a cone was known prior to this and can be found in Euclid's Elements to be 1/3 of the cylinder that just surrounds it. Archimedes then uses this to determine that the volume of a cylinder that just circumscribes a sphere is 3/2 that of the sphere. The ratio of the volume of a triangular prism to that of a prism that just contains it was found in Euclid to 1/3.

Some of these formulas were known even earlier as can be seen from the Moscow papyrus and the Rhind papyrus with various approximations used for the value of π.

Galileo and Archimedes on Mechanics

  Galileo's last work, Mathematical Discourses Concerning Two New Sciences, was published in 1632 a few years before his death and contains four dialogues concerned primarily with a discussion of the resistance of matter to breakage, the mechanics of levers, uniform and accelerated motion and the motion of projectiles. He speaks highly of Archimedes' Mechanics which is mathematical in nature and prefers this to Aristotle who mentions the lever in the last book of his Physics (parts 4 and 6). One of the characters in the dialogue, Simplicius, speaks for Aristotle. Another is Salviati, a deceased friend of Galileo, expounds the author's views and the third, Sagredo, another deceased friend of Galileo's, represents an intermediate position.

Archimedes' work On Floating Bodies contains a discussion which is similar to the cryptic passage in Lagrange's Analytical Mechanics in which the lines of the weights meet at the Earth's center of gravity.

Supplemental (Jan 12): Galileo describes the chain as being very nearly a parabola and uses an inverted chain (arch?) in an argument.

Hero of Alexandria's Mechanics

    The Mechanics of Hero of Alexandria tells us that the ancient Romans had a fairly good understanding of the use of simple machines. Hero discusses "the wheel and axle, the lever, the pulley, the wedge, and the screw." In his Catoptrica he also shows that light reflecting from a mirrored surface takes a minimum path.

  Hero's Mechanics survived only in an Arabic translation. There doesn't appear to be an English translation available but there are French translations and German translations. In French the simple machines are given as "le treuil, le levier, la  moufle (poulie), le coin et la vis sans fin."

Lagrange in his Analytical Mechanics makes a rather cryptic remark about the Center of Force and for heavy bodies subject to gravity it being the center of the Earth. What is peculiar about it is that he equates the ratio of the potential energy to the sum of the forces acting on a mechanism as the distance to the center of the Earth. This would be true if the distances to the Earth's center are the same or the forces acting on the parts of a body are equivalent to a single force acting on a center of mass. The change in direction of the vertical at sea level would also allow us to compute the distance to the Earth's center. Perhaps this is an allusion to the work in Geodesy that was taking place in the early 19th Century.

Supplemental (Jan 6): The key to Lagrange's system of mechanics are the virtual velocities or equivalently small changes in position which produce no change in their scalar product with the corresponding forces. Perhaps Lagrange was suggesting the problems of the chain and the arch for the "student" of analytical mechanics. In 1829 Gauss proposed "A new general Principle of Mechanics," the principle of least constraint based on least squares.

Minimum for a Horizontal Line

  What happens if the path is limited to a horizontal line instead of an ellipse?

In this case y equals a constant value and there is only one independent variable which we can take to be x. We can easily determine a formula for the potential energy and set its derivative equal to zero
to get an equation that can be solve for x.

On simplifying this equation we get a simple expression for the square of x and we have to be careful about the sign when taking the square root. All the factors are positive except for y so we need a minus sign to make the right side of the equation positive and then we can solve for x. The equation for x allows us to determine that the triangles involving the angles are similar and we see that angles are also equal at equilibrium for this problem. A check using the value of y_min for the ellipse shows that we get the same x_min.

What's remarkable is that for both problems the angles are equal at the equilibrium position. We get the same results whether we use the ellipse or the horizontal line which is tangent at the equilibrium point.

Minimum of the Zip Line Ellipse

  The equation for the ellipse only depends on the relative numerical values of the anchor positions so one can solve for the equilibrium position just by finding the lowest point on elllipse. There are no physical "laws" needed. Let's review the equation for the ellipse and those for formulas for the equilibrium found using Lagrange's undetermined multipliers.

What happens if we try to find the minimum value of y for the ellipse? To do that we need to take the derivative of the formula for it and set that equal to zero and solve for the unknown angle.

The cosine of the minimum angle, θ_min, turns out to be equal to the eccentricity times the cosine of the angle of the line through the focal points of the ellipse, θ₀, and we get the same coordinates as the other formulas gave for the equilibrium position.

So the path that the constraints impose the motion of the weight is all that we really need to solve this problem.

Adding a Counter Weight

 One can modify the zip line problem is the previous post by adding a counterweight m. We learned that the two angles were equal for equilibrium so we can simplify the solution making same assumption here.

 The length of the line is now ℓ=a+b+c so we have eliminated one variable and replaced it with another, c. The potential has an additional term for the height of m as well.

We have to modify the potential again to allow an unconstrained variation of the unknown parameters. The constraints here are constants of the variation and we can write them in a form that doesn't change the potential. We again take the derivative of the modified potential, V', and set it equal to zero to find the minimum of V'. If this is true for arbitrary changes in the unknowns their coefficients have to be equal to zero. The first of the three resulting equations gives us the same value for μ as before which allows us to simplify the second equation. We now have two simultaneous equations which allow us to solve for the undetermined multipliers λ and μ. The value of μ allows us to solve for sinθ.

We now have enough information to solve for the values of a and c.

The values of x, y, and sinθ from the previous problem allows us to find the mass m needed to maintain the equilibrium there. The formulas above allow us to solve for the remaining unknowns.

We can plot the data to help visualize the results.

The counterweight allows us to determine that the tension of the line is T=mg for equilibrium. The action of simple machines was all that Lagrange would have needed to develop general methods for mechanics.

A Zip Line Problem

  The section on finding the minimum of a function in a mechanical problem is a little difficult to follow in Lagrange's Analytical Mechanics so an example to illustrate the procedure may be helpful. Let's consider a zip line with a weight suspended on it and see if we can find the equilibrium position. We can set up a coordinate system with the origin at the lowest anchor for the line and the coordinates for the higher anchor will be (L,h). The length of the cable is ℓ=a+b and a point on the line can be designated by the coordinates (x,y) as shown.

Two unit vectors e₁ and e₂ are needed to simplify the specification of the problem and that the sum of the two vectors a and b will always equal the position of the higher anchor. The function that we want to minimize is the potential V. We can use the length of segment b to show that length of segment a is a linear function of the coordinates.

We can also show that the weight moves along an elliptical path by substituting the formulas for x and y as a function of a and θ into the linear equation for a and then solve the result for a.

The variation of the potential function V is subject to two constraints which are the components of the equation for the sum of the two segments. Multiplying the constraints by two undetermined multipliers assumed to be known constants and adding to the potential V we get a modified potential function V' for which we want the minimum. Taking the derivative of V' an setting it equal to zero we get three equations which we need to solve for the three unknowns a, θ, and φ.

The last equation tells us that the dot product of the second unit vector with a vector formed from the Lagrange multipliers is zero so we know it is perpendicular to this vector. Dividing by the square root of the sum of the squares of the multipliers gives us the second unit vector. The same trick works for the second equation and we get the components of the first unit vector in terms of the multipliers. Substituting these results into the first equation allows us to determine the value of μ which in turn allows us to show that the two angles are equal at the equilibrium position.

This result also allows us to simplify the equations for sum of the segments and solve for the unknowns in terms of the known quantities.

We are now in a position to do a numerical problem and find the values of x and y for the equilibrium position,

and plot the results.