Chapter 2

Design a Simple Eave Trough using Optimization

Start with a 30 cm wide, long flat piece of sheet metal, lying in the horizontal plane.  Make sharp bends at 10 cm in from each of the left and right edges as is seen in the diagram following.  Bend each side up the same amount.


There must be some bend that will provide the most trough volume.  If so, what will be the height and width of the trough? 

To answer the question we require an expression that relates a dimension of the bent section to the volume, or equivalently, to the area of the trough's cross section.

We decide to relate the amount of bend to the x coordinate of the tip of the bend on the right. Its range is 5cm. through 15 cm.  The y coordinate of the tip of the bend is given by:

y = 102 - (x -5)2

The area of the trough a is given by a = y*(10+x).  A short spreadsheet table is used to verify our assumption that there is an optimum bend and if so to localize it.  We start with x = 5 cm. and increment it in amounts of 2 cm. as seen next.
Without proceeding further one could choose a bend where x and y were both slightly over 9 cm, but then we would miss the fun part. 

Use a spreadsheet. Start at x = 7 cm, use initial small steps, ∆x = 0.1, and hunt for the maximum by reversing step direction with smaller step size, i.e., ∆x/2, whenever the area is seen to be decreasing with the current step size.


Only the more interesting steps are shown in the table.  Notice that the process seems to be homing in on x = 10 and therefore choose that value as optimum.

Click the cell addresses following to view the cell expressions.

. .
J39 K39 L39 M39

What about the choice of positioning the bends at 10 cm from each edge?  Why don't you use a spreadsheet to see if there is a choice that will provide more area?

In some situations we search for a minimum.  How would the search algorithm change?

Try Polar Coordinates

When a position in a plane is referenced by its distance, r, from a point of origin and its angular offset, θ, from a straight line that passes through that origin, it is being referenced by its polar coordinates r, θ.

Polar coordinates are often most convenient when the problem at hand has some angular symmetry.

When the reference line coincides with the x axis and θ is taken to increase positively with counter clockwise rotation, then one can convert to rectangular coordinates using x = r*cos(θ), y = r*sin(θ).  The inverse expressions are θ = arccos(x/r) and θ = arcsin(y/r). (Sometimes arc is abbreviated to a.)

The unit that is often used for θ is the radian although the degree is sometimes convenient. There are 2π radians to a single 360o rotation around an origin.

Return now to the design of the eave trough.  The right 10 cm. flap of the sheet metal, is bent upward by an angle θ from the angle 0 radians to some angle < π / 2 radians.  The height h of the trough is 10*sinθ.  Its width w is 10*cosθ +10 +10*cosθ. Its area a is 100*sinθ*(1+cosθ).

A rough calculation using steps of 0.25 radians finds a maximum of about 130 cm2 near θ = 1. 

A more precise maximum is then found using an adaptive step size with θ increasing from 0.75 radians.  In this case the change in area, Δa was calculated so that calculation could be terminated when Δa reached ~10-8.

The expressions for the outlined cells in rows 45, 85, and 86 can be seen by clicking the cell addresses that follow.

. .
I45 J45 K45 L45 K85 K86

Whether to employ polar or rectangular coordinates in this optimization problem appears to be six of one and a half dozen of the other.


Optimization and Approximation are closely related topics. Approximation with sinusoids is explored next.

Top Previous Topic Next Topic Topics