COVER AND GRAPHS FOR THE OWNER'S MANUAL

inside the Arch





Former students, John and Scott, took a video camera and me inside the Arch several years ago. An electrician gave us a tour.










It will take a few minutes for the cover and graph, then you can print it.


COVER OF THE MANUAL Figure 1.

THE OWNER'S MANUAL


by William V. Thayer © 1993 PedLog Graphics

The Gateway Arch's hollow inside holds a curve used for basic calculations of Arch construction data. Look at the curve between the inside edge and the outside surface in Figure 1. The middle curve is located one third of the distance from the outside surface and two thirds of the distance to the inside edge of the Arch profile graph. We call the middle curve the centroid curve. The word centroid refers to the center of geometry of the equilateral triangles used in the construction of the Arch. The centroid curve controls the general arch shape of the Arch. The equation for the centroid curve is:

y = 693.8597 - 68.7672 COSH(0.0100333 x) feet for -299.2239 less than or equal to x less than or equal to 299.2239 feet.

Where y feet gives the centroid elevation x horizontal feet from the y axis.

You may have a calculator that gives COSH values meaning HYPerbolic COSine values. Please try the following computations as you read. Take 118.4123 feet for distance x and multiply it by 0.0100333 to get 1.18806613. To find COSH(1.18806613): For scientific calculators press in the number 1.18806613, then press the HYP key and then the COS key. New scientific calculators may have you press the HYP key, then the COS key followed by the number 1.18806613. Graphics calculators have HYPerbolic menu keys under a MATH menu that get COSH(1.18806613) about equal to 1.79277036. Multiply this result by -68.7672. Answer approximately -123.2837979. Now add 693.8597 for an approximate answer of 570.5759021 feet vertical y distance. Congratulations! You have completed the first step in the computation of a construction point for the Arch.

A formula giving the area of an equilateral triangle cross section of the Arch, Figure 1 right corner, with a centroid that is x horizontal feet away from the axis of symmetry is:

Q = 125.1406 COSH(0.0100333 x) square feet.

So while we have our calculator value for COSH(1.18806613) about equal to 1.79277036 corresponding to x=118.4123 feet, we can calculate the area of the equilateral triangle. Multiply 1.79277036 by the coefficient 125.1406 to get an area of 224.3483585 square feet. You completed the second step in the computation of construction data for the arch.

At (118.4123 ft., 570.5759 ft.), the equilateral triangle's area is 224.3483585 square feet.

Use geometry of the equilateral triangle to find the altitude given the triangle has an area of 224.3483585 square feet. A triangle's area

a = hb/2 for altitude h and base b.

The base b is related to the altitude h in an equilateral triangle by

b=2h/(square root of 3). We will write that as 3^0.5

Then the area a=hh/(3^0.5) so h=((3^0.5) a )^0.5.

Using this formula with 224.3483585 square feet gives altitude h about equal to 19.71250252. Congratulations! Step three yields the altitude of the equilateral triangle at the centroid point. From h = 19.71250252 we get one third of the altitude h1 about equal to 6.570834173 and two thirds of the altitude h2 about equal to 13.14166835.

Consider Figure 2. The location of the centroid of the equilateral triangle is (118.4123 ft., 570.5759 ft.) . This point is 6.570834173 feet from the outside surface and 13.14166835 feet from the inside edge of the Arch. The plane of the equilateral triangle is a cross section of the Arch and we need to know the slope of this cross section to determine the coordinates of the inside edge point and the coordinates of the outside surface point.
Figure 2.
CALCULATIONS GRAPH
For the slope we turn to calculus to find the derivative of the centroid equation y:

y' = -0.689961947 SINH(0.0100333 x).

v y' is the slope of the tangent to the centroid curve at x feet. The perpendicular to this tangent at (x, y) is called the NORMAL to the centroid curve at (x, y). Since the tangent is perpendicular to the normal, the normal must have a slope equal to the negative reciprocal of the tangent's slope. Therefore the slope of the normal at a distance of x feet is:

m = 1.449355292/SINH(0.0100333 x)

Using a scientific and/or a graphics calculator to find

SINH(0.0100333}118.4123)

with HYPerbolic and SINe keys or HYPerbolic menu in the MATH menu respectively gives 1.487960202.

Multiplying the reciprocal of this number by 1.449355292 yields

m about equal to .974055146 and we have completed step four!

The inside edge point (xi, yi) and the outer surface point midpoint (xo, yo) can be calculated by using m, h2, h1, and the centroid's coordinates x and y by the following formulas from algebra or trigonometry:

xi = ( x - h2)/((1+ m²)^0.5)

yi = ( y - m h2)/((1+ m²)^0.5)

xo = ( x + h1/((1+ m²)^0.5)

yo = ( y + m h1)/((1+ m²)^0.5)

First we find calculations of the intermediate values related to the slopes above:

1/((1+ m²)^0.5) about equal to 0.716338364

m/((1+ m²)^0.5) about equal to 0.69775307

Then we use x and y with h1 and h2 to get:

xi = 118.4123 - 0.716338364×13.14166835

yi = 570.5759 - 0.69775307×13.14166835

So

xi = 108.9984188

yi = 561.4062606

And

xo = 118.4123 + 0.716338364×6.570834173

yo = 570.5759 + 0.69775307×6.570834173

So

xo = 123.1192406

yo = 575.1607197

This fifth step gives the location of the equilateral triangle with centroid (x, y)!

You may calculate the angle formed between the normal to the centroid equation at the point (x, y) and a horizontal line through this point. The inverse tangent of the slope is the angle.

So m = 0.974055146 makes

TAN 0.974055146 = 44.24700936 degrees.

Again consider Figure 21. and our calculation for the centroid equation point (118.4123 feet, 5570.5759 feet):

OUTLINE OF COMPUTATIONS

• We calculated the area of the normal equilateral triangle 224.3483585 square feet from which we calculated the altitude 19.71250252 feet.

• We split this altitude into thirds 6.570834173 feet and two thirds 13.14166835 feet to locate the centroid of the triangle.

• We found the slope of the normal to the centroid =0.974055146.

• We calculated coordinates for a point on the inside edge, the intrados point, (108.9984188 feet, 561.4062606 feet).

• We calculated coordinates for a point on the outer surface, the extrados point, (123.1192406 feet, 575.1607197 feet).

• We calculated the angle of the normal with any horizontal line = 44.24700936 degrees.

• We could find the length of the side of the equilateral triangle with area 224.3483585 square feet, cut out a model size triangle and begin constructing a model based on the above set of calculations.

Of course, each point on the centroid curve would yield a new set of values for each of the above calculations. The accuracy of the model would then depend on our selection of points on the centroid curve. The blueprints of the Arch contain a large table with each line representing a new point of the centroid curve and each line contains the type of information we calculated for our point.

Whether you calculate many points to build a model of the Arch or just a few to understand the nature of the calculations, you will know and therefore own the basic ideas of your Gateway Arch National Monument. We sold T SHIRTs at local mathematics meetings with Figure 1. on the front and the OWNER'S MANUAL was packaged with the T SHIRT. Help yourself to use the same idea if all of the money is used for student scholarships!

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1. The calculations in Figure 2. were based on 3.0022/299.2239 for this article's use of the 0.0100333 approximation for the coefficient of x in the centroid curve's equation.

inside the Arch





This is one of several elevators in the Arch.

One that takes visitors to the top.













Copyright (c) 1982-1988 William V. Thayer, All Rights Reserved