The Saint Louis Gateway Arch

Latitude/longitude coordinates between coordinates (38.627952,-90.1843880) and (38.62258,-90.186339).

A Shop around the Arch




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Today we use two equations to represent the mathematics of the Gateway Arch:

y = 693.8597 - 68.7672 COSH(0.0100333 x) feet

for x between or equal to -299.2239 and 299.2239 for the general arch shape and

Q = 125.1406 COSH(0.0100333 x) square feet area of a

cross section, that is the local shape. A short paper, "OWNER'S MANUAL FOR THE GATEWAY ARCH, SAINT LOUIS, MO." uses a graphing calculator approach to learning the mathematics of the Arch with the above equations.

The Curve of the Arch




The following presentation of the mathematics of the Arch is based on the history of the construction.




Please enjoy the history approach now.







THE MATHEMATICS AND ARCHITECTURE OF

THE SAINT LOUIS ARCH

by William V. Thayer

Copyright © 1982, © 1988 and © 1998, William V. Thayer, All Rights Reserved


On October 28, 1965 the construction of the GATEWAY ARCH located in St. Louis, Missouri was finished with the top stainless steel section welded in place 630 feet above ground. Dr. Hannskarl Bandel, New York, NY, furnished the equations of the ARCH for Eero Saarinen. These equations gave geometric form to Saarinen's artistic design concepts for this National Monument. Mr. Bruce Detmers, Hamden, CT, and other architects expressed this geometric form for Saarinen in blueprints as:

BASIS OF CALCULATIONS (Subscripts are juxtaposed to variables)

(1) Y = A ( COSH( C X / L ) - 1 ) or

(2) X = ( L / C ) ARGCOSH ( 1 + Y / A )

(3) A = ( fc ) / ( ( Qb / Qt ) - 1 ) = 68.7672

(4) C = ARGCOSH( Qb / Qt ) = 3.0022

(5) fc = MAXIMUM HEIGHT OF CENTROID (IN FEET) = 625.0925

(6) Qb = MAXIMUM CROSS SECTIONAL AREA OF ARCH AT BASE (IN SQ. FEET) = 1262.6651

(7) Qt = MINIMUM CROSS SECTIONAL AREA OF ARCH AT TOP (IN SQ. FEET) = 125.1406

(8) L = HALF WIDTH OF CENTROID AT THE BASE (IN FEET) = 299.2239

(9) El = fc - Y (elevation)

(10) SLOPE = TAN a = (L/C) ( 1/ ( (2AY+Y^2 )^0.5)

(11) Q = ( ( Qb - Qt) / fc) Y + Qt = 1.81977 Y + 125.1406 = CROSS SECTIONAL AREA AT ANY Y

(12) H = ( Q COT 30 )^0.5 = ( 1.73205081 Q)^0.5 = HEIGHT OF THE SECTION

(13) 2W = 2 H TAN 30 = 1.15470054 H = SIDE OF THE SECTION

The blueprint has a data table listing of values found by using these formulas and the following instructions:

INTENT OF GEOMETRY:

1. CENTROID IS DETERMINED BY FORMULA.

2. ALL SECTIONS PERPENDICULAR TO THE CENTROID ARE EQUILATERAL TRIANGLES.

3. THESE SECTIONS VARY IN AREA IN INVERSE PROPORTION TO THE ELEVATION OF THEIR CENTROIDS.

4. ALL EDGES FROM STATION 0 TO STATION 41, X = 230.1923, ARE SMOOTH CURVED LINES SUCH THAT ANY POINT ON THEM WILL SATISFY THE ABOVE GEOMETRY.

5. ALL EDGES FROM STATION 41 TO STATION 71, X = 299.5458, ARE CHORDS BETWEEN ADJACENT STATIONS SUCH THAT ANY POINT ON THE STATION WILL SATISFY THE ABOVE GEOMETRY.

The following figures gives us a graph of the Arch.

The notation is consistent with the following program used to develop data similar to the data given in the blue prints.

Figure A. THE SAINT LOUIS ARCH
Arch Drawing


Figure B. The relationships between a normal cross section and a horizontal cross section viewed from the side.

A normal cross section.




EXERCISES:

1. Use Figure B to show that the coordinates of a point on the intrados of the Arch are given by:

X3 = X - H1 cos a and Y3 = Y + H1 sin a

given that the coordinates of the centroid are (X,Y).

2. Solve for the coordinates of a point on the extrados using the methods found in exercise 1.

3. Use the equations of page 1 to show that:

Q = Qt COSH( C X / L ).

4. (For calculus students.) Start with equation (1) above and show that:

angle a = ARCTAN ( ( L / ( C A ) ) CSCH( C X / L) )

Hint: This time we use the reciprocal instead of the negative reciprocal. Why?

5. (For geometry students) Assume that you know the area of an equilateral triangle is Q. Find the height, H, of this triangle.

What is the height of the triangle at the top of the Arch?

What is the distance, H1, from a side to the centroid in terms of area Q?

Using fc, from (5), above and the value of H1 for the triangle at the top of the Arch, show that the Arch stands 630 feet tall.

6. (For precalculus students) Show that the equation for the elevation, El, of a point on the centroid written in exponential form base two is:

El = 693.8597 - 34.3836 ( 2^(0.014475 X) + 2^( - 0.014475 X) )

7. (For programming students) Find the arc length of the centroid by approximating:

s = 2 times the integral from 0 to L of ( 1+ ((A C / L) SINH( C X / L))^2)^(1/2) with respect to dX

(For calculus students) Show that this formula gives the arc length.

8. (For calculus students) Show that the radius of curvature of the centroid at the top of the Arch is about 144 ft.

9. (For calculus students) Show that the volume enclosed by the surface of the Arch is approximately 924100 cubic feet.

Hint: V = 2 times the integral from 0 to L of ( Q(X) (ds/dX) with respect to dX .

{It is also interesting to find the surface area of stainless steel used in construction of the arch! And the lengths of the intrados and extrados midline.}



BIBLIOGRAPHY:

Boyer, Lee E., Applications Of Mathematics In The New Spirit Of St. Louis Arch, Harrisburg Area Community College, Harrisburg, Pa., 1966.

Thayer, William V., The St. Louis Arch Problem, UMAP Modules 1983, Tools for Teaching, Unit 638, COMAP, Inc. 1983.

Videotape: Monument To The Dream - The Construction Of The Arch, Jefferson National Expansion Historical Association, Inc., 11 North 4th Street, St. Louis, Missouri, 63102.

Owner's Manual For The Gateway Arch, Saint Louis, MO, by William V. Thayer, (c) 1993, for a basic understanding of the mathematics of the Arch using you calculator.


See also: Geometry, St. Louis Style, by William V. Thayer in CONSORTIUM, The Newsletter of Consortium for Mathematics and Its Applications (COMAP), Number 53, Spring 1995.









Copyright © 1982 through © 2001 with all rights reserved by
William V. Thayer, PedLog.