Solving for the Height and Width of St. Louis Arch: A Calculus Problem

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SUMMARY

The discussion focuses on calculating the height and width of the St. Louis Arch using calculus principles. The height of the arch is determined to be 630 feet, derived from the center of the highest triangle and the area of equilateral triangles formed by the arch's cross-section. The calculations utilize the hyperbolic cosine function, specifically the equation 693.8597 – 68.7672 cosh(0.0100333x). However, discrepancies arise in determining the width at ground level, which is expected to be 630 feet, but the calculations yield a width of 307.7239 feet, indicating a potential error in the approach.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically cosh
  • Familiarity with calculus concepts, including area and center of mass
  • Knowledge of equilateral triangle properties and area calculations
  • Ability to solve equations involving inverse hyperbolic functions
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  • Review hyperbolic functions and their applications in geometry
  • Study the properties of equilateral triangles and their centroid calculations
  • Learn about inverse hyperbolic functions and their usage in solving equations
  • Explore calculus applications in real-world structures, focusing on arch design
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Students studying calculus, architects, engineers, and anyone interested in the mathematical principles behind structural design, particularly in relation to the St. Louis Arch.

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Homework Statement
The St. Louis Arch was constructed using the hyperbolic cosine function. The equation used for construction was y=693.8597-68.7672 cosh 0.0100333x, -299.2239<=x<=299.2239, x and y in feet. Cross sections of the arch are equilateral triangles, and (x,y) traces the path of the centers of mass of the cross-sectional triangles. For each value of x, the Area A of the cross-sectional is A = 125.1406 cosh 0.0100333x.
Relevant Equations
y=693.8597-68.7672 cosh 0.0100333x, -299.2239<=x<=299.2239, x and y in feet. (1)
Area A of the cross-sectional is A = 125.1406 cosh 0.0100333 (2)
(a) How high above the ground is the center of the highest triangle? (At ground level, y=0)
Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925

(b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of the triangle and the center of mass of the triangle is located at two-thirds the height of the triangle.)
From equation (2) above, I found the area A of the highest cross-sectional triangle as A = 125.1406 cosh (0) = 125.1406. Then used A = 125.1406 = sqrt (3) c^2 to solve for c and got c=8.5. Then found the height of the highest cross-section triangle as h = tan (60) x (8.5) = 14.72. The height of the arch is then 625.0925 + (1/3) 14.72 = 625.0925 +4.0975 =630 feet. This checks with Arch height on web, so believe (a) and (b) are correct.

(c) How wide is the arch at ground level? Here's where I ran into problems.
I reasoned that the center of mass equation is satisfied for y=0 -> 0 = 693.8597 - 68.7672 cosh 0.0100333 x -> 68.7672 cosh .0100333x = 693.8587 -> cosh .0100333x =693.8587/68.7672 = 10.09. Then I took the inverse cosh of both sides of the equation and got x=2.9941. The substituted back into Area function A = 125.1406 cosh (0.0100333(2.9941) = 125.1406 cosh (1.0045) = 125.1971. Solving for c, 125.71 = sqrt (3) c^2 -> c=8.5. If I add that to 299.2239, I get 307.7239. Web lists the width as 630 feet, so half width is 315. I'm off by over 7 feet. Any help appreciated.
Note: This problem is from Calculus, 8th ed, by Larson et al, that I'm using for home study)
 
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Can you please make it more sparse, to make it more readable? Like breaking it into paragraphs.
 
St. Louis Arch:

The St. Louis arch was constructed using the hyperbolic cosine function. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x. -299.2239<=x<=299.2239. (1) Cross sections of the arch are equilateral triangles, and (x,y) traces the path of the centers of mass of the cross-sectional triangles. For each value of x, the area of the cross sectional triangle is A = 125.1406 cosh .0100333x. (2)

(a) How high above the ground is the center of the highest triangle? (At ground level, y=0) Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925

(b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of the triangle and the center of mass of the triangle is located at two-thirds the height of the triangle.) From equation (2) above, I found the area A of the highest cross-sectional triangle as A = 125.1406 cosh (0) = 125.1406. Then used A = 125.1406 = sqrt (3) c^2 to solve for c and got c=8.5. Then found the height of the highest cross-section triangle as h = tan (60) x (8.5) = 14.72. The height of the arch is then 625.0925 + (1/3) 14.72 = 625.0925 +4.0975 =630 feet. This checks with Arch height on web, so believe (a)and (b) are correct.

(c) How wide is the arch at ground level? Here's where I ran into problems. I reasoned that the center of mass equation is satisfied for y=0 -> 0 = 693.8597 - 68.7672 cosh 0.0100333 x -> 68.7672 cosh .0100333x = 693.8587 -> cosh .0100333x =693.8587/68.7672 = 10.09. Then I took the inverse cosh of both sides of the equation and got x=2.9941. The substituted back into Area function A = 125.1406 cosh (0.0100333(2.9941) = 125.1406 cosh (1.0045) = 125.1971. Solving for c, 125.71 = sqrt (3) c^2 -> c=8.5. If I add that to 299.2239, I get 307.7239. Web lists the width as 630 feet, so half width is 315. I'm off by over 7 feet. Any help appreciated. Note: This problem is from Calculus, 8th ed, by Larson et al, that I'm using for home study)
 
Can you please make it more sparse, to make it more readable? Like breaking it into paragraphs.
 
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gjh said:
St. Louis Arch:

The St. Louis arch was constructed using the hyperbolic cosine function. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x. -299.2239<=x<=299.2239. (1) Cross sections of the arch are equilateral triangles, and (x,y) traces the path of the centers of mass of the cross-sectional triangles. For each value of x, the area of the cross sectional triangle is A = 125.1406 cosh .0100333x. (2)

(a) How high above the ground is the center of the highest triangle? (At ground level, y=0) Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925

(b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of the triangle and the center of mass of the triangle is located at two-thirds the height of the triangle.)

gjh said:
From equation (2) above, I found the area A of the highest cross-sectional triangle as A = 125.1406 cosh (0) = 125.1406. Then used A = 125.1406 = sqrt (3) c^2 to solve for c and got c=8.5. Then found the height of the highest cross-section triangle as h = tan (60) x (8.5) = 14.72. The height of the arch is then 625.0925 + (1/3) 14.72 = 625.0925 +4.0975 =630 feet. This checks with Arch height on web, so believe (a)and (b) are correct.

(c) How wide is the arch at ground level? Here's where I ran into problems. I reasoned that the center of mass equation is satisfied for y=0 -> 0 = 693.8597 - 68.7672 cosh 0.0100333 x -> 68.7672 cosh .0100333x = 693.8587 -> cosh .0100333x =693.8587/68.7672 = 10.09.

gjh said:
hen I took the inverse cosh of both sides of the equation and got x=2.9941. The substituted back into Area function A = 125.1406 cosh (0.0100333(2.9941) = 125.1406 cosh (1.0045) = 125.1971. Solving for c, 125.71 = sqrt (3) c^2 -> c=8.5. If I add that to 299.2239, I get 307.7239. Web lists the width as 630 feet, so half width is 315. I'm off by over 7 feet. Any help appreciated. Note: This problem is from Calculus, 8th ed, by Larson et al, that I'm using for home study)