Expert: Paul Klarreich - 7/17/2007

Question
a stadium is to be built with the stands in the form of two concentric ellipses. The outer ellipse is to be 240 yd. long and 200 yd. wide. The inner ellipse is to be 200 yd. long and 100 yd. wide. A football field of standard dimensions 120 yd. by 160 ft. (convert to yds) is to be laid out in the center of the inner ellipse.

- equation of the 2 ellipses?
- eccentricities of the 2 ellipses?
- how much clearance between the corner of the field and the inner ellipse in the direction of the end line?
- the area of an ellipse is "pi"ab, where "a" and "b" are the semi-axis lenghts. To the nearest square yard, what is the area of the stands? If each seat takes an average of .8yd^2, what is the seating capacity of the stadium?

this one's a bear, any help would be much appreciated.

Answer
Questioner:   Brian
Category:  Calculus

Question:  a stadium is to be built with the stands in the form of two concentric ellipses. The outer ellipse is to be 240 yd. long and 200 yd. wide. The inner ellipse is to be 200 yd. long and 100 yd. wide. A football field of standard dimensions 120 yd. by 160 ft. (convert to yds) is to be laid out in the center of the inner ellipse.

- equation of the 2 ellipses?
- eccentricities of the 2 ellipses?
- how much clearance between the corner of the field and the inner ellipse in the direction of the end line?
- the area of an ellipse is "pi"ab, where "a" and "b" are the semi-axis lenghts. To the nearest square yard, what is the area of the stands? If each seat takes an average of .8yd^2, what is the seating capacity of the stadium?

this one's a bear, any help would be much appreciated.
..............................................
Hi, Brian,

(a small bear, perhaps)

In an ellipse, your basic 'dimensions' are:

a = major semiaxis
b = minor semiaxis
c = distance to either focus from the center.
e = c/a, the eccentricity.
a^2 = b^2 + c^2


For the outer ellipse: (let's convert to FEET for everything)

720 feet is the major axis, so  a = 360 = 60(6)
600 feet is the minor axis, so  b = 300 = 60(5)

c = 60 sqrt(36 - 25) = 60 sqrt(11)

e = c/a = 60 sqrt(11)/(60*6) = sqrt(11)/6

Assuming the center of the field is the origin, and the major axes are East-West (along the x-axes)

WARNING: USE A FIXED-SIZE FONT TO VIEW, SUCH AS COURIER.

 x^2       y^2
------- + ------- = 1
360^2     300^2
...........................................
For the inner ellipse:

You said: The inner ellipse is to be 200 yd. long and 100 yd. wide.

600 feet is the major axis, so  a = 300 = 150(2)
300 feet is the minor axis, so  b = 150 = 150(1)

c = 150 sqrt(4 - 1) = 150 sqrt(3)

e = c/a = 150 sqrt(3)/(150*2) = sqrt(3)/2

Assuming the center of the field is the origin, and the major axes are East-West (along the x-axes)

 x^2       y^2
------- + ------- = 1
300^2     150^2
............................................
- the area of an ellipse is "pi"ab, where "a" and "b" are the semi-axis lengths. To the nearest square yard, what is the area of the stands? If each seat takes an average of .8yd^2, what is the seating capacity of the stadium?

Converting to feet as before, each seat is  0.8 (9) sqfeet = 7.2 sq feet.

I assume that the seating area is simply the area between the two ellipses -- outer - inner.

Outer = pi (720)(600)
Inner = pi (600)(300)

Difference = 600pi(420), which you can work out and divide by 9 for the area in square yards, and you can divide (this) by 7.2 for the seating capacity.

............................................

- how much clearance between the corner of the field and the inner ellipse in the direction of the end line?

This part is not totally clear.  The corner of the field is the intersection between an end line and a side line.  Do you mean:


                      A
+----------------------+--->to the ellipse.
|                      |
|                      |
|                      |
|                      |
|                      |
+----------------------+

In that case, you want the horizontal distance between two points:

A = 1/2 of 120 yards horizontally, 1/2 of 160 feet vertically.
A is the point  (180,80) -- feet, as in the others.

So if we move to the right, where do we intersect the ellipse?  Obviously at a point whose y-coordinate is also 80 and which satisfies the equation of the inner ellipse.

That equation was:

 x^2       y^2
------- + ------- = 1
300^2     150^2

Put in  y = 80:

 x^2      (80)^2
------- + ------- = 1
300^2     150^2

 x^2       8^2
------- + ------- = 1
300^2     15^2

 x^2            8^2
------- = 1 -  -------
300^2          15^2

 x^2     15^2 - 8^2
------- = --------------
300^2       15^2


        300^2(15^2 - 8^2)
 x^2  = -----------------
             15^2

 x^2  = 20^2(225 - 64)

 x^2  = 20^2(161)

x = 20 sqrt(161)

So the distance is that number minus  150 feet.

x = 20 sqrt(161) - 150

My calculator gives about 103.77 feet.