(200, 1000, 650)
SCREAMIN' EAGLE DESIGN AND SPEED
The "Screamin' Eagle" is 3,872 ft. long, and takes 2 minutes and 17
seconds to make one trip. What is the average velocity in Miles Per Hour?
The blueprint indicates that there is a distance of 9 feet between centers
of vertical supports on Eagle Hill. Some horizontal stays are marked with
elevation numbers. Confirm that the distance between centers of the
horizontal stays is 6 feet by completing the following statements.
EAGLE HILL
Elevation of one horizontal stay = ____________________
Elevation of another horizontal stay = ____________________
Distance between these two horizontal stays = _______________
Number of horizontal stays between these two elevations = ____________________
Distance between each horizontal row of stays = _____________
Eagle Hill is the largest hill on the Screamin' Eagle. Use the horizontal and vertical distances between supports and stays as a grid to estimate the vertical distance from the highest point to the lowest point of this section of track.
Vertical Height = _______________
Use the same grid to estimate the horizontal distance from the highest point to the lowest point of the track.
Horizontal Distance = _______________
If we joined the highest point and the lowest point on this hill and valley with a straight line, then we could find the slope of that line and consider it the average slope for the ride.
The Average Slope m = (rise/run) = _______________
If a car fell off the track at the highest point and dropped vertically downward a distance equal to the vertical height then the car's position based on time would be approximated by distance
s = 0.5 g t² .
Where g is 32 feet per second squared in English units or 9.8 meters per second squared in metric units. This equation is Galileo's free fall model with no initial velocity.
Use the free fall model to find how long it would take the car to travel the Vertical Height (above), if any object dropped off the track.
Calculus students can explain how this free fall equation can help to determine the velocity of the car at any point in the time the car is falling. Let them or your teacher show how to derive the velocity equation
v = g t from s = 0.5 g t² .
Use the velocity equation and the time you found in the last exercise to find the velocity of the car if it fell vertically off the track and traveled a distance equal to the Vertical Height. (You may wish to find this answer in miles per hour as well as feet per second.)
Now we may tackle the job of finding an estimation for the position and velocity of the car if it traveled down the Average Slope incline corresponding to the track on Eagle Hill. The free fall equation needs to be adjusted for this trip since 32 feet per second squared is the acceleration constant for free fall and we are now sliding down an incline. Trigonometry students may wish to use the sine function to do this job but algebra students are more familiar with the slope of a line. So we will use Average Slope m (calculated above) to fix up our 32 feet per second squared.
If we run one foot horizontally we would rise m feet vertically and, using the Pythagorean theorem, we would travel the square root of ( 1 + m² ) feet down the Average Slope incline.
The part of 32 feet per second squared acceleration we use to fall m feet is spread over
( 1 + m² )^0.5 feet.
( )^0.5 is just another way of writing "take the square root of ( )" calculation.
Using a proportion for our calculation gives:
Acceleration down the incline is
a = ( m / ( 1 + m² )^0.5) 32 feet per second squared.
For instance, if the average slope were 0.5 then
a = ( 0.5 / ((1 + 0.5² )^0.5))32 = 14.31 feet per second squared.
Calculate a new acceleration value for Eagle Hill's Average Slope.
Eagle Hill's Average Slope Acceleration a = __________
Then the free fall equation would change to the Eagle Hill Average Slope fall equation for distance based on time
s = 0.5 (your value for a) t² ________ (units) or:
Distance s = 0.5 (__________) t² _________ (units),
which simplifies to s = _______ t² ________ (units) and
Velocity v = (your value for a) t which means
v = ______________t _________ (units).
If we use the Horizontal Distance and Vertical Height to find the Distance of the Eagle Hill's Average incline line, we get:
s = ((Horizontal Distance)² + (Vertical Distance)² )^0.5 = ____________ feet.
So the time for this Average Slope incline run would be
t = __________ seconds and
the velocity at the bottom of the Average Slope's incline is
v = __________ (________/sec).
As a comparison, this is __________ miles per hour.
This velocity and the free fall velocity (calculated above) are how related? (and why?)
Some of us wanted to know the value of the acceleration in the steepest part of Eagle Hill. Can you determine that answer and how that would change your estimation for the velocity at the bottom of the hill?
We looked at some video tape in slow motion and found that the lead car traveled the nine foot distance between the vertical posts in about six to seven frames. If the video camera took thirty frames per minute, how slow was the lead car over the hill?
With this initial velocity at the top of Eagle Hill, what is a maximum speed for the car at the next bottom segment of curve?
Some algebra students developed coordinate map equations for sections of track. Can you identify the track near 2x + 5y = 5520?
The Eagle Hill Coordinate Map has the North side on the top and the blueprint is viewed from the North looking South where the Hill is just Northwest of the entrance and exit.
Copyright © 1994 © 1996 with all rights reserved by William V. Thayer, Mathematics Department, St. Louis Community College at Meramec, 11333 Big Bend Blvd., St. Louis, MO 63122-5977, Telephone: 314 984 7866, Email: thayer@stl-online.net
Copyright of all Six Flags' rides and building
names herein belong to Six Flags(R) Theme Park