Neal D. answered • 08/06/16
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30 minutes per revolution clockwise:
A = angle made by positive x-axis and ferris wheel arm
A = 360° / 30 = 12° / min.
Range of seat height is: dm to 135m making the length of arm = 135m - d
Therefor: the radius of arm is (135m - d) / 2
Use a circle to represent the path of the ferris wheel, looking at a right angle
formed by the x-axis and the arm with the car on it,
Assuming the car starts in the bottom most position; then
sin A = H / (135 - d) / 2m; H = distance from x-axis to car, A = -12°t ; t is time in minutes
Multiply both sides of equation by (62.5 - d/2 )
( 62.5 - d/2 )sin (-12t) = H
height above ground = H + (135 - d)/2 + 10 for 7.5 ≤ t ≤ 22.5
and = ( 135 -d ) /2 +10 - H for 0 ≤ t ≤ 7.5 and 22.5 ≤ t ≤ 30
The final equation probably could be done as one, but I had to divide it up for the cars
above the x-axis and those below.
If you have questions, just ask, I will explain.
Neal D.
Jordan,
Let me explain as best I can on this device:
Picture the whole ferris wheel aligned above the x-axis with the y-axis
running down the full length of the diameter (arm)
The x-axis and the (radius) arm running to the car make a right triangle in which
H is the distance from the x-axis to the car
Since the highest car is at 135m and the lowest one is at (d)meters, the length of
the (radius) arm running out to the car = (135 - d ) / 2 ( 62.5 - d/2 )
The ferris wheel completely rotates one full revolution in 30 minutes, which means it
rotates 12° / minute; since the wheel is rotating clockwise these degrees are measured
in negative quantities (counter-clockwise are positive angles)
to get an equation with H in it; using these triangles: Sin (-12t) = H / ( 62.5 - d/2 )
t is time in minutes, the wheel turns -12° every minute; -12t will tell you the angle of
the arm at any time (t minutes)
Multiplying both sides by (62.5 - d/2) yields: (62.5 - d/2) Sin (-12t) = H
For cars above the x-axis, you will have to add the length of the radius (arm) and the
10m the bottom car is above the ground; These would be angles: - 180° ≤ A ≤ - 360°
7.5 min. ≤ t ≤ 22.5 min.
The height above ground for cars at these times would = H + 62.5m + d/2 + 10m
Keep in mind that at t = 0, for the car in the bottom most position, A = -90°
and it is 10m above ground
For cars below the x-axis, you will have to subtract H from the radius and add the
10m above the ground; These would be angles: 0° ≤ A ≤ -180°
22.5 min ≤ t ≤ 30 min.
The height of the cars for this time would be: (62.5m - d/2) - H + 10
I am sure this all could be done in one equation. However, it has been a while since I have
worked a problem like this. I just can't remember how to make it into one equation. Hope this
helps you.
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08/07/16
Jordan H.
Thank you so much! I understand it much better now.
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08/07/16
Neal D.
Jordan,
I have to apologize for one mistake or two:
I refer to finding the height of cars above the x-axis;
This should read: finding height of cars above the horizontal
diameter of ferris wheel; these occur for 7.5 ≤ t ≤ 22.5
and, also
finding height of cars below horizontal diameter of ferris wheel
These heights will occur when 0 ≤ t ≤ 7.5 and when 22.5 ≤ t ≤ 30
Remember they start the initial seat at the bottom most point revolving
in a clockwise rotation
If you need anything explained, let me know.
Neal D.
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08/07/16
Jordan H.
08/07/16