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math portfolio type 2 (eleator)

Extracts from this document...

Introduction

Hl type 2

Math portfolio 2

Designing a freight elevator

image00.png

Candidate name: lokesh baid

Candidate Number:2279-014

Session: 2007-2009

Supervisor: Mrs. Jyoti Uppal

Math’s Portfolio-Designing a Freight Elevator

In this portfolio I will attempt to design a freight Elevator model. Primarily I will study the given model to assess the usefulness and the strengths and weaknesses This will allow me to then form conditions for designing an effective and usefulelevator model.

Therefore I will now study the motion of the elevator by the given displacement equation :image01.png

Displacement

Checking to interpret straight line motion for time (t) =0,1,2,3,4,5,6

Substituting t=0                

image02.png

Substituting t=1

image49.png

Substituting t=2

image60.png

Substituting t=3

image69.png

Substituting t=4

image80.png

Substituting t=5

image90.png

Substituting t=6

image101.png

Displacement/Time Graph

image112.jpg

Velocity

As velocity is the change in displacement at the given time, we differentiate the displacement equation to obtain an equation to give us the velocity of the elevator at a given time.

Therefore :

image123.png

We now substitute r values of  t = 0,1,2,3,4,5,6 to obtain the velocity at the given times

Substituting t=0

image03.png

Substituting t=1

image14.png

Substituting t=2

image25.png

Substituting t=3

image35.png

Substituting t=4

image42.png

Substituting t=5

image44.png

Substituting t=6

image45.png

Velocity/Time graph

image46.jpg

Acceleration

As acceleration is the change in velocity in a given time, we differentiate the velocity equation to obtain an equation to give us the acceleration of the elevator at a given time.Therefore:

image47.png

We now substitute r values of  t = 0,1,2,3,4,5,6 to obtain the velocity at the given times

Substituting t=0

image48.png

Substituting t=1

image50.png

Substituting t=2

image51.png

Substituting t=3

image52.png

Substituting t=4

image53.png

Substituting t=5

image54.png

Substituting t=6

image55.png

Accerleration/Time graphimage56.jpg

Interpretation

...read more.

Middle

as f=0 and g=0

image66.png

Substituting condition 4. t=3, v=0

image67.png

as f=0

image68.png

Substituting condition 5. t=6, s=0

image70.png

image71.png

As f=0

image72.png

Substituting condition 6. t=6, v=0

image73.png

image74.png

As f=0

image75.png

Substituting condition 7. t=6, a=0

image76.png

image77.png

Analysis

As there are 5 variables: a, b, c, d, e and 5 equations with the variables. I have simultaneously equated them using technology.

Variables

Values

a

image78.png

b

image79.png

c

image81.png

d

image82.png

e, f, g

image31.png

After substituting the values found above for a, b, c, d, e, f and g in the general equation. The model that I have final obtained is:

For Displacement (s) as a function of time

image83.png

For Velocity (v) as a function of time

image84.png

For Acceleration (a) as a function of time

image85.png

I will now substitute the values of the variables in the equations below for t=0,1,2,3,4,5,6 to find the desired values for:

  • Displacement
  • Velocity
  • Acceleration

Displacement (s)

When t=0

image86.png

image87.png

When t=1

image88.png

image89.png

When t=2

image91.png

image92.png

When t=3

image93.png

image94.png

When t=4

image95.png

image96.png

When t=5

image97.png

image98.png

When t=6

image99.png

image100.png

image102.jpg

Velocity (v)

When t=0

image103.png

image104.png

When t=1

image105.png

image106.png

When t=2

image107.png

image108.png

When t=3

image109.png

image110.png

When t=4

image111.png

image113.png

When t=5

image114.png

image115.png

When t=6

image116.png

image117.png

image118.jpg

Acceleration (a)

When t=0

image119.png

image120.png

When t=1

image121.png

image122.png

...read more.

Conclusion

Therefore:

Ideal Table

Time

Displacement (s)

Velocity (v)

Acceleration (a)

1

0

0

2

50

0

3

0

0

0

Therefore we will use 6 degree equation as there are 7 variables needed for the 7 equations.

For  

Displacement (s):

image07.png

For

Velocity (v):

image08.png

For

Acceleration (a):

image09.png

I will now use these equations to find out the desired horizontal movement of the elevator.

Substituting condition 1. t=0, x=0

image10.png

image11.png

Substituting condition 2. t=2, x=50

image12.png

image13.png

Substituting condition 3. t=4, x=0

image15.png

image16.png

Substituting condition 4. t=0,v=0

image17.png

image18.png

Substituting condition 5. t=2, v=0

image19.png

image20.png

Substituting condition 6. t=4, v=0

image21.png

image22.png

Substituting condition 6. t=4, a=0

image23.png

image24.pngimage26.png

As f and g are 0 all the equations become 5 variables and I have solved them using technology.

Therefore:

Variables

Values

a

image27.png

b

image28.png

c

image29.png

d

image30.png

e

image31.png

Therefore the general equations of the horizontal model are:

For

Displacement (s):

image32.png

For Velocity (v):

image33.png

For Acceleration (a):

image34.png

By substituting the t=0, 1,2,3,4 into displacement, velocity and acceleration general equations, the value of the displacement, velocity and acceleration is found at different times.

Time

Displacement (s)

Velocity (v)

Acceleration (a)

0

0

0

image26.png0

1

image36.png

image37.png

image38.png

2

image30.png

0

image39.png

3

image36.png

image40.png

image38.png

4

0

0

0

Now the elevator is designed to have vertical motion and horizontal motion. For Dual motion in x and y , the equation will be represented in parametric form as independent equations.

Therefore horizontal motion:

image41.png

Therefore vertical motion:

image05.png

Through this model we see that the lift can move horizontally as well as vertically. The 2 equations have been kept independent as this allows more flexibility in the movement of the Elevator. The horizontal motion of the lift can be used for unloading at 50 m displacement or for picking up people. This model has now become useful for the intended purpose and has also allowed a horizontal motion to be incorporated when needed. Therefore the elevator model is successfully formed.

image43.jpg

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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