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# Designing a Freight Elevator for the Movement of Goods Business Essay

There are many types of elevators like hydraulic elevator, Pneumatic vacuum elevator, mine shaft elevator etc. Freight elevator is an elevator which carries freight. “AA freight elevatorA is used to do just what its name implies: to elevate, or lift,A freight, or goods. It is built to carry goods rather than people”. “The smallest freight elevators are often called dumbwaiters”. “They are typically used in two-story buildings to move household goods such as laundry or dishes up and down. Though older versions were operated by pulling on a rope, modern dumbwaiters include a small electric motor”. “A heavy-duty freight elevator can hold a truck and can handle as much as 100,000 pounds using a dual rope system for support”  In this activity, we will first explore a possible model for the motion of a heavy-duty freight elevator used to raise and lower equipment and minerals in a mineshaft. We will evaluate the model for its strengths and weaknesses and then create a set of specifications to develop a model of our own.

## Analyzing a possible model:

The possible model y = 2.5t3 – 15t2 represents the position of the elevator. Here ‘y’ is in meters and is the vertical displacement. Here y = 0 when the elevator is at ground level. ‘t’ is time in minutes and the starting time t = 0 minutes. The trip up and down the shaft, ignoring time spent at the foot of the shaft, is approximately six minutes and that the depth of the shaft is no more than 100 meters. I will use graphing software “Graph 4.3″ to visualize the motion of the elevator in the shaft which is shown below. displacement.png Graph : Displacement graph using the given model function I have calculated the position of the elevator at different time intervals which I have shown in the table below: Time t (minutes) Displacement (meters) 0 0.0 1 -12.5 2 -40.0 3 -67.5 4 -80.0 5 -62.5 Table : Displacement I have calculated the maximum and minimum displacement during the round trip of the elevator. The maximum / minimum displacement of the freight elevator during the round trip will be when the slope of the displacement function = 0 that is when velocity v = 0 Therefore = 7.5t2 – 30t = 0 A¢”¡’ t = 0 or 4 minutes. Therefore y (0) = 2.5(0)3 – 15(0)2 = 0 y(4) = 2.5(4)3 – 15(4)2 = -80 meters. Hence maximum displacement, ymax = 0 and minimum displacement, ymin = -80 meters. From the above table and graph we can observe the following facts. At time t = 0 and 6 minutes, the displacement is 0. This indicates that the elevator is at ground level at the start and the end of the round trip. Throughout the round trip, from t = 0 to 6 minutes, the displacement is negative. This indicates that the elevator is below the ground level which is evident from the graph also. The elevator descends to a maximum depth of 80 meters which is the ymin value at time t = 4 minutes. The time taken by the elevator to descend the maximum depth of 80 meters is 4 minutes whereas the time taken to ascend the same distance is 2 minutes. Since displacement y(t) = 2.5t3 – 15t2 meters We know that velocity is rate of change of displacement. Therefore Velocity = = y'(t) = 7.5t2 – 30t meters/minute Using graphing software I have graphed the velocity function which is shown below. velocity.png Graph : Velocity graph using the given model function I have calculated the velocity of the elevator at different time intervals which I have shown in the table below: Time t (minutes) Velocity v (meters/minute) 0 0.0 1 -22.5 2 -30.0 3 -22.5 4 0.0 5 37.5 6 90.0 Table : Velocity I have calculated the maximum and minimum velocity during the round trip of the elevator. The maximum / minimum velocity of the freight elevator during the round trip will be when the slope of the velocity function = 0 that is when acceleration a = 0 Therefore =15t – 30 = 0 A¢”¡’ t = 2 minutes. Therefore velocity is minimum at t = 2 minutes. v(2) = 7.5(2)2 – 30(2) = -30 meters / minute Since = 15. Therefore v is minimum at t = 2 minutes. v(2) = 7.5(2)2 – 30(2) = -30 meters/minute Therefore the minimum velocity of the elevator during the round trip is -30 meters/minute. From the above table and graph we can observe the following facts. When t = 0 and 4 minutes, the velocity is 0 meters/minute. It indicates that the elevator is at rest at the bottom of the shaft and at the ground level when it starts descending. During time interval 0 to 4 minutes the velocity is negative. This indicates that the elevator is moving downwards. During time interval 4 to 6 minutes, the velocity is positive. This indicates that the elevator is moving upwards. The minimum velocity is -30 meters/minute at time t = 2 minutes. We know that acceleration is rate of change of velocity. Therefore, Acceleration = = y”(t) = 15t – 30 meters/minute2 Using graphing software I have graphed the acceleration function which is shown below. accn.png Graph : Acceleration graph using the given model function I have calculated the acceleration of the elevator at different time intervals which I have shown in the table below: Time t (minutes) Acceleration (meters/minute2) 0 -30.0 1 -15.0 2 0.0 3 15.0 4 30.0 5 45.0 Table : Acceleration From the above table and graph we can observe the following facts. When t = 2 minutes, the acceleration is 0 whereas before t = 2 minutes it is negative and after t = 2 minutes it is positive. This indicates that the acceleration is changing its direction at time t = 2 minutes. During time interval t = 0 to 2 minutes, the acceleration is negative. This indicates that acceleration is in the downward direction. During time interval t = 2 to 6 minutes, the acceleration is positive. This indicates that acceleration is in the upward direction. I have graphed all the three displacement, velocity and acceleration functions together for comparison which is shown below. displ,vel.accn together.png Graph : Displacement, Velocity and Acceleration graphs using the given model function I have also combined all the values of displacement, velocity and acceleration at different time intervals for a comparative study as shown below. Time t (minutes) Displacement (meters) Velocity v (meters/minute) Acceleration a (meters/minute2) 0 0.0 0.0 -30.0 1 -12.5 -22.5 -15.0 2 -40.0 -30.0 0.0 3 -67.5 -22.5 15.0 4 -80.0 0.0 30.0 5 -62.5 37.5 45.0 6 0.0 90.0 60.0 Table : Displacement,velocity and acceleration. I have analyzed all the three displacement, velocity and acceleration together during the three time intervals as below. Between 0 to 2 minutes. During this time interval at t = 0 minutes, the displacement is 0 meters and at time t = 2 minutes, the displacement is -40 meters. This indicates that the elevator is at ground level at the starting of the time interval and descends 40 meters by the end of this time interval which is also the half of the downward displacement of the elevator. During this time interval at t = 0 minutes, the velocity is 0 meters/minute and at time t = 2 minutes, the velocity is -30 meters/minute which is the minimum velocity of the elevator in the full trip. In this time interval the magnitude of the velocity is increasing. The negative values of velocity indicate that the elevator is moving downwards. During this time interval at t = 0 minutes, the acceleration is -30 meters / minute2 and at time t = 2 minutes, the acceleration is 0 meters / minute2. This indicates that the magnitude of the acceleration is decreasing in this time interval. In this time interval, the direction of both velocity and acceleration is same and downwards hence the acceleration helps the elevator to speed up and the magnitude of the velocity increases. Between 2 to 4 minutes. During this time interval, at t = 4 minutes, the displacement is -80 meters. This indicates that the elevator is at the bottom of the round trip at the end of the time interval. Therefore during this interval of 2 to 4 minutes; the displacement is -40 meters. The elevator descends 40 meters during this time interval which is also the half of the downward displacement of the elevator. During this time interval at t = 2 minutes, the velocity is -30 meters/minute and at time t = 4 minutes, the velocity is 0 meters/minute. Therefore the elevator reaches its lowest point of the trip at time t = 4 minutes. In this time interval the magnitude of the velocity is decreasing. The negative values of velocity indicate that the elevator is moving downwards. During this time interval at t = 2 minutes, the acceleration is 0 meters / minute2 and at time t = 4 minutes, the acceleration is 30 meters/minute2. This indicates that the magnitude of the acceleration is increasing in this time interval. The acceleration is positive during this time interval which indicates that the acceleration is upwards. In this time interval, both velocity and acceleration are in opposite direction. The acceleration is in the upward direction whereas the velocity is in downwards direction and the magnitude of the velocity is decreasing during this time interval. This indicates that, when the acceleration is in the opposite direction of the velocity, the velocity decreases. Between 4 to 6 minutes. During this time interval, at t = 6 minutes, the displacement is 0 meters. This indicates that the elevator is at the bottom of the round trip at the starting of the time interval and is at the ground level at the end of this time interval. Therefore during this interval of 4 to 6 minutes; the displacement is 80 meters. The elevator ascends 80 meters during this time interval which is the full ascends of the elevator. Therefore the elevator takes only 2 minutes to move from the lowest point of the trip to the ground level. During this time interval at t = 4 minutes, the velocity is 0 meters/minute and at time t = 6 minutes, the velocity is 90 meters/minute. The velocity is positive during this time interval and the elevator is moving upwards. Therefore the elevator reaches the ground level again at time t = 6 minutes and completes its trip. In this time interval the magnitude of the velocity is increasing. During this time interval at t = 4 minutes, the acceleration is 30 meters / minute2 and at time t = 6 minutes, the acceleration is 60 meters/minute2. This indicates that the magnitude of the acceleration is increasing in this time interval. The acceleration is positive during this time interval which indicates that the acceleration is upwards. In this time interval, both velocity and acceleration are in the same direction. Both the velocity and acceleration are in the upward direction. The magnitude of the velocity is increasing during this time interval. This indicates that, when the both velocity and acceleration are in the same direction, the magnitude of the velocity increases.

## Conclusion:

When both the velocity and accelerations are of the same sign that is they are in the same direction. The acceleration helps velocity and the magnitude of the velocity increases. When the velocity and the accelerations are of different signs that is they are in the opposite directions. The acceleration acts as retardation and decreases the magnitude of the velocity.

## Usefulness of the model function:

The possible model function y = 2.5t3 – 15t2 models the complete round the trip motion by a single function.

## Problems in the model function:

The main drawback of the given model function is that the elevator does not come to rest at the end of the trip when it returns back to the ground level. The velocity is 90 meters/minute at t = 6 minutes whereas the velocity should be 0 meters/minute. As per the model function, the elevator takes 4 minutes to descend to the bottom and only 2 minutes to ascend back to the ground level. This indicates that it takes double the time in descend compared to the time required to ascend. Since this is a vertical motion and gravitational force provides a gravitational acceleration downwards which always helps any object to move downwards and opposes any object to move upwards. Therefore the gravitational acceleration is helping the elevator while descending and opposing the elevator while ascending but the elevator takes half the time while ascending which means that the source of force required to take the elevator up has to be more powerful. This increases the cost of the equipment and operation. Using this model function the elevator can descend only up to a depth of 80 meters whereas the depth of the mineshaft is 100 meters. So the elevator can never reach the bottom of the mine shaft using this model function.

## Creating our own model:

I tried to gather information for the speeds of freight elevators from the internet. The speeds of the heavy duty freight elevators manufactured by ‘Hitachi Ltd., Japan’ are 30, 45 metres/minute.  I have selected the maximum speed of freight elevator for my model as 45 meters/minute. Therefore the specification for redesigned freight elevator model is: The depth of the mine shaft the freight elevator has to descend is 100 metres. The maximum speed of the freight elevator is 45 metres/minute. The freight elevator takes equal time in descending and ascending from ground level to the depth of the mine shaft. The speed of the freight elevator is 0 meters/minute at the ground level, whether the freight elevator is starting from or reaching the ground, and at the bottom of the shaft. Looking at the above specifications, especially The freight elevator takes equal time during descending and ascending, which means that the displacement function graph is symmetrical during ascend and descend. The elevator velocity is zero at the beginning, middle and at the end of the trip. The function which can suit the above requirements can be a sinusoidal cosine function. The cosine function starts from the maximum value ‘1’ and ends at the maximum value ‘1’. Also the cosine function is at its lowest value ‘-1’ at the middle of its cycle. Also the slope of the cosine function at these three points is zero and is symmetrical; hence it satisfies all the requirements of the model function. Therefore I will try a cosine function. The general cosine function is given by y(x) = A cos (Bx + C) + D. Here, ‘A’ is the amplitude of the cosine function which is the difference between the peak and mean of the maximum and minimum y coordinates. ‘B’ is given by B = where T is the time period and is the distance between the two successive maxima or minima. ‘C’ is the horizontal translation of a normal cosine graph. This is negative when the graph is translated to the right and is positive when the graph is translated to the left. Its value is the horizontal shift of the point cosine (0) from the origin (0, 0). ‘D’ is the vertical translation of the mean of the normal cosine function given by the mean of maxima and minima of ‘y’ value. ‘D’ is negative when the graph is translated downwards and is positive when the graph is translated upwards. Since I have taken the maximum depth of descend of the elevator to be 100 meters from the ground level. Therefore ymax = 0 and ymin = -100, therefore Amplitude, A= = 50 B = Horizontal translation, C = 0 Vertical translation, D = Substituting these values of ‘A’, ‘B’, ‘C’ and ‘D’ in the general equation of the cosine function I get the model displacement function as below: y(t) = = A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦.(1) Therefore, Velocity ‘v’ = = and, A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦..(2) Acceleration ‘a’ = A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦…(3) Now in the above model equations the only unknown is the time period or we can say the time required for the elevator one round trip. In our specification of the model function I have specified that the maximum magnitude of the velocity of the elevator will be 45 meters/minute. Now the velocity ‘v’ can be maximum when sin = -1. Therefore maximum velocity, Vmax = = 45 meters/minute Therefore T = 6.98 minutes. Therefore our model displacement, velocity and acceleration functions will be Displacement, y(t) = meters Velocity v(t) = meters/minute Acceleration a(t) = meters/minute2 I have graphed the above model functions for visual check. cosine displ.bmp Graph : Remodeled displacement graph cosine velocity.bmp Graph : Remodeled velocity graph cosine accn.bmp Graph : Remodeled acceleration graph I have calculated the values of displacement, velocity and acceleration at different time intervals which is shown in the table below. Time ‘t’ (minutes) Displacement 50 cos(0.9t) – 50 (meters) Velocity – 45 sin () (meters/minute) Acceleration – 40.5 cos () (meters/minute2) 0.00 0.00 0.00 -40.50 1.75 -50.00 -45.00 0.00 3.49 -100.00 0.00 40.50 5.24 -50.00 45.00 0.00 6.98 0.00 0.00 -40.50 Table : Displacement, velocity and acceleration as per the remodeled function From the above graphs and table we can see that the velocity is 0 meters/minute at time, t = 0, 3.49 and 6.98 meters when the elevator is at ground, bottom of the shaft and ground level again. The elevator descends a depth of 100 meters at t = 3.49 minutes which is half of the time taken for round trip. Therefore the elevator descends to the bottom of the mine shaft. The displacement graph is symmetrical and the time required for descend and ascend is 3.49 minutes which means that the elevator takes equal time in descend as well as ascend. The remodeled function is single function, which take cares of all the drawbacks of the original model function. Hence I accept this remodeled function as my final model function.

## I will modify my remodel functions for a high speed passenger elevator.

From the same web site of Hitachi limited, I gathered that the speed of passenger elevators is 180, 210 and 240 meters/minute. I will design a displacement, velocity and acceleration function for a passenger elevator with the following specification. The height the passenger elevator has to ascend is 240 meters and the maximum speed of the elevator is 240 meters/minute. Using this specification, I will remodel my function as below. Amplitude, A= = 120 B = Horizontal translation, C = Vertical translation, D = Substituting these values of ‘A’, ‘B’, ‘C’ and ‘D’ in the general equation of the cosine function I get the model displacement function as below: y(t) = = A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦.(1) Therefore, Velocity ‘v’ = = and, A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦..(2) Acceleration ‘a’ = A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦A¢â‚¬A¦…(3) Now in the above model equations the only unknown is the time period or we can say the time required for the elevator one round trip. In our specification of the model function I have specified that the maximum magnitude of the velocity of the elevator will be 240 meters/minute. Now the velocity ‘v’ can be maximum when sin = 1. Therefore maximum velocity, Vmax = = 240 meters/minute Therefore T = 3.14 minutes. Therefore our model displacement, velocity and acceleration functions will be Displacement, y(t) = meters Velocity v(t) = 240 meters/minute Acceleration a(t) = meters/minute2 I have graphed the three functions for visual analysis as below. cosine displ paasenger.bmp Graph : Displacement graph for passenger elevator cosine velocity passenger.bmp Graph : Velocity graph for passenger elevator cosine acceleration passenger.bmp Graph : Acceleration graph for passenger elevator Form the above graphs we can see that with slight modifications of the model cosine function, it meets all the requirements of the high speed passenger elevator. So we can modify the cosine function so that the same can be adapted to any other similar application.

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Designing A Freight Elevator For The Movement Of Goods Business Essay. (2017, Jun 26). Retrieved December 1, 2021 , from

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