Designing A Freight Elevator For The Movement Of Goods Business Essay - Free Essay Example

Sample details Pages: 10 Words: 3059 Downloads: 5 Date added: 2017/06/26 Category Engineering Essay Type Analytical essay Did you like this example? There are many types of elevators like hydraulic elevator, Pneumatic vacuum elevator, mine shaft elevator etc. Freight elevator is an elevator which carries freight. A  freight elevator  is used to do just what its name implies: to elevate, or lift,  freight, or goods. Don’t waste time! Our writers will create an original "Designing A Freight Elevator For The Movement Of Goods Business Essay" essay for you Create order 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  [2] 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 d isplacement 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 ÃÆ' ¢Ãƒ ¢Ã¢â€š ¬Ã‚ ¡Ãƒ ¢Ã¢â€š ¬Ã¢â€ž ¢ 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 min utes 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 ÃÆ' ¢Ãƒ ¢Ã¢â€š ¬Ã‚ ¡Ãƒ ¢Ã¢â€š ¬Ã¢ „ ¢ 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 a t 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 ele vator 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/min ute2. 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 ascen ds 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 t he 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.  [3] 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 symmetri cal 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 transl ated 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) = = ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦.(1) Therefore , Velocity v = = and, ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦..(2) Acceleration a = ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦(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 a t 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) = = ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦.(1) Therefore, Velocity v = = and, ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚ ¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦..(2) Acceleration a = ÃÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦ÃƒÆ' ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒâ€šÃ‚ ¦(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) = mete rs/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.

Theftby Katherine Anne PorterFollow 10 Members - 1550 Words

Theft by Katherine Anne Porter Follow 10 Members Following Summary Themes Analysis More ââ€" » Theft by Katherine Anne Porter Analysis Style and Technique (Comprehensive Guide to Short Stories, Critical Edition) print Print document PDF list Cite link Link â€Å"Theft† is a unique short story in the Porter canon for several reasons. It is the first effort at incorporating autobiographical elements into her work. Porter developed an intense relationship with Matthew Josephson, her literary mentor and lover. His wife, after discovering the affair, told him to choose between them. Josephson chose his wife and wrote Porter a letter detailing the decision and the fervent hope they could continue working together and remain friends. Porter was†¦show more content†¦The curtain of the story rises, and Porter’s protagonist emerges from her bath to see that her â€Å"gold cloth† purse is no longer on the bench where she spread it out to dry the night before. As she recalls the previous evening, trying to discover when it may have gone missing, we learn that she has been robbed several times either â€Å"material[ly] or intangib[ly]†(85), but not of the purse. The subtle yet critical thefts of the night before take place, as so many petty thefts can, under the guise of friendship. As they leave a cocktail party together, her friend, Camilo, insists on walking her through the rain to the Elevated and in doing so ruins his hat; still, she thinks, he will â€Å"associate her with his misery† (79), as if his offer to walk her through rain puts her at fault. Roger, another artist friend, spots her on the steps to the Elevated and offers to take a taxi with her, but then borrows ten cents, a quarter of all the money she possesses, to pay his fare. Once in her apartment building, she runs into a playwright who owes her money for writing the third act of his play, but he won’t give it to her;

Digital Fortress Chapter 18 Free Essays

Standing before the huge plate-glass window in his Tokyo skyrise, Numataka took a long pull on his cigar and smiled to himself. He could scarcely believe his good fortune. He had spoken to the American again, and if all was going according to the timetable, Ensei Tankado had been eliminated by now, and his copy of the pass-key had been confiscated. We will write a custom essay sample on Digital Fortress Chapter 18 or any similar topic only for you Order Now It was ironic, Numataka thought, that he himself would end up with Ensei Tankado’s pass-key. Tokugen Numataka had met Tankado once many years ago. The young programmer had come to Numatech Corp. fresh out of college, searching for a job. Numataka had denied him. There was no question that Tankado was brilliant, but at the time there were other considerations. Although Japan was changing, Numataka had been trained in the old school; he lived by the code of menboko-honor and face. Imperfection was not to be tolerated. If he hired a cripple, he would bring shame on his company. He had disposed of Tankado’s resume without a glance. Numataka checked his watch again. The American, North Dakota, should have called by now. Numataka felt a tinge of nervousness. He hoped nothing was wrong. If the pass-keys were as good as promised, they would unlock the most sought-after product of the computer age-a totally invulnerable digital encryption algorithm. Numataka could embed the algorithm in tamper-proof, spray-sealed VSLI chips and mass market them to world computer manufacturers, governments, industries, and perhaps, even the darker markets†¦ the black market of world terrorists. Numataka smiled. It appeared, as usual, that he had found favor with the shichigosan-the seven deities of good luck. Numatech Corp. was about to control the only copy of Digital Fortress that would ever exist. Twenty million dollars was a lot of money-but considering the product, it was the steal of the century. How to cite Digital Fortress Chapter 18, Essay examples

Art for Heart`s Sake Interpretation Essay Example For Students

Art for Heart`s Sake Interpretation Essay The text under interpretation is a short story Art For Heart`s Sake by Reuben Garrett Lucius Rube Goldberg (July 4, 1883 – December 7, 1970), who was an American cartoonist, sculptor, author, engineer, and inventor. He is best known for a series of popular cartoons depicting complicated gadgets that perform simple tasks in indirect, convoluted ways. Goldberg was a founding member and the first president of the National Cartoonists Society, and he is the namesake of the Reuben Award, which the organization awards to the Cartoonist of the Year. Goldberg produced several series of cartoons all of which were highly popular. Among his best works are Is There a Doctor in the House? , Rube Goldberg`s Guide to Europe and I Made My Bed. The given extract introduces the readers to a story of a wealthy man called Collins Ellsworth who is treated for a disorder which causes irresistible desire of buying things, more commonly referred to as compulsive shopping or oniomania. The man`s obsessive condition leads him to unfortunate results and health problems. Knowing this, Mr Caswell, his doctor, suggests trying some art therapy. This therapy seems to be effective at first, and Caswell`s client even evinces interest to painting and the general drift of affairs in art galleries. subsequently Ellsworth executes a nonsensical unskillful work, which, to everybody`s astonishment, he later exhibits at the Lathrop Gallery. That picture gets the 1st prize of the Exhibition. The story ends with a revelation of the method that helped the old man win the prize. From the point of view of its composition, the story may be divided into four parts. The first part is an introduction starting with a dialogue between the old man and the male nurse. Which annoys the latter very much as the patient is a very disagreeable man, who refuses to follow doctor`s orders. Here we get acquainted with Ellsworth with his inherent disrespect for the people around him. Whether it is a male nurse or a doctor, who are with him in order to help, the businessman pulls no punches. It was not the first outburst of Ellsworth. So, instead of trying to demand an apology, the doctor suggests a new way of getting rid of old man`s problems. That`s where the 2nd part of the story starts. In this part we get to know that doctor considers busying his patient with art to be a way out. The idea of Caswell is to bring a young student Frank Swain to the patient. When Swain arrives to Ellsworth`s house, he starts the therapy, suggesting that the man should try to paint a vase. The old grump starts unwillingly. The first attempts do not meet success, but the practice leads to progress as the vase on the painting gradually develops resemblance to the one on the mantelpiece. Ellsworth asks Swain for more hours and seems to forget about his obsession. He becomes curious about what`s going on in art galleries. Some idea arises in his head. The third part of the extract is connected with the events in the Lathrop Gallery. There was an exhibition in it, which as a lifetime dream of the mature artists. And the newly-made artist Ellsworth aimed at showing his amateurish awkward painting there. The male nurse, Koppel, is sure that if it happens, the old man will become a laughing stock. But the doctor orders him to forbear from interfering in Ellsworth`s business not to ruin their achievements. The painting is accepted, but luckily for the worried fellows, it hangs in a corner where people can barely see it. This part also tells us about the events that took place two days before the closing of an exhibition. Swain, Koppel and the doctor witness a strange follow-up to the story. the old man receives a letter which surprises the people even more: the First Prize of the Lathrop Exhibition has been awarded to Ellsworth`s painting Trees Dressed in White. Euopean Art In The Wake Of World War 1 EssayHe would dwell on the rich variety of colour in a bowl of fruit, he proudly displayed the variegated smears of paint on his heavy silk dressing gown. He would not allow his valet to send it to the cleaner`s. he wanted to show the doctor how hard he`d been working). As Ellsworth gets acquainted with the world of the Museums and Galleries he`s being caught by its charming mysteries. The revolution of the seasons is portrayed with the wonderful metaphor: the spring sun cloaks the fields and gardens with colour. The old man`s idea, as he reveals in his startling announcement, is taking part in the exhibition. In order to do that he creates an ugly picture, which imperfection is described with the epithet god-awful smudge. That exhibition is a lifetime dream of the mature artists, so old man`s participation in it was shocking and out of place. Trees Dressed in white was not just bad. As the simile from the text imparts, it is like a salad dressing thrown violently up against the side of a house. The male nurse wants to stop Ellsworth, not to let him become a laughing stock, but the doctor insists on letting him finish, as they`ve got too far to lose everything. Luckily, the place where the picture hangs is inconspicuous. The metaphor a raucous splash on the wall reminds us about picture`s ugliness. It`s all is followed up by the ironies, which retell that the masterpiece was noticed by the people and as some guys stopped next to the Ellsworth`s strange anomaly, Swain fled in terror. When the letter from the Gallery arrives, Ellsworth requests somebody to read it, as his eyes, ironically, are tired from painting. The news is announced: the Gallery gives Ellsworth the first prize. The reaction of the characters, excluding the old man is quite predictable. While Koppel and Swain, overwhelmed by surprise, try to regain their ability to speak, the doctor congratulates the old grump, which is not easy, or, as the epithet says, is a supreme effort. In the end the reader gets to know that there was no miracle in Ellsworth`s win, that the old man owns a Gallery now. His last remark (Art`s nothing †¦ I bought the Lathrop Gallery last month) crowns the story, proving the point that nothing has changed in him.