Power Scissors Dissection
The primary function of the Power Scissors is to cut through different materials. There are two blades, the bottom one stays stationary, and the top blade moves rapidly, opening and closing. This action allows the tool to cut through things.
2) Another stepper motor, undocks the printer carriage. The dock is used to secure the carriage, and to keep the printer head clean.
3) The printer carriage is operated by a drive belt, which is powered by a third stepper motor.
4) The printer cartridge applies ink onto the paper according to the instructions supplied by the CPU through the copper contacts.
The table belows lists the components for the HP Deskjet 600c printer:
|Part #||Part Name||Category||Function||Material||Picture|
|1||Motor||Input||Serves as Power Supply for blades|
|2||External Casing||Structural Component||Protects internal components and holds them together||Plastic|
|3||Jaws||Ouput||The bottom blade remains stationary, while the top blade moves continually up and down||Steel|
|4||Battery||Input||Provide power to motor||Wrapped in cardboard|
|5||Cam||Motion Conversion||This forces the bracket to move by translating rotation motion of the motor, to translational movement in the bracket. The bracket then causes the upper blade to open and close||Plastic|
|6||Screws||Structural||Holds Casing Together||Metal|
|7||Bracket||Support element||Ball and socket joint limits movement of jaws. The socket is ovular so it allows some movement of the jaws.||Plastic|
|8||Button (spring)||Motion Conversion||Closes switch to start motor by closing circuit.||Button – plastic
Spring – metal Switch - metal
Analysis Of The Belt System
Scope of Analysis
The two engineering specifications that are quantified for the printer belt tensioning system, are the force on the belt required to accelerate the printer head to its maximum speed, and the force to stop the printer carriage and change direction. Both of these specifications pertain to the Dots Per Inch design parameter. The best design is to obtain the maximum DPI rating in the quickest printing time. To achieve this goal, the belt must be able to cope with the forces to accelerate the printer carriage.
Friction force exerted by the slider on the carriage is always constant when moving, and should have been reduced to a minimum by the manufacturer. Lower friction would reduce the force the stepper motor has to supply to move the carriage. Since it is difficult to measure the exact amount of friction force, and it is relatively small compared to the acceleration forces, friction can be neglected.
Since, the actual printer head speed I could not be found, a few assumptions had to be made with the available information. The printer has a 300 DPI rating, which means that it can print 90,000 dots per square inch. It can print 4 pages of text per minute. Assuming that it prints 8.5” x 11” pages with a 1” top and bottom margin, and 1.25” side margins, it leaves a total of 54 square inches of text. This equates to 4.86 x 10^6 dots per page. Multiply the dots per page by the pages per minute, and that results in 2.43 x 10^7 dots per minute. 2.43 x 10^7 dots per minute is the average printing rate for the printer. However, if we assume the format is 12 point, Times New Roman font, the total area per line of text is 1.95 in^2 (6.5” x .3”) 0.54 square inches per page multiplied by 4 pages per minute, divided by 1.95 square inches per pass, and taking the reciprocal, yields .009 minutes per line, which is 0.54 seconds per line. Every line is 6.5 inches long, which means that the printer head moves at 12 inches per second. Since information on the acceleration of the printer carriage could not be obtained, it was instead estimated. After careful observation of other inkjet printers, the carriages reach their top speed almost instantaneously, so it was estimated that it takes 0.2 seconds to reach maximum velocity, and 0.25 seconds to stop and change direction.
Acceleration of Carriage = max velocity/ time = (0.3048 m/s) / 0.2 s = 0.15 m/s2
Mass of Carriage With Ink Cartridge = 0.14 kg
F = ma = (0.14)(0.15) = 0.021 N
Change in velocity = (0.3048+.03048) = 0.6096 m/s
Change in Time = 0.25 s
F = (Mass*Change in Velocity)/ Change in time = 0.34 N
To increase the maximum force that the belt could handle, a few options are available. The cross-sectional area could be increased to decrease the stress on the belt. According the stress equation, stress = load / area , the stress can be reduced by increasing the cross sectional area to compensate for load force. The cross sectional area can be increased by making the belt wider or thicker. Making it wider would be more beneficial because it would provide a greater surface area for the toothed pulley to grip onto. The drawbacks for increasing the surface are the increase in stiffness and cost of manufacturing. Making the belt thicker would reduce its flexibility, making it harder for it to move around the pulley. It would require more torque form the input motor to compensate. Making the belt wider would force the pulleys to be wider, which drives up the cost of the system.
Another method to increase the stress capacity of the belt would be to use a curvilinear design instead of the current trapezoidal design. The curvilinear design looks very similar to the gear sprocket of a bicycle. Instead of a trapezoid shape, the curvilinear design uses a half circle shape. Therefore, the teeth are deeper in the gear which makes it less probable for tooth jumping during high accelerations. Furthermore, there is less material at the edges of the gear, which lowers the moment of inertia, allowing the gear to accelerate faster. 
A final option to consider is to change the material of the belt. Using a more durable and stress resilient material such as steel or composites, may work better. However, it is most likely that these materials are more difficult to manufacture, which increases the cost
a. Blades are sharp enough to cut different materials
i. Have a sharpening function
b. Easy to use
i. Has a low weight ii. Has a small size iii. Comfortable to hold iv. Wireless v. Easily accessible buttom to start
c. Aestheically Pleasing
i. More likely to be bought
d. Long battery life, chargeable e. Changeable blades for different materials to cut f. Doesn’t break if dropped
Engineering Specifications 1. Material for jaws is sharper than cutting materials strength. This corresponds to E. 2. Battery Life is ______ long. This correspsonds to D. 3. Force of scissors is greater than _____. This corresponds to A. 4. Weight is less than _______ (5 lbs?). This corresponds to B. 5. Scissors retain sharpness for _______ (amt of time). This corresponds to A. 6. torque on motor is ______. This corresponds to A. 7. External casing can withstand ______ force, to avoid being dropped. This corresponds to F. 8. A certain percentage of people when surveyed said that the scissors are pleasing. This corresponds to C.