Design Proposal.page11 - closed state. 4.1.2.1 Components...

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P a g e | 11 After the door has been unlocked for 15 seconds, the solenoid 2 will receive power supply and exerted attractive force. The solenoid 2 will no longer push out the door, the springs at the end of the door will bring the door back to the closed state ( figure 9-V ). Since the spring force is only large enough to bring back the door, clients are able to keep the door opening by hold the door without extra work. If the clients want to force the door to be closed within the first 15 seconds, simply push the door to overcome the force from the solenoid 3 and bring the door back to the
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Unformatted text preview: closed state. 4.1.2.1 Components and Installation Instruction of Automatic Open/Close Operation System: Lock solenoid is installed on the door frame while the lock hole is installed on the door since the solenoid will cause the increase of the weight of the door thus the increase of the requirement for the springs is needed. By installing the lock, solenoid and pushbutton on one side of the door beside the empty space of the large outer frame, all the circuit connections can be made through the empty area. 4.2 Circuit Description Fig. 4.2.1 DECODE AND CONTROL CIRCUIT...
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This note was uploaded on 07/12/2011 for the course AER 201 taught by Professor Emami during the Fall '10 term at University of Toronto- Toronto.

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