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Tue Jul 13, 2004 10:21 am
:?: I need help, who can suggest me that how to measure the output of automatic door?. According to Lift regulation, it says that "The kinetics energy of landing door and the mechanical elements which are rigidly connected to it, measured at average closing speed shall not exceed 10J"
Tue Jul 13, 2004 8:05 pm
Joules is a measure of work... How much weight is the door pulling? How much does the door weigh?
Wed Aug 04, 2004 9:25 pm
The formula for kinetic energy is ½mv², where m = mass (in kg) & v = speed (in m/s). This will give you an answer in Joules (J). I would imagine that the mass to use would be that of the door plus "the mechanical elements which are rigidly connected to it".
Thu Aug 05, 2004 4:00 am
Mike Wright wrote:The formula for kinetic energy is ½mv², where m = mass (in kg) & v = speed (in m/s). This will give you an answer in Joules (J). I would imagine that the mass to use would be that of the door plus "the mechanical elements which are rigidly connected to it".
This would allow a 489 lb (222 kg) door to close at one foot per second (0.3 m/s) and deliver less than 10J of energy (9.99J)
1/2(222 kg·(0.3 m/s)²)=9.99
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