System Sensor Strobe Coverage Calculation Examples

File Preview
File Preview
Click below to download for free

File Data

Name system-sensor-strobe-coverage-calculation-examples-5486071293.pdf
Type PDF
Size 822.00 KB
Downloads

Text Preview

Audible Visible Notification Strobe Coverage Calculations Coverage Examples are two examples of the calculations used in determining the correct strobe coverage mearsurements both examples the following variables are applied Height 20 ft Room Size length x width 25 ft x 25 ft of Model 75 cd 1 Distance calculation when the distance the floor is less than 1 2 the room length 2 Distance calculation when the distance along floor is greater than 1 2 the room length 30 UL 1971 Effective Candela 75 x 75cd Candela 56.25 cd 60 UL 1971 Effective Candela 40 x 75cd Candela 30 cd to point in profile to point in profile Height Angle x Room Length COS 90 Angle x ft 30 x 3.14159 180 ft COS 30 x 3.14159 180 ft ft Foot cd ft 2 Foot cd ft 2 lumens square foot lumens square foot 0.106 is greater than 0.0375 lumens square foot UL 1971 requirement 0.144 is greater than 0.0375 lumens square foot UL 1971 requirement Light Intensity following is a description of the mathematics involved in calculating light intensity throughout a rectangular shaped room when the light source is located at the of the ceiling the calculations are simplified by assuming a room first step is to construct the smallest square that encompasses the actual room Then inscribe a circle inside the square room The radius of the circle must be such that all four sides of the square room contact the circle as shown in 1 Distance calculations are then made from 0 degrees right angle to the to 90 degrees parallel to the ceiling in 5 degree increments By assuming room to be square the calculations for one quadrant of the room will apply to all quadrants Utilizing the UL profile requirements Lumens Ft2 can be calculated by the required candela at each angle by the calculated distance at each angle to NFPA 72 the minimum illumination requirement of 0375 lm ft2 at any must be met Figure 1 shows these angles in relationship to the device located the center of the room 07 14

Related Files