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OS:Telesis:Data Calculations

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Telesis
Magnet Motor
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Calculations, Data and Graphs

PES Engineer Preliminary Graph

PES Engineer

This graph represents the measurements of positive and negative force readings in 1/4 of revolution. This graph was produced by a PES Engineer.

Degrees Series Series Series Series

from TDC     1       2       3       4 
    -45    0.00    0.00    0.00    0.00  Your 45 deg position in upper quadrant 
    -40   -0.25   -0.31   -0.31   -0.38 
    -35   -0.44   -0.50   -0.63   -0.56 
    -30   -1.02   -1.04   -1.07   -1.08 
    -27   -1.04   -1.09   -1.06   -1.09 
    -25   -2.01   -2.02   -2.09   -2.06 
    -23   -2.05   -2.07   -2.12   -2.14 
    -20   -3.04   -3.06   -3.12   -3.08 
    -19   -3.01   -3.03   -3.08   -3.06 
    -17   -3.01   -3.07   -3.14   -3.15 
    -15   -4.03   -4.07   -4.08   -4.12  Max negative torque 
    -12   -4.02   -4.02   -4.12   -4.12 
    -10   -3.07   -3.05   -4.00   -4.00 
     -7   -3.00   -3.01   -3.09   -3.12 
     -5   -2.02   -2.10   -2.08   -2.09 
     -2   -2.08   -2.10   -2.11   -2.12 
      0    0.00    0.00    0.00    0.00  Your 90 deg position, Top Dead Center of stator A 
      2    1.08    1.09    1.10    1.14 
      5    2.21    2.07    3.03    3.01 
      7    2.11    2.13    2.13    2.15 
     10    3.03    3.07    4.10    4.08 
     12    4.04    4.02    4.01    4.09 
     15    4.07    4.10    4.11    4.12  Max positive torque 
     17    3.12    3.08    3.10    3.09 
     19    3.06    3.02    3.10    3.08 
     20    3.04    3.04    3.14    3.12 
     23    2.08    2.09    2.12    2.09 
     25    2.01    2.00    2.00    2.02 
     27    1.13    1.11    1.14    1.12 
     30    1.01    1.00    1.02    1.01 
     35    0.50    0.44    0.44    0.44 
     40    0.31    0.25    0.25    0.25 
     45    0.00    0.00    0.00    0.00  Your 45 deg position in lower quadrant

From: [A., Engineer] To: richard Cc: Sterling D. Allan Sent: Friday, June 25, 2004 2:14 AM Subject: Power calculation

Richard, Before multiplying by torque to get power, it is necessary to convert RPM (revolutions/minute) to rad/min (radians/minute).

Since 1 revolution = 2*pi rad (Note 3) = 2*3.14 rad (approx.) = 6.28 rad,

that means 1000 rev/min = 6.28*1000 rad/min = 6,280 rad/min.

For the power calculation, IF you have a constant 24 in-oz of torque throughout the complete 360 degree rotation (note 1), and IF you have that torque while running at 1000 RPM (note 2), then 24 in-oz * 6,280 rad/min = 150,720 in-oz/min (shaft power at 1000 RPM) = 2,512 in-oz/sec (min = 60 sec) = 157 in-lb/sec (lb = 16 oz) = 13.083 ft-lb/sec (ft = 12 in) (Note 3) = 0.02378 hp (hp = 550 ft-lb/sec) = 17.74 Watts (hp = 745.7 Watts) = 5.199 BTU/hr (BTU/hour = 3.4121 Watts) (Note 4)


Note 1: It is not likely that the torque is constant throughout the whole rotation. Note 2: Generally, torque changes with speed. You need a dynamometer or a pony brake. Note 3: "pi" is the ratio of circumference to diameter of any circle, pi = 3.14159... Note 4: BTUs are heat energy, and BTU/hour is heat flow

I hope these calcula Regards, [A., PES Engineer]

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