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General motor power concern...

General motor power concern...

Hello all, I am a bit puzzled about some calculations I have done
regarding motor power. First, I found some formulas relating motor
power, rpm, and torque at http://www.patchn.com/motorformula.htm. Now,
as can be seen from this site,

Power [HP] = Torque [ft-lb] x rpm / 5250,

and simple electronics tells us that

Power [Watts] = I [amps] x V [volts].

My motor (Black & Decker 18V cordless drill motor) is rated at 10
ft-lb of torque and 425 rpm's. The first formula above, then, gives
P = .9 HP. This of course assumes that the given torque and rpm occur
simultaneously. It may be, however, that these values are maximums
which then the max power would occur when these values are halved thus,
P = .2 HP.

Now, based on actual experiments, with no load at 18 volts my motor
draws about 4 amps. The second of the above formulas gives P = 72
Watts = .09 HP.

My problem is the considerable difference in the two calculations. I
understand that the specs on the motor given on the Black & Decker web
page may not be precise, but I still feel my results should be a little
closer than they are. The reason for my concern, I am currently
looking for a larger DC motor, and I hear one good source is old
treadmills. Most of these are rated at 1 to 2 HP. My intuition tells
me that a treadmill motor should blow the socks off of a cordless drill
motor, but how can I compare them if I get power calculations anywhere
from about 1 HP to less than 1 tenth?!?! Any
help/advise/expertise/recommendation would be appreciated.

Thanks, Lucas McGill
 
B

BobG

Jan 1, 1970
0
Now, based on actual experiments, with no load at 18 volts my motor
draws about 4 amps. The second of the above formulas gives P = 72
Watts = .09 HP.
===========================
I'd suggest measuring current at full load. Are you using 746 watts=1HP?
 
G

George

Jan 1, 1970
0
I'm not positive so don't take my word as gospel but, two things I can
think of right off are:

The Black & Decker torque and rpms are probably at the chuck, after the
gear reduction, not the actual motor
and
Your applying AC motor formulas to a DC motor.

George
 
B

BobG

Jan 1, 1970
0
But power is torque x speed. Cross check calcs in SI units. One watt is
nt-m/sec. If the battery pack can hump out 18V at 8 amps or 144 watts,
thats about 1/5th of a HP. The motor is probably 80% efficient or more.
 
B

Bob Eld

Jan 1, 1970
0
General motor power concern...

Hello all, I am a bit puzzled about some calculations I have done
regarding motor power. First, I found some formulas relating motor
power, rpm, and torque at http://www.patchn.com/motorformula.htm. Now,
as can be seen from this site,

Power [HP] = Torque [ft-lb] x rpm / 5250,

and simple electronics tells us that

Power [Watts] = I [amps] x V [volts].

My motor (Black & Decker 18V cordless drill motor) is rated at 10
ft-lb of torque and 425 rpm's. The first formula above, then, gives
P = .9 HP. This of course assumes that the given torque and rpm occur
simultaneously. It may be, however, that these values are maximums
which then the max power would occur when these values are halved thus,
P = .2 HP.

Now, based on actual experiments, with no load at 18 volts my motor
draws about 4 amps. The second of the above formulas gives P = 72
Watts = .09 HP.

My problem is the considerable difference in the two calculations. I
understand that the specs on the motor given on the Black & Decker web
page may not be precise, but I still feel my results should be a little
closer than they are. The reason for my concern, I am currently
looking for a larger DC motor, and I hear one good source is old
treadmills. Most of these are rated at 1 to 2 HP. My intuition tells
me that a treadmill motor should blow the socks off of a cordless drill
motor, but how can I compare them if I get power calculations anywhere
from about 1 HP to less than 1 tenth?!?! Any
help/advise/expertise/recommendation would be appreciated.

Thanks, Lucas McGill

Most likely the rated 10 ft-lb of torque is full load stall torque and the
speed, 425 RPM is the no load speed. They do not occur at the same time. The
simplest way to visulize this is by a linear graph with the torque on the y
axis and the speed on the x axis. A straight line connects 425 RPM at 0
torque to 0 RPM at 10 ft-lb. The actual running torque and speed is along
this line. For example, 212RPM would be in the middle and would have a
torque of 5 ft-lb. The peak power occurs here and would be 0.2HP as you have
stated. This in watts is 0.2 * 746 = 149 watts. The current at 18 volts: I =
149/18 = 8.2 Amps, assuming 100% efficiency. Now, you say the motor draws 4
amps with no load which is due to drag in the motor, gear train, brushes,
etc. This drag is always there but is probably about half value at half
speed. Therefore the drag current is guessed at about two amps at 212 RPM.
This gives a total current of about 8.2 + 2 = 10.2 Amps at full power.

I don't see anything inconsistent in this other than understanding that the
delivered power, speed times torque is ZERO when either the speed or the
torque are at their maximum value. In otherwords when one is at full value
the other is at zero.
Bob
 
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