J
Jon Slaughter
- Jan 1, 1970
- 0
Does anyone know what kinda motors that most drills use? I was thinking it
was a torque motor but not sure.
Also, for DC motors can I get away with changing the voltage to a lower
setting to increase the torque or do I run a huge risk of burning up the
motor?(i.e., are they usually designed for a very narrow operating voltage).
I have this old broken 18v drill and I want to turn it into opening a gate.
I figure that it has enough torque to do it and I can control the speed
quite easily. I plan on using some gears so I can open and close it with a
change of polarity. I'm a little worried that it might have to much torque
or that it might be to fast/slow for the gate(which the gears have a lot to
do with of course).
Just wondering how much I can play around with the voltage to get what I
want if I need too. Also, I assume that P is pretty much constant for
motors? i.e., if the rated power for a drill is P then P = I*V for a large
range of I and V? So if I know its a "100w" drill at 20V then I can
calculate the current at 15 voltages = 100/20 A = 5A and it should be
approximately what the drill will pull? (obviously the load has a lot to do
with it but just curious as to some simple approximation)
Thanks,
Jon
was a torque motor but not sure.
Also, for DC motors can I get away with changing the voltage to a lower
setting to increase the torque or do I run a huge risk of burning up the
motor?(i.e., are they usually designed for a very narrow operating voltage).
I have this old broken 18v drill and I want to turn it into opening a gate.
I figure that it has enough torque to do it and I can control the speed
quite easily. I plan on using some gears so I can open and close it with a
change of polarity. I'm a little worried that it might have to much torque
or that it might be to fast/slow for the gate(which the gears have a lot to
do with of course).
Just wondering how much I can play around with the voltage to get what I
want if I need too. Also, I assume that P is pretty much constant for
motors? i.e., if the rated power for a drill is P then P = I*V for a large
range of I and V? So if I know its a "100w" drill at 20V then I can
calculate the current at 15 voltages = 100/20 A = 5A and it should be
approximately what the drill will pull? (obviously the load has a lot to do
with it but just curious as to some simple approximation)
Thanks,
Jon