Alt-BEAM Archive
Message #13819
To: beam@sgiblab.sgi.com
From: Jeffrey D Spears jspears@engin.umich.edu
Date: Sat, 22 Apr 2000 00:10:24 -0400 (EDT)
Subject: Re: Lift potential for motors?
Greetings Xaphiosis;
Although I am new to BEAM, I am familiar with torque. While I was
taking engineering statics, their came a point in time where I
actually saw (rxF) in my sleep.
The formula for torque (M) is M = rxF. The x is the cross product
of the radius and force. When working in three dimensions, the right
hand rule applies. Luckily, the cross-product reduces to a simple
multiplication when the force is at a right angle to the radius.
Lets say you have a bolt to turn. You stick a wrench on the bolt, and
then you place your hand on the wrench--say 20cm from the center of the
bolt. Next, you apply 2 Newtons of force to the wrench at a right angle
to the bolt. The torque created with this arrangement is radius times
force, or .2m times 2 Newtons, equals .4 Newton-meters.
Now, looking at the motors page in the latest Jameco rag, I see a
stepper motor which claims to produce 144 g-cm. That is, 144 gram
centimeters. This means that if I stick a tiny wrench on this motor
and put my finger on the wrench one centimeter from the center of
the shaft, the motor will exert 144 grams of force on my finger.
Now, I move my finger so that it is 3cm away from the center. The
force on my finger is F = M/r (force equals torque over radius).
Well we see that the force is 48 grams.
Instead of our little pretend wrench, lets say this is the length
of a robot leg attached to this motors' shaft and it is still
3 cm away from the center. The weight this motor, in conjunction
with this *moment arm* is capable of pushing (or lifting) 48 grams.
Think of torque along the same lines as POWER. Where power is
the product of voltage and current, torque is the product of
force and radius. You know how a 2.5V 10mA cell produces 25mWatts.
This power can be expended at 5volts and .5mA, or it can be
expended at 1volt and 2.5mA. Conservation of energy in action.
Now, to put the cross-product into its place, imagine I were trying
to turn that bolt with a wrench. Silly stooge I am, instead of applying
pressure to the wrench at a right angle to the radius, I apply force
directly toward the center of the bolt. If you can imagine this, then
it should generate a chuckle. Anybody knows I will never turn the bolt
if I pull the end of the wrench directly over the top of the bolt. Now
as I figure this out, I gradually change the direction from along the
radius to a point ninety degrees to the radius. As I increase the angle,
then the torque will also increase. The torque resulting from this
constant force (at varying angle) will be at a maximum at ninety
degrees to the radius. To the other extreme, lets say I start with
force on the wrench at ninety degrees to the radius, but then I start
moving out. The torque will decrease as my angle of pull increases
from ninty degrees to 180 degrees. At 180 degrees I will be pulling
the wrench directly away from the bolt, and the bolt will never
turn. If I were a stupid muscle-man, I may manage to shear the top
of the bolt off while trying to turn that bolt 180degrees from
the radius (pulling straight out) -- but I will never turn that bolt
Xaphiosis--I haven't the slightest idea where you are on the
educational ladder. As far as I am concerned, position is not
nearly as important as direction. On your way up the ladder,
take any physics courses that come your way--be it high school
or calc-based--suck em up! The concepts gained will serve you
well!
ok..jef
On Fri, 21 Apr 2000, Xaphiosis wrote:
> Again a silly question ... how do I more or less find out the lifting power
> of a motor
> when connected to an average efficiency propeller?
> Ideally I'd like to know how cheap I can buy a motor that can lift a bit
> more than
> just itself, as my funds are quite limited at the moment :(
>
> Thanks in advance,
>
> X.
>
>
Jeffrey D. Spears
University of Michigan
College of Engineering
``Double-E, can't spell gEEk without it!''
-Captain Gerald M. Bloomfield II, USMC
(my brother)
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