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'OT: Power and Force'
2000\05\12@212958 by PDRUNEN

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Hi Group,

If I have a pump or some device that is holding a 100lb object in the air and
the mass is at rest, and I measure the current and voltage to the pump to get
the pump power input.  How can I relate that to the 100lbs?

Say V=12 and I=10A for 120watts.

The question is how does force relate to power given that the object is not
in motion.

Thanks!

2000\05\12@215342 by Sean Breheny

face picon face
Hi,

Unfortunately, you can't. There is no absolute rule that you need a certain
amount of power to provide a given force (in fact, there are many practical
examples of situations where no power is needed. Think about it, you could
put a 100lb object on to of a bookcase and it would hold it up without
requiring any power).

The reason why a pump like this will require energy is because it is
imperfect (i.e., some of the air or whaever leaks out and more air needs to
be repressurized). If you knew how imperfect it was, you could determine
how much power you need, but that requires more information about the pump.

by the way, power=force times velocity. If you have an object which is
moving and you are applying a force, THEN a minimum amount of power is
involved. It sounds like you may have known this part already.

Sean

At 09:20 PM 5/12/00 EDT, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
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2000\05\12@215558 by Damon Hopkins

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face
PDRUNEN@AOL.COM wrote:
>
> Hi Group,
>
> If I have a pump or some device that is holding a 100lb object in the air and
> the mass is at rest, and I measure the current and voltage to the pump to get
> the pump power input.  How can I relate that to the 100lbs?
>
> Say V=12 and I=10A for 120watts.
>
> The question is how does force relate to power given that the object is not
> in motion.
>
> Thanks!

Not sure if this will help BUT 765W ~ 1 HP ~ 550lbs raised 1Ft in 1
Second

2000\05\13@201012 by Anthony Clay

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I am currently taking an Advanced Placement Physics class at School.  Power
is defined as the amount of work performed per period of time.

Work is the force multiplied by the distance an object is moved.  If there
is no movement or activity, no work is performed, and no Power output.

Please clarify what you want me to calculate, in detail.

2000\05\14@220825 by Fansler, David

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face
Anthony - I use to teach physics, and yes by the physic's definition you are
correct. - however if you were to hold 10 lbs at arms length - by the
physic's definition - you are doing no work!  Try and tell your arm that
after a couple of minutes.

David V. Fansler
Network Administrator
TriPath Imaging, Inc. (formerly AutoCyte, Inc.)
.....DfanslerKILLspamspam@spam@autocyte.com <DfanslerspamKILLspamautocyte.com>
336-222-9707 Ext. 261
Now Showing!  http://www.dv-fansler.com <http://www.dv-fansler.com>
Last updated March 31, 2000
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               {Original Message removed}

2000\05\15@002221 by David Huisman

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face
How is that so ?
F = ma , mass of weight and acceleration = 9.81m/s^2 for gravity.
And Work = energy x time

2000\05\15@004717 by Damon Hopkins

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face
David Huisman wrote:
>
> How is that so ?
> F = ma , mass of weight and acceleration = 9.81m/s^2 for gravity.
> And Work = energy x time

I'm not sure what your asking here, but it looks like you forgot to
include e=mc^2.
MAYBE.... Work = mc^2 * time...

Wow, thats alot of work sitting that's been sitting on my desk for the
last 2 days


                       Damon Hopkins
                       Professional Slacker/Student

2000\05\16@101353 by Roger Morella

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face
It seems to me that you should be able to determine force vs power in your system
experimentally by using different weights or heights. It may be possible to come
up with constants that represent the losses of your system this way.  For
instance, how much power is required to hold an 80lb weight?  what about a 120lb
weight?  Or how much power is required to hold a 100 lb weight at 1/2 the height?
or 1 /12 times the height?  You may find that the losses of the system are
proportional to the force that you need to hold an object at different heights.
Depending on what you are trying to accomplish, you may be able to develop a set
of practical calculations by doing some fairly simple experimentation.   In any
case you need to do some more experimentation to determine the practical
variables in your system.

.....PDRUNENKILLspamspam.....AOL.COM wrote:

{Quote hidden}

2000\05\16@104453 by PDRUNEN

picon face
Thanks for your reply,

Here is some more detail, I have a motor which has a motor current sensor.  I
use a motor to raise weights at a constant velocity.  While the weights are
moving I record the motor current sensor value.

As I add more weight, in increments of 5lbs, I look for a linear increase in
the current sensor value.  The value I am reading is in voltage as it is a
drop across a sense resistor, so I expect it to be somewhat linear.  The
value of the sense resistor voltage is very small so I have to amplify it
with an op-amp which will add some error with offsets, etc.  I am interested
in the error as the weight load increases.

Now I have a relationship between the voltage and current, so I can convert
the voltage back to motor current.  I also have the torque curve of the motor
which I can go back to and see what the torque value is for the measured
current.

But what I would like to do, if from the torque curve of the motor, select a
motor current value and from that determine what my current sensor would read
if I was moving say 15lbs or 20lbs without doing any experimentation.  So, I
am adding a downward force of 15lbs, 20lbs, and looking at the change in
current.  I would expect I could take the torque curve and do the same.

Thanks,

Paul


In a message dated 5/16/00 9:13:58 AM EST, EraseMErjmspam_OUTspamTakeThisOuTBUFFNET.NET writes:

<< It seems to me that you should be able to determine force vs power in your
system
experimentally by using different weights or heights. It may be possible to
come
up with constants that represent the losses of your system this way.  For
instance, how much power is required to hold an 80lb weight?  what about a
120lb
weight?  Or how much power is required to hold a 100 lb weight at 1/2 the
height?
or 1 /12 times the height?  You may find that the losses of the system are
proportional to the force that you need to hold an object at different
heights.
Depending on what you are trying to accomplish, you may be able to develop a
set
of practical calculations by doing some fairly simple experimentation.   In
any
case you need to do some more experimentation to determine the practical
variables in your system. >>

2000\05\17@165148 by Peter L. Peres

picon face
Hi,

now, I can see no more pumps here, and no air.

To measure the weight being lifted at constant speed by a mechanism, using
the input power to the mechanism, you have to consider the fact that the
efficiency of the mechanism+motor system is not constant as the weight
changes. This is more than the torque, it relates to the particular kind
of mechanism, the way friction appears in bearings, and between gear faces
and more. With a small load the efficiency is likely going to be poor on
account of the motor and friction losses, then it will be good for medium
load, and then it will be poor with high load as elastic deformation
begins to increase friction and bearing rolling losses.

The only way out is to calibrate the whole system using a prototype. This
is best a final product taken off the assembly line because there are SO
many things that can change the curves. From experience I can say that the
effects I have quoted above will affect your theoretical read-out (based
on the motor curves) by between 5 and 30% (for low loads).

My experience is based on overload sensing circuitry in equipment that is
probably similar to what you are doing.

hope this helps,

       Peter

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