On Fri, 03 Oct 2003 12:28:57 -0000, Chuck wrote:
The compressor currently has a 3/4 hp motor. At first glance I was kind of
depressed thinking "Gee, just about every compressor I see these days has 3
+ hp. Will this thing even work well?" In asking around I was told that
amperage plays a big part in the actual torque of the motor. This is where
I get confused... I thougth torque was the direct product of the motor's
hp. How does the amperage come into play? Can you have a "strong" or
"weak" 3/4 hp motor? What factors actually determine the torque? Or, am I
looking at this equation in the wrong way?

You're looking at things just fine. Most of the consumer grade compressors
you see today have a rating called "peak HP". What that really means is that
they lock the rotor and measure the peak current drawn as the windings
smoke, then calculate a theoretical HP figure from that. It is totally bogus.
HP is *zero* with a locked rotor.

The figure you want is actual continuous load running HP. That's what
your old motor gives you on its nameplate. (You can trust that figure
because the given maximum current draw, which calculates out to an
input power of 2.2 HP, if it were all real and not reactive, confirms they
aren't doing marketing magic with it.)

Where reading the current spec is useful is to cut through the marketing
bull****. Output power must always be less than input power (no machine
is 100% efficient). Input power is voltage times current, in your case
115 * 14.4, which yields 1656 watts, or 1656/746=2.2 HP. Now since
some of that current is reactive, and the 14.4 figure is a maximum instead
of a sustained value, real continuous load HP will be less than 2.2 HP.
Conservatively, probably based on allowable heat rise in the windings,
they're saying it is 0.75 HP, which is very believeable.

Very few (I'm tempted to say none) of the current crop of consumer
grade compressors will quote you a continuous running HP. They'd
call the compressor you've got *at least* a 2 HP unit, and probably
2 or 3 times that much. They lie. But at least now you have a basis
for understanding their lies, and a way to derate their claims to at
least the input power of their compressors. (Figuring actual efficiency
and allowable temperature rise in the motor windings to arrive at
a true continuous running horsepower figure is a bit tougher.)

Another way to compare is to look at the air delivery. A reasonably
well built single stage compressor will deliver about 3 SCFM @ 90PSI
per horsepower. Pitfalls here are that some advertising doesn't use
SCFM, but rather an ill defined CFM or "free air displacement" figure
instead. Also, some will give the air volume at a lower pressure (40 PSI
is common).

The figure given is rarely the continuous delivery rate, ether. They'll
give a duty cycle of less than 100%. That's because they're running
the system harder than temperature rise in the windings (and pump)
would allow it to be run on a continuous basis. So derate by the
duty cycle to get a true comparison.

Note that your motor can produce more than 0.75 HP for short
periods. So you could do like the consumer compressor marketers
do, and claim it to be larger than it is if you allow it to cool every
few minutes rather than running it continuously. This will allow
you to draw higher volumes of air for short periods (a big tank
lets you do that too).

So conservatively, by consumer marketing standards, you have
at least a 2 HP compressor. You should be able to draw 6 SCFM
at 90 PSI from the tank for at least short periods. That's enough
to run many air tools which are only operated in short bursts.

Gary

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:
Torque bears no relation to any capacity to do work.

Well, it has a relationship. Power is defined as the capacity to do work,
and torque times RPM is power. But power is the independent variable
here. Torque and RPM can be traded off in any fashion as long as their
product equals the input power.

Although the two
quantities are expressed in the same units of force and distance, torque is a
vector quantity and work is scalar one.

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The question of torque is less ambiguous. It is the product of the applied
force, and the length of the lever arm. If the latter is a simple scalar,
a vector times a scalar is always a vector. Also, since the lever arm is at
right angles to the force, if it were treated as a vector, the cross product of
a unit length and a unit force would be zero, which would be inconsistent
with the notions of torque, though it would be consistent with the notion
of torque not being the equivalent of work.

The lever arm's length doesn't fit the definition of the distance used to
calculate work given above (not aligned with the direction of the force).
So we can say that by definition torque is not the same thing as work
regardless of whether we decide work is vector or scalar.

Note too that the torque on a shaft is the same whether the shaft is
turning or stationary. For a torque to do work, the shaft has to turn,
to provide a distance over which the force can act, and then we have
to bring in the notions of RPM and power to give us a rate and capacity
for doing work.

Gary

On Sat, 04 Oct 2003 14:26:36 GMT, "Bob Swinney" wrote:
Richard J Kink sez:

"...No, you said, "Torque is the capacity of an engine to do work", which is
just wrong..."

Uh huh. And would you care to comment on exactly how that statement is
wrong? Torque is a turning moment that defines the force applied to a shaft
in order to do work. Without torque the shaft cannot turn, work cannot be
done.

Mechanical work is defined as the product of the applied force and the
resulting load displacement (distance) *in the direction of the applied force*.

(see Machinery's Handbook, page 92 of the 24th edition, or any good
elementary physics book)

Torque is defined as the product of the applied force and the length of
the lever arm against which it acts. The latter is at *right angles* to the
applied force, and hence the product is *not* work according to the
definition of mechanical work.

You can apply any arbitrary torque and still not do any work, consider
pulling on a wrench applied to a stuck fastener. No matter how hard
or long you pull, you've done no work (in the physics sense) until the
fastener moves.

Power determines the rate of doing work, and over any given period,
the capacity of the system to do work. (We're implicitly assuming a
prime mover fuel tank sufficiently large to maintain that power input
over the given period, of course.)

Note that it is implicit in the definition of power that the shaft *is*
turning. Unlike the case of torque, displacement at some rate is
required, otherwise no work will be done.

Gary

In article , Gary Coffman says...

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too.

Exactly correct. Torque is defined as the vector cross product of
a force and a distance, R X F, that is, R(vector) (cross product) F(vector).

The R(vector) in this case is from the center of motion to the point
at which the force is applied, and the F(vector) of course points
in the direction the force is applied, which is tangential in
most practical cases. The Torque vector as a cross product winds
up pointing along the axis of the shaft by convention. You need
to use one of those right hand rule things, put left hand in pocket,
point fingers in direction of R, curl them around in the direction
of F, and your thumb points in the direction of T.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The question of torque is less ambiguous. It is the product of the applied
force, and the length of the lever arm. If the latter is a simple scalar,
a vector times a scalar is always a vector. Also, since the lever arm is at
right angles to the force, if it were treated as a vector, the cross product of
a unit length and a unit force would be zero, which would be inconsistent
with the notions of torque, though it would be consistent with the notion
of torque not being the equivalent of work.

Although from a dimensional standpoint the units of work and torque
happen to be the same, torque is of course not a unit of work.

Angular velocity has units of inverse seconds (the r in rpm is
not a real unit) so you wind up with units of force x distance x 1/second
when you muliply torque times rpm. Which looks sort of like joules/sec
or watts or hp. IIRC that mulitiplication is actually a dot
product.

The lever arm's length doesn't fit the definition of the distance used to
calculate work given above (not aligned with the direction of the force).
So we can say that by definition torque is not the same thing as work
regardless of whether we decide work is vector or scalar.

Yep.

Note too that the torque on a shaft is the same whether the shaft is
turning or stationary. For a torque to do work, the shaft has to turn,
to provide a distance over which the force can act, and then we have
to bring in the notions of RPM and power to give us a rate and capacity
for doing work.

That's right, similar to the 'stuck wrench' analogy I gave before.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

"Dave Baker" wrote

Just the top of a flat cap and a few stray locks of hair visible over the
lip
of the grave now. At regular intervals a shovel blade appears and another
clod
of earth flies over the edge and lands on the growing pile. As we get
closer
the sound of muttering and grumbling becomes audible from down below; it's
a
very warm day. Crickets chirp....


Nicely done.

Mark

Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 01:22 GMT Daylight Time
Message-id:

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:
Torque bears no relation to any capacity to do work.

Well, it has a relationship. Power is defined as the capacity to do work,
and torque times RPM is power. But power is the independent variable
here. Torque and RPM can be traded off in any fashion as long as their
product equals the input power.

Although the two
quantities are expressed in the same units of force and distance, torque is
a
vector quantity and work is scalar one.

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to

wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never zero.
The equation has to be balanced by the total expenditure of energy consumed in
doing the work.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.

On Sat, 04 Oct 2003 18:00:01 -0400, Gary Coffman
wrote something
.......and in reply I say!:

I am not so sure about the extremity of the problem of power claims,
unless the US is now getting even worse than here in OZ (and these
things all come from Asia anyway, right? G).

I have a "2.5HP" direct drive comp, quite cheap, that delivers about
7CFM at 90 PSI. This gels with the 3 CFM/HP claim. So it is _not_
claiming to be a 2.5 HP machine, when actually it's only say 1/2 HP
usable, with a stalled current to make it 2.5.

So I would be careful about assuming that this 3/4 HP motor will
deliver as much as a modern "2HP" motor or whatever.

On Fri, 03 Oct 2003 12:28:57 -0000, Chuck wrote:
The compressor currently has a 3/4 hp motor. At first glance I was kind of
depressed thinking "Gee, just about every compressor I see these days has 3

You're looking at things just fine. Most of the consumer grade compressors
you see today have a rating called "peak HP". What that really means is that
they lock the rotor and measure the peak current drawn as the windings
smoke, then calculate a theoretical HP figure from that. It is totally bogus.
HP is *zero* with a locked rotor.


************************************************** ****************************************
Those who can, do. Those who can't, teach.
The rest sit around and make snide comments.

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/

On Sun, 05 Oct 2003 00:56:09 -0400, Tom Quackenbush wrote:
On Sat, 04 Oct 2003 20:22:54 -0400, Gary Coffman
wrote:
SNIP
Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.
SNIP
I don't know.

The way I learned, with w=mv^2, work must be scalar because we're
squaring the velocity.

That's not obvious. Ke = (mv^2)/2, and it isn't obvious that work should
always be twice kinetic energy. Kinetic energy and work aren't the same
thing, though they share the same units. A meteor in orbit around the Sun
has kinetic energy, but is doing no work by simply traversing its orbit.

Work is defined as the product of a force and the displacement it causes.
So

w = F * d

And

F =m * a

So we can substitute and get

w = m * a * d

in units, that equates to

kg * m/s^2 * m

which can be simplified to

kg * (m/s)^2

which has the same units as your formulation.

It arises from a different cause, and may differ numerically,
ie the 'm' in 'a' and the 'm' in 'd' may have different values
(they almost certainly will in most calculations), so you can't
just square a single numerical value for 'm' in reality.

This is like the case of torque, where the units are the same,
but we already know torque isn't work, and that torque *is*
a vector. Kinetic energy also has the units of work, but a
body with kinetic energy most certainly has a direction, so
is it vector or scalar? I'm afraid we can't tell by the way the
units work out.

Gary

On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 01:22 GMT Daylight Time
Message-id:

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:
Torque bears no relation to any capacity to do work.

Well, it has a relationship. Power is defined as the capacity to do work,
and torque times RPM is power. But power is the independent variable
here. Torque and RPM can be traded off in any fashion as long as their
product equals the input power.

Although the two
quantities are expressed in the same units of force and distance, torque is
a
vector quantity and work is scalar one.

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never zero.
The equation has to be balanced by the total expenditure of energy consumed in
doing the work.

That *sounds* good, but I'm not sure it holds up to analysis. For example,
if I raise a 1 kilogram mass 1 meter, I do work, requiring energy to accomplish.
But if I then let the mass return to its original position, I regain the energy.
So we have a situation where there has been a zero net change in energy.
Doesn't that also require that negative work was done so that net work is
also zero? Otherwise, we seem to have a violation of the conservation laws
because we'd have net work with no net energy expenditure.

Gary

Gary Coffman wrote:

On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
... However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units ...

... negative work was done so that net work is
also zero? ...

The problem here is that work is a RATE (of energy). You can't, or
don't want to, talk about the conservation of a rate (work). The "...
work to the object by moving it ..." is meaningless. You want to talk
about the ENERGY to move the object. I could use 10,000 HP to move it
one direction and .01 HP to move it back.

Bob

Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 08:53 GMT Daylight Time
Message-id:

On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 01:22 GMT Daylight Time
Message-id:

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:
Torque bears no relation to any capacity to do work.

Well, it has a relationship. Power is defined as the capacity to do work,
and torque times RPM is power. But power is the independent variable
here. Torque and RPM can be traded off in any fashion as long as their
product equals the input power.

Although the two
quantities are expressed in the same units of force and distance, torque
is
a
vector quantity and work is scalar one.

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it

a vector too. If I recall my elementary physics correctly, it doesn't
matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer
is
the same, but one resultant is vector and the other scalar. So we are left
to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each
to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same
direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never
zero.
The equation has to be balanced by the total expenditure of energy consumed
in
doing the work.

That *sounds* good, but I'm not sure it holds up to analysis. For example,
if I raise a 1 kilogram mass 1 meter, I do work, requiring energy to
accomplish.
But if I then let the mass return to its original position, I regain the
energy.
So we have a situation where there has been a zero net change in energy.
Doesn't that also require that negative work was done so that net work is
also zero? Otherwise, we seem to have a violation of the conservation laws
because we'd have net work with no net energy expenditure.

I was anticipating that reply and if it hadn't been so late last night I was
going to mention it in advance. The reason being is you had alluded to it in
the bit about energy states at the top of your earlier post. Potential energy
states and work musn't be confused. You don't do work to an object when it
falls against gravity, the earth does. Whether you arrange to have that energy
recovered, hydroelectricity etc, doesn't affect the amount of work that got
done during the process. Work isn't something that gets conserved, only energy
is.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.

Subject: Compressor Motor: HP v.s. Amps?
From: Bob Engelhardt
Date: 05/10/03 14:47 GMT Daylight Time
Message-id:

Gary Coffman wrote:

On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
... However, If we do 10
units of work to the object by moving it in a particular direction and
then
another 10 moving it back again the net work is always 20 units ...

... negative work was done so that net work is
also zero? ...

The problem here is that work is a RATE (of energy).

Work is not a rate of energy. There is no time element in the measure of work.
It is simply force times distance. HP is a rate of doing work.

You can't, or
don't want to, talk about the conservation of a rate (work). The "...
work to the object by moving it ..." is meaningless. You want to talk
about the ENERGY to move the object.

You can talk about either. Work done is a valid measure as is energy consumed
in doing that work. They are not necessarily the same thing though depending on
the efficiency of the process.

I could use 10,000 HP to move it
one direction and .01 HP to move it back.

So it would take longer to move it back than it did to move it there. I'm not
quite sure what your point is.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.

In article , Gary Coffman says...

... Kinetic energy also has the units of work, but a
body with kinetic energy most certainly has a direction, so
is it vector or scalar?

The velocity is defined as a vector.

The Ek is a scalar. The V(vector) times V(vector) in
1/2 mV(sq) is a dot product.

Kinetic energy of a moving object and the work
done to produce that motion are defined as the
same thing - and yes the units are the same
as they should be. This comes under the general
heading of 'conservation of energy.'

Sure there are other hidden, parasitic losses, friction,
etc, but when you account for them all, the
balance sheet works out.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , Tom Quackenbush says...

The way I learned, with w=mv^2, work must be scalar because we're
squaring the velocity.

Right, and that product is a *dot* product of two
vectors. The answer then by definition is a scalar.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , Old Nick wrote:
On Sat, 04 Oct 2003 18:00:01 -0400, Gary Coffman
wrote something
......and in reply I say!:
I am not so sure about the extremity of the problem of power claims,
unless the US is now getting even worse than here in OZ (and these
things all come from Asia anyway, right? G).

I have a "2.5HP" direct drive comp, quite cheap, that delivers about
7CFM at 90 PSI. This gels with the 3 CFM/HP claim. So it is _not_
claiming to be a 2.5 HP machine, when actually it's only say 1/2 HP
usable, with a stalled current to make it 2.5.

So I would be careful about assuming that this 3/4 HP motor will
deliver as much as a modern "2HP" motor or whatever.

i have read the thread but want to state here that _any_ two hp motor
needs 220-240vac to run. you cannot get it with 110-120 in _normal_
environments. special wiring/breakers might get you close, but that
would require planning (in which case you would plan 220) or retrofit.

--Loren

In article , Dave Baker says...

Work is not a rate of energy. There is no time element in the measure of work.
It is simply force times distance. HP is a rate of doing work.

Of course this is true.

Work and energy have the same units. In MKS this would be joules.

The time rate of work is Power. Here that would be joules/sec,
or watts, or hp.

I think some of the confusion in this thread is from
mixing up the nomenclature.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

Dave Baker wrote:

Work is not a rate of energy. ...

Oh, right ... never mind. Mouth/fingers engaged before brain up to
speed.

Old Nick wrote ...

... My figuring. ...

Great analysis and advice Nick. Thanks much.

Daniel

On 6 Oct 2003 11:28:38 -0700, (Marcus)
wrote something
.......and in reply I say!:

Nurries. Hope it's _right_! G

Old Nick wrote ...

... My figuring. ...

Great analysis and advice Nick. Thanks much.

Daniel

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On 5 Oct 2003 07:45:54 -0700, jim rozen wrote:
Kinetic energy of a moving object and the work
done to produce that motion are defined as the
same thing - and yes the units are the same
as they should be. This comes under the general
heading of 'conservation of energy.'

But we have just been told (in the parts you snipped) that
work is w = mv^2 and kinetic energy is Ke = (mv^2)/2. So
they aren't the same, by a factor of 2. If both equations
are correct, we need an explanation for the factor of 2
difference.

(Explanation of that could go a long way toward resolving
this issue.)

But back to the original question, why is mechanical work
calculated with the dot product w = F dot D instead of the
cross product w = F X D? I've heard some handwaving
about the conservation laws (and I'm a firm believer in
them), but no specific derivation which requires dot over
cross for this calculation.

However, I did a bit of digging and came up with another
defining equation for mechanical work in Van Nostrand's
Scientific Encyclopedia. (Sorry for the kludged nomenclature,
I don't have an integral sign on this keyboard.)

W = [definite integral from a to b] F cos(theta) ds

Were W is work, a and b are the endpoints for the
distance over which work is done, F is force, cos(theta)
is the angle of the force with the direction of work (ie
yields the component of the force in the direction work is
being done), and s is unit displacement.

Note that W is capitalized, which is the normal way to
denote a vector when an over-arrow is not available in
the typeface used. Note also that an integral is really
a repeated sum taken to a limit. The sum of any number
of vectors is still a vector.

So, while I *know* energy isn't supposed to be a vector
quantity, though an energy flux *is*, I still don't have a
satisfactory mathematical answer why mechanical work,
which certainly has to have a direction while it is being
done, is not.

Note, I *believe* it to be true that work is not a vector
quantity, what I'm asking for is some form of at least
semi-rigorous derivation showing why that must be so.

The dog ate my homework is not a satisfactory response.

Gary

I don't think my previous "Thank You" post made it into the thread. Thanks
to all that replied.

Gary, your answer (and the answer of a couple of other people) was the
'simple' answer I was looking for.

Much of the physics dissertation now ongoing in the thread is way above me.
grin Glad others are enjoying it. g

Thanks!
Chuck

Gary Coffman wrote in
:

Where reading the current spec is useful is to cut through the
marketing bull****. Output power must always be less than input power
(no machine is 100% efficient). Input power is voltage times current,
in your case 115 * 14.4, which yields 1656 watts, or 1656/746=2.2 HP.
Now since some of that current is reactive, and the 14.4 figure is a
maximum instead of a sustained value, real continuous load HP will be
less than 2.2 HP. Conservatively, probably based on allowable heat
rise in the windings, they're saying it is 0.75 HP, which is very
believeable.

In article , Gary Coffman says...

On 5 Oct 2003 07:45:54 -0700, jim rozen wrote:
Kinetic energy of a moving object and the work
done to produce that motion are defined as the
same thing - and yes the units are the same
as they should be. This comes under the general
heading of 'conservation of energy.'

But we have just been told (in the parts you snipped) that
work is w = mv^2 and kinetic energy is Ke = (mv^2)/2. So
they aren't the same, by a factor of 2. If both equations
are correct, we need an explanation for the factor of 2
difference.

I missed that. Work done on an object is 1/2 mv(sq).
But that's a conservation of energy thing, from a
force vector standpoint, its F(vector) x D(vector) x cos (theta)
where the cosine of the angle is there because only
the component of displacement in the direction of the
force counts.

But back to the original question, why is mechanical work
calculated with the dot product w = F dot D instead of the
cross product w = F X D?

Because it's only the component of the work in the
direction of the displacement that adds to the energy
of the particle.

The dot product is just the magnitute of each vector
times the cosine of the angle between them. For
example, if you push down really hard on a roller
skate, but it only moves slowly along the floor, it's
only the component of your hand's force *parallel*
to the floor that is actually pushing the skate to
speed it up. There the value of the dot product is
small as it should be.

I've heard some handwaving
about the conservation laws (and I'm a firm believer in
them), but no specific derivation which requires dot over
cross for this calculation.

OK, I managed to get a copy of halliday and resnick and
looking it over, it looks like they use the term "speed"
when calculating 1/2 mv(squared), not the vector velocity.

They proceed to derive an example of the "work - energy
theorem" for the case of a constant force acting on an
object. But they don't use vector velocity, but rather
the scalar speed instead. So where I put Vs below, they're
scalars, not vectors.

begin quote:
======================================

The simplest situation to consider is that of a constant
force F(vector). Such a force F(vector) will produce a
constant acceleration a(vector). Choose the x-axis to be
the common direction of F(vector) and a(vector). What is the
work done by this force on the particle in causing a
displacement x? For constant acceleration, the relations are

a= (V - Vo)/t

and

x= (t) X (V + Vo)/2

where Vo is the particle's *speed* [emphasis mine] at
t = 0 and V is its speed at time t.

Then the work done is:

W = (F)(x) = (ma) (x) [newton's law]

W = (m) [(V - Vo)/t] [(V + Vo)/2] (t)

W = 1/2 mV(sq) - 1/2 mVo(sq)

We call one half the product of the mass of a body and the
square of its *speed* [again, emphasis mine] the kinetic
energy of the body. If we represent kinetic energy by the
symbol K then

K = 1/2 mV(sq)

The work done by the result force acting on a particle is
equal to the change in kinetic energy of the particle.

Although this is proved for a constant force only, it holds
whether the resultant force [again, the dot product between
F(vector) and D(vector)] is constant or variable.

================================================== ===========
(end quote)

They go on to give some examples, where the force is not
in the direction of accelleration (circular motion, for
example, where cos(theta) is equal to zero) and where the
force varies in magnitude as well.

So basically according to H&R I was wrong in saying that
work is proportional to the dot product of two vectors,
it's not. It's proportional to the square of the *speed*
which is not a vector.

Sorry for the confusion.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

On Tue, 07 Oct 2003 09:42:29 -0400, Tom Quackenbush wrote:
SNIP

The conclusion I'm coming to is that I was wrong when I said w =
mv^2. : (

If KE = .5mv^2 (and I'm sure it is) and w = delta KE (Work -
Energy Theorem - had to look that one up), I don't see I how my
original statement could be correct.

Sorry 'bout that.

Well, no problem, you found the correct answer. w = delta Ke.
Since we know Ke is a scalar, then it follows that a change in
Ke will also be a scalar. QED.

Gary

On 05 Oct 2003 14:36:23 GMT, a (Dave Baker) wrote:
From: Gary Coffman
On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
From: Gary Coffman
Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.
snip
The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same
direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never
zero. The equation has to be balanced by the total expenditure of energy
consumed in doing the work.

That *sounds* good, but I'm not sure it holds up to analysis. For example,
if I raise a 1 kilogram mass 1 meter, I do work, requiring energy to accomplish.
But if I then let the mass return to its original position, I regain the energy.
So we have a situation where there has been a zero net change in energy.
Doesn't that also require that negative work was done so that net work is
also zero? Otherwise, we seem to have a violation of the conservation laws
because we'd have net work with no net energy expenditure.

I was anticipating that reply and if it hadn't been so late last night I was
going to mention it in advance. The reason being is you had alluded to it in
the bit about energy states at the top of your earlier post. Potential energy
states and work musn't be confused. You don't do work to an object when it
falls against gravity, the earth does. Whether you arrange to have that energy
recovered, hydroelectricity etc, doesn't affect the amount of work that got
done during the process. Work isn't something that gets conserved, only energy
is.

Well, Feynman disagrees (at least partly). He gives examples in his
"Lectures on Physics" of cases where work must be negative. One
of those cases is an orbiting body, another is a frictionless pendulum.
The equation w = F dot D turns out not to be general enough to cover
all cases. The more general equation is:

w = [definite integral from a to b] F cos(theta) dS

Where w is work, a and b are the end points of the displacement,
F is a force vector, theta is the angle between the force and the
direction of displacement, and S is the unit displacement vector.

Now for straight line displacements, and where theta = 0,
cos(theta) = 1, and we have the simple degenerate case of
w = F dot D. But consider the case where theta = 180.
Cos(theta) is then -1, and the resulting work is -w, negative
work.

Another way of looking at this is that work = delta Ke or
Ke(b) - Ke(a). In cases where Ke(a) is larger in magnitude
than Ke(b), you get a result which is negative work.

This is *why* bodies in orbit, or frictionless pendulums,
show zero net work. The positive work done during one
part of the motion is canceled by the negative work done
through another part of the motion, resulting in zero
net work being done.

Feynman shows that for any closed conservative system
which is in principle reversible, net work will always be zero,
no matter how torturous the path taken to close the loop.

Gary

On 7 Oct 2003 08:19:48 -0700, jim rozen wrote:
In article , Gary Coffman says...
But back to the original question, why is mechanical work
calculated with the dot product w = F dot D instead of the
cross product w = F X D?

Because it's only the component of the work in the
direction of the displacement that adds to the energy
of the particle.

The dot product is just the magnitute of each vector
times the cosine of the angle between them. For
example, if you push down really hard on a roller
skate, but it only moves slowly along the floor, it's
only the component of your hand's force *parallel*
to the floor that is actually pushing the skate to
speed it up. There the value of the dot product is
small as it should be.

Yes! In 3 space, it would be over each of the three
vector components, X, Y, and Z. But we can simplify
that by contriving the force to be orthogonal to two
of those dimensions by choosing an appropriate
reference frame, leaving us with a single cosine
factor.

OK, I managed to get a copy of halliday and resnick and
looking it over, it looks like they use the term "speed"
when calculating 1/2 mv(squared), not the vector velocity.

I don't think so, see below.

They proceed to derive an example of the "work - energy
theorem" for the case of a constant force acting on an
object. But they don't use vector velocity, but rather
the scalar speed instead. So where I put Vs below, they're
scalars, not vectors.

begin quote:
======================================

The simplest situation to consider is that of a constant
force F(vector). Such a force F(vector) will produce a
constant acceleration a(vector). Choose the x-axis to be
the common direction of F(vector) and a(vector). What is the
work done by this force on the particle in causing a
displacement x? For constant acceleration, the relations are

a= (V - Vo)/t

For 'a' to be a vector, as you correctly say it is, then you need
at least one vector on the opposite side of the equation, if you
are to obey the rules of vector math. That can't be 't', but it can
be the difference between *vectors* V and Vo. So they can't
correctly be treating V and Vo as *speeds* here.

and

x= (t) X (V + Vo)/2

where Vo is the particle's *speed* [emphasis mine] at
t = 0 and V is its speed at time t.

Ok, I don't understand this. If your 'X' is supposed to be
the cross product, that doesn't make sense, because 't'
isn't a vector. If it is meant to be '*' (simple multiplication),
then the equation makes sense, but 'x' is total displacement
*only if both Vo and V are in the same direction*. The math
can't enforce that if they are treated as scalar speeds. Only
if they are vectors, can it require that by resolving their
components.

Then the work done is:

W = (F)(x) = (ma) (x) [newton's law]

W = (m) [(V - Vo)/t] [(V + Vo)/2] (t)

W = 1/2 mV(sq) - 1/2 mVo(sq)

We call one half the product of the mass of a body and the
square of its *speed* [again, emphasis mine] the kinetic
energy of the body. If we represent kinetic energy by the
symbol K then

K = 1/2 mV(sq)

The work done by the result force acting on a particle is
equal to the change in kinetic energy of the particle.

Although this is proved for a constant force only, it holds
whether the resultant force [again, the dot product between
F(vector) and D(vector)] is constant or variable.

================================================= ============
(end quote)

They go on to give some examples, where the force is not
in the direction of accelleration (circular motion, for
example, where cos(theta) is equal to zero) and where the
force varies in magnitude as well.

So basically according to H&R I was wrong in saying that
work is proportional to the dot product of two vectors,
it's not. It's proportional to the square of the *speed*
which is not a vector.

I don't think that they've showed that at all, except in
special contrived cases. Feynman says that work is
invariant over all reference frames for all forces and
displacements, not just for the ones that conveniently
line up. The only way that can be true is if work is the
dot product of F and D.

Dot products have the vector math property that they
are invariant over all reference frames. That's not true
for cross products.

The reason Feynman says work is invariant is, if it
were not true, then you could contrive a perpetual
motion machine from which you could eternally extract
net energy by simply changing reference frames. The
conservation of energy law forbids that.

(This isn't a terribly obvious point, it takes Feynman
three chapters to work up to it. But it yields a result
which is rigorously correct in both the vector math
sense and the physics sense.)

I've been trying to get people to discover this by
using the Socratic method, but as is often the case
here, that hasn't worked very well. So now I've just
spelled it out.

Gary

In article , Gary Coffman says...

I've been trying to get people to discover this by
using the Socratic method,

To really prove this rigorously, you apparently
need to use special relativity.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , Gary Coffman says...

Well, Feynman disagrees (at least partly). He gives examples in his
"Lectures on Physics" of cases where work must be negative.

But 'negative work' is trivial, the example that I
gave from H&R showed that the value calculated is
really the difference between an initial velocity
and a final velocity of the body. If the final
velocity is less than the initial, the "work" is
negative, ie the body has supplied some energy
to the thing that deaccelerated it.

Basically the limits of integration can be from
a small number to a large number, or from a large
to a small. For the second case the integral
winds up being negative in sign. This isn't
anything fancy, it just tells you what way the
work 'went.'

This is *why* bodies in orbit, or frictionless pendulums,
show zero net work.

Not really. The real reason is that the dot product
of the force and the displacement has cos(theta) = 0
because the force is radial, and the displacement is
at 90 degrees to that.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , Gary Coffman says...

OK, I managed to get a copy of halliday and resnick and
looking it over, it looks like they use the term "speed"
when calculating 1/2 mv(squared), not the vector velocity.

I don't think so, see below.

Yes, this is how it's derived for a 101 level
physics course. They don't go into detail of
why they do it this way, and they don't discuss
the rapid shift from vectors to scalars, but
this is how it is taught.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:

Would it be accurate in the case of the compressor to say that torque
would be related to the PSI that can be produced, and HP would relate
to the airflow (cfm) at that PSI.

Subject: Compressor Motor: HP v.s. Amps?
From: "Bob Swinney"
Date: 04/10/03 02:58 GMT Daylight Time
Message-id: vPpfb.489338$Oz4.334706@rwcrnsc54

"Richard J Kinch" wrote in message
1...
Bob Swinney writes:

Torque is the capacity of an engine to do work ...

This, and many of your other comments here elided, are quite wrong.


So, Richard - we would like to hear your definition of torque. And could
you elucidate a bit on the many "elided" comments? Please, give us your
"take" on the correct facts.

Torque bears no relation to any capacity to do work. Although the two
quantities are expressed in the same units of force and distance, torque is a
vector quantity and work is scalar one. Torque merely defines an instantaneous
twisting force about an axis. It could be expressed as a capacity to overcome a
given load applied to that axis but not to do a given amount of work. We need
to also know speed to calculate that and thus horsepower is what defines
capacity to do work.

Inherent in the use of the word "capacity" is a time element. Any engine could
theoretically do any amount of work if left running for long enough. The use of
the term capacity without recognising the inherent time element makes the term
meaningless.

For example, the question "can an engine producing X amount of torque lift
those bricks to the top of that building?" is meaningless. Appropriately geared
any engine could do that but only the horsepower defines how fast it could do
it.

All torque therefore tells one is the shaft speed at which an engine produces a
given amount of horsepower and the load the engine can overcome at that shaft
speed without further gearing.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.

On 8 Oct 2003 10:30:46 -0700, jim rozen wrote:
In article , Gary Coffman says...

Well, Feynman disagrees (at least partly). He gives examples in his
"Lectures on Physics" of cases where work must be negative.

But 'negative work' is trivial, the example that I
gave from H&R showed that the value calculated is
really the difference between an initial velocity
and a final velocity of the body. If the final
velocity is less than the initial, the "work" is
negative, ie the body has supplied some energy
to the thing that deaccelerated it.

Basically the limits of integration can be from
a small number to a large number, or from a large
to a small. For the second case the integral
winds up being negative in sign. This isn't
anything fancy, it just tells you what way the
work 'went.'

This is *why* bodies in orbit, or frictionless pendulums,
show zero net work.

Not really. The real reason is that the dot product
of the force and the displacement has cos(theta) = 0
because the force is radial, and the displacement is
at 90 degrees to that.

No it isn't. Analyze pendulum motion and come back
and tell us that the force is radial. The force, mg, is
always *down*, but pendulum motion is radial around
the pivot.

Gary

In article , Gary Coffman says...

The force, mg, is
always *down*, but pendulum motion is radial around
the pivot.

I was talking about circular motion.

But for a pendulum, the displacement
and the force are still orthogonal.

The dot product is still zero.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , Gary Coffman says...

The force, mg, is
always *down*, but pendulum motion is radial around
the pivot.

I was talking about circular motion.

But for a pendulum, the displacement
and the force are still orthogonal.

The dot product is still zero.

For long pendulums, that is. That
makes is a lot easier to solve for
the motion.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

On Wed, 08 Oct 2003 23:15:31 -0400, Robert Salasidis wrote:
Would it be accurate in the case of the compressor to say that torque
would be related to the PSI that can be produced, and HP would relate
to the airflow (cfm) at that PSI.

No, not in the general case. Depending on the design, you can trade
between pressure and flow in a fairly arbitrary manner, same as you
can trade between RPM and torque to get power.

It is true that for a given piston head area, crank throw, etc, maximum
pressure will be torque limited, and volume will be RPM limited. But you
can't say that either is proportional to either in a general sense away
from the limit cases.

Gary

On Fri, 10 Oct 2003 13:22:30 -0400, Gary Coffman
wrote:

I would disagree - as foir a given horsepower motor, the PSI vs the
CFM is the determinant of the HP.

Therefore one can achieve a greater PSI at a lower CFM and vise versa.

The same applies to car engines - the Torque x RPM = HP

(One can say Volt x Amps = Watts follows the same rules.)

In th case of design - many variables would come into play - such as
torque vs RPM curves - which are different depending on the type of
motor/engine in question

On Wed, 08 Oct 2003 23:15:31 -0400, Robert Salasidis wrote:
Would it be accurate in the case of the compressor to say that torque
would be related to the PSI that can be produced, and HP would relate
to the airflow (cfm) at that PSI.

No, not in the general case. Depending on the design, you can trade
between pressure and flow in a fairly arbitrary manner, same as you
can trade between RPM and torque to get power.

It is true that for a given piston head area, crank throw, etc, maximum
pressure will be torque limited, and volume will be RPM limited. But you
can't say that either is proportional to either in a general sense away
from the limit cases.

Gary

On Mon, 13 Oct 2003 22:17:29 -0400, Robert Salasidis wrote:
I would disagree - as foir a given horsepower motor, the PSI vs the
CFM is the determinant of the HP.

So what are you disagreeing about?

Gary

Therefore one can achieve a greater PSI at a lower CFM and vise versa.

The same applies to car engines - the Torque x RPM = HP

(One can say Volt x Amps = Watts follows the same rules.)

In th case of design - many variables would come into play - such as
torque vs RPM curves - which are different depending on the type of
motor/engine in question

On Fri, 10 Oct 2003 13:22:30 -0400, Gary Coffman
wrote:
On Wed, 08 Oct 2003 23:15:31 -0400, Robert Salasidis wrote:
Would it be accurate in the case of the compressor to say that torque
would be related to the PSI that can be produced, and HP would relate
to the airflow (cfm) at that PSI.

No, not in the general case. Depending on the design, you can trade
between pressure and flow in a fairly arbitrary manner, same as you
can trade between RPM and torque to get power.

It is true that for a given piston head area, crank throw, etc, maximum
pressure will be torque limited, and volume will be RPM limited. But you
can't say that either is proportional to either in a general sense away
from the limit cases.

Gary

On 9 Oct 2003 05:44:20 -0700, jim rozen wrote:
But for a pendulum, the displacement
and the force are still orthogonal.

No it isn't. The pendulum swings down
and in, or up and out. The up and down
components are in line with the force.
So work is done in each half of the
swing. It is just that the positive work
while the pendulum is swinging down
is offset by the negative work done
as the pendulum is swinging up. So
net work is zero.

The dot product is still zero.

Depends on over what part of the swing you calculate
it. If you picture the pendulum end points as a and c,
and call the midpoint b. Then from a to b positive work
is done, and from b to c negative work is done. If you
calculate directly from a to c, of course, the result is
zero. That's because you're adding the work done in
the two halves of the swing together.

Wac = Wab + (- Wbc)

Since, ignoring friction, Wab and Wbc have the same
magnitudes, but opposite signs, Wac should always
equal zero.

Gary

Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 14/10/03 23:33 GMT Daylight Time
Message-id:

On 9 Oct 2003 05:44:20 -0700, jim rozen wrote:
But for a pendulum, the displacement
and the force are still orthogonal.

No it isn't. The pendulum swings down
and in, or up and out. The up and down
components are in line with the force.
So work is done in each half of the
swing. It is just that the positive work
while the pendulum is swinging down
is offset by the negative work done
as the pendulum is swinging up. So
net work is zero.

The dot product is still zero.

Depends on over what part of the swing you calculate
it. If you picture the pendulum end points as a and c,
and call the midpoint b. Then from a to b positive work
is done, and from b to c negative work is done. If you
calculate directly from a to c, of course, the result is
zero. That's because you're adding the work done in
the two halves of the swing together.

Wac = Wab + (- Wbc)

Since, ignoring friction, Wab and Wbc have the same
magnitudes, but opposite signs, Wac should always
equal zero.

I'm not convinced this recent line of argument about zero net work is actually
valid in the first place. All a pendulum does is convert potential energy into
kinetic and vice versa. The only work that is done is against the friction in
the pivot and that is a scalar quantity and irreversible.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.

In article , Gary Coffman says...

But for a pendulum, the displacement
and the force are still orthogonal.

No it isn't.

OK, I was thinking of the force on the wire, holding up
the ball.

The gravitational force that causes it to oscillate
is pretty much vertical though, and pretty much
cancelled out by the force of the wire holding the
ball up.

The analysis of a pendulum that resolves into simple
harmonic motion *depends* on the wire being long,
so the restoring force away from the bottom of the
swing looks nearly proportional to the displacement
of the ball.

If the wire is short, or if the swing becomes long,
then really the restoring force looks proportional
to the *sine* of the displacement angle, and the
ball no longer displays simple harmonic motion.

Jim

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In article , Dave Baker says...

I'm not convinced this recent line of argument about zero net work is actually
valid in the first place. All a pendulum does is convert potential energy into
kinetic and vice versa. The only work that is done is against the friction in
the pivot and that is a scalar quantity and irreversible.

That's right. Of course there is some tiny amount of energy lost
in the knife edge pivots or whatnot. Also in air friction for
example. So real examples have some way of adding that tiny
extra energy back into the system.

Gary's example of course has the provision that 'the thing
runs in vacuum, and has perfect, non-lossy pivots.' Or
something to that effect.

Jim

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On 14 Oct 2003 22:41:51 GMT, a (Dave Baker) wrote:
Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 14/10/03 23:33 GMT Daylight Time
Message-id:

On 9 Oct 2003 05:44:20 -0700, jim rozen wrote:
But for a pendulum, the displacement
and the force are still orthogonal.

No it isn't. The pendulum swings down
and in, or up and out. The up and down
components are in line with the force.
So work is done in each half of the
swing. It is just that the positive work
while the pendulum is swinging down
is offset by the negative work done
as the pendulum is swinging up. So
net work is zero.

The dot product is still zero.

Depends on over what part of the swing you calculate
it. If you picture the pendulum end points as a and c,
and call the midpoint b. Then from a to b positive work
is done, and from b to c negative work is done. If you
calculate directly from a to c, of course, the result is
zero. That's because you're adding the work done in
the two halves of the swing together.

Wac = Wab + (- Wbc)

Since, ignoring friction, Wab and Wbc have the same
magnitudes, but opposite signs, Wac should always
equal zero.

I'm not convinced this recent line of argument about zero net work is actually
valid in the first place. All a pendulum does is convert potential energy into
kinetic and vice versa. The only work that is done is against the friction in
the pivot and that is a scalar quantity and irreversible.

That's not what the definition of work says, though. The definition contains
no requirement of irreversibility. In fact it is easy to show that net work can
be done in a system which is nondissipative.

Feynman analyzes pendulum motion as I have above precisely to show
why a closed reversible system must exhibit both positive and negative work
in order to be consistent with both the definition of work and the principle
of conservation of energy.

Gary