Accurate plane performace?

[R]

I love FS2004 -- hell of a value for $50. :) I've recently been
downloading and playing with the slew of fan-designed aircraft. I
love speed, and so I've been gravitating to some of the supersonic
military aircraft -- Mig 29, F-18, F-16, F-15, F-14, SR-71, etc.

However, I've yet to find one such aircraft that performs in speed
and altitude the way that they're supposed to. The F-16, for example,
struggles to break mach 1.2 at 40,000 feet -- this is a plane capable
of Mach 2 at that altitude. The SR-71 can sustain Mach 3.3 for an
hour at 80,000 feet. The SR-71 I downloaded stalls out at around
70,000 feet -- at 60,000 feet, it's struggling along at 200 kts. (The
U-2 stalled out at 55,000 feet -- that got a good laugh from me.) All
of these planes can tear the MSFS world up at low altitudes -- I had
the SR-71 popping Mach 3 at 5,000 feet (which is also unrealistic; the
plane would fly apart at that speed at that altitude). But get them
to their cruise ceiling, and suddenly they perform subsonically.

Am I doing something wrong? Yes, gear is up, flaps are up, spoilers
are not deployed. Is MSFS not that good at rendering supersonic
flight? Or are the makers of these add-on planes not building them
correctly to interface with MSFS 2004? I've even tried going into the
..air file and doubling or tripling the engine thrust... doesn't seem
to help much at the higher altitudes. (Which are not really "high" to
these aircraft.) I'm puzzled. I'd love to fly a SR-71 mission 80,000
feet over Russia, but that's not going to happen at 100 kts. ;)

-R

[Peter Duniho]

"Jeffrey Voight" > wrote in message
...
> I don't know if this is what you're experiencing in the FS, but I'd
> check the GPS ground speed before filing a bug report to M$.

Your guess as to what the person posting as "R" is complaining about is
probably right on the money.

For the non-pilot crowd, often the best solution is simply to change the
setting so that true airspeed is displayed rather than indicated. The
option is under the Realism settings. They usually want to see a specific
number on the airspeed indicator, and don't care about the difference
between true and indicated.

Of course, if they are displaying true, they need to remember that the stall
speed will be higher at higher altitudes. They should not be surprised if
one cannot slow down very much from max cruise before stalling, when flying
at 50000 feet (for example).

Pete

[Lars Møllebjerg]

"R" > wrote in message
...
> ...
> However, I've yet to find one such aircraft that performs in speed
> and altitude the way that they're supposed to. The F-16, for example,
> struggles to break mach 1.2 at 40,000 feet -- this is a plane capable
> of Mach 2 at that altitude. The SR-71 can sustain Mach 3.3 for an
> hour at 80,000 feet. The SR-71 I downloaded stalls out at around
> 70,000 feet -- at 60,000 feet, it's struggling along at 200 kts. (The
> U-2 stalled out at 55,000 feet -- that got a good laugh from me.) All
> of these planes can tear the MSFS world up at low altitudes -- I had
> the SR-71 popping Mach 3 at 5,000 feet (which is also unrealistic; the
> plane would fly apart at that speed at that altitude). But get them
> to their cruise ceiling, and suddenly they perform subsonically.
> ...

I can't say much about the max speed at different altitudes, but I did
notice one point where you are probably not maintaining the correct climb
method.
You should NEVER EVER allow the SR-71 to slow down to 200 doing a climb. As
you slow down, less air move over the wings - meaning they produce less
lift. To compensate, you have to rise the nose a bit which increase the lift
but this produce more drag, hence slowing the plane further and the cycle
continues - you are heading streight for a stall no matter what kind of
plane you are flying - Cessna or SR-71 - it does not matter the physics is
the same. Of course if the model is wrong it won't work, but you forgot to
say which model we are talking about?? As far as I recall, I manage to get
Kirk Olsson's F-16 over 60000 using this method, and apparently the correct
number is 61500 feet.

If you at full power start loosing speed in a climb, lower the nose
immidiatly to build up speed. Run it as close to the max speed at the
current altitude as you can to get maximum lift. When you are flying at
maximum speed and the plane no longer is able to climb 100 feet per minute
(I think that's the number, but I'm not too sure - might be 300 or another
random number) you have reached the planes maximum altitude (and this is the
configuration where you should compare it to the real world aircraft to see
how well the model is done).

You can play around and rise the nose to see how much higer you can get, but
you should not expect to maintain the altitude gained this way. Again, from
memory I think it was 65000 I managed to get Kirk Olsson's F-16 to for a few
seconds. I used the flaps just before it stalled to buy another couple of
hundred feets, but using the flaps to maintain altitude is like peeing in
your pants to keep warm - it might help for a few seconds, but then you are
worse of than you where before. :)

Regards
Lars

[Peter Duniho]

"Lars Møllebjerg" > wrote in message
...
> I can't say much about the max speed at different altitudes, but I did
> notice one point where you are probably not maintaining the correct climb
> method.
> You should NEVER EVER allow the SR-71 to slow down to 200 doing a climb.

This part might be true. Not having flown either the real or simulated
SR-71, I don't know. The rest of the post leaves a lot to be desired.

> [...] If you at full power start loosing speed in a climb, lower the nose
> immidiatly to build up speed. Run it as close to the max speed at the
> current altitude as you can to get maximum lift.

This is very wrong advice.

There are a few interesting airspeeds one can use for climbs. Two in
particular have special names, "Vx" and "Vy". "Vx" is the airspeed at which
the airplane will gain the most altitude over a given distance; it is the
steepest climb angle. "Vy" is the airspeed at which the airplane will gain
the most altitude over a given time; it produces the largest vertical speed.
These airspeeds are what one should choose for "best" climb performance;
which one to choose depends on whether you want a steeper angle or a faster
rate of ascent.

Vx is always slower than Vy, and both are WELL below normal cruise speed,
never mind "max speed at the current altitude". As far as "to get maximum
lift" goes, that is also incorrect. In unaccelerated flight (e.g. a steady
state climb), lift is equal to weight. Always. There's no minimum or
maximum...lift is simply always the same as weight. The "Vx" and "Vy"
airspeeds correspond to the airspeeds at which the airplane has the most
excess thrust and the most excess power, respectively. Lift keeps an
airplane in the air. Excess power is what makes an airplane climb.

A more general airspeed is known as the "cruise climb" airspeed. This is
going to be somewhat higher than Vy, but still will be nowhere near the
maximum airspeed for a given altitude, except for normally aspirated piston
engine airplanes near their ceiling.

> When you are flying at
> maximum speed and the plane no longer is able to climb 100 feet per minute
> (I think that's the number, but I'm not too sure - might be 300 or another
> random number) you have reached the planes maximum altitude

Actually, what you're referring to is called "service ceiling" and in the US
that's established at the altitude at which the climb is 100 feet per
minute. An airplanes "absolute ceiling" is the altitude at which the
airplane simply will not climb any higher. As it happens, the "Vx" and "Vy"
airspeeds are equal at this altitude (and converge all the way from their
sea level values to their single absolute ceiling value).

> (and this is the
> configuration where you should compare it to the real world aircraft to
see
> how well the model is done).

Well, it's certainly one data point one ought to be looking at. However, a
simulation model that gets the service ceiling correct may or may not get
any of the rest of the simulation details correct.

Your general advice -- one cannot simply climb at a higher rate by blindly
raising the nose -- is well-intentioned and the basic idea is correct.
However, you took things too far by ignoring the fact that the best climb
performance does still occur in the lower region of the airplane's airspeed
range, and the nose will in fact be pitched relatively high.

Claiming that one should climb at the maximum airspeed possible for a given
altitude is just plain wrong, and in fact that advice will NEVER work. The
maximum airspeed possible for a given altitude (assuming no descent) is in
straight and level, *zero* vertical speed flight. If you are climbing, you
can *always* go a little faster by lowering the nose and not climbing.

Pete

[Lars Møllebjerg]

Yes, I should probably have made it clearer that I was only refering to
climbing close to the ceiling. Would have saved you a lot of typing. :)

Regards
Lars

[Lars Møllebjerg]

"Lars Møllebjerg" > wrote in message
...

By the way, saying that lift is equal to weight is a bit wierd as the lift
is a force generated, while weight isn't a force, but a number calculated
from the mass and gravity. But I guess it's one of those simplification
making it easier for people to understand. :)

/Lars

[Peter Duniho]

"Lars Møllebjerg" > wrote in message
...
> By the way, saying that lift is equal to weight is a bit wierd as the lift
> is a force generated, while weight isn't a force, but a number calculated
> from the mass and gravity.

The number you calculate from mass and gravity is a force. Gravity is an
acceleration, and mass is, well...mass is mass. F=ma.

In other words saying that lift is equal to weight makes perfect sense.
They are both forces.

> But I guess it's one of those simplification
> making it easier for people to understand. :)

I know you mean well, but frankly you really have not been helpful in the
"understanding" department.

Pete

[Chris]

> By the way, saying that lift is equal to weight is a bit wierd as the lift
> is a force generated, while weight isn't a force, but a number calculated
> from the mass and gravity. But I guess it's one of those simplification
> making it easier for people to understand. :)

The above comment is wrong. First of all weight *is* a force. Then
"saying that lift is equal to weight is a bit wierd" is not wierd but
it is a simplification. This will be true only in straight and level
flight. In a turn or descent/climb, lift will not be equal to weight.

In fact in a steady climb lift is *smaller* than weight! This is
because weight is tilted backward relative to the flight path (weight
always point to the centre of Earth). Therefore it adds a component to
the drag, which must then be balanced by an increase in thrust (the
"excess thrust") but the weight component that must be balanced by the
lift is now less so lift is smaller.


Chris

[R]

On Sun, 7 Dec 2003 01:27:39 -0800, "Peter Duniho"
> wrote:

>"Lars Møllebjerg" > wrote in message
...
>> I can't say much about the max speed at different altitudes, but I did
>> notice one point where you are probably not maintaining the correct climb
>> method.
>> You should NEVER EVER allow the SR-71 to slow down to 200 doing a climb.
>
>This part might be true. Not having flown either the real or simulated
>SR-71, I don't know. The rest of the post leaves a lot to be desired.

Actually, I wasn't even climbing at that speed. I'd get it to
around 65,000, level off, and see how fast it would fly. 200kts was
about average. Even with "true airspeed" selected, I wasn't coming
near Mach 1 in a plane that should pull Mach 3.3 at 80,000 feet. I
had similar speed issues with most fighter aircraft, as well -- the
F-16 barely chugging along at 400 kts at 50,000 feet. This is a plane
that has enough thrust to climb from takeoff to 20,000 feet at a *90
degree angle*.

With the SR-71, I mostly climb with the autopilot. Granted, I
usually set at the default 1,100 fpm climb rate, but I didn't think
that would be too much of a problem for the fastest (known) plane in
the world, a plane that pulls Mach 3.3 at 80,000 feet. At some point
well before 80,000 feet, the autopilot more or less fails, pulling the
nose up into a stall.

I don't have stalling problems with fighter aircraft, but they don't
go anywhere near their advertised speed capability at high altitude.
Oddly, they come closer to approaching their max speed at low
altitude, which is kind of silly, as a plane flying mach 2 at 500 feet
would promptly break apart or begin to melt.

I did find one interesting speed glitch in the game. With the
air-stress option untoggled, it is possible to get nearly any jet
aircraft out of the sim reality's "envelope" and send it into a
hyperspeed journey. Take any jet aircraft (fighters work well) up to
a high altitude, say, 65,000 feet. Bring the nose to 90 degrees down.
Do some barrel rolls on the way down. At some point, your plane gets
a tremendous speed boost, up to around mach 5. Pull out of it.
Better have your wing leveler on, as, at this speed, the plane bounces
all over the place with turblence. Once you do this, it is impossible
to slow down. Cut the throttle, turn the engines off, it doesn't
matter. You are now God's guided missile. I made a trip from Seattle
to LAX in around 15 minutes this way.

I never owned FS2002. Did supersonic aircraft perform correctly in
that package? I'm tempted to think that most of the 2004 planes are
just 2002 conversions, and that perhaps the two versions handle high
speed dynamics differently. If the developers didn't change anything
before porting to 2004 (or, in the case where I just load 2002 planes,
like the SR-71), the planes may perform incorrectly.

-R

[H M]

> With the SR-71, I mostly climb with the autopilot. Granted, I
> usually set at the default 1,100 fpm climb rate, but I didn't think
> that would be too much of a problem for the fastest (known) plane in
> the world, a plane that pulls Mach 3.3 at 80,000 feet. At some point
> well before 80,000 feet, the autopilot more or less fails, pulling the
> nose up into a stall.

is that autopilot capable of maintaining a certain IAS during a climb?
if so, i'd try 300 knots or so and see how it goes. if not, try keeping
an eye on the IAS and lowering the 1100 fpm if IAS goes below those 300
knots...

[Quilljar]

> yap yap yap ypa
>
> The Speedbyrd :>


Yes I have always admired your helpful and intelligent inputs here. Don't
bother replying as I have an excellent killfile :-)


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[Lars Møllebjerg]

Sorry, I was indeed sleeping a bit with the weight thing. Too many years
since I had to mess too much with it. :)

But please notice that I never EVER said that I was saying how to climb the
fastest (either covering the least amount of distance when climbing, nor
gaining altitude as fast as possible, nor reach altitude using as little
fuel as possible). Basically your long statement of how wrong I was is based
on trying to acheive the most efficiant climb - which I never stated it was.
Just like you trid to make it look like I thought a flight model can be
judged from one single messurement. Obviously I never stated that, and
obviously I never meant it, so why comment on it as if I made a mistake?

I can see you might interprete my "maximum lift" statement as "maximum
climb" - it is not how I read it, but I agree - it is a bit "on the edge", I
will give you that.

If you take this into account, along with the fact that I was trying to
explain a simple way of reaching max speed, I still claim my original
information is good enough to reach the ceiling - and that was what this
question was about, the rest you made up.

/Lars

[Peter Duniho]

"Lars Møllebjerg" > wrote in message
...
> But please notice that I never EVER said that I was saying how to climb
the
> fastest [...] Basically your long statement of how wrong I was is based
> on trying to acheive the most efficiant climb - which I never stated it
was.

No. You said that to climb, one needed to fly at the maximum speed for the
given altitude. This is simply wrong. The maximum speed for any given
altitude will NOT produce a climb.

> Just like you trid to make it look like I thought a flight model can be
> judged from one single messurement. Obviously I never stated that, and
> obviously I never meant it, so why comment on it as if I made a mistake?

I didn't "try to make it look like" anything. You said "this is the
configuration where you should compare it to the real world aircraft to see
how well the model is done". In my reply, I even refrained from saying that
you were flat out wrong. I simply agreed that the "data point" you
described was useful, but pointed out it was not sufficient.

If anything, I gave way more benefit of the doubt than your post deserved.

> I can see you might interprete my "maximum lift" statement as "maximum
> climb" - it is not how I read it, but I agree - it is a bit "on the edge",
I
> will give you that.

The point is that the phrase "maximum lift" is meaningless. You get the
same amount of lift at any airspeed for normal, unaccelerated flight.

> If you take this into account, along with the fact that I was trying to
> explain a simple way of reaching max speed, I still claim my original
> information is good enough to reach the ceiling - and that was what this
> question was about, the rest you made up.

"The rest you made up"? Funny. Bottom line: your advice, to fly at maximum
speed, won't allow the airplane to climb at all, never mind reach the
airplane's ceiling.

I understand your desire to save face, but revisionist history just doesn't
cut it on Usenet. Anyone can go back to your previous post to see what you
really said.

Pete

[Andreas Maurer]

On Mon, 08 Dec 2003 02:32:58 -0600, R > wrote:

> With the SR-71, I mostly climb with the autopilot. Granted, I
>usually set at the default 1,100 fpm climb rate, but I didn't think
>that would be too much of a problem for the fastest (known) plane in
>the world, a plane that pulls Mach 3.3 at 80,000 feet. At some point
>well before 80,000 feet, the autopilot more or less fails, pulling the
>nose up into a stall.

Checck WWW for somew real-life SR-71 procedures.

Get her supersonic at 30.000 ft (this means an IAS of about 600
kts...!!!!)... and keep that IAS during a shallow climb. Adjust the
climb rate accordingly - most important is to keep IAS. The higher you
climb, the higher gets your MACH - until you level off at 70.000 ft
with an IAS of abpout 600 kts at Mach 3.x.


Bye
Andreas

[David CL Francis]

On Sun, 7 Dec 2003 at 12:41:21 in message
>, Lars Møllebjerg
> wrote:

>By the way, saying that lift is equal to weight is a bit wierd as the lift
>is a force generated, while weight isn't a force, but a number calculated
>from the mass and gravity. But I guess it's one of those simplification
>making it easier for people to understand. :)

I presume you are referring to Newton's laws. However if you want to be
pedantic then you are wrong. Weight is a force, Mass isn't. We had a
famous popular scientist who used to ask "how much does a satellite
weigh in orbit?" His answer is zero although its mass remains the same.
Two forces are cancelling out in orbit.

Weight _is_ a force, although it is a restricted case that refers only
to the force that is generated by gravity.

In what way does this 'simplification' make any difference to the
equations? Of course lift only equals weight in steady straight and
level flight.
--
David CL Francis

[Peter Duniho]

"David CL Francis" > wrote in message
...
> Its mass is the same; its weight differs. You are still being confused
> between weight and mass.

It is true that its weight differs. It's farther from the Earth's center of
gravity, thus the weight is necessarily less. However, I think what Jeffrey
was trying to point out is that the satellite still does *weigh* something.
And in fact, its weight is almost as great as it would be sitting on the
surface of the Earth.

> Weight is the measure that you find if you weigh something on a spring
> balance.

I think this view of "weight" is what's tripping you up. The satellite in
freefall would appear to weigh nothing if weighed on a weighing scale that
is also in freefall with the satellite. However, that doesn't mean that the
satellite weighs nothing. In fact, if it weren't for its weight, it would
fly off at a tangent to its orbit.

The satellite's weight is what keeps it in orbit. It's just not true that
the satellite weighs zero in orbit. It's my impression that this is what
Jeffrey was saying in his post.

> Mass is a measure of the total quantity of matter in an object. If you
> are floating in deep space in free fall, then you cannot detect any
> weight.

Detecting weight and the existence of weight are two different things.
Consider the folks riding the "Vomit Comet", the jet used to create freefall
conditions without going into orbit. The occupants of the aircraft during
its parabolic flight cannot detect their weight. However, it is their very
weight that keeps them accelerating toward the planet, as it always does
during the non-parablic phases of flight or even while standing on solid
ground.

> However the _mass_ is the same and if a force (perhaps from a
> rocket motor} is applied then the acceleration depends on the force
> exerted by the rocket and the mass of the object.

I'm not sure what this has to do with the so-called "weightless satellite".

> Some of the confusion arises because in the imperial system of units
> there is no obvious distinction in the measurement of them.

I'm not sure that explains your confusion regarding whether a satellite in
orbit is weightless or not.

> For ordinary everyday, stuck on the surface of earth, people the
> distinction is subtle. To engineers, physicists and applied
> mathematicians the distinction is essential.

Which is why it's odd you seem to think that a satellite in orbit is
weightless. It's not.

Pete

[Nick Coleman]

Peter Duniho wrote:

[confused stuff on weight snipped]

Sorry, but you are confusing mass and weight. There is no such thing as
weight. Weight is a nebulous term that non-physicists use to describe
gravitational attraction between two bodies, or so it seems to me.

A body has mass and velocity; that's all it has, it doesn't have weight. No
reflection on PD, since we all did different things at school, but this is
plain-vanilla high-school physics.

A satellite has no weight in orbit; it has mass and velocity (and therefore
momentum). The reason it doesn't fly off on a tangent is because of the
gravitational attraction between the mass of the Earth and the mass of the
satellite, exactly counter-balancing the satellite's momentum which would
act to keep it going in a straight line.

Similarly, in the Vomit Comet it is not weight that is attracting the people
towards the ground, it is the gravitational attraction between the mass of
the Earth and the mass of the people.

--
Nick

[David CL Francis]

On Sun, 14 Dec 2003 at 13:08:32 in message
>, Peter Duniho
> wrote:
>"David CL Francis" > wrote in message
...
>> Its mass is the same; its weight differs. You are still being confused
>> between weight and mass.
>
>It is true that its weight differs. It's farther from the Earth's center of
>gravity, thus the weight is necessarily less. However, I think what Jeffrey
>was trying to point out is that the satellite still does *weigh* something.
>And in fact, its weight is almost as great as it would be sitting on the
>surface of the Earth.
>
That is mass you are talking about. I defined weight earlier. But why
'almost' in your world?

>> Weight is the measure that you find if you weigh something on a spring
>> balance.
>
>I think this view of "weight" is what's tripping you up. The satellite in
>freefall would appear to weigh nothing if weighed on a weighing scale that
>is also in freefall with the satellite. However, that doesn't mean that the
>satellite weighs nothing. In fact, if it weren't for its weight, it would
>fly off at a tangent to its orbit.
>

It is not a 'view' it is definition to help to try and help you
understand the difference between force and mass. I am not the list bit
'tripped up'.

>The satellite's weight is what keeps it in orbit. It's just not true that
>the satellite weighs zero in orbit. It's my impression that this is what
>Jeffrey was saying in his post.
>
Once more, it is the satellite's MASS that keeps it in orbit and the
MASS of the earth..

>> Mass is a measure of the total quantity of matter in an object. If you
>> are floating in deep space in free fall, then you cannot detect any
>> weight.
>
>Detecting weight and the existence of weight are two different things.
>Consider the folks riding the "Vomit Comet", the jet used to create freefall
>conditions without going into orbit. The occupants of the aircraft during
>its parabolic flight cannot detect their weight. However, it is their very
>weight that keeps them accelerating toward the planet, as it always does
>during the non-parablic phases of flight or even while standing on solid
>ground.
>
You are talking mass again. Weight is a force, Mass is the _quantity_ of
material in an object.

>> However the _mass_ is the same and if a force (perhaps from a
>> rocket motor} is applied then the acceleration depends on the force
>> exerted by the rocket and the mass of the object.
>
>I'm not sure what this has to do with the so-called "weightless satellite".
>

>> Some of the confusion arises because in the imperial system of units
>> there is no obvious distinction in the measurement of them.
>
>I'm not sure that explains your confusion regarding whether a satellite in
>orbit is weightless or not.
>
For the last time - I am not confused.

>> For ordinary everyday, stuck on the surface of earth, people the
>> distinction is subtle. To engineers, physicists and applied
>> mathematicians the distinction is essential.
>
>Which is why it's odd you seem to think that a satellite in orbit is
>weightless. It's not.
>
I give up. Get a book on physics or applied mechanics. Perhaps someone
else might be able to help. You seem to be stuck with your preconceived
idea as to what weight and mass are, or to put it another way as to what
a force and a mass are.

If I mention that force is a vector quantity (weight is a force) and
that mass is a scalar quantity, I suppose that will mean nothing to you?

A quote from an A level physics book:
~~~~~~begin quote~~~~~~~~~~
The weight of a body is the force of gravity acting on it towards the
centre of the earth. Weight is thus a _force_ , not to be confused with
mass which is independent of the presence or absence of the earth.
~~~~~~~~~~~~~end quote~~~~~

The Gravitational force F between two particles of masses m1 and m2 , a
distance r apart ,is given by;

F=(G*m1*m2)/r^2 where G is the Gravitational constant.
--
David CL Francis

[Peter Duniho]

"David CL Francis" > wrote in message
...
> That is mass you are talking about. I defined weight earlier. But why
> 'almost' in your world?

That is not mass I'm talking about. It's weight, and the reason it's
"almost" the same as on the ground is that weight depends both on the masses
of the two objects being considered as well as the distance between them.
Weight decreases with the square of the distance.

> It is not a 'view' it is definition to help to try and help you
> understand the difference between force and mass. I am not the list bit
> 'tripped up'.

So you say.

> >The satellite's weight is what keeps it in orbit. It's just not true
that
> >the satellite weighs zero in orbit. It's my impression that this is what
> >Jeffrey was saying in his post.
> >
> Once more, it is the satellite's MASS that keeps it in orbit and the
> MASS of the earth..

The mass only keeps it in orbit inasmuch as mass near another mass causes a
force. This force is weight, and the satellite in orbit has this force
called weight. Without the weight, the satellite would fly off in a
straight line. The acceleration due to weight is the only thing that allows
the satellite to follow a curved path.

> > [...] However, it is their very
> >weight that keeps them accelerating toward the planet, as it always does
> >during the non-parablic phases of flight or even while standing on solid
> >ground.
> >
> You are talking mass again. Weight is a force, Mass is the _quantity_ of
> material in an object.

No, I'm talking weight. Mass is not a force. The acceleration toward Earth
is caused by a force. What force? Weight.

> For the last time - I am not confused.

Yes, you keep saying that. And yet...

> I give up. Get a book on physics or applied mechanics. Perhaps someone
> else might be able to help. You seem to be stuck with your preconceived
> idea as to what weight and mass are, or to put it another way as to what
> a force and a mass are.

So you say. And yet, that's not really the problem here.

> If I mention that force is a vector quantity (weight is a force) and
> that mass is a scalar quantity, I suppose that will mean nothing to you?

I know the difference between a vector and a scalar. So?

> A quote from an A level physics book:
> ~~~~~~begin quote~~~~~~~~~~
> The weight of a body is the force of gravity acting on it towards the
> centre of the earth. Weight is thus a _force_ , not to be confused with
> mass which is independent of the presence or absence of the earth.
> ~~~~~~~~~~~~~end quote~~~~~

How does that contradict my statement that the satellite in orbit DOES have
weight? The satellite in orbit is affected by the force of gravity acting
on it towards the center of the Earth. Your quote defines this as "weight".
By your own quote, the satellite DOES have weight.

> The Gravitational force F between two particles of masses m1 and m2 , a
> distance r apart ,is given by;
>
> F=(G*m1*m2)/r^2 where G is the Gravitational constant.

I am quite aware of that. You'll notice that this gravitational force is
the same as weight. You'll also notice that nowhere in that equation is
there any term that would differentiate between a satellite in orbit and a
satellite sitting motionless at the same distance from the planet. Both
satellites wind up with the same "F", and that "F" is their weight.

So, please...I'd love to hear you try to explain again why it is the
satellite in orbit has no weight.

Pete

[David CL Francis]

On Mon, 15 Dec 2003 at 17:52:06 in message
>, Peter Duniho
>
wrote:

>No, I'm talking weight. Mass is not a force. The acceleration toward Earth
>is caused by a force. What force? Weight.

A large amount of quote is not necessary here. The object in orbit or in
any trajectory around a massive object is in free fall. The path of the
object caused by gravitational attraction means that there is no force
measurable on the object (If you happened to be part of it.) If we want
to be very precise then you might detect the gravitational gradient
across the object if it had significant dimensions. Otherwise, except by
external observations, you would be unable to detect acceleration to
provide 'weight'.

No one denies the force of gravity except in the sense that other
theories claim that the effect of a large mass is to distort space.

I see now at last what you are talking about, but to call the force of
gravity 'weight' seems curious to me. Weight cannot be detected except
when a body is not in free fall. So in orbit an object has no weight.
Gravity enables 'weight' on objects that are on the surface of a body to
be measured. Astronauts cannot weigh things in orbit.
--
David CL Francis

[Dashi]

"David CL Francis" > wrote in message
...
> On Mon, 15 Dec 2003 at 17:52:06 in message
> >, Peter Duniho
> >
> wrote:
>
> >No, I'm talking weight. Mass is not a force. The acceleration toward
Earth
> >is caused by a force. What force? Weight.
>
> A large amount of quote is not necessary here. The object in orbit or in
> any trajectory around a massive object is in free fall. The path of the
> object caused by gravitational attraction means that there is no force
> measurable on the object (If you happened to be part of it.) If we want
> to be very precise then you might detect the gravitational gradient
> across the object if it had significant dimensions. Otherwise, except by
> external observations, you would be unable to detect acceleration to
> provide 'weight'.
>
> No one denies the force of gravity except in the sense that other
> theories claim that the effect of a large mass is to distort space.
>
> I see now at last what you are talking about, but to call the force of
> gravity 'weight' seems curious to me. Weight cannot be detected except
> when a body is not in free fall. So in orbit an object has no weight.
> Gravity enables 'weight' on objects that are on the surface of a body to
> be measured. Astronauts cannot weigh things in orbit.

But can they lay things in orbit?

> --
> David CL Francis

[Peter Duniho]

"David CL Francis" > wrote in message
...
> I see now at last what you are talking about, but to call the force of
> gravity 'weight' seems curious to me.

What else would you call it? And perhaps more interestingly, what is
"weight" if not the force of gravity? I'm curious: if "weight" is not the
force of gravity, what is it? How would you describe "weight" without
noting that it's a force resulting from gravity?

> Weight cannot be detected except
> when a body is not in free fall.

Of course it can. Simply note your acceleration, multiply by your mass, and
you get your weight. The acceleration itself is proof of the force called
"weight". Without a force, there is no acceleration. In free fall, the
*only* force acting on the object is weight.

> So in orbit an object has no weight.

Again, yes it does.

> Gravity enables 'weight' on objects that are on the surface of a body to
> be measured. Astronauts cannot weigh things in orbit.

Of course they can. As I described above, all they need to do is note their
acceleration, multiply by their mass, and they get their weight.

Just because you do not have a handy fixed object against which to measure a
force, that does not mean that the force does not exist. It simply means
that you need something other than a spring attached to a fixed object to
measure the force.

By your definition, if I were to jump from the roof of a two-story building,
for the period of time before I struck the ground, I would be "weightless".
Yet, by any reasonable definition of "weight", a "weightless" object would
not fall to the ground at all.

Pete

[R]

On Wed, 17 Dec 2003 21:40:51 -0800, "Peter Duniho"
> wrote:


>> Weight cannot be detected except
>> when a body is not in free fall.
>
>Of course it can. Simply note your acceleration, multiply by your mass, and
>you get your weight. The acceleration itself is proof of the force called
>"weight". Without a force, there is no acceleration. In free fall, the
>*only* force acting on the object is weight.
>
>> So in orbit an object has no weight.
>
>Again, yes it does.

However, the spacecraft is moving in a straight line along the
geodisc. There is no way to distinguish a "weightless" (freefall)
object moving in a straight line along the geodisc of a gravitational
field from a "weightless" (non-accelerating) object that is nowhere
near a gravitational field. Both travel in straight lines along the
geodisc. What you're saying is dead-on in the Newtonian sense, but
as far as relativity is concerned, an object in freefall is synonymous
with an object not affected by a gravitational field. An object not
affected by a gravitational field by definition has no weight.

The Earth is "accelerating" towards the sun in the Newtonian sense
every day, just as the Sun is "accelerating" towards the center of the
galaxy, but I'm guessing all but the most insufferable graduate
student is willing to leave these non-negligable forces out when they
report their weight on their driver's licence.

-R

[Gene Nygaard]

"Peter Duniho" > wrote in message >...
> "David CL Francis" > wrote in message
> ...
> > I see now at last what you are talking about, but to call the force of
> > gravity 'weight' seems curious to me.
>
> What else would you call it? And perhaps more interestingly, what is
> "weight" if not the force of gravity? I'm curious: if "weight" is not the
> force of gravity, what is it? How would you describe "weight" without
> noting that it's a force resulting from gravity?
>
> > Weight cannot be detected except
> > when a body is not in free fall.
>
> Of course it can. Simply note your acceleration, multiply by your mass, and
> you get your weight. The acceleration itself is proof of the force called
> "weight". Without a force, there is no acceleration. In free fall, the
> *only* force acting on the object is weight.
>
> > So in orbit an object has no weight.
>
> Again, yes it does.
>
> > Gravity enables 'weight' on objects that are on the surface of a body to
> > be measured. Astronauts cannot weigh things in orbit.
>
> Of course they can. As I described above, all they need to do is note their
> acceleration, multiply by their mass, and they get their weight.
>
> Just because you do not have a handy fixed object against which to measure a
> force, that does not mean that the force does not exist. It simply means
> that you need something other than a spring attached to a fixed object to
> measure the force.
>
> By your definition, if I were to jump from the roof of a two-story building,
> for the period of time before I struck the ground, I would be "weightless".
> Yet, by any reasonable definition of "weight", a "weightless" object would
> not fall to the ground at all.
>
> Pete

Weight is an ambiguous word--it has several different meanings.

Francis Weston Sears and Mark W. Zemansky, University Physics,
Addison-Wesley, 4th ed., 1970 used your definition, Pete, but said:

There is no general agreement among physicists as
to the precise definition of "weight." Some prefer to
use this term for a quantity we shall define later and
call the "apparent weight" or "relative weight." In
the absence of a generally accepted definition we shall
continue to use the term as defined above.

Most physicists today use David's definition, however.

Note that in Pete's definition the term "weightless" would never have
entered our language. Sure, there are many places where you would be
weightless in Pete's definition, some closer to the Earth than the
moon is, but we don't go there except in science fiction.

Of course, that only scratches the surface of the ambiguity in the
word weight. Consider all those labels at the supermarket and the
hardware store: "net weight" is not a physics term, and weight in
this context doesn't have either of the definitions you two are using.
That weight does not vary with the strength of the local
gravitational field--and it should not.

Same goes for "atomic weight" and "molecular weight" and "dry weight"
and "carat weight" and "troy weight" and "thrust-to-weight ratios" and
lots of other applications.

--
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
"It's not the things you don't know
what gets you into trouble.

"It's the things you do know
that just ain't so."
Will Rogers

[Gene Nygaard]

On Thu, 18 Dec 2003 02:23:00 -0600, R > wrote:

>On Wed, 17 Dec 2003 21:40:51 -0800, "Peter Duniho"
> wrote:
>
>
>>> Weight cannot be detected except
>>> when a body is not in free fall.
>>
>>Of course it can. Simply note your acceleration, multiply by your mass, and
>>you get your weight. The acceleration itself is proof of the force called
>>"weight". Without a force, there is no acceleration. In free fall, the
>>*only* force acting on the object is weight.
>>
>>> So in orbit an object has no weight.
>>
>>Again, yes it does.
>
> However, the spacecraft is moving in a straight line along the
>geodisc. There is no way to distinguish a "weightless" (freefall)
>object moving in a straight line along the geodisc of a gravitational
>field from a "weightless" (non-accelerating) object that is nowhere
>near a gravitational field. Both travel in straight lines along the
>geodisc. What you're saying is dead-on in the Newtonian sense, but
>as far as relativity is concerned, an object in freefall is synonymous
>with an object not affected by a gravitational field. An object not
>affected by a gravitational field by definition has no weight.
>
>The Earth is "accelerating" towards the sun in the Newtonian sense
>every day, just as the Sun is "accelerating" towards the center of the
>galaxy, but I'm guessing all but the most insufferable graduate
>student is willing to leave these non-negligable forces out when they
>report their weight on their driver's licence.
>
>-R

Bad example. All forces, of whatever kind, are totally irrelevant to
the weight on that driver's license.

http://physics.nist.gov/Pubs/SP811/sec08.html

Thus the SI unit of the quantity weight used in this
sense is the kilogram (kg) and the verb "to weigh" means
"to determine the mass of" or "to have a mass of".

Examples: the child's weight is 23 kg

American Society for Testing and Materials, Standard for Metric
Practice, E 380-79, ASTM 1979.

3.4.1.2 Considerable confusion exists in the use
of the term weight as a quantity to mean either force
or mass. In commercial and everyday use, the term
weight nearly always means mass; thus, when one
speaks of a person's weight, the quantity referred to
is mass.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

[R]

On Thu, 18 Dec 2003 15:30:26 GMT, Gene Nygaard >
wrote:

>Bad example. All forces, of whatever kind, are totally irrelevant to
>the weight on that driver's license.

This is definately the case with most of the women that I know.


> 3.4.1.2 Considerable confusion exists in the use
> of the term weight as a quantity to mean either force
> or mass. In commercial and everyday use, the term
> weight nearly always means mass; thus, when one
> speaks of a person's weight, the quantity referred to
> is mass.

Oh, I don't know. Maybe it's because I live in a country using
english-based units; when someone says they weigh 170 pounds, they
mean weight, as their weight was determined by a device created to
measure weight, not mass. But that's moot anyhow; while you are
correct in that we are entangled in a difference in semantics, the
real difference is not the one that you are referring to. Peter, for
example, is referring to weight in the Newtonian sense. In this
context, weight is a force, and acceleration is proof of that force.
Peter is correct within the framework that he's talking about. I am
referring to weight in the Relativistic sense. In this context,
weight is not a force, it is part of the geometry of spacetime.
Philosophically, Relativity probably holds more weight in this
argument. The whole problem with the Newtonian perspective in this
case is "what perspective do you approach this from"; i.e., what
gravitational fields do you reference when calculating "weight". Do
you leave out the sun? Do you leave out the galaxy? Heck, the galaxy
itself is accelerating. Relativity is not bothered by these
questions. Your own point of view is the only one that matters; if
you are in freefall, if you are moving along the geodisc, you are
"weightless". If you're not, you're not.

Under this definition, the Earth itself has no weight.

-R

[Gene Nygaard]

On Thu, 18 Dec 2003 19:41:06 GMT, R > wrote:

>On Thu, 18 Dec 2003 15:30:26 GMT, Gene Nygaard >
>wrote:
>
>>Bad example. All forces, of whatever kind, are totally irrelevant to
>>the weight on that driver's license.
>
> This is definately the case with most of the women that I know.
>
>
>> 3.4.1.2 Considerable confusion exists in the use
>> of the term weight as a quantity to mean either force
>> or mass. In commercial and everyday use, the term
>> weight nearly always means mass; thus, when one
>> speaks of a person's weight, the quantity referred to
>> is mass.
>
> Oh, I don't know. Maybe it's because I live in a country using
>english-based units; when someone says they weigh 170 pounds, they
>mean weight,

How can that make any difference, when those pounds are legally
defined as 0.45359237 kg, all around the world where pounds were used?
It wouldn't make any difference if the NIST example had used
pounds--the only reason it used kilograms is because SP811 is the
official NIST "Guide for the Use of the International System of Units
(SI)."

What is this country which uses the "licence" spelling and has pounds
on the driver's licence? Does Canada, or some of its provinces, have
pounds on the licence? Anyway, if it matters to you, you shouldn't be
the mystery woman without a name and without an email address, or you
should put that country info into a signature or something.

Sure, there are also pounds force, a recent *******ization, something
never well defined before the 20th century. There are also kilograms
force too--they used to be quite acceptable units. The only
difference is that the metric system is still fully supported and
updated, and those kilograms force are not a part of the modern
version of the metric system, the International System of Units.
Nonetheless, we still see many vestiges of their use--but not in the
use of kilograms for body weight. They are units of mass just as the
pounds are, when used for this purpose in medicine or in sports.


>as their weight was determined by a device created to
>measure weight, not mass.

Nonsense. Sure, we often accept the measurements of a cheap spring
scale as a substitute for what we want to measure. Those scales
aren't very accurate for the measuring either force or mass--if your
mother's scale shows you to be 5 lb more than your scale home showed
an hour earlier, you don't congratulate yourself on a successful
weight loss program, do you?

But what about when you get serious about your weight, and weigh
yourself on one of those platform type beam balances at the doctor's
office or the gym? Those are mass-measuring devices, as are the ones
which promise

HONEST WEIGHT
NO SPRINGS

>But that's moot anyhow; while you are
>correct in that we are entangled in a difference in semantics, the
>real difference is not the one that you are referring to. Peter, for
>example, is referring to weight in the Newtonian sense. In this
>context, weight is a force, and acceleration is proof of that force.
>Peter is correct within the framework that he's talking about. I am
>referring to weight in the Relativistic sense.

No. As I had already pointed out in my earlier reply to Peter,
quoting from Sears and Zemansky, your differences are at a more
fundamental level than differences in how you would define it in
Newtonian physics or using relativity. As S&Z said, "There is no
general agreement among physicists as to the precise definition of
"weight."

The differences they are talking about hinge on things like whether or
not "centrifugal force" is accounted for in your definition of weight.

> In this context,
>weight is not a force, it is part of the geometry of spacetime.
>Philosophically, Relativity probably holds more weight in this
>argument. The whole problem with the Newtonian perspective in this
>case is "what perspective do you approach this from"; i.e., what
>gravitational fields do you reference when calculating "weight". Do
>you leave out the sun? Do you leave out the galaxy?

That applies whether you are using Newtonian physics or relativity.

>Heck, the galaxy
>itself is accelerating. Relativity is not bothered by these
>questions. Your own point of view is the only one that matters; if
>you are in freefall, if you are moving along the geodisc, you are
>"weightless". If you're not, you're not.
>
> Under this definition, the Earth itself has no weight.
>
>-R

Yet Cavendish was successful in "Weighing the Earth" (the title of his
paper, IIRC)--in the same meaning used in commerce and for body
weight, not the different force definitions you and Peter have been
arguing about.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

[R]

On Thu, 18 Dec 2003 23:22:10 GMT, Gene Nygaard >
wrote:

>On Thu, 18 Dec 2003 19:41:06 GMT, R > wrote:
>
>>On Thu, 18 Dec 2003 15:30:26 GMT, Gene Nygaard >
>>wrote:
>>
>>>Bad example. All forces, of whatever kind, are totally irrelevant to
>>>the weight on that driver's license.
>>
>> This is definately the case with most of the women that I know.
>>
>>
>>> 3.4.1.2 Considerable confusion exists in the use
>>> of the term weight as a quantity to mean either force
>>> or mass. In commercial and everyday use, the term
>>> weight nearly always means mass; thus, when one
>>> speaks of a person's weight, the quantity referred to
>>> is mass.
>>
>> Oh, I don't know. Maybe it's because I live in a country using
>>english-based units; when someone says they weigh 170 pounds, they
>>mean weight,
>
>How can that make any difference, when those pounds are legally
>defined as 0.45359237 kg, all around the world where pounds were used?
>It wouldn't make any difference if the NIST example had used
>pounds--the only reason it used kilograms is because SP811 is the
>official NIST "Guide for the Use of the International System of Units
>(SI)."

Point noted; you are correct that legally pounds are measuring lbm,
or pounds of mass.

>What is this country which uses the "licence" spelling and has pounds
>on the driver's licence? Does Canada, or some of its provinces, have
>pounds on the licence? Anyway, if it matters to you, you shouldn't be
>the mystery woman without a name and without an email address, or you
>should put that country info into a signature or something.

On the assumption that you're not trying to open a trolling front,
I'll say that it doesn't really matter who I am or where I am from.
That information is not difficult to divine, anyway, if you know how
to trace header information.

Bitching about spelling is bad form, as well.

>>as their weight was determined by a device created to
>>measure weight, not mass.
>
>Nonsense. Sure, we often accept the measurements of a cheap spring
>scale as a substitute for what we want to measure. Those scales
>aren't very accurate for the measuring either force or mass--if your
>mother's scale shows you to be 5 lb more than your scale home showed
>an hour earlier, you don't congratulate yourself on a successful
>weight loss program, do you?
>
>But what about when you get serious about your weight, and weigh
>yourself on one of those platform type beam balances at the doctor's
>office or the gym? Those are mass-measuring devices, as are the ones
>which promise
>
> HONEST WEIGHT
> NO SPRINGS

They don't measure mass in the absence of a gravitational field,
though I agree that they will give consistent readings any time that
they are oriented against any field of any strength. That is, a scale
the likes that you are talking about (no springs involved) that
returns a value of "100 pounds" on the earth will return a value of
"100 pounds" on the moon as well. As such, you're right -- those
scales measures mass, they just fail to work in a zero-g environment.
But they still measure mass.


>>But that's moot anyhow; while you are
>>correct in that we are entangled in a difference in semantics, the
>>real difference is not the one that you are referring to. Peter, for
>>example, is referring to weight in the Newtonian sense. In this
>>context, weight is a force, and acceleration is proof of that force.
>>Peter is correct within the framework that he's talking about. I am
>>referring to weight in the Relativistic sense.
>
>No. As I had already pointed out in my earlier reply to Peter,
>quoting from Sears and Zemansky, your differences are at a more
>fundamental level than differences in how you would define it in
>Newtonian physics or using relativity. As S&Z said, "There is no
>general agreement among physicists as to the precise definition of
>"weight."
>
>The differences they are talking about hinge on things like whether or
>not "centrifugal force" is accounted for in your definition of weight.

Yes, but, again, centrifugal force is not a force in relativistic
physics, it is only a property of the geometry of spacetime. In the
debate between Peter and David (which I have unceremoniously butted
into), this is important. It demonstrates that both of them may be
correct, depending on which variant of physics you use to view the
problem, and explains a bit why the two of them are so far from
agreeing. In Newtonian physics, what you are saying is important, but
in relativistic physics, "centrifugal force", "gravitational force",
"acceleration" -- these are all the same thing.

>> In this context,
>>weight is not a force, it is part of the geometry of spacetime.
>>Philosophically, Relativity probably holds more weight in this
>>argument. The whole problem with the Newtonian perspective in this
>>case is "what perspective do you approach this from"; i.e., what
>>gravitational fields do you reference when calculating "weight". Do
>>you leave out the sun? Do you leave out the galaxy?
>
>That applies whether you are using Newtonian physics or relativity.

No, really, it doesn't, given the problem at hand. The Newtonian
approach states that one must measure the effects of gravitational
fields to determine weight, even if one is in freefall (orbit). This
can be difficult, as one must take into account forces which one may
not even be aware of or able to measure. Relativity assumes that
anything in freefall is moving in a straight, non-accelerated line
along the geodisc, and thus has no weight. If something is not being
accelerated, it has no weight, as there is no "force" acting on it.

How does Newtonian physics in this instance account for the fact
that we are *accelerating* away from distant galaxies at near the
speed of light? Why not factor in this "repulsive" force to find
one's weight? Certainly the force pushing us away is as valid as the
force that draws us towards massive objects. But then where do you
begin to measure the force? The force is huge if you measure out to
the edge of the observable universe, but virtually nonexistent if you
measure across the solar system. Relativity doesn't have this
problem. I tend to like Relativity to answer this question of weight
because with Newtonian physics the answer to this "weight" question is
arbitrary to the frame of reference that you choose to include at
possible the exclusion of other frames of reference; be it earth, sun,
galaxy, or universe. Relativity assumes that all frames of reference
are equally valid; or, more specifically, the only frame of reference
that is of any matter to you is your own.

>>Heck, the galaxy
>>itself is accelerating. Relativity is not bothered by these
>>questions. Your own point of view is the only one that matters; if
>>you are in freefall, if you are moving along the geodisc, you are
>>"weightless". If you're not, you're not.
>>
>> Under this definition, the Earth itself has no weight.
>>
>>-R
>
>Yet Cavendish was successful in "Weighing the Earth" (the title of his
>paper, IIRC)--in the same meaning used in commerce and for body
>weight, not the different force definitions you and Peter have been
>arguing about.

Henry "weighed the earth" using Newtonian laws and did so long
before the discovery of Relativity. He was correct within the
Newtonian framework; however, often, when dealing with matters of
stellar physics, relativistic physics is the often only way to fly. I
don't really care what commerce and those who divine national
standards think about how "weight" should be described; they have no
use for Relativity (let alone Quantum gravity), as Newtonian physics
is almost completely accurate in measuring their widgets. I'm
interested in the philosophical question of which approach is
"correct" (so far as we can tell) and most useful in describing
reality. Given that relativistic physics is correct in every instance
that Newtonian physics is correct and is also correct in instances
where Newtonian physics breaks down, I suspect that relativistic
physics is the right course to take in answering this "weight"
question. Relativity doesn't ask one to believe in an imaginary
"weight" that can't be directly measured by someone in a closed room
with no windows or external sensors.

-R