Can I set my own bindings?

On Feb 21, 4:07 pm, (Jeff Davis) wrote:
In article ,

klaus wrote:

No. Kinetic energy is defined as 1/2 mv^2. The slab can lose
elasticity (stiffen) without moving. A moving slab does not have zero
elasticy except at the interface, which is no longer a slab. Therefore
kinetic energy is zero in both cases. Quoting from a dictionary does
not prove your point. It just makes you look like you don't lnow what
you are talking about.

The slab creeps before it releases and stores elastic energy. It most
certainly has kinetic energy due to your own equation. Nice to have an
intelligent conversation relevant to skiing.
--
According to John Perry Barlow, "Jeff Davis is a truly gifted trouble-maker."

"Elasticity" and "elastic energy" are not the same thing.

Jeff Davis wrote:
In article ,
klaus wrote:

No. Kinetic energy is defined as 1/2 mv^2. The slab can lose
elasticity (stiffen) without moving. A moving slab does not have zero
elasticy except at the interface, which is no longer a slab. Therefore
kinetic energy is zero in both cases. Quoting from a dictionary does
not prove your point. It just makes you look like you don't lnow what
you are talking about.

The slab creeps before it releases and stores elastic energy.

Yes. But this has little to do with kinetic energy. Creep is quasi-static. velocities by difinition are negligible.

It most
certainly has kinetic energy due to your own equation. Nice to have an
intelligent conversation relevant to skiing.

Only when it begins to move. Not due to it's change of elasticty as
you initially asked. Change (increase) of elasticity of a slab does
*NOT* change a slab's kinetic energy. Period. You're playing a
game. It's like saying, does a change of a spring's spring constant
change its kinetic energy, and then when the responder says no.. you
say.. "Wrong, the end broke free and it contracted." It's bull**** and
nonsensical..

-klaus

In article .com,
Richard Henry wrote:
On Feb 21, 4:07 pm, (Jeff Davis) wrote:

The slab creeps before it releases and stores elastic energy. It most
certainly has kinetic energy due to your own equation. Nice to have an
intelligent conversation relevant to skiing.

"Elasticity" and "elastic energy" are not the same thing.

No **** Sherlock. Where did I type that? Thanks for the hot ****ing tip
Dickie.
--
According to John Perry Barlow, "Jeff Davis is a truly gifted trouble-maker."

In article ,
klaus wrote:
Jeff Davis wrote:
In article ,

The slab creeps before it releases and stores elastic energy.

Yes. But this has little to do with kinetic energy. Creep is
quasi-static. velocities by difinition are negligible.

We're measuring that movement. It's observable. We can calculate a
value for v and plug it in to your equation and calculate a real value.

It most certainly has kinetic energy due to your own equation. Nice to
have an intelligent conversation relevant to skiing.

Only when it begins to move. Not due to it's change of elasticty as
you initially asked. Change (increase) of elasticity of a slab does
*NOT* change a slab's kinetic energy. Period. You're playing a
game. It's like saying, does a change of a spring's spring constant
change its kinetic energy, and then when the responder says no.. you
say.. "Wrong, the end broke free and it contracted." It's bull**** and
nonsensical..

We're discussion a dynamic, constantly changing snowpack. It seems to
me when the slab slides, it has no elasticity. It can't regain its
original shape. Nice anallogy on the spring breaking. You picked right
up on what I was questioning.
--
According to John Perry Barlow, "Jeff Davis is a truly gifted trouble-maker."

Jeff Davis wrote:
In article ,
klaus wrote:
Jeff Davis wrote:
In article ,

The slab creeps before it releases and stores elastic energy.

Yes. But this has little to do with kinetic energy. Creep is
quasi-static. velocities by difinition are negligible.

We're measuring that movement. It's observable. We can calculate a
value for v and plug it in to your equation and calculate a real value.

No you can't. As long as the slab is in place, a change of elasticity
does not imply motion. Internal stress may increase, but that doesn't
mean it moves. Only when the boundary conditions change does motion
occur, and this is independent of elasticity other than reaching thee
breaking point of internal stress and the brittle qualities of the
slab..


It most certainly has kinetic energy due to your own equation. Nice to
have an intelligent conversation relevant to skiing.

Only when it begins to move. Not due to it's change of elasticty as
you initially asked. Change (increase) of elasticity of a slab does
*NOT* change a slab's kinetic energy. Period. You're playing a
game. It's like saying, does a change of a spring's spring constant
change its kinetic energy, and then when the responder says no.. you
say.. "Wrong, the end broke free and it contracted." It's bull**** and
nonsensical..

We're discussion a dynamic, constantly changing snowpack. It seems to
me when the slab slides, it has no elasticity.

Wrong. The interface has infinite elasticty. The slab remains unchanged.

It can't regain its
original shape.

If it is a slab, it never lost it. Much. Only where it breaks does the
elasticity change. Elasticity does not change instantaneously. Ask
your dad about crystals and if they instantaneously change elasticity
without a phase change occuring.

Nice anallogy on the spring breaking. You picked right
up on what I was questioning.

The interface of the spring to the anchor does not effect the spring
constant of the spring.

-klaus

In article ,
(Jeff Davis) wrote:

In article ,
Alan Baker wrote:
In article ,
(Jeff Davis) wrote:

Oh. How does an airplane wing change mass?

It doesn't. A wing encounters m mass of air each t unit of time. To that
air it imparts downward velocity.

F = dm/dt * v

I still don't see where the mass changes. Lift is a function of the
ideal gas law in respect to pressure.

We'll take this one step at a time, OK? Just tell me if you agree or
disagree with each statement.

The airplane experiences a downward force due to gravity.

In level flight (to use the most straightforward case) that force must
be balanced by a force that the air (since it is all that is touching
the plane) exerts on the airplane.

Since Newton's laws tell us that any force exerted on a body is also
returned, the plane must also exert a force on the air.

Since the plane only momentarily encounters any particular parcel of
air, the force the plane exerts on the air cannot be accelerating a
fixed mass as the usual, F = ma, formulation suggests.

Instead, it is encountering a *flow* of air. A constant rate of mass
with respect to time is being changed by a fixed velocity.

Hence F = dm/dt * v, instead of F = m * dv/dt


--
According to John Perry Barlow, "Jeff Davis is a truly gifted trouble-maker."


--
"The iPhone doesn't have a speaker phone" -- "I checked very carefully" --
"I checked Apple's web pages" -- Edwin on the iPhone and how he missed
the demo of the iPhone speakerphone.

In article ,
(Jeff Davis) wrote:

In article ,
Alan Baker wrote:

Nope. You've yet to provide a cogent explanation of your statement.

What the **** is this **** dumbass? I asked klaus a question. We have
delusion, do you want to ad narcissism as well to your list of defects?

Your question was not clear.

Your explanation was lacking.

And now, out comes the vitriol (you have learned to spell it by now,
right?)

You can add anything you like to the list you keep in your head of my
supposed "defects".

Just as I can believe what I want to believe.

g

--
According to John Perry Barlow, "Jeff Davis is a truly gifted trouble-maker."


"-- "

--
"The iPhone doesn't have a speaker phone" -- "I checked very carefully" --
"I checked Apple's web pages" -- Edwin on the iPhone and how he missed
the demo of the iPhone speakerphone.

In article ,
VtSkier wrote:

frankenskier wrote:
On Feb 19, 4:46 pm, VtSkier wrote:
Walt wrote:
VtSkier wrote:
Walt wrote:
I cannot find a single definition of torque that
doesn't require motion that is either happening
or is about to happen.
That's odd, since I've only posted it about 5 times. Here it is again:
T = r x F
where F is the force and r is the moment arm vector.
Now, what, exactly, is in motion here? A force, acting on a moment arm
produces torque. We agree that it's possible for a force to exist
without motion. The above definition shows that a stationary force will
produce torque.
I really can't make this any clearer.
If I didn't know you better I'd say that you were just trolling.
//Walt
It goes to the definition of VECTOR. My reading, which I
posted, it that a vector has magnitude and direction.
Those are the qualities which create a vector, no?

Magnitude is usually expressed as a unit of length.

Force is expressed as a unit like pounds or newtons.
Force, by this definition is static. It takes
movement to make force into work. Or torque.

If you multiply a unit by another unit, you have
created yet a third unit with the first two units
as components of the third.

If you multiply a force (weight) unit by a length
unit you have a third unit that has both weight
and length as components.

if the first two units were pounds and feet the
third unit will be pound-feet. This is a unit
that requires that a force be moved a distance.
If you specify the direction of the distance
and/or add leverage that unit is called TORQUE
as opposed to simply WORK. But the force still
has to move a distance.- Hide quoted text -

- Show quoted text -

On this one, sorry but you're just wrong. Consistently wrong. As in,
you've been wrong every single time, on every post in this thread.

Except for the few instances where you've agreed with Walt.

Oh? Well so far nobody has been able to show me where I'm wrong.

You included. You've simply said that I'm wrong and I'll discount
your statement until you can back it up.

Walt at least has been very patient in trying to show me by
his words and others (cites) where I'm wrong. And I STILL
haven't gotten it yet.

Klaus chimed in to try to shed a little light with his
explanation of component torque which may well be where I've
been trying to go.

It's very hard for me to say that the following is a false
statement:

"Total torque is zero, therefore there is no torque."

Component torque, WTF does that mean? A few more words
would be helpful here. Does it mean that there is a
component OF torque being applied? Since there are
only two parts to torque (at least in the case of
tightening or loosening a nut), force and vector,
does it mean that force only is being applied to the
lever arm which creates the vector until the nut moves
and the vector exists?

Torque is analogous to force in linear motion.

Would you say that you are exerting no force on an object if you're
pushing against it, but it doesn't move? No.

You'd be exerting a force, but something else must be exerting a force
in the opposite direction to balance the total force on the object.

--
"The iPhone doesn't have a speaker phone" -- "I checked very carefully" --
"I checked Apple's web pages" -- Edwin on the iPhone and how he missed
the demo of the iPhone speakerphone.

Walt wrote:

The Real Bev wrote:
Walt wrote:

...And someday I plan to get the other two volumes
of Feynman....

There are THREE? I thought there were only two.

Yup. Three. The first volume is on mechanics & thermo , the second on
E&M and the third on quantum.

See http://en.wikipedia.org/wiki/The_Fey...res_on_Physics

I've got volume 2, which I picked up at a garage sale for a buck a
decade or two ago. I've been looking for other underpriced stray
volumes ever since. I call this the Bev method of library acquisition.

We cleaned out a pile of books. We filled a 100-gallon recycling container
with stuff like old DOS manuals and other books that nobody will ever want
to read or buy or even be given. We gave (while they were closed, of
course) an equivalent amount to the library. Books proliferate beyond all
possibility of utility. I am intensely grateful for the space. Only about
200x that amount left to go...

Allen has Vol 1. Friend has Vol 2. Both are keeping theirs. This has
caused a certain amount of friction.

A friend told me a Feynman story yesterday.

Caltech has Freshman Camp up in the mountains for all new students a week or
so before school starts. The friend, just arrived from Shanghai, brought a
dufflebag full of his stuff, including his sleeping bag. Unfortunately, he
had left the key to the padlock home. Somebody said "Don't worry, Feynman
is here." After dinner, Feynman opened the lock with a screwdriver and a
paperclip. It took only a few seconds.

Too bad he turned his accent on and off depending on the circumstances.

--
Cheers, Bev
---------------------------------------------------
I have no idea what you're talking about, so here's
a bunny with a pancake on his head:
http://www.ebaumsworld.com/forumfun/misc15.jpg

Richard Henry wrote:

On Feb 21, 10:18 am, Walt wrote:
The Real Bev wrote:
Walt wrote:

...And someday I plan to get the other two volumes
of Feynman....

There are THREE? I thought there were only two.

Yup. Three. The first volume is on mechanics & thermo , the second on
E&M and the third on quantum.

Seehttp://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics

I've got volume 2, which I picked up at a garage sale for a buck a
decade or two ago. I've been looking for other underpriced stray
volumes ever since. I call this the Bev method of library acquisition.

I've been looking for the books of the MIT Radiation Lab series. So
far, I have one.

(Maybe I should look on eBay...)

How about the Amazon resellers? We just bought an older tech book from a
guy in India at a not-unreasonable price.

--
Cheers, Bev
---------------------------------------------------
I have no idea what you're talking about, so here's
a bunny with a pancake on his head:
http://www.ebaumsworld.com/forumfun/misc15.jpg