Force and displacement are graphically depictable with _curved_
vectors;
which may curve variously to show changes in speed and direction,
whereas
velocity and the one way passage of time, from past to present, toward
the future, require depicting graphically with _straight_ vectors.

This puts a new slant on acceleration, which heretofore has been
depicted as a series of straight lines; where the resultant depicts the
displacement. Now a curved vector can be used directly, to show the
involuted displacement.

"Don1" <[Only registered users see links. ]> wrote in message
news:1109341391.123038.246280@f14g2000cwb.googlegr oups.com...

Hi

Believe one can use a curved vector to present a straight vector and of course
vice versa. (Using straight vectors to show a curved vector or a curve.)

It's just how the formulas involved are written.

Or just compare polar versus rectangular vectors; r, "angle" versus (x, y).
These forms are the curved and the straight forms of any 2 dimensional vector.

So we see that it's not the question of curved versus straight vector really that
is the main problem, but the amount of dimension(s). Expressed by any curved
or straight vector formula.

I think not. However, I see the source of your confusion. There are
really two kinds of physics vectors:

1. Difference vectors, which connect two points. Displacement and
average velocity are two examples of difference vectors. In difference
vectors, the tail is fixed at a starting point and the tip of the arrow
is fixed at the ending point. Thus, if I'm on a 90-degree exit ramp on
the highway, turning from north-bound to east-bound, the displacement
vector for my displacement on the exit ramp is a *straight vector*
starting at the beginning of the ramp and ending at the end of the
ramp, and so pointing 45 degrees northeast.

2. Instantaneous vectors, which apply at a *single* point. Force,
instantaneous velocity, and electric field are all examples of
instantaneous vectors. In instantaneous vectors, the tail is fixed at
the single *point* of application of the quantity represented by the
vector. (Thus, for velocity, the tail sits at the *point* where the
object at that moment.) The direction of the arrow tells you the
direction of that quantity at the *point* where the tail sits, and only
where the tail sits. The length of the vector tells you the size of the
quantity. In this kind of vector, the head (arrow tip) of the vector is
"disengaged" in the sense that it doesn't have to land on anything that
represents the object's location or path or destination. For example,
if I'm slowing down on the exit ramp, the velocity vector at the start
of the ramp is relatively long and points due north, even though only
the tail of the vector is actually touching the ramp. At the midway
point on the ramp, the velocity vector is straight and shorter than
before and pointing northeast. At the end of the ramp, the velocity
vector points due east and is shorter still.

Observe the difference between the vectors and the path. For example,
the displacement in the first example is a straight line pointing
northeast, even though I took a straight path. Also note that if I
strung together the three velocity vectors in the second look at the
example, I would not end up with circular path (because the vectors are
getting shorter) though I know the path is indeed circular. The path is
curved, the vectors are not. Trying to get the vectors to represent the
curved path will lead to trouble.

A similar problem will be encountered when trying to associated a
curved "vector" with a force. I'll give you an example. Take a comet in
an elliptical orbit around the sun. Note that the comet is never headed
directly at the sun -- the sun lies inside the ellipse. However, the
force on the comet that makes it take this path is *always* pointed
*directly* at the sun. The force does *not* lie on the path the comet
takes. You'd have to explain to me how to draw the force as a curved
arrow for this comet. I can see how to draw a curved path for the
comet. I can't see how to draw a curved force for the comet. Which
direction does that curve go?

One final note. The direction of a force vector does not tell you what
direction the object is going to go. It tells you how the "go" is going
to *change*. A simple example is braking in a car to a stop. The force
on the car is *backwards*, though the car never goes backwards during
the braking.

> Believe one can use a curved vector to present a straight vector and of

Vectors are not curved. They have magnitude and direction. Magnitude has no
curve (duh). Direction has no curve either. A path can be curved, but a path
is not a direction. A path, at any point has a tangent that indicates its
direction at that point. A curved path has a direction that changes from one
point to another.

What formulas?

Polar coordinates are not curved. Just as with cartetian coordinates, polar
coordinates are represented as am ordered set of numbers. These numbers are
not curved (duh). In particular, an angle is not curved, it is just a
number.

The amount of dimension(s)? Did you lick a cane toad?

"Don1" <[Only registered users see links. ]> wrote in message
news:1109341391.123038.246280@f14g2000cwb.googlegr oups.com...
You must be referring to a parametric curve. Each parametric curve could be
considered to have a multitude of straight vectors in it. Vectors are
straight. Collections of vectors need not be straight. Planets would
travel in straight lines if it were not for the gravity of the sun pulling
them into a circular or rather elliptical orbit.

"ošin" <oš[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

Hi

The topic is "Curved vectors".

No "vectors" are curved by math definition.

So, what does the topic implement?

To me it implement the non linear path.

A carrier can has a linear path, a orbital or a mix of these two paths. OK.

So, what's the fuss about?

As I see it all depends on what you take as a STANDARD path? - The mathematical
linear path or the mathematical orbital path? You don't take the mix by logical reasons.

Don is right - to my point of view - by asking a good question.

The scientific community has taken the linear path as a mathematical STANDARD.
But you can by all means take the orbital path as a STANDARD.

Both STANDARDS are mathematically correct, but different by the stock you choose.

What can you do with curved vectors that you can't do with straight
vectors?

Perhaps you should whip up a little textbook developing your ideas and
showing all the applications and advantages of your curved vectors.
With math you really have to show people how it's done before they can
grasp it and see the usefulness of it.

Curved vector analysis might turn out to be something novel and
interesting, like Spaceman's algebra.

Just because people have always thought of vectors as straight, doesn't
mean they always are.

Alan Keyes used to always think his daughter was straight too!