Path Integral question

[Lea-disabled]

Hi All,

Please see the enclosed diagram: -
- Take a cube with dimensions 6m x 6m x 6m. ( m = meters )
- I am trying to setup a function that describes the length of the path line at any point in time as the line begins its decent from the top of the side of the cube traveling around all 4 side of the cube.

- I think the cylindrical co-ordinates are the most appropriate for this function.... because...

- � = starting from the top - the decent is a constant rate downwards to the bottom of the cube @ 6m units height per 1 unit of time ( say 1 hour )

- ϕ = the angular velocity is also constant @ 1 complete rotation per 1 unit of time ( say 1 hour )

- r = the radius from the center of the cube to the outside of the cube is, however, constantly increasing & decreasing according to the position on the 360 circle.
please see diagram. The radius increases & decreases in each quadrant as the path line descends around the outside of the cube.

..................................................
I hope the enclosed makes sense.

Please could anyone help me find the correct function to solve this problem.

Many thanks for your help & attention.

Best regards,
Lea Rebanks

[Lea-disabled]

Dear ALL,

Many thanks for all your replies & effort to my Path line question, again.

- It seems that I have caused quite a stir & confusion from my inability to describe my problem correctly.
- For jmG - the problem is not from a book or a physical problem. But is actually my simplified conceptual attempt at a more complex proof which would certainly cause much greater confusion if I tried to explain.
- I thought jmG�s comment of "Path line" is strictly not English or American, a pure farmer expression ... "the length of line". Absolutely hilarious & I deserve that for my inability to state my problem correctly within the letter of the law of maths in the correct contractual form.
- Thanks to jmG for the 0_3D_CUBE scatter & the Path Integral Example_final(1) files which are really appreciated, but unfortunately not what I am looking for here.
- There are no cylinders involved or helix spiral wanted here. Only a cube with a straight line moving around the four sides at constant angular velocity. ( See Fred�s point in a moment.)
- >>> I am looking for a straight line around the edge of a cube descending from the top to the bottom of the length as Lou pointed out - The path length is that of the hypotenuse = sqrt(6^2+24^2) = 24.74.
- >>> However as Fred said � �If the angular velocity is constant and the vertical (dz/dt) velocity is constant, then the horizontal velocity will vary with the distance from the axis of rotation. You can keep a "straight line" and solve for the horizontal velocity required to keep it.� THIS IS WHAT I WANT.
- I want to solve for the horizontal velocity required to keep the straight line. Can someone show me this with a 3D diagram.
- As Richard said � �I think Lou is assuming a constant downward velocity and a straight path, in which case the angular velocity cannot be constant. You are looking at the case where the angular velocity is constant and the path is straight, in which case the downward velocity is not constant. I solved for the case where the angular velocity and downward velocity are constant, but the path is not straight. Pick any two.� --- I am asking here - If I have to pick one constant to keep the straight line then I want to retain the constant angular velocity.
- Can someone show me this with a 3D diagram showing the straight line on the cube.

Many thanks for help & attention.
Best regards,


Lea Rebanks �

[RichardJ]

OK. That's not a big change. You know z must be linearly related to x and y (otherwise the path cannot be straight). It's just some very basic geometry to work out that relationship.

Richard

[TomGutman]

With the path a straight line the horizontal, vertical, and along path displacements are all proportional to one another. Thus the length of the path at any time is proportional to the vertical displacement. Having calculated the Z coordinate as a function of time, one need only multiply this by a suitable constant (the total length of the line by geometry divided by the Z distance of 6) to get the length of the line at that time. No calculus, fast or slow implementation, needed.
__________________
� � � � Tom Gutman

[RichardJ]

True, but that would make it too easy 😉

Richard

[Lea-disabled]

Dear Richard,

Perfect! That�s exactly what I was after with the whole calculation & 3D diagram.
And thanks so much for the additional notes from jmG, really appreciated, cheers�

Best regards,
Lea �

[ptc-1368288]

On 1/26/2009 3:37:43 AM, Lea wrote:
>Dear Richard,
>
>Perfect! That�s exactly what I
>was after with the whole
>calculation & 3D diagram.
>And thanks so much for the
>additional notes from jmG,
>really appreciated, cheers�
>
>Best regards,
>Lea �
___________________________

Good thing: we all found the same travelled distance. I disagree with Richard to cut and paste differential elements. Mathcad differentiates and integrates discontinuous functions, as demonstrated.
Finally, you wanted the "cumulative arc length function of the angular position of a punctual source descending the Z axes at constant velocity".
I saved the work sheet, cleaned it and tutored adequately for future collaboration. It does not have much use as such, but good didactic.

Obviously, the "calculus method" attached is slower than the "cut and paste" . The later is not technical and should be rejected w/o dispute.

The project can be resumed:
Calculate the "angular Pyhagoras in the X, Y, Z space"

jmG

[ptc-1368288]

Lea,

I understand Fred proposal or interpretation, i.e: the path of travel is straight, but the speed is not. Now you are integrating the "time line" which is a line integral in the general sense of line integral. That is a bit different story, but as simple,. The function that will produce the line is now the sec(x) over the 8 quarters, each varying 0... 45� . Them sum them as Richard did in a discontinuous function (piecewise) and you plug that in the phi(t) in the last work sheet. But the length will not change. At this point, you can size both the discontinuous and the straight line to either one of them by scaling, scaling Chebyshev type but more general Cheby is � 1.

A line has lot of mathematical definitions, thus the unique "line integral", which is the definition of the integral . The path of integration may not be over from end to end, so the path of integration describes the line. I understand that by nature english is not very analytical and american english is worst with their habit of making expressions like 10 lines long german words. "Path integral" means strictly nothing, but "path OF integration" has now full meaning. In fact, the best english is the one from non native english speaking countries. Curiously enough, the best "reference english" is the Canadian one as well as the Canadian french. Historically speaking, these two speakers were abandoned by their kings and left alone. That is only true for Canadians not too "Americanised".

I had the discontinuous sec(x) done but zapped it, being sure it didn't purport your project. I might give it a go again, but don't trust me 100% yet. Drawing on the cube seems feasible. It would require drawing the 4 planes and project each sec(x) segment. But I can tell you in advance you will hardly see more than 4 straight lines.

jmG

[ptc-1368288]

Things are all upside down. Just to demonstrate the principle of wrapping a function over a plane .

jmG

[AlvaroDíaz]

Lea: "Actually I don't know the difference between Line Integral / Path Integral. Are they the same thing?"

In spanish books the difference is this: line integrals refers a geometrical curve (and in geometry there are not times), and path integral a 'physical' one, and in this case the 'point' (a particle with mass) can pass several times for the same point.

In the parametrization of a 'geometrical' curve then, you must restrict the parameter to some interval and not cover the same point more than one time (with punctual exceptions).

In 'physical' curves the parameter is usually the time, and there are not restrinctions in the parameter interval.

The trayectory is defined as the geometrical figure, this is, the geometrical curve described for the particle, and the path name is reserved as the actual position of the particle as function of the time.

For example, a segment in the geometrical sense is very distinct from a physical segment, and the point can cover this trayectory in stranges forms.

That, in spanish literature, and in the curve theory context.

Alvaro.

[ptc-1368288]

That's right Alvaro:

The line integral is a definition and the "path integral" defines the application (so to speak). In better words, "line integral is a scalar function" unlike the "path of integration" that defines the "geometry of the line" so to speak again).


jmG

[ptc-1368288]

... back to what you commented in

http://collab.mathsoft.com/~Mathcad2000/read?119741,16

The "line integral" over the path will be integrating the feston of the 8 segments. You may have to slope the feston but you mentioned only "angular speed" or something of the sort. I made the feston quick, discretized to be integrated by finite differences. I you have the patience to put it in discontinuous form, that would be more elegant and a bit more accurate.

So, at any time (t) in the Cartesian X you will have on the Cartesian Y the "cumulative line integral" over the path at this point in time.

After all, quite interesting ! if of any use ?

jmG

[ptc-1368288]

... the way I understand the problem as applied is like this. A gun is shooting paint from the center of a cube at constant "angular velocity", starting at the top face of the cube. For the spots to be equally spaced, the jet will have the unit velocity at right angle to the side face of the cube, then as the gun rotates, the spot will have higher speed as per sec(,) [secant(,)] up to 45� (pi/4). Then decreasing speed again down to unit speed ... and so on until the 2 pi rotation is complete. The path is the sec(x) feston. The cumulative integral will mean something if for each "elemental travel" something is associated to it like $. So, the cumulative integral is a scalar.
That is incomplete whereas the gun travels a longer distance than simply "rotating flat". So, you will have to add the cumulative integral based on the Pythagore distance already calculated in the previous sheet(s). I didn't bother making a function for the path of the sec(x). It would have to go to the numerical integrator and would be probably very close or lot worst than the finite differences method attached.

Don't misinterpret, Lea: very interesting but confusing. If the problem could have been done before computers, the same method is today faster in man/hour but not different in the principles.

jmG