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

[TomGutman]

Your description of a constant vertical velocity and a constant angular velocity is inconsistent with your diagram of the unfolded cube showing a straight line from corner to corner. Please specify which two of these three descriptions are accurate and which is wrong.
__________________
� � � � Tom Gutman

[RichardJ]

Assuming your statements are correct, but not your diagram, here's the (or a) way to get the solution.

Addendum: I just noticed the horrendous file size. One case where XMCDZ format is much smaller, so I've attached that too.

Richard

[ptc-1368288]

On 1/16/2009 3:58:57 PM, rijackson wrote:
>Assuming your statements are
>correct, but not your diagram,
>here's the (or a) way to get
>the solution.
>
>Richard
_______________________________

I don't understand the 2nd diagram and the bottom one is totally incorrect. The trajectories add together, they have unequal length that must be calculated individually then summed. Thus the total path is a straight line of the form L = a + b*x. It does not involve any integral at all.

I bet there is an obscure subtility.
In literature "line integral" is that what you call "path int'l" ?

jmG

[ptc-1368288]

A much smaller size sheet, same quality .

jmG

[RichardJ]

On 1/16/2009 11:38:38 PM, jmG wrote:
> The
>trajectories add together, they have
>unequal length that must be calculated
>individually then summed.

I'm not sure what you mean by "trajectories". The path along each side of the cube? If so, they must all have equal length, because the angular velocity is constant.


>Thus the total
>path is a straight line of the form L =
>a + b*x. It does not involve any
>integral at all.

If the angular velocity and the rate of change in the z direction are both constant, then the path cannot be a straight line.

>In literature "line integral" is that
>what you call "path int'l" ?

As far as I'm aware path and line integrals are the same thing.

Richard

[ptc-1368288]

On 1/17/2009 9:12:20 AM, rijackson wrote:
>On 1/16/2009 11:38:38 PM, jmG wrote:
>> The
>>trajectories add together, they have
>>unequal length that must be calculated
>>individually then summed.
>
>I'm not sure what you mean by
>"trajectories". The path along each side
>of the cube? If so, they must all have
>equal length, because the angular
>velocity is constant.
>
>
>>Thus the total
>>path is a straight line of the form L =
>>a + b*x. It does not involve any
>>integral at all.
>
>If the angular velocity and the rate of
>change in the z direction are both
>constant, then the path cannot be a
>straight line.
>
>>In literature "line integral" is that
>>what you call "path int'l" ?
>
>As far as I'm aware path and line
>integrals are the same thing.
>
>Richard
___________________________________

"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".
"Z = starting from the top - the decent is a constant rate downwards to the bottom of the cube ..."
-"f = the angular velocity is also constant @ 1 complete rotation per 1 unit of time ( say 1 hour )".
_________________________

I don't understand what the angular velocity is doing there ?
Angular velocity of what ?
Then ignore angular velocity as it may later on represent torque of some kind if under a force of some kind too. The total length is the sum of the 4 segments, for which there are no side coordinates. Therefore, the cartesian representation will be a + b*x where x spans over the 4 corner lines. The line integral of the a + b*x will just be a number. It seems Lea wants the cumulative integral, i.e: the value of the line integral at each point of the path. It will then be a parabola, which parabola (a function) can then be further exploited ... for instance : how much petrol to push the truck starting from the bottom up to a certain point.
That's my understanding from the imagined above statements of the project.

"Path Integral" is probably as good as "Line Integral" . To me , "Line Integral" is purely scalar, unlike "Path Integral" suggestive of a relation between the "scalar" and other functions ... etc.

The project is not very explicit in sizable words, and the picture it prints in my mind is as exposed, I may be wrong but not guilty .

jmG

[RichardJ]

On 1/17/2009 1:04:37 PM, jmG wrote:

>I don't understand what the angular
>velocity is doing there ?
>Angular velocity of what ?

Take an axis up the center of the cube (the z-axis). Then take a line that is perpendicular to the axis, parallel with the top face of the cube, and intersecting the axis and one vertex on the top face. Now rotate the line about the axis, and let the intersection with the z-axis move so that in one full revolution of the line the line drops the height of the cube. The rotation rate of the line (i.e. angular velocity) and the rate of movement of the intersection of the line and the z-axis are both constant. The intersection of the rotating line and the faces of the cube forms a 3D path. That path is not a set of straight lines.

> It seems Lea wants the
>cumulative integral, i.e: the value of
>the line integral at each point of the
>path.

That's true. A minor modification to the worksheet. See the attached.

Richard

[TomGutman]

Repeated evaluation of an integral tends to be slow. Solving an ODE is much faster. Here I've increased the resolution of the length plot to a thousand points (the default for quickplots) to accentuate the speed difference.
__________________
� � � � Tom Gutman

[ptc-1368288]

I understand XX(1406), yours too , but one way or another the length of the path is 3*6+(6�+6�)^� = 26.4853 ... not 24.XXX. The error results from not vectorizing Z(theta).
Like said before, no integral is involved as the last value of the "cumulative path integral" is the same as the previous path integral. The travelling is in straight line, therefore the "cumulated travelled distance is a straight line too.

The problem is surely not stated this way and we don't know what is what in there. Interesting to follow.

jmG

[TomGutman]

>>The travelling is in straight line<<

If one believes the text (rather than the picture) the travelling is not in a straight line, neither overall nor on any individual face.

Nor does your expression reflect travelling in a straight line (on the unrolled cube). Rather it show travelling at a fixed Z (contradicting the constant dZ/dt in the conditions) for three faces and then down the diagonal of the fourth face. The matches nothing in the problem statement, neither the stated conditions nor any of the diagrams.

Thud. Thud. Thud. Over and out.
__________________
� � � � Tom Gutman

[ptc-1368288]

On 1/18/2009 3:01:11 AM, Tom_Gutman wrote:
>>>The travelling is in straight line<<
>
>If one believes the text
>(rather than the picture) the
>travelling is not in a
>straight line, neither overall
>nor on any individual face.
>
>Nor does your expression
>reflect travelling in a
>straight line (on the unrolled
>cube). Rather it show
>travelling at a fixed Z
>(contradicting the constant
>dZ/dt in the conditions) for
>three faces and then down the
>diagonal of the fourth face.
>The matches nothing in the
>problem statement, neither the
>stated conditions nor any of
>the diagrams.
>
>Thud. Thud. Thud. Over and
>out.
>__________________
>� � � � Tom Gutman
______________________________

"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".

...traveling around all 4 side of the cube"

That is the statement of the problem.

>Thud. Thud. Thud. Over and out.<<br>
There is nothing specifying more than what is says. The cube is a solid, there is no cylinder involved. If the cube is a parallelepiped, same applies: by Pythagore. The problem might look different if there would be work done by a particle ... etc.
"...the length of the path line a any point in time". Your terminal "arc length" is < the Pythagorean length, it can't be so.

jmG

[Lea-disabled]

Dear Richard, Tom & jmG,

Many thanks for all your replies & effort to my Path line question.
- Actually I don't know the difference between Line Integral / Path Integral. Are they the same thing?
- I did my drawing to try to explain what I am trying to do & it still was ambiguous & caused confusion. So apologies for that.
- Your replies have caused me to reconsider my explanation as I still haven't made myself clear. ( However, you all may be tied of this problem by now .)
- Actually Richard's reply, many thanks Richard, most closely resembles what I am looking for� I.E. The LengthPathIntegral(t) = an increasing undulating line.
- However I failed to convey that I want the physical path taken from the top of the cube to the bottom to be a straight line. This is an important part of what I am trying to do.
However the length of this line LengthPathIntegral(t) increases at the undulating rate due to the fact that is moving around the side of a cube @ constant Ang vel. From the center of the cube.)
- That was interesting that Richard & everyone was inclined to provide the function as a Cartesian parametric function. Because I thought the problem lent it's self better to the cylindrical system. Because to the z = constant descent.
Theta = angular velocity with which the path travels around the cube, as it descends from top to bottom, is constant.
Therefore the radius = constantly increasing & decreasing so that z & the Ang vel. remain constant & the line path on the surface of the cube remains straight....surely this is possible!

Many thanks for your help & attention.
Best regards,
Lea Rebanks �

[TomGutman]

>>Therefore the radius = constantly increasing & decreasing so that z & the Ang vel. remain constant & the line path on the surface of the cube remains straight....surely this is possible!<<

It is very much not possible. As I said in my first post, you have to choose two of the three conditions. The three conditions are inconsistent with each other.
__________________
� � � � Tom Gutman

[ptc-1368288]

"Dear Richard, Tom & jmG,"

Lea,

A path integral would be a double integral, suggesting that along the arc length cumulative integral, another physical something would operate. Read more:
1. Line integral independent of the path. Line integral of Gradients
2. Line integral in the space.
3. Line integral in the plane.
4. Line integral in complex plane
5. Cauchy, Green, Stokes ... theorems.

Line integral would be the inner cumulative integrand of the 2nd cumulative integrand of a double integral. All what that means is that a line integral is not synonymous of path integral.
That something moves at different inclination at constant speed: that I understand. But you should give the entire problem it was stated. Words, providing they are correct and meaningful have legal (contract) values, unlike drawings that are just drawings. The Cartesian Y seems to be the "time" (t) as the inner cumulative integrand, then associated with the time is something unknown.

Resume: the Cartesian X spans 0... to the Pythagoran total length of the travelled distance in 4 segments corresponding to each corner of the cube. On each of the segment, then the Cartesian Y which is (t). That will give a discontinuous plot of linear segments, that discontinuous function you will integrate. The overall task is via Odesolve of two functions, but you have not provided the 2nd function neither the points in the cube. But for the points in the cube, you can generalise the discontinuous function for a general solution.

I don't understand "angular velocity", as you say it is constant. If you mean "angular velocity around the cube", sure it it angular as a cube is rather angular but does not involve angle or functions of angle(s) !

Back to the design board.
It's all there, under the fingers.
Get inspired from the attached.

jmG

[RichardJ]

On 1/18/2009 9:51:12 PM, Lea wrote:

>- That was interesting that
>Richard & everyone was
>inclined to provide the
>function as a Cartesian
>parametric function. Because I
>thought the problem lent it's
>self better to the cylindrical
>system.

You wanted the path on the faces of a cube, and anything on the faces of a cube is very easily described in a Cartesian coordinate system.

>Theta = angular velocity with
>which the path travels around
>the cube, as it descends from
>top to bottom, is constant.
>Therefore the radius =
>constantly increasing &
>decreasing so that z & the Ang
>vel. remain constant & the
>line path on the surface of
>the cube remains
>straight....surely this is
>possible!

No. As Tom says, it is not possible. Forget the changing z position, and just think of a simpler case in which only the angle changes. The path traced out by the intersection of the line and the cube is then obviously a set of straight, horizontal, lines. The position on the cube face is not linearly related to angle, however, so if the angle of the line changes at a constant rate the length of the lines do not increase at a constant rate. If you now add in a z component that changes at a constant rate it should be obvious that the resulting line cannot be straight.

Richard

[ptc-1368288]

>The path traced out by the intersection of the line and the cube is then obviously a set of straight, horizontal, lines<<br> ___________________________

No, the time (t) is the Cartesian X and the travelled distance plots on the Cartesian Y, therefore ... like a saw tooth going up. Where all Y's values will the Pythagore distance deduced from the points at the 3 corners + the last one SQRT(_�+_�)

jmG

[ptc-1368288]

Adapt to your project.

jmG

[RichardJ]

On 1/19/2009 10:17:41 PM, jmG wrote:

>No, the time (t) is the
>Cartesian X and the travelled
>distance plots on the
>Cartesian Y, therefore ...

No. The time is angle. Both x and y change with time.

Richard

[ptc-1368288]

On 1/20/2009 8:29:49 AM, rijackson wrote:
>On 1/19/2009 10:17:41 PM, jmG wrote:
>
>>No, the time (t) is the
>>Cartesian X and the travelled
>>distance plots on the
>>Cartesian Y, therefore ...
>
>No. The time is angle. Both x and y
>change with time.
>
>Richard
______________________

I didn't see that in the statement of the project.

jmG

[RichardJ]

On 1/20/2009 9:24:36 AM, jmG wrote:
>On 1/20/2009 8:29:49 AM, rijackson
>wrote:
>>On 1/19/2009 10:17:41 PM, jmG wrote:
>>
>>>No, the time (t) is the
>>>Cartesian X and the travelled
>>>distance plots on the
>>>Cartesian Y, therefore ...
>>
>>No. The time is angle. Both x and y
>>change with time.
>>
>>Richard
>______________________
>
>I didn't see that in the statement of
>the project.

It seems clear enough to me. The path tracks down the faces of the cube, with the z-axis up the center of the cube. So on two of the faces x is constant and y changes with time. On the other two faces y is constant and x changes with time. In general, both x and y change with time.

Richard

[LouP]

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

Yes, different words for the same concept.


Lea: "However I failed to convey that I want the physical path taken from the top of the cube to the bottom to be a straight line. This is an important part of what I am trying to do."


I'll take two conditions to start, and I'll skirt the issue of coordinate systems per se, since that is not the core of the problem.

1. You have repeated the statement that the path is a straight line along the unfolded cube, per your diagram, so I'll assume no ambiguity here.

2.In your file, you have another condition: "Z = starting from the top - the decent is a constant rate downwards to the bottom of the cube..."

These two statements constrain the motion along the path to be at constant speed, since a constant downward velocity implies a constant horizontal component of velocity (look at the triangle of the unfolded cube). In this case, the length of the path travelled is simply proportional to time. The path length is that of the hypotenuse = sqrt(6^2+24^2) = 24.74. If the total time to traverse the four sides is T, then the length of the path at an intermediate time t is simply given by 24.74*(t/T).

Now let's look at the constant angular velocity with respect to the center axis of the cube you have also stated as a fact which may be used in the analysis. As Tom has pointed out twice, the "decent is a constant rate downwards" and "the angular velocity is also constant @ 1 complete rotation per 1 unit of time" are inconsistent with each other. The solution depends on which condition you choose to impose. I suspect it is the "constant rate downwards," since I think you inferred(incorrectly) that this would also imply constant angular velocity - a secondary condition. However, you will need to tell us which condition you really want.

In your plan view, pick a point P on the path along the left side of the cube (say between the horizontal radius and the diagonal radius). Call the center of the cube O, and let C be thepoint at the left end of the horizontal radius (on the cube face). In triangle OCP, let Y be the distance CP (actually the horizontal distance component of the path along the face. Let A be center angle COP. the distance OC is half the cube side = 3. The relation between Y and A is Y = 3*tan(A). Taking time derivatives, we get dY/dt = 3*[sec(A)^2]*dA/dt.

dY/dt is the horizontal velocity of a point on the path. dA/dt is the angular velocity. Since sec(A) is not a constant with time (a varies), the two velocities cannot both be constant with time.

If the path is a straight line as you have reconfirmed, then the horizontal(dY/dt) and vertical(dZ/dt) velocity components are either both constant or both changing. Therefore, constant dZ/dt is not consistent with constant dA/dt(angular velocity). You need to pick one or the other.

Lou

[ptc-1368288]

Interesting, Lou

Though I don't devote time on that. Everybody is right and wrong in there as the issue is "contractual" (per say), meaning that the wording of the statement of the problem is not able to separate the "dispute" [Dispute in the legal meaning, i.e; not a war flames !]

I understand all arguments, but "line integral" is "line integral". Path integral is suggestive of a double integral, but the statement does not say so. Interesting to follow, reading some companion books while waiting Lea tells all what it is about.

jmG

[Fred_Kohlhepp]

On 1/20/2009 10:08:53 AM, lpoulo wrote:
>Lea: "However I failed to
>convey that I want the
>physical path taken from the
>top of the cube to the bottom
>to be a straight line. This is
>an important part of what I am
>trying to do."
>
>
>I'll take two conditions to
>start, and I'll skirt the
>issue of coordinate systems
>per se, since that is not the
>core of the problem.
>
>1. You have repeated the
>statement that the path is a
>straight line along the
>unfolded cube, per your
>diagram, so I'll assume no
>ambiguity here.
>
>2.In your file, you have
>another condition: "Z =
>starting from the top - the
>decent is a constant rate
>downwards to the bottom of the
>cube..."
>
>These two statements constrain
>the motion along the path to
>be at constant speed, since a
>constant downward velocity
>implies a constant horizontal
>component of velocity (look at
>the triangle of the unfolded
>cube).

This is not true. 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.

In this case, the
>length of the path travelled
>is simply proportional to
>time. The path length is that
>of the hypotenuse =
>sqrt(6^2+24^2) = 24.74. If the
>total time to traverse the
>four sides is T, then the
>length of the path at an
>intermediate time t is simply
>given by 24.74*(t/T).
>
>Now let's look at the constant
>angular velocity with respect
>to the center axis of the cube
>you have also stated as a fact
>which may be used in the
>analysis. As Tom has pointed
>out twice, the "decent is a
>constant rate downwards" and
>"the angular velocity is also
>constant @ 1 complete rotation
>per 1 unit of time" are
>inconsistent with each other.
>The solution depends on which
>condition you choose to
>impose. I suspect it is the
>"constant rate downwards,"
>since I think you
>inferred(incorrectly) that
>this would also imply constant
>angular velocity - a secondary
>condition. However, you will
>need to tell us which
>condition you really want.
>
>In your plan view, pick a
>point P on the path along the
>left side of the cube (say
>between the horizontal radius
>and the diagonal radius). Call
>the center of the cube O, and
>let C be thepoint at the left
>end of the horizontal radius
>(on the cube face). In
>triangle OCP, let Y be the
>distance CP (actually the
>horizontal distance component
>of the path along the face.
>Let A be center angle COP. the
>distance OC is half the cube
>side = 3. The relation between
>Y and A is Y = 3*tan(A).
>Taking time derivatives, we
>get dY/dt =
>3*[sec(A)^2]*dA/dt.
>
>dY/dt is the horizontal
>velocity of a point on the
>path. dA/dt is the angular
>velocity. Since sec(A) is not
>a constant with time (a
>varies), the two velocities
>cannot both be constant with
>time.
>
>If the path is a straight line
>as you have reconfirmed, then
>the horizontal(dY/dt) and
>vertical(dZ/dt) velocity
>components are either both
>constant or both changing.
>Therefore, constant dZ/dt is
>not consistent with constant
>dA/dt(angular velocity). You
>need to pick one or the other.
>
>Lou



Fred Kohlhepp
fkohlhepp@sikorsky.com

[ptc-1368288]

Sorry Fred but this is the statement [the contract !].

"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".

"Path line" is strictly not english or american, a pure farmer expression ... "the length of line".

The function is then a 3D vector representation, and the time to travel is the Pythagore length (already posted) divide by the speed. OK, the 3D vector representation is over a cube and at this point there is not trigonometry involved. There is no dv/dt because the speed is constant.

If there is an original problem to be solved for real or exercise, at least we must read it but not so yet. BTW, the statement does not mention any "line integral", does it ?

jmG

[RichardJ]

On 1/20/2009 1:06:10 PM, fkohlhepp wrote:

>This is not true. 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.

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.

Richard

[ptc-1368288]

What about statement that has meaning ?



jmG

[Fred_Kohlhepp]

On 1/20/2009 2:17:04 PM, rijackson wrote:
>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.


Thank you for saying it much more clearly than I did.



Fred Kohlhepp
fkohlhepp@sikorsky.com

[ptc-1368288]

Lea,

I'm done with that one project.

As you mentioned a cylinder inscribed within the cube, the path over the cylinder is a spiral. I have dawn a cube of 2 and the spiral sized to view correct in the 3D, as well as the real path of the spiral. But all that has nothing to do with your line integral. The line integral is the integral of the Pythagore line, which is a straight line sloped, the word is sloped. You then integrate the line 0+ m*t, which is a parabola and the all lot is scalar. What that means is that you just make the slope anything you want for a new parabola This is in perfect agreement with the double integral for the center of gravity of a triangle, except that the "line to integrate" has proportionally different values.

Bloody messy, but this is only a long breakfast project, friday style at - 20� and snow to shovel, but not to shovel in my bed !

Now, the other point is about the rotation: this is for one rotation. If there are two rotations the system will extend to two and if the system is fractional, the Pythagore line will be in relation to the radian(s) of the rotation.

The cube and the spiral is for the view and meat under the skin, but not needed. The line integral around a cube is a parabola. If that line integral is producing some work for instance or in other relation with another entity in the system, and linear with it, the second integral is a parabola again. but if the unknown relation is not linear, the second integral will be anything the Odesolve will solve for.

What do you think ?

At any point in time, the value of the line integral is the ordinate of the cumulative integral, i.e: the parabola... and that was your main question "... at any point in time...". Time and rotation is same monkey business. If you say one rotation per unit of time, then for any fractional time you have the corresponding fractional rotation, is it ? The spiral projects straight Pythagore segments on the cube and they simply sum over and over.

jmG

[ptc-1368288]

Same worksheet as the one posted in the morning, but cleaned . I'm still puzzled by something, but the maths are maths and welcome for collab to contribute/contradict.

jmG