Functions, Volume of Revolutions

TZK203 shared this question 7 years ago
Answered

I'll attach my file.


You'll probably initially notice that it has an abundant amount of points. I'm only doing this for accuracy.


Now, that said, all the fit functions make any manifested functions go on forever instead of around the points encompassing the function.


So, I ask, how do I get this to work out?


OR


Is there a way to take a set amount of points and somehow make a volume of revolution around the x-axis? Anything?


Thanks!

https://ggbm.at/563647

Comments (15)

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Hi,


Maybe this is a correct aproximation for the volume (not for a function), I'm not really sure.


https://ggbm.at/563649


Raymond


addendum:

more nice, more correct but not more warranty


https://ggbm.at/563655

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Hi

A file to help for your problem

try with this file

Your file has too many points so it is very long to open your file. Here only 18 dots

Volume is made with intégral of the the Adjustpoly[liste1,18] between X(A) and x(V)

See picture (call message in a bottle) the blue area is the integral of the function

4f09439a25ee4040e73ba8a85b040a3a


https://ggbm.at/563651


Enjoys

Daniel

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Hi,


Maybe this is a correct aproximation for the volume (not for a function), I'm not really sure.

[attachment=1]Forum_ 33205_A_Water Bottle Project02.ggb[/attachment]

Raymond


addendum:

more nice, more correct but not more warranty

[attachment=0]Forum_ 33205_B_Water Bottle Project02.ggb[/attachment]


Ok, care to answer a few questions for me?


First of all, how did you create a list without any points to use? I thought lists needed points (list2 - 5).


Second, what is the variable L that is spoken of in list2?


Also, as for volume, I thought it might be better doing the disk method to flip it across the x-axis... pi * (f(x))^2.

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I'm talking about the second file, btw.

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I'm talking about the second file, btw.


Also, I just want to ask if you can try to make sense of this to me.


So far, I can understand the fact of the showpoint boolean value. I have the list of values as list1.


I just don't quite understand how you got to the steps you did; any clarification would be nice.

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First of all, how did you create a list without any points to use?

I create a lot of points but all points are in a list (more exact: a matrix). You can see this points if you activate visible for list2.

The difference between list1 and list 2 is, that all points have minimum 1 point more with the exact the same Y-value.


I thought lists needed points (list2 - 5).

No.

polyLine need the points of list1.

polyLine is used in list2 to generate new points (in a Matrix).

list2 is used in list3 this generate segments

list3 is used in list4 this keep some segments

list4 is used in list5 this calculate voluminas


list2 generate points in a matrix:

    KeepIf[Length[L] > 1, L, Sequence[RemoveUndefined[{Intersect[y = n, polyLine]}], n, 0, -y(A), stepY]]


Sequence[....] generate a lot horicontal lines y=n (from n=0 to the top of picture (-y(A)) in a distance of stepY)

Each of this lines intersect with polyLine = PolyLine[list1] and for each of this lines is generated a list of point with {Intersect[y = n, polyLine]}

(Note the curly braces. This mean: make a list of points)

All this point-lists are a part of a list generated with Sequence[]. That mean all generated points are a part of a Matrix.

Each line of the matrix with 2 or more elements (points) is keep with KeepIf[Length[L] > 1, L, <matrix>

L is the free chosen variable name of 1 line in the matrix (= 1 element = 1 list) in the command KeepIf[] (I chose L to remember me it's a list).

Also in Sequence[] od Zip[] exists free chosen variables (for examample 'L' and 'S' in list4)


Also, as for volume, I thought it might be better doing the disk method to flip it across the x-axis... pi * (f(x))^2.

I not use a disk methode.

I cut the bottle into nested tubes/cylinder with a selectable wall/shell thickness.

See the following file with a very rough cuted bottle with cylindrical (include volume-formula):


https://ggbm.at/563673


This is necessary because of the solid of revolution can not be represented by a function.

The problem is the bottom of the bottle. It is curved inward (toward + X).

What would be possible with a function: a straight, vertical bottle bottom

Then you can use the disc method or the definite integral of a function.

But the shape is too complex to generate a function by using points.

If the disc-methode is what you want to convey to your students, then both my method and your bottle is inappropriate.

Then eg a wineglass would be more suitable.

This is the reason why the Example of Daniel Mentrard has another bottle bottom and a simple form/silhouette with match less points.


Raymond

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Hi

And why not with the polygone of the points and the area of this polygone by using séquence of segments for the représentation

00fbb6bb8f07903829ab66713338310b

See the gg file


https://ggbm.at/563677


Daniel

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Hi,

A cylinder as equivalent to the volume of the bottle has estimated the dimensions of

radius ≈ 2.5

length ≈ 19

This results in an estimated volume between 350 to 390 cm ^ 3

I think: 286 cm^3 is definitely too small.


Raymond

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Guldinus theorem[/url]"]The second theorem

The second theorem states that the volume V of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d traveled by its geometric centroid.

  • gif

This leads me to:

    vol = poly2 * abs( y( Centroid[ poly2 ] ) ) * 2 * pi

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Thanks Loco :D

Now I can give a small garantie for the pipe-methode.

I have slightly adjusted the volume formula in list5.


Raymond

https://ggbm.at/563679

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Many thanks Loco :D :laughing: :laughing:

This theorem allows me and us to use polygones rather than functions to evaluate a volume in many case.

http://www.geogebratube.org...

That works fine

Daniel

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Further investigations could involve the way to compute the bottle volume partially filled, by means (if possible) of dynamic lists and varying x, as shown underneath. So ... terrific for me!

Cheers

Philippe35cb8304947ff45a3939c03f301c0fc3

Files: 23.png
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Thank you guys for the polyline suggestions. They actually work well for an extremely accurate answer.


Nevertheless, I have attached the same file, only this time with fewer points.


The issue I'm running into is finding good integral functions for this--specifically it seems that it is quite difficult making vertical curves.


Also, some horizontal curves go through the bottle picture, when all I want them to do is go through the points that compose their specific function.

I'm trying to make multiple functions so I can do the volume calculations by myself, on paper, just for reference and error calculations. One big function would be nice, but for accuracy, it might be helpful to get me on the path I stated above.

https://ggbm.at/563733

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I still want to account for the ridges, btw.

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The issue I'm running into is finding good integral functions for this


The problem is the bottom of the bottle. It is curved inward (toward + X).

What would be possible with a function: a straight, vertical bottle bottom

Then you can use the disc method or the definite integral of a function.

But the shape is too complex to generate a function by using points.

If the disc-methode is what you want to convey to your students, then both my method and your bottle is inappropriate.

Then eg a wineglass would be more suitable.

This is the reason why the Example of Daniel Mentrard has another bottle bottom and a simple form/silhouette with match less points.


The maximum (to draw the exact profile) wath is possible:

https://ggbm.at/563751

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