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- Thread starter Superposed_Cat
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- #2

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I tried endY-beginY. in case your wondering.

- #3

Mark44

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This is very confusing.

If w = f(x, y, z), then ##∇w = <\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}>##. What function are you working with?

I don't know what this means.I tried endY-beginY. in case your wondering.

- #4

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I don't know what this means.

I meant divide not subtract sorry. what's w again?

- #5

Mark44

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In the equation w = f(x, y, z), w is the dependent variable in what I'm showing as a function, written using normal notation. I'm trying to get you to write something meaningful.

- #6

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this is for a program im writing, in it there is a cylinder consisting of rows of circles, its essentailly a plane wrapped in a circular shape using trig. now i need to get that cylinder to stretch between two points, so im using the straight line equation to do this. so there isnt a function where I can use partial derivatives. im trying to extend the 2d line equation to 3d in non parametric form (delta x over delta y). I have gotten the z and x shift values using what I said earlier but I cant figur out how to do the y shift. If you can think of another way to do what i need to do im open to change in the method.Reason I said End.Y/Begin.Y was because i tried to get the gradient relative to the xz plane, so y2-0/y1-0.

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- #7

Mark44

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Still not clear. Mathematically, a cylinder extends infinitely far along its central axis. For example, xthis is for a program im writing, in it there is a cylinder consisting of rows of circles, its essentailly a plane wrapped in a circular shape using trig. now i need to get that cylinder to stretch between two points, so im using the straight line equation to do this.

[Superposed_Cat said:so there isnt a function where I can use partial derivatives. im trying to extend the 2d line equation to 3d in non parametric form (delta x over delta y). I have gotten the z and x shift values using what I said earlier but I cant figur out how to do the y shift. If you can think of another way to do what i need to do im open to change in the method.Reason I said End.Y/Begin.Y was because i tried to get the gradient relative to the xz plane, so y2-0/y1-0.

BTW, the word "gradient" has different meanings in American English vs. British English. Here in America we talk about the slope of a line, and reserve gradient to mean the vector of partial derivatives of a function of two or more variables.

- #8

Chestermiller

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Chet

- #9

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Thats what im trying to do, what I did initially as a temporary fix was to add offset=gradient*(cylinder vertex height) to each vertex in the cylinder. which worked for x and z shifting. Is there a better way to do this if the y gradient cant be obtained?

- #10

Chestermiller

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Chet

- #11

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like the deck of cards, thats just the way im trying to do it, is there a better way?

- #12

Chestermiller

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Is this a solid cylinder or a thin tube?like the deck of cards, thats just the way im trying to do it, is there a better way?

chet

- #13

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I said earlier a plane wrapped cylindricly therefor thin tube.

- #14

Chestermiller

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You are trying to determine the deformation gradient tensor for the deformation you illustrated in post #9, correct?I said earlier a plane wrapped cylindricly therefor thin tube.

Chet

- #15

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- #16

Chestermiller

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Let x

Chet

- #17

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CylinderPointPos = OriginalCylinderPointPos + ((positon2 - positon1) / RingCount * RingHeight);

- #18

Chestermiller

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Well, this is not what I have in mind, but if you are happy with this solution, then I am.

CylinderPointPos = OriginalCylinderPointPos + ((positon2 - positon1) / RingCount * RingHeight);

Chet

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- #20

Chestermiller

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[tex]x=x_0+tan\theta z_0[/tex]

[tex]y=y_0[/tex]

[tex]z=z_0[/tex]

So, if two neighboring material points in the original configuration of the body are located at x

[tex]dx=dx_0+\tanθ dz_0[/tex]

[tex]dy=dy_0[/tex]

[tex]dz=dz_0[/tex]

So,

[tex]\left( \begin{array}{c}dx\\dy\\dz\end{array} \right)=\left( \begin{array}{ccc}1&0&tanθ\\0&1&0\\0&0&1\end{array}\right)\left( \begin{array}{c}dx_0\\dy_0\\dz_0\end{array} \right)[/tex]

The 3x3 matrix is the deformation gradient tensor components.

Chet

- #21

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Okaty a standard skew matrix? did not think of that, funny because my entire program is based on matrices. Am I correct in thinking that I can convert this to 4x4 (the software uses only 4x4 so rotation and translation can be supported in a single matrix) by going:

\begin{array}{ccc}

1 & 0 & tan(θ) & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \end{array}

this seems to flatten the cylinder when I apply the matrix. Is there a version that wont do that?

\begin{array}{ccc}

1 & 0 & tan(θ) & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \end{array}

this seems to flatten the cylinder when I apply the matrix. Is there a version that wont do that?

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- #22

Chestermiller

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I really don't understand what you are referring to here. Maybe you learned this stuff differently from me. I don't understand what your 4x4 matrix is supposed to represent. The transformation I gave maps an arbitrary differential position vector between two material points in the initial undeformed configuration of the cylinder into the corresponding differential position vector between the same two material points in the deformed configuration. From this, you can determine all the strains, which I assume, is your objective.Okaty a standard skew matrix? did not think of that, funny because my entire program is based on matrices. Am I correct in thinking that I can convert this to 4x4 (the software uses only 4x4 so rotation and translation can be supported in a single matrix) by going:

\begin{array}{ccc}

1 & 0 & tan(θ) & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \end{array}

this seems to flatten the cylinder when I apply the matrix. Is there a version that wont do that?

Chet

- #23

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