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In summary, the poster is working on a project involving 3D gradients and is looking for a way to obtain the Y gradient. They are trying to stretch a cylinder between two points using the straight line equation and have already obtained values for the Z and X gradients. They are open to suggestions for other methods and are also trying to determine the deformation gradient tensor. They provide additional information about the problem, including the use of matrices and tensors, and the angle of the axis of the cylinder in the deformed configuration.

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Mathematics news on Phys.org

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I tried endY-beginY. in case your wondering.

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This is very confusing.Superposed_Cat said:

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.Superposed_Cat said:I tried endY-beginY. in case your wondering.

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Mark44 said:I don't know what this means.

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

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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.

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this is for a program I am 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 I am using the straight line equation to do this. so there isn't a function where I can use partial derivatives. I am 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 can't figur out how to do the y shift. If you can think of another way to do what i need to do I am 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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Still not clear. Mathematically, a cylinder extends infinitely far along its central axis. For example, xSuperposed_Cat said:this is for a program I am 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 I am using the straight line equation to do this.

[Superposed_Cat said:so there isn't a function where I can use partial derivatives. I am 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 can't figur out how to do the y shift. If you can think of another way to do what i need to do I am 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.

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Chet

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Thats what I am 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 can't be obtained?

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Chet

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like the deck of cards, that's just the way I am trying to do it, is there a better way?

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Is this a solid cylinder or a thin tube?Superposed_Cat said:like the deck of cards, that's just the way I am trying to do it, is there a better way?

chet

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I said earlier a plane wrapped cylindricly therefor thin tube.

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

Chet

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Let x

Chet

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

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

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

Chet

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

[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

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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 won't 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 won't do that?

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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.Superposed_Cat said:

\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 won't do that?

Chet

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