Help with math formula to maintain distance along the x-axis between an Outer cube and an inner cube of 0.88 while scaling

Question

Help with math formula to maintain distance along the x-axis between an outer cube and an inner cube of 0.88 while scaling.

Outer cube dimensions = x=4.4, y=11.4, z=11
Inner cube dimensions = x=2.64, y=9.48, z=16.8

My thought process was to scale everything along the x-axis then reduce the outer cubes faces so it will maintain the 0.88 distance between the two. I'm just having an issue with the formula required to do this.

Example:

If I scale the object along the x-axis 3 times it gives me this see image below:

But I'm trying to get something like this scale / adjust the outer faces of the cube along the x-axis so the distance maintains the 0.88 distance.

When I do this graphically it comes out the scale I should use is 0.7339...How can I do this using a formula?

PS: I know I can select the outer faces and scale in the x-axis direction till it reaches 0.88.

I'm looking for a formula due to the fact that this is going to be used in Geometry Nodes to make different magnet holders parametric which will then be 3D printed to test different Halbach array setups.

Answer 1

Here's an answer for the x dimension. The others should be obvious.

Let $w$ be the width distance you want to separate the inner and outer box by.

Let $x_0$ be the original width of the outside box.

let $x_1 = x_0 - 2w$ be the width of the inside box.

let $s_0$ be the factor you want to scale the outside box by.

The new outside width is $x_0*s_0$. Call this $X_0$

The new inside width then should be $X_0 -2w$. Call this $X_1$.

$X_1 = X_0 - 2w = s_0*x_0 - 2w$

But we want $X_1 = s_1 * x_1$ so we need to solve for $s_1$.

$s1 = X_1 / x_1 = (s_0 *x_0 - 2w) / (x_0 - 2w)$

Because your original inner width depends on your original outer width and your original width difference, you can calculate the new scale factor for the inner box in terms of the original outer width, the width difference, and the scale factor for the outer box.

Answer 2

What I had help getting....

Suppose you want scale the $\ x$-dimension of the inner object by a scale factor $\ s\ $, and keep the $\ x$-distance between the surfaces of the inner and outer objects fixed at $\ d\ $.

Let $\ w\ $ be the $\ x$-width of the inner object before scaling. Then the $\ x$-width of the outer object will be $\ w+2d\ $. After scaling, the $\ x$-width of the inner object will be $\ sw\ $, and the $\ x$-width of the outer object will have to be $\ sw+2d $. The outer object will thus have been scaled in the $ x$-direction by the factor $$ f=\frac{sw+2d}{w+2d}\ , $$ and if you've already scaled the outer object instead by the factor $\ s\ $, then you'll have to rescale it by the factor $\ \frac{f}{s}\ $ to obtain the right dimensions.

In your example, $\ w=2.64,$$\,d=0.88\ ,$ and $\ s=3\ $, and if we plug these into the above formula, we get $$ f=\frac{3\times2.64+2\times0.88}{2.64+2\times0.88}=\frac{11}{5} $$

Thus, after scaling the outer box up by a factor of $\ 3\ $, you need to rescale it by the factor $\ \frac{11}{15}\approx0.7333\ $ to get the right dimensions. I assume the digit $9$ appearing in the fourth decimal place of the scaling factor $\ 0.7339\dots\ $ you give is either a typo or rounding error.

Answer 3

Maybe make this all with GeoNodes. Make a cube, delete the top face and with extrude node set to 0.88 and some selection nodes you can isolate the faces that you want to extrude and then you can use this as a group to instance everything.

Answer 4

There is actually a tool in Blender to do exactly that which is called push/pull https://docs.blender.org/manual/en/latest/modeling/meshes/editing/mesh/transform/push_pull.html#:~:text=Push%2FPull%20will%20move%20the,center%20by%20the%20same%20distance.