A new version of Cineversity has been launched. This legacy site and its tutorials will remain accessible for a limited transition period

Visit the New Cineversity
   
 
Creating this animation
Posted: 15 October 2020 12:11 AM   [ Ignore ]  
Total Posts:  98
Joined  2018-05-13

Hello Sassi,

I am trying to work out how to create a similar animation in C4D as per the link below:

https://en.wikipedia.org/wiki/Rhombicuboctahedron#/media/File:P2-A5-P3.gif

I know you can move the polygon faces in the Normal direction but how to create new the faces like the animation?

Any insight is appreciated.

Thanks,
David

Profile
 
 
Posted: 15 October 2020 12:21 AM   [ Ignore ]   [ # 1 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

Hi David,

Please have a look here:
https://www.amazon.com/clouddrive/share/eR1Y8WaJaLcO57qa0pFtnIHiuifLrFwTrGX9MkQOxqX

If you have a different geometry, let me know. I’m happy to look into it.

All the best

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
Posted: 15 October 2020 12:37 AM   [ Ignore ]   [ # 2 ]  
Total Posts:  98
Joined  2018-05-13

Well there you go - the fillet radius, I do make life unnecessarily difficult for myself.

As always, thanks for your grounding and enlightening reply.

-David

Profile
 
 
Posted: 15 October 2020 12:44 AM   [ Ignore ]   [ # 3 ]  
Total Posts:  98
Joined  2018-05-13

One more Sassi please. A variation of this, not animated, is this:

https://en.wikipedia.org/wiki/Elongated_square_gyrobicupola

I’d like to see how this can be modelled where the above link states “The solid can also be seen as the result of twisting one of the square cupolae (J4) on a rhombicuboctahedron” which is under the heading “Construction and relation to the rhombicuboctahedron”

Thanks!
David

Profile
 
 
Posted: 15 October 2020 01:14 AM   [ Ignore ]   [ # 4 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

You’re welcome, David.

Enlightenment is easy when the question is clear.

Complex things are easy; to get it simple is what we are aiming for, right?

Always a pleasure to look into your challenges.

So, I hope I got this one right…
https://www.amazon.com/clouddrive/share/H6yYguaGZoZCi2PpIRK5fozKExKTzXFSL4ezElHzD9c

Enjoy.

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
Posted: 15 October 2020 02:47 AM   [ Ignore ]   [ # 5 ]  
Total Posts:  98
Joined  2018-05-13

Thank you Sassi, that’s not quite what I was looking for but very interesting to see nonetheless thank you.

It is this shape here on the wiki page that is based on the rhombicuboctahedron, but then the top third “layer” of the geometry is rotated 45 degrees, so the square and triangular polygons on the top and bottom alternate. Can you see the subtle difference?

https://en.wikipedia.org/wiki/Elongated_square_gyrobicupola#/media/File:Elongated_square_gyrobicupola.png

If possible, I’d like to see a short video on how this can be modelled or what techniques can be employed to create this “cleanly”...

Thanks, David

Profile
 
 
Posted: 15 October 2020 04:22 AM   [ Ignore ]   [ # 6 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

David,

Got it; the 45º rotation was not targeting the eight tilted squares. I was caught in the idea of animation, as that was the initial question (animate…). No problem. I even refined it:
https://www.amazon.com/clouddrive/share/0Q74zQfQfml3kiZ35j3E4Vj2maFy2zV9mGhwJ1hy8t2

There are many ways to do that. I have recorded two, so you can choose. The more complex seems too much work, but I shared this, as the procedure is more important here. It s a hybrid polygon modeling approach, with the use of the Array.

The simple version needs a little bit of math. (Thinking in X, Y, and Z values only, given that the squares are either straight or 45 degrees.) With 45º, we can say that the short sides compared to the hypotenuse is square root out of one, if the hypotenuse is one. Pythagoras.

This allows us to set up a cube quickly. Then turn its “head” and connect it back on again (Optimize).

Here are the two screen captures
https://www.amazon.com/clouddrive/share/Gwc93Hzb7dGovHkKs7Gr1TYkpQKoQm038yoC3xPr0mL

If you need anything else, I’m just a few clicks away.

My best wishes

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
Posted: 15 October 2020 05:32 AM   [ Ignore ]   [ # 7 ]  
Total Posts:  98
Joined  2018-05-13

Fantastic! Thanks for showing both ways Sassi, interesting use of replicating a section with the array to create the shape.

I should learn the math and impressed to see your level of understanding on display again. I’m always spending my spare time learning the myriad features in C4D, let alone dusting off my school days of trigonometry.

I did try an alternative way but it didn’t create the size of the side squares equally. Please see my video below, I was hoping the second round of beveling the points would have created the side squares of equal size….but it didn’t. Close smile

https://www.dropbox.com/s/l6s5yxxp5zrubkb/Rhombicuboctahedron.mp4?dl=1

Thanks again for your help, David

Profile
 
 
Posted: 15 October 2020 06:02 AM   [ Ignore ]   [ # 8 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

Very nice, David.

Exploring and “messing” from time to time with all options, just to see what happens: this fills your creative library, certainly for modeling.
Just keep going, you are on the right path, and perseverance is the underlaying key!

Please have a look here:
https://www.amazon.com/clouddrive/share/L3IonaTUbGQkAU5WHQFQTdErQ95wsT2fMZ6zbXSnnER

Just a hint: The Bevel offset is based on Pythagoras. To showcase that it is working, I placed three splines in it.

Going with your set up, start with the Edges, all selected, while the cube is 200cm, use 58.579cm for the offset in the bevel tool. (The number is rounded on the last digit.)

Enjoy.

Midnight here, it was a long day!

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
Posted: 15 October 2020 06:53 AM   [ Ignore ]   [ # 9 ]  
Total Posts:  98
Joined  2018-05-13

Thanks for your reply and encouragement, it means a lot. Agreed, exploring is rewarding and something I have done a lot more of in C4D, it’s therapeutic in many cases.

Just to close off, can you explain again how you arrived at the 58.579cm offset for the bevel tool based on the last setup.

Thanks and have a good rest.

Profile
 
 
Posted: 15 October 2020 04:30 PM   [ Ignore ]   [ # 10 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

Hi David,

Let’s say we have an 8-sided-spline with two vertical and two horizontal sides, i.e., it doesn’t stand on a corner.

If that spline covers an area of 200 by 200 cm, then a quarter of it would cover 100 by 100cm. This quarter is of interest.

With the size of a hundred by hundred, cm or % are equal.

We also know that all sides are equal in length before cutting it into the quarter, which means the vertical and horizontal line is now half.

So we can just set this up, measure the distance in y, for example, and take the distance between total size and the first point. There you have the number you are after for the offset.

Reading along might feel too much at first, but it will be as simple as opening a door with a key once you see it.

In math, it takes a little bit longer. We are going back to the full 8 sided one. We know that the sides are equal. We also know that a side rotated 45 degrees will have a horizontal or vertical projection of sqrt(2)/2 compared to the full length.


Pythagoras (a*a)+(b*b)= (c*c)
Since the two shorter sides are equal 2* (a*a)= (c*c)
If c is considered one: 2*a*a=1*1
or a*a=1*1/2
or a*a=0.5
or a = sqrt(0.5)
0.707…
Since we have two of those and only one vertical
(2*0.707…)+ one = 2.414…
So if we divide 100 by 2.414, we get half of the vertical (or horizontal), and this number 1- the short side is the offset. Since this is a symmetrical shape, we don’t have to divide all numbers by two, as the ratio is the key. 58.579… is the number we can use directly in this case. So we are back to the quarter, which is all we care about for the bevel.

Please have a look at the practical example.
https://www.amazon.com/clouddrive/share/vzTi0WBDg6ILhtTL9sqa8pUIXtWfPTBQih71b0BuuJd

All the best

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
Posted: 15 October 2020 10:32 PM   [ Ignore ]   [ # 11 ]  
Total Posts:  98
Joined  2018-05-13

Many thanks for taking the time to write up your explanation, it’s starting to sink in.

Have a good day/night.

-David

Profile
 
 
Posted: 16 October 2020 12:01 AM   [ Ignore ]   [ # 12 ]  
Administrator
Avatar
Total Posts:  12043
Joined  2011-03-04

Hi David,

Yes, the wonderful feeling when something becomes clear and then feels as simple as using a pencil. I enjoy those moments a lot.

It is certainly one of the numbers I apply the most often: sqrt(2). wink

Thanks for the wishes.
Have a great weekend.

 Signature 

Dr. Sassi V. Sassmannshausen Ph.D.
Cinema 4D Mentor since 2004
Maxon Master Trainer, VES, DCS

Photography For C4D Artists: 200 Free Tutorials.
https://www.youtube.com/user/DrSassiLA/playlists

NEW:

NEW: Cineversity [CV4]

Profile
 
 
   
 
 
‹‹ popcorn      How can I mirror an object ? ››