making cubes with unit rods and joints

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Hey. I’m having trouble with this work I’ve been set about structures. I understand some of it, which I’ll explain later, but I really need your help with it!

My task is to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes. For example, 2 by 2 by 2 cube, 3 by 3 by 3 cube etc etc etc…

I desperately need help on getting a formula to work this out. I know already, as my teacher gave us these results, that a 1 by 1 by 1 cube has the following:

8 Three-Joints ( as do all cubes)
0 Four Joints
0 Five Joints
0 Six Joints

and a 2 by 2 by 2 has:

8 three joints
12 four joints
2 five joints
1 six joints

I really need a formula in order to calculate how many joints a 3 by 3 by 3 has, and with that formula to use it on lets say a 55 by 55 by 55 cube.

All help would be appreciated!
 
A "joint" is usually defined as something that bends, such as an elbow. What do you mean by "joint", that a cube would have one?

Please reply with the exact text of and instructions for the exercise, along with appropriate definitions. Also, since you posted this to the "Advanced" (that is, post-calculus) category, rather than to "Geometry", please include the topic under study, along with whatever techniques, formulas, etc, you believe you are expected to bring to bear.

Thank you.

Eliz.
 
Sorry, I was vague.
---
Rigid Structures are constructed using unit rods.

An example is shown below:

* Pic opf a 2 by 2 by 2 cube*

The individual rods in the structures are held together using different types of joints. (Just think of making cubes using matchsticks and playdough*.

joints3lz.jpg


A three joint would be on the vertices of each cube no matter what the size and so therefore there'll be 8 three-joints.
--

3 joints - found on the vertices of the cube and connect three different rods
4joints - found on the edges of the cube and connect four different rods
 
Hello, carlito!

My task is to investigate different sized cubes, made up of single unit rods
and justify formulae for the number of rods and joints in the cubes.
For example, 2 by 2 by 2 cube, 3 by 3 by 3 cube etc etc etc…

I desperately need help on getting a formula to work this out.

I know already, as my teacher gave us these results:

a 1-cube has: 8 three-joints, 0 four-joints, 0 five-joints, 0 six-joints

a 2-cube has: 8 three joints, 12 four joints, 6 five joints, 1 six joints

I really need a formula in order to calculate how many joints a 3-cube has,
and with that formula to use it on, let's say, a 55-cube.

All help would be appreciated!
Click to expand...
After making dozen of messy sketches, here's what I found . . .


On an \(\displaystyle n\)-cube , each face looks like this:
Code: 
      3 4 4 ... 4 4 3
      4 5 5 ... 5 5 4
      4 5 5 ... 5 5 4
      4 5 5 ... 5 5 4
      : : : ... : : :
      4 5 5 ... 5 5 4
      3 4 4 ... 4 4 3
There is a three-joint at each vertex: \(\displaystyle \,8\text{ three-joints}\)

There are \(\displaystyle n-1\) four-joints on each of the 12 edges: \(\displaystyle \,12(n-1)\text{ four-joints}\)

There are \(\displaystyle (n-1)^2\) five-joints on each of the 6 faces: \(\displaystyle \,6(n-1)^2\text{ five-joints}\)

Inside the cube, there is an \(\displaystyle (n-1)^3\) cube comprised of six-joints: \(\displaystyle \,(n-1)^3\text{ six-joints}\)


This checks out since: \(\displaystyle \,8\,+\,12(n-1)\,+\,6(n-1)^2\,+\,(n-1)^3 \;= \;(n\,+\,1)^3\)
\(\displaystyle \;\;\)the expected number of joints in an \(\displaystyle n\)-cube.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

A 55-cube has:
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\:\:\:\:\:8\text{ three-joints}\)
\(\displaystyle 12\,\times\,54\,=\:\;\;\:\,648\text{ four-joints}\)
\(\displaystyle 6\,\times\,54^2\:=\:\;17,496\text{ five-joints}\)
\(\displaystyle \;\:\:\,54^3\:=\:157,464\text{ six-joints}\)

and: \(\displaystyle \,8\,+\,648,\,+\,17,496\,+\,157,464\:=\:175,616\:=\:56^3\)
 
Very impressive, Soroban.

My head starts to hurt from the 2by2 on... :shock:
 
Thanks ever so much.

So can you confirm these formae for 4 joints, 5 joints and 6 joints please?

Also, if I wanted to work out how many 4 joints, 5 joints and 6 joints in a 3 by 3 by 3 cube how would I go about it? And also, how would I go about 4 by 4 by 4 up to lets say 12 by 12 by 12?

3 joints is easy as it's always 8.
 
What work have you done on testing and confirming the formula that has been provided to you? (Most of the tutors don't do all the work for you, and Soroban might not be back for a while.)

Eliz.
 
I've drawn cubes but I'm honestly very stuck and need help ... I hope you understand... I'm very greatful,.
 
carlito said:
I've drawn cubes but I'm honestly very stuck and need help ... I hope you understand... I'm very greatful,.
Click to expand...
I don't understand . . .

Why are you still drawing cubes?

Why aren't you using my formulas?
If you're not going to use them, I'll delete my posts . . .
 
carlito said:
So can you confirm these formae for 4 joints, 5 joints and 6 joints please?
Click to expand...
Geezzzzz carlito; Soroban CLEARLY gave you formulas for these:
you seem to be quite lost, me boy :shock:
 
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