ASCII TURNED SHAPE

How can we depict a turned shape simply and nicely using ascii characters in 3D space?

I suggest this method :

Instead of a circle we use a square rotated by 45° for the cutting section so that we only need '/' and '\' characters to draw it.

/\
/  \
\  /
\/

And we use '_' character for the profiles: upper, lower and median.

_________
/\        \
/  \________\
\  /        /
\/________/

Isn't it Turning complete? Well, if you agree, write a full program or a function taking an unsigned integer value N , representing a number of steps , producing a 3d shape as described below.

The profile of this turned shape has the form of a stair step curve raising from 0 to N steps and lowering back to 0 where each step is 2 char('/') high and 5 char long ('_').

Maybe some examples describes it more clearly.

For N = 0 you may output nothing but it's not mandatory to handle it.

N = 1

_____
/\    \
/  \____\
\  /    /
\/____/
.
N = 2

_____
/\    \
___/_ \    \__
/\    \ \    \ \
/  \____\ \____\_\
\  /    / /    / /
\/____/ /    /_/
\  /    /
\/____/
.
N = 3
_____
/\    \
___/_ \    \__
/\    \ \    \ \
___/_ \    \ \    \ \__
/\    \ \    \ \    \ \ \
/  \____\ \____\ \____\_\_\
\  /    / /    / /    / / /
\/____/ /    / /    / /_/
\  /    / /    / /
\/____/ /    /_/
\  /    /
\/____/

Rules :
- Margins are not specified.
- Standard loopholes are forbidden.
- Standard input/output methods.
- Shortest answer in bytes wins.

• What if I don't agree ? I may be wrong, but I don't have any task in that case. Oct 15 '19 at 7:32
• Oct 15 '19 at 7:37

Charcoal, 70 69 bytes

ＮθＦθ«Ｊ⊕×⁷⊕ι⁰≔⊗⊕ιι↙ι↑←×⁵_Ｐ↖²→↗ι×⁴_↖ι←×⁵_↓Ｐ↙²→↘ιＪ⁻×⁹θ⊗ι⊖ι__↗ι←Ｐ←__↖ι←__

Try it online! Link is to verbose version of code. Explanation:

ＮθＦθ«

Input N and loop that many times.

Ｊ⊕×⁷⊕ι⁰

≔⊗⊕ιι

Get the size of the slice.

↙ι↑←×⁵_Ｐ↖²→↗ι×⁴_↖ι←×⁵_↓Ｐ↙²→↘ι

Draw the front slice.

Ｊ⁻×⁹θ⊗ι⊖ι

Jump to the bottom (that we can see) of the back slice.

__↗ι←Ｐ←__↖ι←__

Draw the back slice.

JavaScript (ES8),  322 282 278  272 bytes

Builds the output line by line.

f=(n,y=0,Y=(k=y>2*n)?4*n-y:y,S='___/_25__,/\\    \\25,\\/____/14_/,\\ 14,\\/____/,_____, \\____\\, /    /, \\    \\,_\\, /, \\'.split,)=>~Y?(y-2*n?''.padEnd(n*5-(Y>>1)*5-3-y%2-k)+S[y?Y?k*2+y%2:4:5]:'/ 03').replace(/\d/g,n=>S[+n+6].repeat(Y-(n>2||-k)>>1))+
+f(n,y+1):''

Try it online!

How?

For each row $$\0 \le y \le 4n\$$, we define:

$$k=\cases{ 0,&\text{if y \le 2n}\\ 1,&\text{if y > 2n} }$$ $$Y=\cases{ y,&\text{if k=0}\\ 4n-y,&\text{if k=1} }$$

Each row is first converted into a main pattern. A main pattern may contain digits: they are placeholders for repeated sub-patterns that are expended afterwards.

The middle row (when $$\y=2n\$$) is a special case which is processed separately. For all other rows, we compute the main pattern ID $$\p\$$ with:

$$p=\cases{ 5,&\text{if y=0}\\ 4,&\text{if y \neq 0, Y=0}\\ 2k+(y \bmod 2),&\text{otherwise}\\ }$$

There are no leading spaces for the middle row. For all other rows, the number $$\s\$$ of leading spaces is given by:

$$s=5n-5\left\lfloor\frac{Y}{2}\right\rfloor-3-(y\bmod 2)-k$$

There are inner sub-patterns (marked with a digit $$\\le2\$$) and outer sub-patterns (marked with a digit $$\>2\$$), which are repeated $$\n_1\$$ and $$\n_2\$$ times respectively:

$$n_1=\left\lfloor\frac{Y+k}{2}\right\rfloor\\ n_2=\left\lfloor\frac{Y-1}{2}\right\rfloor$$

The above formulae apply to the middle row as well, but are irrelevant for the first and last rows, which do not have sub-patterns.

Below is what we get for $$\n=3\$$:

y Y k |  p  |  s  |   before .replace()  | n1 | n2 |      after .replace()
--------+-----+-----+----------------------+----+----+-----------------------------
0 0 0 |  5  | 12  | ............_____    |  0 | -1 | ............_____
1 1 0 |  1  | 11  | .........../\    \25 |  0 |  0 | .........../\    \
2 2 0 |  0  |  7  | .......___/_25__     |  1 |  0 | .......___/_ \    \__
3 3 0 |  1  |  6  | ....../\    \25      |  1 |  1 | ....../\    \ \    \ \
4 4 0 |  0  |  2  | ..___/_25__          |  2 |  1 | ..___/_ \    \ \    \ \__
5 5 0 |  1  |  1  | ./\    \25           |  2 |  2 | ./\    \ \    \ \    \ \ \
6 6 0 | n/a | n/a | / 03                 |  3 |  2 | /  \____\ \____\ \____\_\_\
7 5 1 |  3  |  0  | \ 14                 |  3 |  2 | \  /    / /    / /    / / /
8 4 1 |  2  |  1  | .\/____/14_/         |  2 |  1 | .\/____/ /    / /    / /_/
9 3 1 |  3  |  5  | .....\ 14            |  2 |  1 | .....\  /    / /    / /
10 2 1 |  2  |  6  | ......\/____/14_/    |  1 |  0 | ......\/____/ /    /_/
11 1 1 |  3  | 10  | ..........\ 14       |  1 |  0 | ..........\  /    /
12 0 1 |  4  | 11  | ...........\/____/   |  0 | -1 | ...........\/____/