I need to know what function to use here?

As an architect working at a firm, you receive an email from a returning client asking for a floor plan for a postmodern installation:

“Dear Jamie Doe,

We are very happy with your design for our new gallery and we thought that an additional installation might compliment it. Can you please design it following the requirements below:

(i) The ground floor should be octagonal and all sides should be 4 metres from the centre; (ii) 8 smaller octagonal floors where all sides are 1 metre away from their centre; (iii) These 8 floors must be equidistant from the ground floor’s centre; (iii) These 8 floors’ centres must be 3 metres away from the ground floor’s centre; (iv) These 8 floors’ walls must align with the borders of the ground floor;

(v) A final square shaped floor to be placed in the centre of the ground floor where all the sides are 1 metre away from the centre.

We just need to see the shapes of the floor plan from a top-view, there is no need for taking into account

the support beams.

Kind regards,

OtherFirm PLC”

Design the floor plan as required by your client on MATLAB.

[Bob Thompson]

Could you post your code, or how you think things should start, so we can help you from there?

[Mark Sherstan]

MATLAB Answers is a place you can get help from the user community on specific MATLAB questions. It is not intended as a place to have strangers do your homework for you.

Desiring help with homework is fine, but:

• show what you have done so far (post some code and explain what you are thinking about how to solve the problem).
• ask a specific question about either specific MATLAB syntax/constructs, or general guidance (such as "what functions should I look at to achieve [result]")
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If you need help with the very basics of MATLAB, please read the getting started documentation.

Answer 1

[Guillaume]

Certainly, the problem is easy to solve using nsidedpoly. 5 lines of code (1 line for the anonymous function that converts wall distance to circumscribed radius, 3 lines for the 3 kind of floors, 1 line for calling plot) were all that was required to produce:

Do I get the job?

Note that you haven't asked an actual question. You just reposted your assignment verbatim.

[dd62]

thank you a lot , can you please tell me what you did exactly to get the shapes?

[dd62]

or can you please send the code?

[Guillaume]

As I wrote, I used nsidedpoly (and a call to polyshape.rotate for the smaller octagons)

[dd62]

okay thanks

[sally z]

can someone provide the code

[Guillaume]

As per Mark's comment to the question, we're not going to provide the solution to a homework problem. That's cheating and you don't learn anything from it. Now, if you've made an attempt at solving the problem and are struggling with a specific point, we can help. But you'll have to show us what you've done and ask a specific question.

[sally z]

t = (1/16:1/8:1)'*2*pi;
x1 = cos(t)+3.5;
y1 = sin(t);
 
x2 = cos(t)+2.5;
y2 = sin(t)-2.5;
 
x3 = cos(t);
y3 = sin(t)+3.5;
 
x4 = cos(t);
y4 = sin(t)-3.5;
 
x5 = cos(t)+2.5;
y5 = sin(t)+2.5;
 
x6 = cos(t)-3.5;
y6 = sin(t);
 
x7 = cos(t)-2.5;
y7 = sin(t)+2.5;
 
x8 = cos(t)-2.5;
y8 = sin(t)-2.5;
 
x = 4.8*cos(t);
y = 4.8*sin(t);
plot(x,y,x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x8,y8)
fill(x,y,'c',x1,y1,'w',x2,y2,'w',x3,y3,'w',x4,y4,'w',x5,y5,'w',x6,y6,'w', x7,y7,'w',x8,y8,'w')
axis square
r = rectangle('Position',[-1 -1 2 2]')
plot(r)

[sally z]

the problem we are facing is that notice the height is not exactly 8

[Bob Thompson]

The simplest way to determine what your constant multiplications are is to work backwards from the values of t. For example you know that the outermost octagon must have edges at 4. So if we consider x = C * cos(t), then we know t(4) is part of the left most edge of the octagon. So if we want x = 4, then set 4 = C * cos(t(4)) and solve for C. You should get an answer of approximately 4.33. This same method will work for all of the other constants.

Also, before anybody else mentions it, I find that indexing my variables is much smoother than having a series of numbered variables. It just helps keep things more organized, and makes accessing the elements simpler.

x(1) = cos(t)+3.5;
y(1) = sin(t);
 
x(2) = cos(t)+2.5;
y(2) = sin(t)-2.5;
 
x(3) = cos(t);
y(3) = sin(t)+3.5;
 
x(4) = cos(t);
y(4) = sin(t)-3.5;
 
x(5) = cos(t)+2.5;
y(5) = sin(t)+2.5;
....

[dd62]

the big octagon has to have its points from 0,4 and 0,-4 at the bottom

[dd62]

we are trying to do that but it's not working

[Bob Thompson]

See my edited response.

[dd62]

thank you

[Guillaume]

As I said, use nsidedpoly. The only thing you have to work out is the radius of the circumscribed circle which is a trivial bit of trigonometry. For the 8 inner octagon, you also need to rotate them. Again working out the rotation angles is trivial.