Stuck designing a diamond pattern leaded window.

[Rob Edwards]

Hello everyone.

This is my first post - apologies if I don't follow correct forum etiquette.

I am a traditional woodworker designing a window in CAD. My maths is very rusty (30 years since I did A-Levels in the UK)

I attach a picture of the problem I'm trying to solve and repeat the equations below.

I have been trying to solve these all day..

A = n.w + (n-1)d

B = m.h + (m-1)e

tan(a) = h/w

tan(a) = e/d

d = c / sin(a)

e = c / cos(a)

Find w, h and a in terms of n, m, c, A and B?
* the B is missing in the image

By substitution I got as far as

c(m - n) = n.B.cos(a) - m.A.sin(a)

which proved to be a dead end for me.

I enjoy maths and can normally get by, but this problem is well beyond me.

Any pointers would be great.

 

[Dr.Peterson]

Your work agrees fully with mine; good work. I assume B is also given.

Here's the rest of the work the way I do it:

[MATH]nB\cos(\alpha) - mA\sin(\alpha) = c(m - n)[/MATH]​

​

[MATH]\frac{nB}{\sqrt{m^2A^2+n^2B^2}}\cos(\alpha) - \frac{mA}{\sqrt{m^2A^2+n^2B^2}}\sin(\alpha) = \frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

Let [MATH]\beta = \arctan\frac{-mA}{nB}[/MATH]; then the equation becomes

[MATH]\cos(\alpha - \beta) = \frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

so

[MATH]\alpha = \beta + \arccos\frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

You can probably finish. You should also definitely check against a simple case to find any mistakes I made ...

 

[Rob Edwards]

Your work agrees fully with mine; good work. I assume B is also given.

Here's the rest of the work the way I do it:

[MATH]nB\cos(\alpha) - mA\sin(\alpha) = c(m - n)[/MATH]​

​

[MATH]\frac{nB}{\sqrt{m^2A^2+n^2B^2}}\cos(\alpha) - \frac{mA}{\sqrt{m^2A^2+n^2B^2}}\sin(\alpha) = \frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

Let [MATH]\beta = \arctan\frac{-mA}{nB}[/MATH]; then the equation becomes

[MATH]\cos(\alpha - \beta) = \frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

so

[MATH]\alpha = \beta + \arccos\frac{c(m - n)}{\sqrt{m^2A^2+n^2B^2}}[/MATH]​

You can probably finish. You should also definitely check against a simple case to find any mistakes I made ...

THANKYOU! THANKYOU! THANKYOU!

Absolutely Fantastic.

If you like you can see it working here

I don't understand what magic spell you cast, but it works an absolute treat!

If you don't mind where can I find out more? Is there a name for what you just did?

That's the problem with Google - if you don't know a name it don't help.

Just as you posted I found this http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf

a cos x + b sin x = R cos(x − α)

Was I on the right track?

 

[Dr.Peterson]

Yes, that page shows just what I did, and explains it very nicely. They don't give it a memorable name, and I don't know of anything better than "adding a sine and a cosine with the same period". But I know it when I see it.

 

[Rob Edwards]

Excellent - I will put it on my reading list for when I've finished this project.
I have a few more ideas in the pipe that will challenge my mathematical limitations, so I'm sure I'll be back.
But for now Thankyou once again.