# Need Help Solving a problem for work

• Nov 2nd 2012, 08:09 AM
macleodjb
Need Help Solving a problem for work
Hi All,

I've been tasked at work for create an excel based product library for my company which will product parts. I'm working on a product that needs to be parametric. Please see the attached image
Attachment 25525
If the lengths of the angled lines were a fixed length i would know how to calculate all of this. But since they lengths are relative to the front radius I'm kinda at a loss on how to generate the x and y points for each line. Any help would be appreciated. I'm sure this is a snap for most of you but I suck at math especially trig.

Joe
• Nov 2nd 2012, 11:09 PM
chiro
Re: Need Help Solving a problem for work
Hey macleodjb.

How much calculus do you know?

The reason is that you will know the tangent for the start and ends of the circle since you know the angles and you you have a formula regarding what the derivatives are for a given circle with an equation x^2 + y^2 = r^2 and the dy/dx will correspond to the gradient as a function of the angle.

So you have two gradients so you will essentially get a relation to where the (x,y) points are for the circle part (you get two solutions and discard one for each tangent) and this will give you the relative points for that circle which gives the (x,y) points at the two positions.
• Nov 3rd 2012, 05:02 AM
macleodjb
Re: Need Help Solving a problem for work
Sorry I don't know any calculus. I googled derivatives but the math jargon is just way over my head. Would you mind breaking it down in laymens terms for me. Like Step 1 Angle1/Radius+Whatever.
• Nov 3rd 2012, 09:23 AM
Soroban
Re: Need Help Solving a problem for work
Hello, Joe!

This takes a lot of work.
I have a start on it.
Maybe you or someone else can finish it.
(I assume the lower-left corner is a right angle.)

Code:

      |       |                      o P       |                  *  *       |              *    *       |          *        *       |      *          *     B|  * β            *  (0,b)o - - - - -      *       |                *     b |              *       |              * α   - - * - - - - - - o - - - - - - -       O      a    (a,0)                     A
The line $\displaystyle AP$ has slope $\displaystyle \tan\alpha.$
Its equation is: .$\displaystyle y \:=\:\tan\alpha(x-a) \quad\Rightarrow\quad y \:=\:x\tan\alpha - a\tan\alpha$ .[1]

The line $\displaystyle BP$ has slope $\displaystyle \tan\beta.$
Its equation is: .$\displaystyle y \:=\:x\tan\beta + b$ .[2]

Equate [1] and [2]:

. . $\displaystyle \begin{array}{c}x\tan\alpha - a\tan\alpha \:=\:x\tan\beta + b \\ x\tan\alpha - x\tan\beta \:=\: a\tan\alpha + b \\ x(\tan\alpha - \tan\beta) \:=\:a\tan\alpha + b \\ x \:=\:\dfrac{a\tan\alpha + b}{\tan\alpha - \tan\beta} \end{array}$

Substitute into [2]:

. . $\displaystyle y \:=\:\left(\frac{a\tan\alpha + b}{\tan\alpha - \tan\beta}\right)\tan\beta + b \quad\Rightarrow\quad y \:=\:\frac{\tan\alpha(a\tan\beta + b)}{\tan\alpha - \tan\beta}$

We know the coordinates of point $\displaystyle P\left(\frac{a\tan\alpha + b}{\tan\alpha - \tan\beta},\;\frac{\tan\alpha(a\tan\beta + b)}{\tan\alpha - \tan\beta}\right)$

We are given $\displaystyle r$, the radius of the circle.

Find the equation of the line parallel to $\displaystyle AP$ (to the left of $\displaystyle AP$)
. . and at a distance $\displaystyle r$ from $\displaystyle AP.$

Find the equation of the line parallel to $\displaystyle BP$ (and below $\displaystyle BP$)
. . and at a distance $\displaystyle r$ from $\displaystyle BP.$

The intersection of these lines is the center of the circle.