Help finding length of perpendiculars in a box of known dimension

[ttk]

For a research problem, I'd like a way to find the distance of each of 4 lines perpendicular to one of 4 walls connected to a point that is within a box of known dimension. I know the distance from the center of each wall to the point of interest (C1 to C4), but I do not know the angle this line makes relative to the wall (usually it will not be perpendicular). I'm attaching an image of the problem. C1 to C4 are known, as is the dimension of the square box. What I want to know is B1 to B4. My overall goal is to understand the distance of the point of interest (which could be anywhere within the box) from the nearest wall along a perpendicular to that wall (it's closest distance from the wall).

View attachment 5915

 

[I like Serena]

For a research problem, I'd like a way to find the distance of each of 4 lines perpendicular to one of 4 walls connected to a point that is within a box of known dimension. I know the distance from the center of each wall to the point of interest (C1 to C4), but I do not know the angle this line makes relative to the wall (usually it will not be perpendicular). I'm attaching an image of the problem. C1 to C4 are known, as is the dimension of the square box. What I want to know is B1 to B4. My overall goal is to understand the distance of the point of interest (which could be anywhere within the box) from the nearest wall along a perpendicular to that wall (it's closest distance from the wall).

Hey ttk! Welcome to MHB! ;)

Let's pick the center of the square to be our origin.
And let's pick the coordinates of the point of interest to be $(x,y)$.
Oh, we already had an $x$ for the side of the square.
Well, let's discard that one, and let's pick $h$ to be half the side of the square, if you don't mind.
I just like $x$ and $y$ to be my unknowns, and use other letters for known values.

With those choices, we have the following system of equations:
\[\begin{cases}
x^2 + (h-y)^2 = C_1^2 \\
(h-x)^2 + y^2 = C_2^2 \\
x^2 + (h+y)^2 = C_3^2 \\
(h+x)^2 + y^2 = C_4^2
\end{cases}
\Rightarrow\begin{cases}
x^2 + y^2 - 2hy = C_1^2 - h^2\\
x^2 + y^2 - 2hx= C_2^2 - h^2 \\
x^2 + y^2 +2hy = C_3^2 - h^2 \\
x^2 + y^2 +2hx = C_4^2 - h^2
\end{cases}
\Rightarrow\begin{cases}
4hx= (C_4^2-h^2) - (C_2^2 - h^2) \\
4hy = (C_3^2 - h^2) - (C_1^2 - h^2) \\
\end{cases}
\Rightarrow\begin{cases}
x= \frac{C_4^2 - C_2^2}{4h} \\
y = \frac{C_3^2 - C_1^2}{4h} \\
\end{cases}
\]

Does that satisfy your needs? (Wondering)

 

[ttk]

You cracked it! Thanks so much. With x,y, I can easily determine the distance to the walls. Can't thank you enough. Take care,

Terry