# Determining distance across face of bars following a curve

So I'm not much of a mathematician, and I've been trying to figure out how one would solve this real world problem I have.

I have 2"x4" wood cut into squares (2x4x4).

I am trying to figure out at what distance to place each board on top of each other so that the top edges would follow the curve (-x^2)+25 from -5 to 0 where each increment of 1 is 1 foot.

I hope this makes sense, please let me know if i need to clarify.

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When you say "on top of", are you stacking them so that the square faces meet, or so that the "edges" (i.e. the 2 inch sides) meet? –  Zev Chonoles Nov 29 '11 at 18:23
@Zev: so that the square faces meet. –  kylex Nov 29 '11 at 18:27

It's a bit easier to solve this problem by flipping the curve on its side; i.e., instead of looking at $y=-x^2+25$:
$\hskip2.7in$
looking at $y=\sqrt{25-x}$:

$\hskip1.4in$

Using the latter setup, what you want to find first (if I have understood your question correctly) are the non-negative values of $x$ where $$y=\sqrt{25-x}=\frac{n}{3}$$ where $n$ is an integer (all my numbers will be in feet, hence $4$ inches is $\frac{1}{3}$). These are just the numbers $$25-\frac{n^2}{9}$$ as $n$ ranges from $1$ to $15$, i.e.

224/9, 221/9, 24, 209/9, 200/9, 21, 176/9, 161/9, 16, 125/9, 104/9, 9, 56/9, 29/9, 0


or, to use approximate values,

24.89, 24.56, 24.00, 23.22, 22.22, 21.00, 19.56, 17.89, 16.00, 13.89, 11.56, 9.000, 6.222, 3.222, 0


This shows how the number of wood pieces you can fit within the boundary increases by one at each of these values of $x$:

As you can see, for $x\leq25-\frac{n^2}{9}$, we can fit in $n$ wood pieces.

Now, the thickness of each wood piece is 2 inches, or $\frac{1}{6}$, so the top of the $m$th row (counting the row whose bottom is the $y$-axis as row 1) is at $x=\frac{m}{6}$. Thus, you will be able to fit $n$ wood pieces in on the $m$th row if and only if $$\frac{m}{6}\leq25-\frac{n^2}{9},$$ or $$n\leq \sqrt{225-\frac{3m}{2}},$$ so the number of wood pieces you can fit in on the $m$th row is $$\left\lfloor\sqrt{225-\frac{3m}{2}}\right\rfloor.$$ Using this formula, we can generate a side-on view of what the final result of your project will look like:

Is this what you had in mind?

Mathematica code for that final image:

f[m_] := Floor[Sqrt[225 - (3 m/2)]]

Show[ParametricPlot[Table[{m/6, u*f[m]/3}, {m, 1, 150}], {u, 0, 1}],
Plot[Sqrt[25 - x], {x, 0, 25}, AspectRatio -> 1/5],
ParametricPlot[Table[{u Floor[150 - 2 n^2/3]/6, n/3}, {n, 1, 15}], {u, 0, 1}],
PlotRange -> {0, 5}]

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Thank you! Unfortunately I did not specify that the 2" side would be facing the curve, not the 4" side, but that's a simple substitution. What you've provided solves my dilemma. –  kylex Nov 29 '11 at 19:27
@kylex: Ah, I see - glad I could help! –  Zev Chonoles Nov 29 '11 at 19:40
+1 for the nice pictures. –  TonyK Nov 29 '11 at 20:57
Thanks, @TonyK :) –  Zev Chonoles Nov 30 '11 at 4:22

You can make a table using your equation: $\begin {array} {c c c}x&h&boards\\ -5 &50&25\\ -4& 41&20.5\\ -3&34&17\\ -2&29&14.5\\ -1&26&13\\ 0 &25&12.5 \end {array}$

Then at each $x$ position pile up enough boards to get the height you want: $25$ of them at $5$ feet. As your boards are only $1/3$ foot in size, maybe you need more lines in the table. Is this what you were looking for?

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