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Minding the Gaps of Picket Fencing

Date: 03/17/2012 at 22:31:31
From: David
Subject: Fence Construction. Fence Gap Length.

I'm trying to build some picket fencing, but the math is frustrating me.

I know the length between fence posts: 81.5 inches. Each picket is
5.5 inches wide. I'd like a 3-inch gap between each picket (or 
fence post).

How many 5.5 inch-wide pickets can fit into a 81.5 inch space, allowing a
3-inch gap in between each picket?

I got as far as dividing 81.5 by 5.5, and getting 14.81 -- but that's not
a whole number of pickets, and it does not take into account the 3-inch
gaps in separating them.

David



Date: 03/17/2012 at 22:57:18
From: Doctor Peterson
Subject: Re: Fence Construction. Fence Gap Length.

Hi, David.

Just to make sure I'm interpreting your plans correctly, here's a smaller
example viewed from above:

         S   P
       |<->|<->|
   +---+                                           +---+
   |   |   =====   =====   =====   =====   =====   |   |
   +---+                                           +---+
       |<------------------- F ------------------->|

What I'm calling F (the distance between fence posts) would be your 
81.5 in; I'm measuring it between the near sides, not center to center.

What I'm calling P (picket width) and S (space width) are your 5.5 in and
3 in, respectively. I'm assuming you want the same width at the ends as
between pickets.

If that's all correct, the key to working this out is to notice that there
is one more space than picket.

In my sketch, for example, there are 5 pickets and 6 spaces:

   F = 5P + 6S

In general, n pickets require n + 1 spaces:

   F = nP + (n + 1)S

We want to solve for your n. 

Rather than do this the way we'd normally do it in algebra (simplifying
and solving), I'll take the common-sense approach I use in real life.

Subtract one space from each side of the equation, so that we are now
trying to fill a distance of F - S with n pairs of P + S:

   F - S = nP + nS

   F - S = n(P + S)

   n = (F - S)/(P + S)

             P + S
           |<----->| repeated n times
   +---+                                           +---+
   |   |   =====   =====   =====   =====   =====   |   |
   +---+                                           +---+
           |<--------------- F - S --------------->|

(With this picture in mind, I wouldn't actually use the formula; I'd just
divide the reduced fence distance F - S by the picket-and-space width, 
P + S, to get the number of the latter that fit in the former.)

So for your example,

       F - S   81.5 - 3   78.5
   n = ----- = -------- = ---- = 9.235
       P + S    5.5 + 3    8.5

Now, as you said, you need a whole number of pickets (n). I'll suppose
that F and P are really fixed, but you can adjust the spacing S. You'd
like the actual space (which I'll call A) to be no more than S. That means
you will round n UP, increasing the number of pickets to the next whole
number (10, in this case), and then find the appropriate space A by solving

   nP + (n + 1)A = F

Doing the actual algebra this time gives

   A = (F - nP)/(n + 1)

For your case, this works out to

   A = (81.5 - 10*5.5)/11 = 2.41 inches

You might also try rounding n down to 9 instead, which would give a larger
space, but closer to 3 inches:

   A = (81.5 - 9*5.5)/10 = 3.2 inches

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Geometry
High School Practical Geometry
Middle School Algebra
Middle School Geometry

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