Equal Spacing Layout
Evenly space balusters, pickets, studs, shelf pins, or layout marks in a run.
Example
You enter
- Total run (in) 60
- Item width (in, 0 for marks) 1.5
- Solve for max-gap
- Maximum gap (in; just under the 4 in sphere) 4
You get
- Count 11
- Gap (in) 3.625
- Center to center (in) 5.125
No published worked example is on file for this calculation. It is checked against a hand-worked derivation of its method and an independent reference test.
Details, formula, and sources
ScopeN items of width w have N+1 equal gaps of (run - N x w)/(N+1) and a center-to-center pitch of gap + w. Solve for the count from a maximum gap (the smallest N with gap at or below the limit, e.g. the IRC 4-inch-sphere guard rule) or the gap from a desired count, with the mark positions (first-principles layout arithmetic).
Formula or tableN items of width w in a run R have N+1 equal gaps of gap = (R - N x w)/(N+1) and a center-to-center pitch of gap + w. Max-gap mode: smallest N with gap at or below gmax is N = ceil((R - gmax)/(w + gmax)). Item-center positions are measured from the start of the run.
Source and editionEqual-spacing (baluster / picket / stud / division-point) layout - first-principles arithmetic; public domain.
Free accessPure layout arithmetic, public; the run, item width, and gap limit or count are user-supplied.
What governsEstimate. AHJ and licensed professional govern.
API field namesrun_in, item_width_in, mode, max_gap_in, count, gap_in, center_to_center_in
- Equal end gaps the N items sit in a run with N+1 equal gaps, including the two end gaps (the post-to-baluster case); width 0 lays out division pointslayout arithmetic