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Lawn Mower Curvature

  1. Jul 25, 2012 #1
    "Lawn Mower Curvature"

    Given a curve C in the plane, if you pick a consistent perpendicular direction, you can construct a new curve by moving out a fixed distance ε from the curve in that direction. For small values of ε, the area between these two curves will be approximately equal to ε times the length of C, but there will be a certain shrinkage (or expansion). For example, if C is a circular arc with length s and radius of curvature r (choosing the perpendicular direction inwards), the area between the curves is given by εs - ε2s/(2r). If C consists of two straight line segments of length s/2 connected by a single angle with measure θ, again taking the perpendicular direction inwards, the area is given by εs - ε2cot(θ/2). A good measure of this shrinkage or expansion might be lim(ε→0)(ΔA - εs)/ε2. For lack of a better term, I'll call this the "lawn mower curvature," since thinking about the area swept out by a lawn mower as it follows a curved path was what got me thinking about this in the first place.

    This line of reasoning leads me to believe that the lawn mower curvature of any smooth, closed, non-self-intersecting plane curve, taken in the inward direction, is -π: Imagine using n-sided polygons with vertices on the curve to approximate it. The sum of the exterior angles of the polygon, with angles (near concavities) that point the "wrong way" being taken as negative, must be equal to 2π. In terms of the exterior angle θi, the lawn mower curvature at each vertex is -tan(θ/2), so the total lawn mower curvature of the polygonal chain is -Ʃitan(θi/2) As the number of vertices of the polygon increases, the measure of each exterior angle must decrease, and in the limit as the number of vertices becomes infinite, the tangent of each exterior angle becomes equal to its value in radians, so the total lawn mower curvature becomes -Ʃiθi/2 = (-1/2)Ʃiθi = (-1/2)(2π) = -π.

    Is there a better way to prove this? Is it true in general that the lawn mower curvature of a curve is equal to half its total curvature in the ordinary sense? Moreover, is what I call "lawn mower curvature" actually used in established mathematical theories, perhaps as an alternative way to define curvature? On the other hand, perhaps I made an error in reasoning at some crucial step, and what I claimed above isn't actually true.
  2. jcsd
  3. Jul 28, 2012 #2


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    Re: "Lawn Mower Curvature"

    Need to define what the area is if the parallel curve backtracks, i.e. when ε exceeds the radius of curvature of C inside a bend. If you treat it as a signed entity, I believe your result stands. Just consider a short section ds turning through angle dψ. Radius of curvature r = ds/dψ. Area swept out by margin width ε is r2dψ/2 - (r-ε)2dψ/2 = (rε - ε2/2)dψ. (Note that if ε > 2r then this goes negative.) The 'straight line' sweep would be εds = εrdψ. Subtracting this leaves - (ε2/2)dψ. Integrating wrt ψ etc...
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