On the curvature extrema of special cubic Bézier curves
aa r X i v : . [ c s . C G ] J a n On the curvature extremaof special cubic Bézier curves
Kenjiro T. Miura, Péter SalviJanuary 21, 2021
In this document we are going to prove that a cubic Bézier curve C ( t ) = X i =0 P i (cid:18) i (cid:19) t i (1 − t ) − i (1)with the special control point configuration P = Q , P = (1 − a ) Q + aQ , P = aQ + (1 − a ) Q , P = Q (2)has at most one local extremum of curvature when t ∈ (0 , and a ∈ ( , . We need to treat two special cases first. When Q = Q , the curve degeneratesto a line segment, which has a kink (a point with infinite curvature) at t = ,so it clearly has exactly one local curvature extremum.With this handled, let us set, without loss of generality, Q = ( − , , Q = ( b, h ) , Q = (1 , , (3)where b, h ≥ . The second special case is when h = 0 . Once again, the curvedegenerates to a line segment. When b ∈ ( − , , its curvature is always ,otherwise it has a single kink.In the following we will assume h > . The signed curvature of a planar polynomial curve ( x ( t ) , y ( t )) is given [1] as κ = x ′ y ′′ − x ′′ y ′ ( x ′ + y ′ ) , (4)1here x ′ is the first derivative of x with respect to t etc. The derivative of itssquare is ( κ ) ′ = 2 κκ ′ = 2 ( x ′ y ′′ − x ′′ y ′ ) ( x ′ y ′′′ − x ′′′ y ) (cid:0) x ′ + y ′ (cid:1) ( x ′ + y ′ ) − ( x ′ y ′′ − x ′′ y ′ ) (cid:0) x ′ + y ′ (cid:1) (2 x ′ x ′′ + 2 y ′ y ′′ )( x ′ + y ′ ) (5)so (cid:0) x ′ + y ′ (cid:1) κκ ′ = ( x ′ y ′′ − x ′′ y ′ ) (cid:2) ( x ′ y ′′′ − x ′′′ y ) (cid:0) x ′ + y ′ (cid:1) − x ′ y ′′ − x ′′ y ′ ) ( x ′ x ′′ + y ′ y ′′ )] . (6)Here x ′ y ′′ − x ′′ y ′ = 0 means that the curvature is equal to , and it correspondsto inflection points. Consequently, ( x ′ y ′′′ − x ′′′ y ) (cid:0) x ′ + y ′ (cid:1) − x ′ y ′′ − x ′′ y ′ ) ( x ′ x ′′ + y ′ y ′′ ) = 0 (7)corresponds to curvature extrema. Let N ( t, a ) denote the left-hand side of Eq. (7), applied to our curve, withcontrol points given as in (3). We need to show that N ( t, a ) has at most one0-crossing when t ∈ (0 , and a ∈ ( , .First note that N (0 , a ) = − a h (cid:2) (1 + b )(12 + 3 a (5 + b ) − a (7 + b )) + a ( − a ) h (cid:3) , (8)which is positive for a = . Let us denote the expression in brackets with f ( a ) .We have f ( ) < , f (1) < and df ( a ) da = 6( b + 6 b + 5 + h ) a − b + 8 b + 7 + h ) , (9)from which we can see that df ( a ) /da is negative at a = , and increases with a . We can conclude that as a goes from to , f first decreases, and then itmay increase, but since f (1) < , it always remains negative, so N (0 , a ) > . (10)The derivative of N with respect to t is given as ∂N ( t, a ) ∂t = 1296 ah · f ( t, a ) · f ( t, a ) , (11)2here f ( t, a ) = 2 t − t + 1 − (cid:0) t − t + 1 (cid:1) a, (12) f ( t, a ) = − a ( b + b (10 − t ) + h + 60( t − t + 13)+ 4 a (5 b (1 − t ) + 60( t − t + 11) + 80(1 − t ) t − . (13)It is easy to see that f ( t, a ) ≥ , since its derivative with respect to a is negative,and f ( t, is positive. Consequently, the signs of ∂N ( t, a ) /∂t and f ( t, a ) arethe same.As for f ( t, a ) , note that it is a quadratic function of t , and the coefficient of t is − a − a + 4) , which is negative for a ∈ ( , . We can also prove f (0 , a ) < , (14)by ascertaining that f (0 , ) < , ∂f (0 , ) /∂a < and ∂ f (0 , a ) /∂a < , asthis means that ∂f (0 , a ) /∂a decreases as a goes from to , starting from anegative value, and thus f (0 , a ) itself is always negative, as well. b ≤ − a The maximum of f ( t, a ) for a fixed a value is found by solving ∂f ( t, a ) /∂t = 0 ,giving t = ab + 3 a − a − . (15)When b ≤ − a , this will be in the [0 , interval, and f ( t , a ) = 8 − a + 6 a − a h + 2 a b (16) < − a + 6 a − a h + 2 a (cid:18) − a (cid:19) (17) = (24 − h ) a − a + 16 =: f ( a ) . (18)Since f is quadratic in a , and also f ( ) < , f (1) < and df ( ) /da < ,we can see that as a goes from to , the value of f decreases, starting froma negative value, and while it may start to increase, it remains negative.In summary, we have shown that in this case the maximum of f is negative,so ∂N ( t, a ) /∂t < , i.e., the curvature decreases monotonically. b > − a In this case f takes its maximum over the [0 , interval at t = 1 . We can alsostate the following (these will be proved below): N (1 , a ) < when f (1 , a ) > , (19) ∂f (0 , a ) ∂t > , (20) ∂ f ( t, a ) ∂t < . (21)3quation (21) shows that ∂f ( t, a ) /∂t decreases monotonically as t goes from to , while by Eq. (20) it starts from a positive value, so it may have at mostone 0-crossing. Consequently f first increases, starting from a negative value(Eq. 14), and then it may decrease.When f (1 , a ) < , since this is its maximal value, it means that f is alwaysnegative, so the curvature decreases monotonically.Otherwise f has exactly one 0-crossing, so N ( t, a ) first decreases, startingfrom a positive value, and then increases, as t goes from to . Since in thiscase N (1 , a ) < (Eq. 19), there is exactly one curvature extremum. Since f (1 , a ) = − a (cid:18) a − a − b (cid:19) − a h + 36 (cid:18) a − (cid:19) (cid:18) a − (cid:19) , (22)it can be seen that f (1 , a ) can only be positive when a > . Under theseconstraints it holds that a − > a , so f (1 , a ) | b< < f (1 , a ) | b =1 < . (23)Consequently, the assumption f (1 , a ) > implies b > .Restructuring N (1 , a ) = − a h (cid:2) b −
1) + 4 a (7 + ( b − b + h ) − a (5 + ( b − b + h ) (cid:3) (24) = − a (4 − a ) h "(cid:18) b − − a + 16 a − − a ) a (cid:19) − a − (4 − a ) a + h (25)shows that N (1 , a ) is negative when ( b, h ) is outside the circle with center (cid:16) − a +16 a − − a ) a , (cid:17) and radius a − (4 − a ) a . But ( b, h ) will be outside for any b > ,which proves Eq. (19). We have ∂f (0 , a ) ∂t = 20 [ a (3 a − b + 3 a (3 a −
4) + 4] , (26)and a (3 a −