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In hot pursuit of the athletic dollar
China is licking its chops at the prospect of hundreds of thousands of tourists swarming across its borders to watch the 2008 Summer Olympics.
Cash registers will be ringing up sales throughout the stifling month of August.
Korea, a couple of hours away by plane, looks on with envy. With so many tourists in the region, it comes as no surprise that tourism officials are dreaming up schemes to get a slice of the action.
This country has bold plans for developing its tourism, and the Olympics is a vast cash cow.
Attracting visitors this summer would boost the Korea Tourism Organization's chances of meeting its target of $5.8 billion in tourism revenue and 7 million visitors, unveiled two weeks ago.
The grand plan is to lure international athletes bound for the Beijing Games to Korea with promises of discounts and freebies.
"Of course these athletes will be in serious training for the summer events and might not have time for sightseeing, so we don't expect a surge," said Ji Hae-nam, an official at Seoul city government who is responsible for attracting teams.
"But we're looking long term. If these athletes come to Korea to train and rest prior to the games and if they have a good impression of our country, they might return in the future to train or as tourists. And they might spread the word in their home country, which is a kind of indirect promotion," Ji said.
The Korea Tourism Organization has stepped up its marketing for the Olympics by signing an agreement to strengthen cooperation with the Korea Sports Council.
"I only regret that we didn't unite sports and tourism much earlier," said Kim Jung-kil, chairman of the sports council during the agreement signing ceremony, which was held at the tourism organization's office in Seoul.
"This is a significant moment as we are introducing Korea through sports," Kim continued.
Oh Jee-chul, president of the tourism organization, said his institution along with the Korea Sports Council will expand cooperation with local governments. They hope to promote Korea as a training ground to international teams.
Joung Jin-soo, leader of the strategic product development team at the tourism organization, said, "We are hoping to attract approximately 2,000 people before the Olympics begin."
According to the tourism organization, as of Monday local city governments including Seoul have secured agreements with 425 athletes and coaches of 30 national teams from 14 countries.
They include Canada, Denmark, Poland and Japan. Negotiations with Egypt, Algeria, Russia and Bulgaria are underway.
"We're lucky," the official said.
"Even if China has the most advanced sports facilities the world has seen, there won't be enough accommodation for the number of athletes that need to rent them. If they get the chance to rent them, it will only be for short periods, such as stretching and warmups before a competition. Korea has a geographical advantage and we have the facilities to meet the needs of the international teams," he said.
The official said Japan is Korea's only rival.
"Hong Kong doesn't stand a chance because it is a city-state and its sporting facilities are limited compared with Japan and Korea. Although Japan's facilities are far better than Korea's, we're geographically closer," he said.
The tourism organization began its aggressive marketing activities through its overseas branch offices last year.
"Korea's facilities are well-equipped and meet international standards," said Kim of the sports council, adding that the food and climate are also similar to Beijing.
Jeju has been the most successful local government so far, securing more than 250 athletes from nine countries competing in six sporting events.
Hwang Seong-hwa, in charge of sports for the Jeju Special Self-Governing Province, said the island is offering hotel discounts ranging between 20 percent and 60 percent.
Discount flights and free use of sport facilities including gymnasiums have also been put on the table.
"Some couldn't believe the offers and called up several times to confirm the deals," said Hwang.
"Free use of the gym is what has really set us apart from Japan, which admittedly has better facilities than us," he added.
Interestingly, the Japanese national swimming team and the synchronized swimming team have made reservations to practice in pools on Jeju Island.
"The Japanese team is a bit superstitious. They won gold medals in the past after practicing on an island," Hwang said.
The Japanese synchronized swimming team has booked for May and August; the swim team has booked the use of Jeju facilities in July.
Jeju Island is expecting 1.2 billion won ($1.27 million) in revenue from this sporting enterprise, cash that will be put to good use.
Jeju is hoping to host foreign teams at events in Asia such as the World Championships in Athletics in 2011 in Daegu and the Asian Games scheduled for 2014 in Incheon.
"We tried to attract mostly domestic athletic teams to train in Jeju in the past," said Hwang. "The Beijing Olympics have allowed us to expand to the international market."
Gyeonggi has already designated which cities will accommodate which sport. Suwon and Ansan have been assigned football teams, while Seongnam and Pyeongtaek are concentrating on hockey.
Hwaseong has been assigned fencing, and Pocheon badminton.
Suwon and Yongin are trying to lure international taekwondo teams, both cities having many years' experience working with the top taekwondo fighters in Korea.
The American national diving team trained earlier this month in Suwon for three days in preparation for the FINA Diving World Cup held between Feb. 19 and 25 in Beijing.
The city competed against Hong Kong and Macao before finalizing the agreement with the U.S. team.
The Seoul city government earlier this month announced its own promotional campaign.
Its campaign, "Visit Korea 50," focuses on international athletes and tourists, offering significant discounts on tourism services, including shopping, eateries and accommodation discounted at hotels approved by the Seoul city government by up to 70 percent.
The campaigns will be held 50 days prior to major international events in the region.
To further attract overseas visitors, the Seoul city government is planning entertainment via cultural festivals before, after and even during the Olympics. Transportation is likely to be free.
However, the promotional campaign is extremely disorganized.
Some figures provided by the Korea Tourism Organization and local government don't match.
Data from the tourism organization suggests that Jeju has attracted 250 athletes from nine countries, but Jeju claims a higher number.
"Ten countries with 410 athletes, coaches and trainers have made reservations to use our facilities in Jeju this year," said Hwang. "We're hoping to host more than 1,000 this year."
What's more, the Seoul city government said it has secured more than just the Canadian taekwondo team to its sports facilities, but the tourism organization has different information.
"A number of countries are interested in training here before the Olympics," said Ji of the Seoul city government.
"Poland's national team has agreed to train in Seoul and the Australian and Cuban teams have shown interest," Ji claimed.
Joung of the tourism organization said figures from local governments may differ as non-national teams competing in the Olympics and figures from last year might have been included in the different tallies.
But there is clearly an organizational problem at work: too many organizations and local governments are working independently and not communicating enough with each other.
Ji of the Seoul city government agrees.
"Some international teams even visit Seoul without our knowledge because they are invited by some other organization," he said, adding that the Ministry of Culture and Tourism wants to unify the system.
Another problem facing these ambitious plans to bring in overseas teams are the outdated facilities, despite what Kim of the sports council claims.
Ji said Seoul has attracted fewer athletes than Jeju partly because it is easier to schedule facilities at regional governments like on Jeju because the population is less concentrated.
"The track and the infield at the Olympic Stadium in Jamsil haven't been updated and most of the recently built facilities are open to the public. It's not easy to ban Seoul residents from using the facilities because we want to rent them to international teams," Ji said.
Organizers are also struggling with language issues.
"Organizers from the host cities should be competent in English and another language, like Spanish for example, but there's a shortage of specialized manpower in this area," Kim admitted.
Korea may have its geographical advantages, bags of discounts on offer and enough enthusiasm to motivate an entire Olympic squad, but questions remain.
Is there a will to organize the different authorities into a more cohesive unit so Korea can take advantage of the opportunities ahead in the summer?
At the moment it looks like every man for himself in this latter-day gold rush.
By Lee Ho-jeong Staff Reporter [ojlee82@joongang.co.kr]


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\(\epsilon \kappa \)-Curves: controlled local curvature extrema
Kenjiro T. Miura ORCID: orcid.org/0000-0001-9326-31301,
R. U. Gobithaasan ORCID: orcid.org/0000-0003-3077-87722,
Péter Salvi ORCID: orcid.org/0000-0003-2456-20513,
Dan Wang ORCID: orcid.org/0000-0003-1906-49981,
Tadatoshi Sekine ORCID: orcid.org/0000-0003-1813-822X1,
Shin Usuki ORCID: orcid.org/0000-0002-4363-63461,
Jun-ichi Inoguchi ORCID: orcid.org/0000-0002-6584-57394 &
Kenji Kajiwara ORCID: orcid.org/0000-0002-0543-93845
The Visual Computer volume 38, pages 2723–2738 (2022)Cite this article
The \(\kappa \)-curve is a recently published interpolating spline which consists of quadratic Bézier segments passing through input points at the loci of local curvature extrema. We extend this representation to control the magnitudes of local maximum curvature in a new scheme called extended- or \(\epsilon \kappa \)-curves.\(\kappa \)-curves have been implemented as the curvature tool in Adobe Illustrator® and Photoshop® and are highly valued by professional designers. However, because of the limited degrees of freedom of quadratic Bézier curves, it provides no control over the curvature distribution. We propose new methods that enable the modification of local curvature at the interpolation points by degree elevation of the Bernstein basis as well as application of generalized trigonometric basis functions. By using \(\epsilon \kappa \)-curves, designers acquire much more ability to produce a variety of expressions, as illustrated by our examples.
Avoid the most common mistakes and prepare your manuscript for journal editors.
The \(\kappa \)-curve, proposed recently by [28], is an interpolating spline which is curvature-continuous almost everywhere and passes through input points at the local curvature extrema. It has been implemented as the curvature tool in Adobe Illustrator® and Photoshop® and is accepted as a favored curve design tool by many designers (see, e.g., [4, 6]).
We consider the reasons for the success of \(\kappa \)-curve to be:
Information along contours is concentrated at local maxima of curvature.
Curves of low degree have smooth distribution of curvature.
\(G^2\)-continuous curves tend to look fairer than only \(G^1\)-continuous ones.
Attneave [1] suggested, based on his empirical study, that information along contours is concentrated in regions of high magnitude of curvature, as opposed to being distributed uniformly along the contour, and it is further concentrated at local maxima of curvature (see also [33]). Although Attneave never published the details of his methods, [20] conducted a similar experiment and obtained the same results. Levien and Séquin [12] argue similarly and assert that points of maximal curvature are salient features.
The curvature of a polynomial curve is given by a relatively complicated rational function [7], and its distribution might not be globally smooth. However, if the curve is of a low degree, the curvature distribution is more uniform and the curve is fairer, thus more suitable for illustration. The quadratic polynomial curve has the nice property that its curvature has only one local maximum, and its location is easily computable [28], which makes the handling of curvature extrema much easier.
Graphic designers often accept \(G^1\) continuity as good enough for illustration. However, discontinuity remains; for example, if you join a straight line and a circular arc with \(G^1\) continuity, the rhythm of the curve will be broken at the joint. For this reason, we give preference to \(G^2\)-continuous curves.
Nonetheless, \(\kappa \)-curves are not perfect, and their further investigation is necessary [29]. The following are two important shortcomings of \(\kappa \)-curves:
They are not curvature-continuous everywhere: at inflection points only \(G^1\) continuity is guaranteed.
Since the degree of freedom (DoF) of the quadratic segments is limited, it is impossible to control the magnitudes of local maximum curvature at the input points.
For the first shortcoming, Wang et al. [26] provided a solution through the use of log-aesthetic curves [15, 17] instead of polynomial curves. Log-aesthetic curves have a shape parameter (\(\alpha \)), which can be utilized to control the curvature distribution as shown in Fig. 1. Their method guarantees \(G^2\) continuity everywhere, including inflection points, since log-aesthetic curves with negative \(\alpha \) values can represent S-shaped curves with \(G^2\) continuity. (Note that these cannot be represented by quadratic Bézier curves.)
These curves, however, are defined by a Cesàro equation and thus take extra time to evaluate, making the interpolation method impractical for real-time design purposes.
Because of the second shortcoming, if the designer wants to increase or decrease the magnitude of the curvature extremum, she needs to add extra input points, as shown in Fig. 2. Yan et al. [27] proposed a piecewise rational, quadratic, interpolatory curve that is able to reproduce circles and other elliptical or hyperbolic shapes. Although their main intention was to reproduce circles, their method could also control the magnitude of local maximum curvature. However, only rational quadratic curves are applicable, and it is not possible to extend their method for other types.
In this paper, we propose a new method to solve the second shortcoming by degree elevation of the Bernstein basis functions, giving an extra DoF to each quadratic curve segment, providing control over the magnitudes of local maximum curvature at input points as shown in Fig. 3. In order to increase the designers' possible choices, we also introduce a new trigonometric basis for which we can perform degree elevation. In addition, we propose a general method for bases with extra shape parameters. By adding one more parameter to each of the curve segments, the designers obtain more expressive power for their illustration. The family of this new curve is denoted as \(\epsilon \kappa \)-curves.
Log-aesthetic curves with various \(\alpha \) values [26]. Input points are depicted by black boxes, and green points correspond to positions of local maximum curvature. The blue curves show the normal curvature. These curves are \(G^2\)-continuous everywhere
Addition of extra input points to control the magnitude of curvature extrema: b is the original \(\kappa \)-curve, a, c shows deformed curves with large and small curvature extrema by adding two extra input points
The left- and rightmost curves are \(\kappa \)-curves; the others are \(\epsilon \kappa \)-curves with gradual changes in the global shape parameter a. The face part (for \(a=0.95\) and \(a=2/3\)) is zoomed in for better comparison. When a=2/3, \(\kappa \)- and \(\epsilon \kappa \)-curves are identical
\(\epsilon \kappa \)-curves preserve all of the appealing properties of \(\kappa \)-curves, i.e., point interpolation, \(G^2\) continuity (except at inflection points), continuous modification (changes smoothly when the input points move), local influence, and real-time generation. With a small processing overhead, \(\epsilon \kappa \)-curves offer the ability to control the magnitude of local maximum curvature.
We have implemented \(\epsilon \kappa \)-curves in MATLAB® and Julia [11]. The source of the Julia code is available online [22].
The rest of this paper is organized as follows. Section 2 reviews the related work. Section 3 presents a method to control the magnitudes of local maximum curvature by degree elevation of the Bernstein basis functions. Section 4 introduces a new trigonometric basis and proposes a method with degree elevation similar to the one proposed in the previous section. Finally, we end with conclusions and discussion of future work.
In this section, we first review [28] and their underlying strategy developing \(\kappa \)-curves. Next, we discuss related researches on various kinds of basis function formulations for curve design.
\(\kappa \)-Curve
The basic framework of our method is adopted from [28], generating curves controlled by interpolation points. \(\kappa \)-curves have stimulated the field of interpolatory curve generation, resulting in works such as [5], which proposes a method for good control over the location and type of geometric feature points (e.g., cusps and loops).
Yan et al. [28] create a sequence of quadratic curves with \(G^2\) continuity almost everywhere. They derived explicit formulae for the point between two quadratic segments to guarantee \(G^2\) continuity and for the additional condition that input points should be interpolated at maximum curvature magnitude positions. \(\kappa \)-curves are determined by the locations of the middle control points of quadratic Bézier segments—the rest is easily derived from the continuity constraints. Hence, the variables are the locations of these middle control points.
Their basic strategy to determine these locations is to adopt a local/global approach [13, 23]: \(G^2\)-continuous connection is performed locally, while the interpolation at maximum curvature magnitude positions is done globally.
Regarding control of the magnitude of local curvature, the most common technique is to change the weight of a control point of a rational curve [7]. A larger weight attracts the curve to its control point, which makes local curvature larger. However, this technique is not applicable for interpolatory curves. Another technique is to introduce extra parameters called bias and tension to the B-spline formulation for controlling local curvature [2], but this is also not applicable to interpolatory curves.
Basis functions
As mentioned in [28], curve modeling has a long history, especially in computer-aided geometric design, as well as computer graphics. In CAGD and applied mathematics, to extend the expressive power of curves, many researchers have been trying to develop new bases with extra shape parameters. The following is a (nonexhaustive) list of such bases:
C-Bézier curve [31]
Cubic alternative curve [10]
Cubic trigonometric Bézier curve (T-Bézier basis) [8]
\(\alpha \beta \)-Bernstein-like basis [32]
Quasi-cubic trigonometric Bernstein basis [30]
Trigonometric cubic Bernstein-like basis [24]
Our method proposed in the next section can be applied for curves based not only on polynomials, but also other bases such as a trigonometric basis with degree elevation property which we will introduce in Sect. 4. All of the representations listed above have extra parameters for shape control, and we can utilize these parameters to control the magnitudes of local maximum curvature. (Not all curve types are applicable, however, as explained in Appendix C.)
It is interesting to note that most researchers to date have attempted to develop new bases using four control points, based on cubic polynomials or quadratic trigonometric functions. They prefer using four control points out of concern for connections at both ends of the curve. In order to control the magnitude of curvature at the two ends independently, at least two control points are necessary at each end. This differs fundamentally from Yan et al.'s (and our) approach, which uses only three control points.
The importance of [28] is their proposed paradigm shift for curve generation, by considering the local maximum curvature in the middle, instead of focusing on the endpoints. If we can assume that the curvature has just one local maximum in each segment, then only one extra parameter per segment is adequate to control the magnitude of its local maximum curvature.
To our best knowledge, no trigonometric basis family for arbitrary degree has been published yet. Our novel generalized trigonometric basis functions range from linear, using three control points, to any higher degree n, using \(2n+1\) control points. The curve can be evaluated by a recursive method , similar to de Casteljau's algorithm, as explained in Appendix B. Since the curve uses trigonometric functions as blending functions, it can represent a circular arc exactly, without using a rational form.
Cubic Bernstein polynomials
In this section, we extend \(\kappa \)-curves in a direct manner, by elevating the degree of quadratic Bézier segments to cubic. \(\epsilon \kappa \)-curves retain the following properties of \(\kappa \)-curves:
Interpolate all input points (control points).
All local maximum curvature points are the same as the input points.
\(G^2\) continuity is guaranteed almost everywhere (except for inflection points).
In the following, we discuss only closed curves, but it is straightforward to extend our methods to open curves as has been demonstrated for \(\kappa \)-curves.
If we elevate the degree of a planar Bézier curve, we obtain an additional control point, which has two DoFs (the x and y coordinates). To reduce these to one, we add a geometric constraint on the location of the second and third control points of the cubic Bézier curve, as shown in Fig. 4. Here, a is an internal division ratio, where the larger a is, the closer the control points \(P_1\) and \(P_2\) are to the control point \(Q_1\). We make the restriction \(2/3 \le a < 1\) because the curve should not have a complicated curvature distribution. Using \(Q_i\), the curve \(C(t;a)\) is expressed by
$$\begin{aligned} C(t;a)&= (1-t)^3 Q_0 + 3(1-t)^2t \left[ (1-a)Q_0+a Q_1 \right] \nonumber \\&\qquad + 3(1-t)t^2 \left[ a Q_1+(1-a)Q_2 \right] + t^3 Q_2. \end{aligned}$$
Note that if \(a=2/3\), the curve degenerates to quadratic.
Constrained cubic Bézier curve. If \(a=2/3\), the curve becomes quadratic
We have proved that by constraining the construction of the cubic polynomial curve as in Eq. (1), using only three control points instead of four, the curvature in one curve segment has at most one local maximum for \(2/3 \le a < 1\); see details in [16], as well as Appendix A for a general discussion on the curvature extrema of cubic polynomial curves, and a high-level summary of the proof. In addition, we have made MATHEMATICA simulation available in [9] to compute the number of curvature extrema for this curve using both classical approach and Sturm's theorem. Hence, we can safely assume that the curvature in one curve segment has at most one local maximum, and a single extra parameter for each segment is enough to control the magnitudes of local maximum curvature.
Geometric constraints
We assume that \(\epsilon \kappa \)-curves consist of a sequence of constrained cubic polynomial curves
$$\begin{aligned} c_i(t;a_i)&= (1-t)^3c_{i,0} + 3(1-t)^2t \left[ (1-a_i)c_{i,0}+a_ic_{i,1}\right] \nonumber \\&\quad + 3(1-t)t^2 \left[ a_ic_{i,1}+(1-a_i)c_{i,2}\right] + t^3c_{i,2}, \end{aligned}$$
parameterized by t and also \(a_i\), which is an extra shape parameter. The control points are given by \(c_{i,0}\), \(c_{i,1}\) and \(c_{i,2} \in {\mathbb {R}}^2\), corresponding to \(Q_i\), \(i=0,1,2\) in Fig. 4. The \(a_i\)'s are reserved for designers and can be manipulated independently.
The curvature of this curve \(c_i(t;a_i)\) is given by
$$\begin{aligned} \begin{aligned}&\kappa _i(t; a_i) = {\det \left( \frac{\partial c_{i}(t;a_i)}{\partial t}, \frac{\partial ^2 c_{i}(t;a_i)}{\partial t^2} \right) }\ \big /\ {\left\| \frac{\partial \,c_{i}(t;a_i)}{\partial t } \right\| ^3} \\&\quad = \frac{4}{3}\cdot \frac{\triangle (c_{i,0},c_{i,1},c_{i,2})\left[ a_i^2(1-t)t + a_i(1-a_i)\left( (1-t)^2+t^2\right) \right] }{\Vert (1-t)^2 a_i r_i +2(1-t)t(1-a_i)(r_i+s_i)+ t^2 a_i s_i \Vert ^3 },\nonumber \end{aligned}\\ \end{aligned}$$
where \(\triangle \) indicates the area of the triangle specified by its arguments, and \(r_i = c_{i,1}-c_{i,0}\), \( s_i = c_{i,2}-c_{i,1}\).
In the quadratic case (i.e., \(\kappa \)-curves), as the curvature has such a simple formula, we can express the parameter \(t_i\) at the point of maximal curvature explicitly, in terms of the Bézier coefficients of the \(i^\mathrm {th}\) quadratic Bézier curve as
$$\begin{aligned} t_i = \frac{\langle r_i,r_i-s_i\rangle }{\Vert r_i-s_i \Vert ^2}, \end{aligned}$$
where \(\langle a,b\rangle \) means the scalar product of vectors a and b. Then, we add the condition
$$\begin{aligned} c_i(t_i)=p_i, \end{aligned}$$
where \(p_i\) is the \(i^\mathrm {th}\) input point. Solving for \(c_{i,1}\) and substituting into Eq. (4), we get a cubic equation in \(t_i\) that depends only on the endpoints \(c_{i,0}\) and \(c_{i,2}\), and the input points \(p_i\).
Unfortunately, we cannot obtain an explicit formula like Eq. (4) for the parameter \(t_i\) in the cubic case, because of its high degree (see details in Appendix A.1), but this is not a problem. Solving Eq. (5) for \(c_{i,1}\), we arrive at
$$\begin{aligned} \begin{aligned} c_{i,1}&= \left[ p_i-(1-t_i)^3 c_{i,0}\right. \\&\quad -3(1-t_i)t_i(1-a_i)((1-t_i)c_{i,0}+t_ic_{i,2}) \\&\quad \left. -t_i^3 c_{i,2}\right] \big / (3 a_i (1-t_i) t_i). \end{aligned} \end{aligned}$$
Substituting this into the derivative of Eq. (3), and letting it equal 0, we obtain (after some simplification) a polynomial equation of degree 9 in \(t_i\). This equation can be derived by the Maxima [14] code in Fig. 13 (Appendix A). We solve this equation and select a real root in [0, 1]. Note that we have proved that there is one and only one solution for the polynomial equation of degree 9 in \(t_i \in [0,1]\) as in the case of \(\kappa \)-curves; see details in [16], as well as Appendix A.2.
However, when we use other types of curves with more complicated representations (see examples in Appendix C), this kind of formula may be hard to derive. In these cases, we can use the relaxed Newton's method to compute the maximum curvature. For this, we need to be able to compute the curvature and its derivative; we do this using a quadratic Taylor series approximation around the last value of \(t_i\).
The leftmost curve is a \(\kappa \)-curve. The other curves are \(\epsilon \kappa \)-curves: the \(a_i\) values are equal to 2/3 except for one, two and three input points, respectively, where \(a_i=0.85\)
The a values of the two wing tips in the second figure from the left are 0.95, and those of the other input points are 2/3; note the sharpening of the wings. In the third figure, the roles are reversed. The left- and rightmost curves in red are \(\kappa \)-curves with the same input points
We introduce the constant \(\lambda _i\) (\(0< \lambda _i < 1\)) according to the construction method of \(\kappa \)-curves and set
$$\begin{aligned} c_{i,2} = c_{i+1,0} = (1-\lambda _i) c_{i,1} + \lambda _i c_{i+1,1}. \end{aligned}$$
Let the curvatures at the endpoints of the curve segment be denoted by \(\kappa _i(0;a_i)\) and \(\kappa _i(1;a_i)\), then from Eq. (3)
$$\begin{aligned} \kappa _i(1;a_i)&= \frac{4}{3}\cdot \frac{(1-a_i) \triangle ^+_i}{ a_i^2 \lambda _{i}^2\Vert c_{i+1,1} - c_{i,1} \Vert ^3},\nonumber \\ \kappa _{i+1}(0;a_{i+1})&= \frac{4}{3}\cdot \frac{(1-a_{i+1}) \triangle ^-_{i+1}}{ a_{i+1}^2 (1-\lambda _{i})^2\Vert c_{i+1,1} - c_{i,1} \Vert ^3}, \end{aligned}$$
introducing the notations \(\triangle ^+_i=\triangle (c_{i,0}, c_{i,1}, c_{i+1,1})\) and \(\triangle ^-_{i}=\triangle (c_{i-1,1}, c_{i,1}, c_{i,2})\).
The leftmost curve is a \(\kappa \)-curve; the other curves are \(\epsilon \kappa \)-curves: a is equal to 0.75, 0.85 and 0.95, respectively. Note the curvature at the inflection points
By adopting the local/global approach, we treat \(c_{i,0}\) as fixed for the computation of \(\kappa _i(1;a_i)\), although it depends on \(\lambda _{i-1}\) (similarly for \(c_{i+1,2}\)).
In order to guarantee \(G^2\) continuity at the joint of two consecutive segments, the following equations should be satisfied:
$$\begin{aligned} \kappa _i(1;a_i) = \kappa _{i+1}(0;a_{i+1}). \end{aligned}$$
Hence,
$$\begin{aligned} \lambda _i = \frac{\sqrt{(1-a_i)\triangle ^+_i}}{\sqrt{(1-a_i)\triangle ^+_i} + \frac{a_i}{a_{i+1}}\sqrt{(1-a_{i+1})\triangle ^-_{i+1}}}. \end{aligned}$$
Since \(0< a_i, a_{i+1} <1\), \(\lambda _i\) is real and \(0< \lambda _i <1\).
In the global phase, we calculate the positions of the middle control points \(c_{i,1}\) by solving a linear system of equations. We treat the current values of \(\lambda _i\) (internal division ratios of \(c_{i,1}\) and \(c_{i+1,1}\)) and \(t_i\) (parameters of local maximum curvature) as fixed.
Substituting Eq. (7) into Eq. (5), we get
$$\begin{aligned} p_i&= (1-t_i)^3 \left[ (1-\lambda _{i-1}) c_{i-1,1} + \lambda _{i-1} c_{i,1} \right] \nonumber \\&\quad + 3 (1-t_i)^2 t_i \left[ (1-a_i) \left[ (1-\lambda _{i-1}) c_{i-1,1} + \lambda _{i-1} c_{i,1} \right] \right. \nonumber \\&\quad \left. + a_i c_{i,1} \right] \nonumber \\&\quad + 3 (1-t_i) t_i^2 \left[ (1-a_i) \left[ (1-\lambda _{i}) c_{i,1} + \lambda _i c_{i+1,1} \right] \right. \nonumber \\&\quad \left. + a_i c_{i,1}\right] \nonumber \\&\quad + t_i^3 \left[ (1-\lambda _{i}) c_{i,1} + \lambda _i c_{i+1,1} \right] , \end{aligned}$$
which can be solved for \(c_{i,1}\).
The optimization process is summarized in Algorithm 1.
Figure 5 shows examples of closed \(\epsilon \kappa \)-curves along with the original \(\kappa \)-curve. The input points are located at the same positions. The \(a_i\) values of these curves are equal to 2/3, except for one, two and three input points, respectively, where \(a_i\) is set to 0.85. If the \(a_i\) of all input points are 2/3, the \(\kappa \)-curve on the left is generated. Since we specify a larger value for some input points, the magnitudes of the corresponding local maximum curvature increase, as we expected.
Figure 6 shows another example of local curvature control. From the left to right, the first drawing shows a bird using \(\kappa \)-curves. In the second, we set a at the wing tips to 0.95, while leaving all others at the default 2/3. This has the effect of sharpening the wing tips. In the third figure, we reversed the role of the input points, giving \(a=2/3\) to those at the wing tips and 0.95 to all other points. Here, the wings are rounded, while other parts of the bird get sharper.
Notice that the bird's beak resembles a cusp, but is actually the start and end points of an open curve located at the same position. In our implementation, we limit \(2/3 \le a \le 1\) to make a curve with smooth curvature distribution, which disallows the generation of a cusp even at \(a=1\). In cases where the designer wants to use a cusp, the curve should be cut in two, or the input points should be relocated to form a cusp as explained in [28].
Figure 7 shows examples of global curvature control. There are three \(\epsilon \kappa \)-curves, with a set to 0.75, 0.85 and 0.95, respectively, along with the original \(\kappa \)-curve (\(a=2/3\)) for comparison. By increasing a, the magnitudes of local maximum curvature increase. As the close-up windows indicate, at the inflection point \(G^2\) continuity is violated for \(\kappa \)-curves, and only \(G^1\) continuity is guaranteed. However, for \(\epsilon \kappa \)-curves with a larger a, the magnitude of curvature at inflection points, and consequently the \(G^2\) error, becomes smaller. Note that although these curves are almost \(G^2\)-continuous everywhere, they are quite different from those in Fig. 1 (generated using the same input points).
Figures 3, 8 and 9 show the effect of changes of the global shape parameter a on various designs. As discussed above, larger a values generally induce larger local curvature extrema and steeper curvature variation. The resulting curves look more sharp at the input points and more flat between them.
Changing the global shape parameter in the bear model
Changing the global shape parameter in the elephant model
Generalized trigonometric basis
In this section, we describe our new generalized trigonometric basis. This is based on the trigonometric cubic Bernstein-like basis [24], which we are going to review first.
The trigonometric cubic Bernstein-like basis functions have an extra shape parameter \(\alpha \) and are defined by
$$\begin{aligned} f_0&= \alpha S^2 - \alpha S + C^2 = 1 + (\alpha -1) S^2 - \alpha S, \nonumber \\ f_1&= \alpha S(1-S), \nonumber \\ f_2&= \alpha (S^2+C-1) = \alpha C(1 - C), \nonumber \\ f_3&= (1-\alpha ) S^2 - \alpha C + \alpha = 1 +(\alpha -1) C^2-\alpha C, \end{aligned}$$
where \(S=\sin \frac{\pi t}{2}\), \(C=\cos \frac{\pi t}{2}\), for \(\alpha \in (0,2)\), \(t \in [0,1]\). Note that these functions satisfy partition of unity, i.e., \(\sum _{i=0}^3 f_i(t) = 1\) for any \(\alpha \). When \(\alpha =1\), the above functions are simplified to
$$\begin{aligned} f_0&= 1 - S, \nonumber \\ f_1&= S(1-S), \nonumber \\ f_2&= C(1-C), \nonumber \\ f_3&= 1 - C. \end{aligned}$$
If we add the second and third functions together and rename them to u, v and w, we obtain blending functions \(\{ u, v, w \}\) as follows:
$$\begin{aligned} \begin{aligned}&u = 1-S, \\&v = S(1-S) + C(1-C) = S+C-1, \\&w = 1 - C. \end{aligned} \end{aligned}$$
It is straightforward to define a curve by these blending functions with three control points, which we can regard as a "linear" trigonometric curve since the highest degree the trigonometric functions are in is one.
One interesting relationship among these functions is
$$\begin{aligned} v^2 = 2 u w, \end{aligned}$$
which enables
$$\begin{aligned} (u+v+w)^2 = u^2 + 2 u v + 4 u w + 2 v w + w^2, \end{aligned}$$
and yields the five blending functions \(\{u^2\), 2uv, 4uw, 2vw, \(w^2 \}\), associated with five control points. We can define a curve using these blending functions and regard it as a "quadratic" trigonometric curve since the highest power of each blending function is now degree two.
In a similar way, we can extend blending functions of "degree" n with \(2n+1\) control points. As explained in Appendix B, we can perform a recursive procedure to evaluate a curve of any degree similar to de Casteljau's algorithm avoiding the overhead of trigonometric function evaluation. This means that it is not necessary to calculate the coefficients of blending functions, or keep a coefficient table.
We formulate the \(\epsilon \kappa \)-curve in this basis using a strategy similar to that in the previous section, i.e., using a sequence of quadratic trigonometric curves with a constraint on the positions of their control points, as shown in Fig. 10. Note the location of the control point \(P_2=\left[ (1-a)Q_0+2 a Q_1+(1-a) Q_2\right] /2\). The curve \(c(t;a)\) is defined by
$$\begin{aligned} \begin{aligned} c(t;a) =&\;u^2 Q_0 + 2 u v \left[ (1-a) Q_0+a Q_1 \right] \\&+ 2 u w \left[ (1-a)(Q_0+Q_2)+2 a Q_1 \right] \\&+ 2 v w \left[ a Q_1+ (1-a) Q_2 \right] + w^2 Q_2. \end{aligned} \end{aligned}$$
When a is equal to 1/2, the curve degenerates to a linear trigonometric curve.
Constrained quadratic trigonometric curve. When \(a=1/2\) (bottom), the curve becomes linear
Geometric constraints and optimization
First, we analyze the linear trigonometric curve since it corresponds to the original \(\kappa \)-curve. Let \(c_i(t)\) be a linear trigonometric curve with control points \(c_{i,0}\), \(c_{i,1}\) and \(c_{i,2}\) and defined by
$$\begin{aligned} c_i (t) = (1 - S) c_{i,0} + (S+C-1) c_{i,1} + (1-C) c_{i,2}, \end{aligned}$$
where \(S=\sin \frac{\pi t}{2}\), \(C=\cos \frac{\pi t}{2}\) and \(t \in [0,1]\). Its curvature is given by
$$\begin{aligned} \kappa _i(t) = \frac{2 \Delta (c_{i,0}, c_{i,1}, c_{i,2})}{ \left( C^2 \Vert r_i \Vert ^2 + 2 C S \langle r_i, s_i\rangle + S^2 \Vert s_i \Vert ^2 \right) ^\frac{3}{2} }, \end{aligned}$$
where \(r_i = c_{i,1}-c_{i,0}\) and \( s_i = c_{i,2}-c_{i,1}\). The numerator of the above formula does not depend on t, so the extrema of the following \(f_i(t)\) corresponds to those of \(\kappa _i(t)\):
$$\begin{aligned} f_i(t) = C^2 \Vert r_i \Vert ^2 + 2 C S \langle r_i, s_i\rangle + S^2 \Vert s_i \Vert ^2, \end{aligned}$$
and its derivative with respect to t is given by
$$\begin{aligned} \begin{aligned} \frac{d f_i(t)}{d t} =&\;{\pi }(-CS \Vert r_i \Vert ^2+(-S^2+C^2)\langle r_i, s_i\rangle \\&+ SC \Vert s_i \Vert ^2 ). \end{aligned} \end{aligned}$$
By assuming \({d f_i(t)}/{d t}=0\) with \(S, C \ne 0\), we obtain
$$\begin{aligned} S^2 - \gamma SC - C^2=0, \end{aligned}$$
where \(\gamma =( \Vert s_i \Vert ^2-\Vert r_i \Vert ^2)/\langle r_i, s_i\rangle \). We can solve the above equation and obtain
$$\begin{aligned} C = \frac{-\gamma + \sqrt{\gamma ^2+4}}{2} S = \beta S. \end{aligned}$$
Since \(0 \le S, C \le 1\), we have the unique solution
$$\begin{aligned} S = \frac{1}{\sqrt{\beta ^2+1}}. \end{aligned}$$
$$\begin{aligned} t = \frac{2}{\pi } \arcsin \frac{1}{\beta ^2+1}. \end{aligned}$$
Note that when \(r_i\) and \(s_i\) are perpendicular to each other, if \(\Vert r_i\Vert = \Vert s_i \Vert \), then the curve becomes a circular arc, and no local maximum curvature exists. If \(\Vert r_i \Vert > \Vert s_i \Vert \), then the curvature at \(t=1\) will be maximum, and if \(\Vert r_i \Vert < \Vert s_i \Vert \), the curvature at \(t=0\) will be maximum in this curve segment.
The two curves on the top left (red) are \(\kappa \)-curves; the two on the top right (brown) are \(\epsilon \kappa \)-curves using cubic Bernstein basis functions with \(a=0.75\) and 0.9. The bottom row shows \(\epsilon \kappa \)-curves using quadratic trigonometric basis functions with \(a=0.55\), 0.6, 0.75 and 0.9
A Christmas tree drawn with \(\epsilon \kappa \)-curves using the cubic Bernstein basis functions (left) and the quadratic trigonometric basis functions (right). Note that the latter has more rounded forms and—in this case—preferable
For a quadratic trigonometric curve, the curvatures \(\kappa _i(1;a_i)\) and \(\kappa _{i+1}(0;a_i)\) at the endpoints of the constrained quadratic trigonometric curve \(c_i\) are given by
$$\begin{aligned} \kappa _i (1;a_i)&= \frac{1-a_i}{a_i^2}\cdot \frac{ \triangle ^+_i}{ \lambda _i^2 \Vert c_{i+1,1}-c_{i,1} \Vert ^3}, \nonumber \\ \kappa _{i+1} (0;a_{i+1})&= \frac{1-a_{i+1}}{a_{i+1}^2}\cdot \frac{\triangle ^-_{i+1}}{(1-\lambda _i)^2 \Vert c_{i+1,1}-c_{i,1} \Vert ^3}. \end{aligned}$$
We can calculate \(\lambda _i\) by guaranteeing \(G^2\) continuity at the joint of \(c_i(1;a_i)\) and \(c_{i+1}(0;a_{i+1})\):
$$\begin{aligned} \lambda _i = \frac{\sqrt{(1-a_i)\triangle ^-_i}}{\sqrt{(1-a_i)\triangle ^-_i} + \frac{a_i}{a_{i+1}}\sqrt{(1-a_{i+1})\triangle ^+_{i+1}}}. \end{aligned}$$
As before, we get a linear system of equations for \(c_{i,1}\):
$$\begin{aligned} p_i&= u_i^2 ( (1-\lambda _{i-1}) c_{i-1,1} + \lambda _{i-1} c_{i,1} ) \nonumber \\&\quad + 2 u_i v_i ( (1-a_i)(1-\lambda _{i-1}) c_{i-1,1}\nonumber \\&\quad + ((1-a_i)\lambda _{i-1} + a_i )c_{i,1} ) \nonumber \\&\quad +2 u_i w_i ( (1-a_i) (1-\lambda _{i-1}) c_{i-1,1} +((1-a_i)\lambda _{i-1} \nonumber \\&\quad +2 a_i+(1-a_i)(1-\lambda _{i}) ) c_{i,1} + (1-a_i) \lambda _{i} c_{i+1,1} ) \nonumber \\&\quad + 2 v_i w_i ( (a_i+(1-a_i)(1-\lambda _{i}) ) c_{i,1}\nonumber \\&\quad + (1-a_i) \lambda _{i} c_{i+1,1} ) \nonumber \\&\quad + w_i^2 ( (1-\lambda _{i}) c_{i,1} + \lambda _{i} c_{i+1,1}), \end{aligned}$$
where \(u_i=1-\sin \frac{\pi t_i}{2}\), \(v_i=\sin \frac{\pi t_i}{2} + \cos \frac{\pi t_i}{2} -1\) and \(w_i = 1-\cos \frac{\pi t_i}{2}\).
Figures 11 and 12 show examples of \(\epsilon \kappa \)-curves using the quadratic trigonometric basis functions explained in this section. In the first figure, the top left two curves (red) are \(\kappa \)-curves. The top right two curves (brown) are \(\epsilon \kappa \)-curves, using cubic Bernstein basis functions with \(a=0.75\) and 0.9. The bottom row (green) shows \(\epsilon \kappa \)-curves using quadratic trigonometric basis functions with \(a=0.55\), 0.6, 0.75 and 0.9. The curves in the bottom row are more rounded than those of the Bernstein basis. By increasing a, the differences between the two types of curves become smaller as they approach a polyline generated by connecting the input points.
The second figure shows a case where these more rounded forms are clearly preferable.
Use of built-in shape parameters
So far our strategy made use of degree elevation—from quadratic to cubic in the polynomial case, and from linear to quadratic in the trigonometric case. Yet another strategy is to use extra shape parameters built into the basis functions.
As an illustrative example, take the trigonometric cubic Bernstein-like basis functions reviewed in the previous section. In the framework of \(\epsilon \kappa \)-curves, we have to degenerate the curve by relocating the positions of its control points, essentially reducing its degree.
The trigonometric cubic Bernstein-like basis functions need four control points to define a curve. To construct "quadratic" curves corresponding to the quadratic Bézier segments of \(\kappa \)-curves, we make the second and third control points collocate. Hence, the blending functions become
$$\begin{aligned} b_0(t;\alpha )&= 1+(\alpha -1)S^2-\alpha S,\nonumber \\ b_1(t;\alpha )&= \alpha (S+C-1),\nonumber \\ b_2(t;\alpha )&= 1+(\alpha -1)C^2-\alpha C, \end{aligned}$$
where \(S=\sin \frac{\pi t}{2}\), \(C=\cos \frac{\pi t}{2}\), for \(\alpha \in (0,2)\), \(t \in [0,1]\). Incidentally, this will result in the same curve as Eq. (17), if we substitute 2a for \(\alpha \).
Note that several curve types—such as the cubic alternative curve and the \(\alpha \beta \)-Bernstein-like basis functions, for specific extra parameters—have zero curvature at the endpoints, so we cannot obtain the internal division ratio \(\lambda _i\). For these curves, the method shown in this section cannot be applied. See details in Appendix C.
We have proposed two types of \(\epsilon \kappa \)-curves as extensions of \(\kappa \)-curves, for controlling the magnitudes of local maximum curvature. Our methods use degree elevation of the Bernstein basis functions and a new family of trigonometric basis functions.
In line with Yan et al.'s paradigm shift for curve generation, we consider the local maximum curvature in the middle, instead of focusing on the endpoints. The new curves preserve all of the nice properties of \(\kappa \)-curves, i.e., point interpolation, \(G^2\) continuity (except at inflection points), continuous modification (changes smoothly when the input points move) and local influence. Computing \(\epsilon \kappa \)-curves is quite fast, and designers can manipulate them interactively, acquiring much more expressive power for curve design, as illustrated by our examples. The processing time for generating \(\epsilon \kappa \)-curves is similar to that of \(\kappa \)-curves, especially for \(\epsilon \kappa \)-curves with cubic Bernstein basis. For example, \(\kappa \)-curves consume 0.07 sec to draw Fig. 8 (bear). Under similar conditions, cubic Bézier \(\epsilon \kappa \)-curves consume 0.08 sec and generalized trigonometric \(\epsilon \kappa \)-curves takes 0.37 sec on average.
Future work includes the development of \(\epsilon \kappa \)-curve plug-ins for Adobe Illustrator® and Photoshop®. Another possible research direction is to apply the proposed method to different types of aesthetic curves, e.g., log-aesthetic curves [26], \(\sigma \)-curves [18] or \(\tau \)-curves [19].
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This work was supported by JST CREST (No. JPMJCR1911); JSPS Grant-in-Aid for Scientific Research (B, No. 19H02048); JSPS Grant-in-Aid for Challenging Exploratory Research (No. 26630038); Solutions and Foundation Integrated Research Program; ImPACT Program of the Council for Science, Technology and Innovation; and the Hungarian Scientific Research Fund (OTKA, No. 124727). The authors acknowledge the support by 2016, 2018 and 2019 IMI Joint Use Program Short-term Joint Research "Differential Geometry and Discrete Differential Geometry for Industrial Design" (September 2016, September 2018 and September 2019). The second author acknowledges University Malaysia Terengganu for approving sabbatical leave which was utilized to work on emerging researches, including this work.
Graduate School of Science and Technology, Shizuoka University, Hamamatsu, Shizuoka, Japan
Kenjiro T. Miura, Dan Wang, Tadatoshi Sekine & Shin Usuki
Faculty of Ocean Engineering Technology & Informatics, University Malaysia Terengganu, Kuala Nerus, Terengganu, Malaysia
R. U. Gobithaasan
Department of Control Engineering and Information Technology, Budapest University of Technology and Economics, Budapest, Hungary
Péter Salvi
Institute of Mathematics, University of Tsukuba, Tennodai, Tsukuba, Japan
Jun-ichi Inoguchi
Institute of Mathematics for Industry, Kyushu University, Fukuoka, Fukuoka, Japan
Kenji Kajiwara
Kenjiro T. Miura
Dan Wang
Tadatoshi Sekine
Shin Usuki
Correspondence to Kenjiro T. Miura.
Below is the link to the electronic supplementary material.
Supplementary material 1 (mp4 6667 KB)
Local maximum curvature of cubic polynomial curves
General case
The signed curvature \(\kappa (t)\) of a cubic polynomial curve \(c(t) = ( x(t), y(t) )\) is given by [3] as
$$\begin{aligned} \kappa (t) = \frac{x' y''-x'' y' }{\left( x'^2 + y'^2\right) ^{\frac{3}{2}}}, \end{aligned}$$
where \(x' = {\mathrm dx(t)}/{\mathrm dt}\), \(x'' = {\mathrm d^2x(t)}/{\mathrm dt^2}\), and higher derivatives are expressed in a similar way.
At the extremum \({\mathrm d\kappa (t)}/{\mathrm dt} = 0\) and \({\mathrm d\kappa (t)^2}/{\mathrm dt} = 2 \kappa (t) \cdot {\mathrm d\kappa (t)}/{\mathrm dt}\). Hence, if we exclude points where \(\kappa (t)=0\), we can obtain t values at the extrema.
By differentiating
$$\begin{aligned} \kappa (t)^2 = \frac{(x'(t) y''(t)-x''(t) y'(t) )^2}{( x'(t)^2 + y'(t)^2 )^3} \end{aligned}$$
with respect to t, we obtain
$$\begin{aligned} \left( x'^2 + y'^2\right) ^4 \kappa \frac{\mathrm d\kappa }{\mathrm dt}&= (x''y' - x'y'')\bigl ((x'''y' - x'y''')(x'^2 + y'^2) \nonumber \\&\quad -3(x'x'' + y'y'')(x''y' - x'y'')\bigr ), \end{aligned}$$
where \(x'y''-x''y'=0\) means the curvature is equal to 0, and it corresponds to inflection points. Consequently,
$$\begin{aligned}&(x'''y' - x'y''')(x'^2 + y'^2)\nonumber \\&\qquad -3(x'x'' + y'y'')(x''y' - x'y'')= 0 \end{aligned}$$
corresponds to curvature extrema. The above equation is regarded as (quadratic \(\times \) quartic) − (cubic \(\times \) cubic), but the coefficient of sextic terms vanishes and it becomes quintic. Therefore, a cubic polynomial curve can have at most 5 curvature extrema. Please refer to [25] for more details. Since Eq. (33) is quintic, it is generally not possible to have analytical solutions, so we must use numerical approach to obtain the points on the curve where curvature extrema occurs.
Here, we give a high-level summary of the proof [16] that the curvature of a cubic curve of the form shown in Eq. (1) has at most one local extremum in the (0, 1) parameter interval. As a consequence, the degree-9 polynomial in Sect. 3.1, computed by the Maxima [14] program in Fig. 13, has at most one real solution in the (0, 1) interval. When it has none, this means that no local extremum is present in the curvature, so the extremum occurs at
$$\begin{aligned} t_0=\mathop {{{\,\mathrm{arg\,max}\,}}}\limits _{t=0,1}|\kappa (t)|. \end{aligned}$$
Maxima [14] code to calculate the degree-9 polynomial in Sect. 3.1
Without loss of generality, place the control points as follows:
$$\begin{aligned} Q_0&=(-1,0),&Q_1&=(b,h),&Q_2&=(1,0), \end{aligned}$$
where \(b\ge 0\) and \(h>0\). (The special cases where \(Q_0=Q_2\) or \(h=0\) are also handled in [16].)
Let N(t, a) denote the left side of Eq. (33) applied to this curve. When \(b\le 3-2/a\), it is easy to see that \(\partial N(t,a)/\partial t<0\) for any \(t\in (0,1)\) and \(a\in [2/3,1]\), so N is decreasing. Since N(0, a) is always positive, this means N has at most one 0-crossing.
In the following, let us also assume that \(b>3-2/a\). Then, the following statements can also be proven:
$$\begin{aligned} N(1,a)&<0&\text { when }\partial N(1,a)/\partial t>0, \end{aligned}$$
$$\begin{aligned} \partial N(0,a)/\partial t&<0,\end{aligned}$$
$$\begin{aligned} \partial N(t,a)/\partial t&<0&\text { when }\partial N(1,a)/\partial t<0,\end{aligned}$$
$$\begin{aligned} \partial ^2N(0,a)/\partial t^2&>0,\end{aligned}$$
$$\begin{aligned} \partial ^3N(t,a)/\partial t^3&<0. \end{aligned}$$
From the above,, it is easy to prove that \(N(t,a)=0\) has exactly one solution—and thus, the curvature has at most one local extremum—in (0, 1).
Generalized trigonometric basis functions
Recursive evaluation
For our new trigonometric basis, we can derive a recursive algorithm similar to de Casteljau's algorithm . For simplicity we explain only the quadratic case, but it can be extended to a general degree n by induction. To shorten expressions, we use \(u=1-S(t)\), \(v=S(t)+C(t)-1\) and \(w=1-C(t)\), where \(S(t) = \sin \frac{\pi t}{2}\) and \(C(t) = \cos \frac{\pi t}{2}\). Note that \(v^2 = 2 u w\), and
$$\begin{aligned} \begin{aligned}&(u+v+w)^2 = \\&\qquad u(u+v+w) + v(u+v+w)+w(u+v+w). \end{aligned} \end{aligned}$$
For a quadratic curve with this basis, five control points \(P_i\) (\(i=0\dots 4\)) are used, and the curve point at t is evaluated as
$$\begin{aligned} \begin{bmatrix} u&v&w \end{bmatrix} \begin{bmatrix} P_0 &{} \quad P_1 &{} \quad P_2 \\ P_1 &{} \quad P_2 &{} \quad P_3 \\ P_2 &{} \quad P_3 &{} \quad P_4 \end{bmatrix} \begin{bmatrix} u &{} \quad &{} \\ &{} v \quad &{} \\ &{} &{} \quad w \end{bmatrix}. \end{aligned}$$
Hence, the algorithm repeats a simple blending of three points \(u P_{i-1} +v P_{i} +w P_{i+1}\) to generate a point on the given curve.
Triangle method
We can also construct a triangle using the coefficients of trigonometric basis functions, similarly to Pascal's triangle. Below is a table of degree elevation, from the first row representing degree 1 to the sixth row representing degree 6:
$$\begin{aligned} \begin{array}{ccccccccccccc} &{} \quad &{} \quad &{} \quad &{} \quad &{} \quad 1 &{} \quad 1 &{} \quad 1&{} \quad &{} \quad \quad &{} \quad &{} \quad &{}\\ &{} \quad \quad &{} \quad &{} \quad &{} \quad 1 &{} \quad 2 &{} \quad 4 &{} \quad 2&{} \quad 1&{} \quad &{} \quad &{} \quad &{} \\ &{} \quad &{} \quad &{} \quad 1 &{} \quad 3&{} \quad 9 &{} \quad 8 &{} \quad 9 &{} \quad 3&{} \quad 1 &{} \quad &{} \quad &{}\\ &{} \quad &{} \quad 1&{} \quad 4 &{} \quad 16 &{} \quad 20 &{} \quad 34 &{} \quad 20 &{} \quad 16 &{} \quad 4 &{} \quad 1 &{} \quad &{} \\ &{} \quad 1&{} \quad 5&{} \quad 25 &{} \quad 40 &{} \quad 90 &{} \quad 74 &{} \quad 90 &{} \quad 40 &{} \quad 25 &{} \quad 5 &{} \quad \quad 1 &{} \\ 1&{} \quad 6&{} \quad 36 &{} \quad 70 &{} \quad 195 &{} \quad 204 &{} \quad \quad 328 &{} \quad 204 &{} \quad 195 &{} \quad 70 &{} \quad 36 &{} \quad 6 &{} \quad 1 \end{array} \end{aligned}$$
Various basis functions
Here, we check the applicability of the bases listed in Sect. 2, except for the trigonometric cubic Bernstein-like basis functions [24], since that was already discussed in the paper.
The basis of the C-Bézier curve is \(\{ \sin t, \cos t, t, 1 \}\), and the curve is defined by the following formula:
$$\begin{aligned} B_\alpha (t)&= Z_0(t) q_0 + Z_1(t) q_1 + Z_2(t) q_2 + Z_3(t) q_3 \\&= \frac{1}{\alpha -S} \left[ \begin{array}{c} \sin t \\ \cos t \\ t \\ 1 \end{array} \right] ^\top \left[ \begin{array}{cccc} C &{} \quad 1-C -M &{} \quad M &{} \quad -1 \\ -S &{} \quad (\alpha -K)M &{} \quad -KM &{} 0 \\ -1 &{} \quad M &{} \quad -M &{} \quad 1 \\ \alpha &{} \quad -(\alpha - K)M &{} \quad KM &{} \quad 0 \end{array} \right] \\&\quad \times \left[ \begin{array}{c} q_0 \\ q_1 \\ q_2 \\ q_3 \end{array} \right] . \end{aligned}$$
Here, \(\alpha \) is a built-in shape parameter satisfying \(0 < \alpha \le \pi \), and \(S=\sin \alpha \), \(C = \cos \alpha \). The parameter of the curve is \(t \in [0,\alpha ]\), and
$$\begin{aligned} K&= \frac{\alpha - S}{1-C}, \\ M&= {\left\{ \begin{array}{ll} 1 &{} \quad \text{ if } \alpha = \pi , \\ \frac{S}{\alpha - 2K} = \frac{S(1-C)}{2 S - \alpha -\alpha C} &{} \quad \text{ if } 0< \alpha < \pi . \end{array}\right. } \end{aligned}$$
Blending functions of the degenerated (quadratic) C-Bézier curve
We degenerate the curve by adding its second and third basis functions, i.e., placing the second and third control points at the same position. The curve is then defined by three control points. However, even if we vary \(\alpha \) from 0 to \(\pi \), the blending functions do not vary much, as shown in Fig. 14. Therefore, this type of curve is not suitable for changing the magnitude of local maximum curvature.
The cubic alternative curve is similar to the cubic Bézier curve, and is defined by
$$\begin{aligned}&Z(t) = F_0(t) P_0 + F_1(t) P_1 + F_2(t) P_2 + F_3(t) P_3, \\&\quad 0 \le t \le 1, \end{aligned}$$
where the basis functions \(F_i(t)\), \(t=0\dots 3\) are
$$\begin{aligned} F_0(t)&= (1-t)^2 (1+(2-\alpha )t), \\ F_1(t)&= \alpha (1-t)^2 t, \\ F_2(t)&= \beta t^2 (1-t), \\ F_3(t)&= t^2 (1+(2-\beta )(1-t)). \end{aligned}$$
When \(\alpha =\beta =2\), the curve becomes the cubic Ball curve; for \(\alpha =\beta =3\), the classical cubic Bézier curve; and for \(\alpha =\beta =4\), the cubic Timmer curve. The basis functions are nonnegative when \(0 \le \alpha \le 3\).
We assume that \(\beta =\alpha \), and the curve is degenerated by adding the second and third blending functions. Hence,
$$\begin{aligned} A_0(t;\alpha )&= (1-t)^2(1+(2-\alpha )t), \\ A_1(t;\alpha )&= \alpha (1-t)t \\&= 1- (1-t)^2(1+(2-\alpha )t) -t^2(1+(2-\alpha )(1-t)), \\ A_2(t;\alpha )&= t^2(1+(2-\alpha )(1-t)). \end{aligned}$$
We define the curve \(c(t;\alpha )\) by
$$\begin{aligned} c(t;\alpha )&= A_0(t;\alpha ) P_0 + A_1(t;\alpha ) P_1 + A_2(t;\alpha ) P_2 \\&= A_0(t;\alpha ) (P_0-P_1) + P_1 + A_2(t;\alpha ) (P_2-P_1). \end{aligned}$$
$$\begin{aligned} \frac{d c(t;\alpha )}{d t}&= (1-t)(-3\alpha t+6t+\alpha )(P_1-P_0) \\&\quad + t(3\alpha t - 6 t-2\alpha +6) (P_2-P_1), \\ \frac{d^2 c(t;\alpha )}{d t^2}&= 2 \big [ (3 \alpha t-6t-2 \alpha +3) (P_1-P_0) \\&\quad + (3 \alpha t-6t-\alpha +3) (P_2-P_1) \big ]. \end{aligned}$$
Hence, the curvature \(\kappa _i(t;\alpha )\) for each curve segment is given by
$$\begin{aligned}&\kappa _i(t;\alpha ) =\\&\quad \ \ \frac{2 \alpha _i(3(\alpha -2)(1-t)t-\alpha _i+3) (P_1-P_0)\times (P_2-P_1) }{\Vert (1-t)(-3\alpha _it+6t+\alpha _i)(P_1-P_0)+t(3 \alpha _i t-6 t-2 \alpha _i+6)(P_2-P_1) \Vert ^3}. \end{aligned}$$
When \(\alpha =3\), this degenerates to the classical cubic Bézier curve with collocated second and third control points, and at the endpoints we get \(\kappa _i(0,3)=\kappa _i(1,3)=0\), since the directions of the first and second derivatives are the same. Hence, the proposed method is not applicable in this case.
When \(\alpha \ne 0\),
$$\begin{aligned} \kappa _i(1;\alpha _i)&= \frac{4(3-\alpha _i)\triangle ^+_i}{ \alpha _i^2 \lambda _i^2 \Vert c_{i+1,1}-c_{i,1} \Vert ^3}, \\ \kappa _{i+1}(0;\alpha _i)&= \frac{4(3-\alpha _{i+1})\triangle ^-_{i+1}}{ \alpha _{i+1}^2 (1-\lambda _i)^2 \Vert c_{i+1,1}-c_{i,1} \Vert ^3}. \end{aligned}$$
To guarantee \(G^2\) continuity at the joint of the segments, when \(0< \alpha _i, \alpha _{i+1} < 3\), we can compute \(\lambda _i\) (\(0< \lambda _i < 1\) ) by
$$\begin{aligned} \lambda _i = \frac{\sqrt{ (3-\alpha _i) \triangle ^+_i}}{\sqrt{(3-\alpha _i)\triangle ^+_i} + \frac{\alpha _i}{\alpha _{i+1}}\sqrt{ (3-\alpha _{i+1}) \triangle ^-_{i+1}}}. \end{aligned}$$
When \(\alpha _i, \alpha _{i+1}>3\),
$$\begin{aligned} \lambda _i = \frac{\sqrt{ (\alpha _i-3) \triangle ^+_i}}{\sqrt{(\alpha _i-3)\triangle ^+_i} + \frac{\alpha _i}{\alpha _{i+1}}\sqrt{ (\alpha _{i+1}-3) \triangle ^-_{i+1}}}. \end{aligned}$$
In other cases, such as \(\alpha _i>3\), \(0< \alpha _{i+1} < 3\) or \(0< \alpha _i < 3\), \(\alpha _{i+1} > 3\), \(\lambda _i\) can be determined by careful handling of the signs of \(3-\alpha _i\) and \(3-\alpha _{i+1}\).
By assuming a local maximum curvature at \(t_i\), we can express the input point \(p_i\) as
$$\begin{aligned} p_i&= (1-t_i)^2(1+(2-\alpha _i)t_i) \left[ (1-\lambda _{i-1}) c_{i-1,1} + \lambda _{i-1} c_{i,1} \right] \\&\quad + \alpha _i (1-t_i)t_i c_{i,1} \\&\quad + t_i^2(1+(2-\alpha _i)(1-t_i)) \left[ (1-\lambda _{i}) c_{i,1} + \lambda _i c_{i+1,1} \right] . \end{aligned}$$
Cubic trigonometric Bézier curve [8]
For \(\lambda \), \(\mu \in [-2,1]\), \(t \in [0, 1]\), the cubic trigonometric Bézier denoted as T-Bézier basis functions are defined by
$$\begin{aligned} b_0(t)&= (1 - S)^2 (1 - \lambda S), \\ b_1(t)&= S (1 - S)( 2+ \lambda - \lambda S ), \\ b_2(t)&= C (1 - C )( 2+ \mu - \mu C ), \\ b_3(t)&= (1 - C)^2 (1 - \mu C), \end{aligned}$$
where \(S=\sin \frac{\pi t}{2}\) and \(C=\cos \frac{\pi t}{2}\).
We assume \(\lambda =\mu =\alpha \), and the curve is degenerated by adding the second and third blending functions. Hence,
$$\begin{aligned} A_0(t;\alpha )&= (1-S)^2(1-\alpha S ), \\ A_1(t;\alpha )&= 1-(1-S)^2(1-\alpha S )- (1-C)^2(1-\alpha C), \\ A_2(t;\alpha )&= (1-C )^2(1-\alpha C). \end{aligned}$$
$$\begin{aligned} c(t;\alpha )&= (1-S)^2(1-\alpha S ) P_0 \\&\quad +(1-(1-S)^2(1-\alpha S)- (1-C )^2(1-\alpha C) )P_1 \\&\quad + (1-C)^2(1-\alpha C) P_2 \\&= (1-S)^2(1-\alpha S ) (P_0-P_1) + P_1 \\&\quad + (1-C)^2(1-\alpha C) (P_2-P_1). \end{aligned}$$
$$\begin{aligned} \frac{d c(t;\alpha )}{d t}&= \frac{\pi }{2} \big [ (1-S)C (2+\alpha -3\alpha S)(P_1-P_0) \\&\quad +(1-C) S (2+\alpha -3\alpha C) (P_2-P_1) \big ], \\ \frac{d^2 c(t;\alpha )}{d t^2}&=\frac{\pi ^2}{4} \big [ (S-1)( (2S+1)( 2+\alpha - 3 \alpha S) + 3 \alpha C^2) (P_0-P_1) \\&\quad + (1- C)( (2C+1)( 2+\alpha - 3 \alpha C) + 3 \alpha S^2) (P_2-P_1) \big ]. \end{aligned}$$
Hence, \(d c/dt \times d^2 c/dt^2\) is given by
$$\begin{aligned}&\frac{d c}{dt} \times \frac{d^2 c}{dt^2}= (1-S)(1-C) \\&\quad \big [ C (2+\alpha -3 \alpha S) ((2C +1)(2+\alpha -3\alpha C)+3 \alpha S^2) \\&\quad + S(2+\alpha -3\alpha C) ((2 S +1)(2+\alpha -3\alpha S+3 \alpha C^2 ) \big ] \\&\quad (P_1-P_0)\times (P_1-P_0). \end{aligned}$$
When \(\alpha =0\), this curve is the same as the trigonometric cubic Bernstein-like curve with \(\alpha =2\), and the curvatures at the start and end points \(\kappa (0)\) and \(\kappa (1)\) are generally not equal to zero. For other \(\alpha \) values, \(\kappa (0)=\kappa (1)=0\), so our method is not applicable. This is because the directions of the first and second derivatives are the same.
\(\alpha \beta \)-Bernstein-like basis functions [32] and quasi-cubic trigonometric Bernstein basis curves [30]
For arbitrary \(\alpha \), \(\beta \in [2,+\infty ]\), and \(t \in [0,1]\), the \(\alpha \beta \)-Bernstein-like basis functions are defined by
$$\begin{aligned} A_0(t;\alpha )&= (1-t)^\alpha , \\ A_1(t;\alpha )&= 1-3t^2 + 2 t^3 -(1-t)^\alpha , \\ A_2(t;\beta )&= 3 t^2 - 2t^3 - t^\beta , \\ A_3(t;\beta )&= t^\beta . \end{aligned}$$
When \(\alpha =\beta =2\), these degenerate to cubic Said-Ball basis functions [21]. When \(\alpha =\beta =3\), these become cubic Bernstein basis functions.
Let \(\beta =\alpha \) and we degenerate the curve by adding the second and third blending functions. Hence,
$$\begin{aligned} A_0(t;\alpha )&= (1-t)^\alpha , \\ A_1(t;\alpha )&= 1-(1-t)^\alpha -t^\alpha , \\ A_2(t;\alpha )&= t^\alpha . \end{aligned}$$
We define the curve \(c(t)\) by
$$\begin{aligned} c(t;\alpha )&= (1-t)^\alpha P_0 + (1-(1-t)^\alpha -t^\alpha ) P_1 + t^\alpha P_2 \\&= (1-t)^\alpha (P_0-P_1) + P_1 + t^\alpha (P_2-P_1). \end{aligned}$$
$$\begin{aligned} \frac{d c(t;\alpha )}{d t}&= \alpha \left[ (1-t)^{\alpha -1}(P_1-P_0) + t^{\alpha -1} (P_2-P_1) \right] , \\ \frac{d^2 c(t;\alpha )}{d t^2}&= \alpha (\alpha -1) \left[ (1-t)^{\alpha -2}(P_0-P_1) + t^{\alpha -2} (P_2-P_1) \right] . \end{aligned}$$
Hence, its curvature \(\kappa (t;\alpha )\) is given by
$$\begin{aligned} \kappa (t;\alpha ) = \frac{\alpha -1}{\alpha }\cdot \frac{(1-t)^{\alpha -2} t^{\alpha -2} (P_1-P_0)\times (P_2-P_1) }{\Vert (1-t)^{\alpha -1} (P_1-P_0)+t^{\alpha -1} (P_2-P_1) \Vert ^3}. \end{aligned}$$
When \(\alpha =2\) (cubic Said-Ball curve), the curvatures at the start and end points \(\kappa (0;\alpha )\) and \(\kappa (1;\alpha )\) are generally not zero. However, when \(\alpha > 2\), we get \(\kappa (0;\alpha )=\kappa (1;\alpha )=0\). Therefore, in this case our method is not applicable.
In the case of quasi-cubic trigonometric Bernstein basis curves, the directions of the first and second derivatives at the ends are the same; thus, our method is not applicable.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
Miura, K.T., Gobithaasan, R.U., Salvi, P. et al. \(\epsilon \kappa \)-Curves: controlled local curvature extrema. Vis Comput 38, 2723–2738 (2022). https://doi.org/10.1007/s00371-021-02149-8
Issue Date: August 2022
Interpolatory curves
Curvature continuity
Control of curvature magnitude
Degree elevation

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Home / Movie / News: John Wick 2 has received critical acclaim
News: John Wick 2 has received critical acclaim
07.02.2017	Movie 131 Views
Thriller "John Wick 2" has earned high praise from critics. At the time of this writing, his approval rating on the website-aggregator of professional reviews from Rotten Tomatoes, his approval rating was 91 per cent, i.e. the majority of the reviews were positive.
"The biggest compliment I can make "John Wick 2″, is that I've lost count of the corpses within the first 15 minutes," wrote the author of the book Entertainment Weekly Chris Nashawaty.
See also: The first poster of Thriller GLASS
"Sit back, enjoy the stunts, speed and style" – echoes reviewer for Time Out's Tom Huddleston.
"This is one of the most stylish, elegant, well choreographed and creative performances in the genre of action," – said the author of the Forbes Scott Mendelson.
"John Wick 2″ is as close to a true cinema as possible for projects of this genre", adds To Kai from Den of Geek.
See also: In the Network laugh at the angry pedestrian who had received by merit. Video
The film starred Keanu Reeves, Thomas of Sadoski, ruby rose, Bridget Moynahan, Peter Stormare, Ian McShane, John Leguizamo, Laurence Fishbourne, common, David Patrick Kelly, and other actors. Recall that the first series, released in 2014, earned rentals of $ 130 million when the production budget of only $ 20 million.
Source: RottenTomatoes
Tags acclaim critical received
The composer of "Astral" will release an album with tracks that are not included in the film
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CTB film companyinvolved in production of the Franco-British mystical Thriller Sonata (The Sonata), introduced Russian …


There's a full lineup of author/speakers, panel discussions, tutorials and more at Storyfort 2019.
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Google and NORAD have a couple of swell sites for Santa-spotting.
A staged reading of the children's play is free to attend.
On Thursday, Dec. 6, Brewer will read a draft of "Relic" at Surel's Place backed by projections of her drawings, but before that, she will teach a poetry comics workshop on Saturday, Dec. 1.
The choir will the kids will perform at Grace Lutheran Church in Caldwell on Wednesday, Nov. 21, at 7 p.m.
Today's the last day to cast an in-person early vote, the nation's jobless report holds at 3.7 percent, you'll get an extra hour of sleep this weekend, it looks like there will be a sequel to Gladiator and Finland celebrates National Jealousy Day by disclosing every citizens' personal income.
Gov. C.L. "Butch" Otter endorses Medicaid expansion, a geothermal line breaks in downtown Boise, a nonprofit says it will bring pizza to long lines at U.S. polling places, severe flooding puts more of Venice underwater, and horrible TV ratings for the World Series.
In a few seconds, this thimble-sized device screws onto any household faucet.
Kinner has been charged with numerous felonies, including first-degree murder for the killing of 3-year-old Ruya Kadir.
"America's weird. We're the richest country with the worst poverty."
"We will always focus on nutritious meals, but if students don't participate with us, the preparation of nutritious meals doesn't mean anything."
Over the course of a week, Camp Musical takes a group of 60 students, ages 9-18, and coaches them through performances of three Broadway and contemporary songs.
Now is the perfect time to indulge the little Bob (or Bobbette) the Builder in your life with some fun LEGO merch for the new year.
Starting at 8 p.m., a slate of storytellers and poets with DACA ties, including Boise poet Hannah Q. Rodabaugh, Elizabeth Almanza, Ben Stein and Maria Andrade, will share stories of immigration.
The Hello Kitty Cafe Truck, a project debuted by Hello Kitty Parent company Sanrio in 2014, will make a stop at the Boise Towne Square Mall on Saturday, Aug. 18.
Rediscovered Publishing, an arm of Rediscovered Books, is set to release its first title, A Kid's Guide to Boise, by Rick Just.
Tragedy at Lake Lowell, a new "free-range parenting" law is about to go into effect in Utah, robocalls continue to flood our phones, Giuliani says Trump may plead the Fifth, and the Avengers still rule the box office.


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Provost – University of Delaware
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Differential charge
Differential Charge for Engineering, Nursing and Business & Economics Students
The University of Delaware, as part of its commitment to provide a high-quality education and attract the best students and faculty, is implementing a differential charge beginning with the 2018-19 academic year in three areas—the College of Engineering, the Alfred Lerner College of Business & Economics and the School of Nursing. This new financial structure will more closely align the student charges with the higher cost of delivery for these programs while supporting UD's continued strategic growth, academic excellence and affordability.
Q: Why is UD assessing a differential charge?
A: The differential charge will more closely align the student rate with the higher cost of delivery for certain programs while supporting UD's continued strategic growth, academic excellence and affordability. In this way, the University can remain competitive with its peer institutions, continuing to provide students with the very best academic experience, tools and life preparation possible. The funds generated from the differential charge will be used to increase financial aid, hire new faculty, broaden course access, expand specialized career services and continue providing the transformational learning environments needed to keep pace with advancing technologies.
Many higher education institutions have already implemented similar charges. In fact, 60 percent of public research universities today assess tuition based on students' year of study, major or both, according to research by Gregory C. Wolniak, director of the Center for Research on Higher Education Outcomes at New York University.
Revenues from these charges will stay entirely within the respective college or program.
Q: Which students will be impacted?
A: Beginning with the 2018-19 academic year, students in the Alfred Lerner College of Business & Economics, the College of Engineering and the School of Nursing will have a differential charge added to their bill each semester.
Q: Why are only the Lerner College, College of Engineering and the School of Nursing affected by this change?
A: While all colleges and University programs remain competitive at UD, these three highly ranked areas have been assessed for specific criteria to warrant greater necessary funding and investment to sustain and build competitive advantage for graduates in these industries. Such criteria include high cost of instruction, high demand by students, high national standing, high placement rate and salary upon graduation, and finally, high economic impact on the state and beyond.
Q: What is the amount of the differential charge?
A: To enable students impacted by the differential to plan in advance, the total charges will be phased in incrementally. For the 2018-19 academic year, the differential charge will be $500 per semester for full-time students registered for 12 or more credits. For part-time students, the charge will be $250/semester. The table below provides the planned maximum differential charge by semester for the next three academic years. However, to assist in funding this new charge, students enrolled at UD as of the spring 2018 semester will receive a credit until it is fully phased in (spring 2021). Current full-time students will receive a credit of $250 per semester, while current part-time students will receive a credit of $125/semester during this time period.
Projected Differential Phased Increases
Fall '18/Spring '19* Fall '19/Spring '20* Fall '20/Spring '21*
Nursing $500/Semester $1,000/Year $750/Semester $1,500/Year $750/Semester $1,500/Year
Lerner College of Business & Economics $500/Semester $1,000/Year $750/Semester $1,500/Year $1,000/Semester $2,000/Year
College of Engineering $500/Semester $1,000/Year $1,250/Semester $2,500/Year $2,000/Semester $4,000/Year
* For current UD students, those enrolled as of spring, 2018, a credit will apply. See above.
Q: Will any additional financial aid or funding be available to cover the cost of the differential charge?
A: Yes. Support will be provided in the following forms:
We guarantee no net increase for all students with the greatest financial need, our Pell-eligible students, as determined by the Free Application for Federal Student Aid (FAFSA). This includes covering this charge under the Commitment to Delawareans financial aid model.
All students attending UD on full scholarships (Unidel Eugene du Pont Memorial Distinguished Scholars, Veterans Benefits or Athletics) will receive funding to offset these costs.
Q: What will my total cost of attendance be?
A: The University each year thoughtfully considers how to ensure outstanding academics at an affordable price. In fact, UD has kept tuition increases below the rate for public universities nationwide and has been ranked a "best value" by publications like Money, Forbes and Kiplinger's. Annually the Board of Trustees considers the cost of tuition, fees and room and board, and the rates are approved and published in June or July. Annual changes in the tuition, fees, and room and board will be in addition to the projected increases in the differential charge, as the differential supports the distinctive offerings provided by the College of Engineering, Lerner College, and the School of Nursing.
Q: Does this differential charge apply to both in-state and out-of-state students?
A: Yes. Students are charged the same differential charge regardless of whether they are Delaware residents or non-residents. As with any other program, however, financial aid support will be made available as applicable.
Q: Will there be a differential charge for Winter Session and Summer Session?
A: No. The differential charge will be billed for the regular fall and spring semesters only.
Q: Will the tuition benefits/funding, such as tuition remission and tuition exchange, cover the differential charge?
A: Students who are eligible for tuition exchange will be responsible for the differential charge as these funding sources only cover tuition. University employees who are eligible for tuition remission or fee waiver will not be responsible for the differential charge.
Q: Who can I contact if I have further questions?
A: The best way to contact us with questions is by emailing the dean's offices for each college as follows.
College of Engineering: Email engr-contacts@udel.edu
Lerner College of Business and Economics: Email lerner@udel.edu
College of Health Sciences/School of Nursing: Email CHS-info@udel.edu
For additional information on differential charge billing, please visit the Student Financial Services website. This UDaily story has additional information on the University's plan to advance student success.
Contact the Provost
Priority Initiatives
Diversity at UD


Eighteen months ago, I was so proud of myself. I had lost 90 pounds and looked and felt fabulous. I was never going back. Except I did. I went about 25 pounds the other way. Twenty of them in the six months since we moved. And it feels miserable. Sure, there are reasons for it. Stress. Stress eating. Minimal exercise. Bronchitis. Bad weather for running. But, eventually, the reasons become excuses. And, once again, I'm done with excuses.
Two weeks ago, I ran my second marathon. It was a miserable experience. I ran the first 14 miles and felt pretty good. But then it got hot. And I started to walk – which tightened up all my muscles, making it difficult to do anything other than walk. Even the walking was excruciating. But, I finished. It took me 29 minutes longer to finish than my last one did – perhaps because it was 77 degrees. Perhaps because I weighed 20 pounds more than I did last May.
So, even though I am proud of myself for finishing the marathon, I am incredibly disappointed in myself. I made choices that led to a bad performance. I didn't train well – no running for six weeks because of bronchitis, skipping long runs because of bad weather and the holidays. Eating when I knew I shouldn't have.
But I'm done with it. I'm done with my favorite clothes not fitting and feeling embarrassed by myself again. I'm finished with eating food that only makes me feel terrible about myself. And, perhaps most importantly, I am over not exercising properly and giving my body the opportunity to excel the way that I know it can. So, I'm holding myself accountable. At least thirty minutes a day for myself. Yoga, running, walking. Pushing myself toward the me I know I want to be and can be. Are you?
This post will perhaps seem to be a complete contradiction to the one I just wrote… We arrived at school on Tuesday morning for an emergency faculty meeting. The time-stamp on the email was 12:48 a.m. It had to be bad news. And it was. Gayle Gay, a much beloved educator at Crawford, had passed away the night before. For five minutes, I'm not sure where I was – unable to fully process the news, I watched others around me. Heads buried in hands, tears streaming down faces, confusion on the faces of the teachers walking in late.
Listening to the students as they received the news was even more difficult. Kids who seemed so tough, like nothing could break them, doubled over at their desks, their bodies racked with sobs. The others eerily silent – still processing, perhaps shocked by the reactions of their classmates. This was even more difficult than hearing the news for the first time. It was the second time some of them had confronted the death of someone close to them in four months. And for a little while, we were all just people hurting together.
But I wish I'd known more. I wish I'd stopped to talk more, to ask personal questions, to get to know Gayle outside the hallway. She was one of my favorite people at work; she always put a smile on my face. And I never stopped to talk for long. Always somewhere to go, something to copy, someone to find. So perhaps my lesson is just to slow down. Take an extra 30 seconds. Or five minutes. In the end, won't it have been worth it?
I know… It's part of the wedding vows, not the parenting vows. But I think it applies. After all, aren't we supposed to love our children in sickness and in health? But surely there's an exception to this, where maybe we don't have to like them very much? Particularly if one parent is out of town and the remaining parent is also sick with the same virus and the child is just as sick and tired of you after three days together? Because if that's the case, I'm right on track.
As soon as I saw that Justin would be out of town for two full weeks in a row (home on the weekend between, thank God!), I knew that we were destined for illness. It seems obvious, doesn't it? So, it came as no surprise to get a text from Justin Tuesday afternoon that Robbie had a fever of 102.5 degrees. I headed out of school, knowing that I would be gone for Tuesday and Wednesday. I never imagined Thursday, too. Nor did I imagine that I would have the same virus, knocking me out for a day and a half as well.
Robbie and I only left the bedroom to get food yesterday. And, since neither of us was eating, that was rare. It was constant together time. We went to sleep last night planning on going to school today. We were both ready, each of us over our respective virus. Then Robbie's fever shot up to 101 this morning and I was still horribly nauseous. We headed back to the doctor, since Robbie complained about his chest and throat hurting. Another nose swab and throat culture gave us no definitive answers; we just had different versions of the same virus.
Home we came, prepared for an afternoon of movies and cuddling. At least that's what I was prepared for. Robbie was prepared for disaster. We made it through The Cat in the Hat, which Robbie watched in his chair while I napped fitfully on the couch. Then, determined to be somewhat productive, I went to the bedroom to put away the laundry my sister had generously folded yesterday when she came to bathe Robbie and got sucked into our laundry pile. It was very quiet, so I asked Robbie what he was doing. His response left me struggling for words.
And he had. Into four pieces. He had also spilled the entire box of oyster crackers onto the floor and done some sort of dance on them. It was all I could do to maintain my composure. OK. I didn't maintain my composure. I yelled. I cried. Robbie cried. But my glasses were still broken.
The night continued to improve… Robbie dropped an entire pizza face-down on the floor – I have no idea how that happened. And, as we finally cuddled up in bed (we ARE going to school tomorrow), the dog jumped up with us. And puked all over my newly cleaned duvet cover. And did I mention that Justin is out of town until tomorrow?
Stunned, there wasn't much more I could say. Even Justin, who was following my comment with, "Well, I helped make you, Rob. You're half mine," was speechless. Because Robbie's right. God did make him. And I'm not sure how or why, but I sure am glad that when God made Robbie, he entrusted us to be his parents.
There were no words. I squeezed my child a little tighter and kissed the top of his head as tears welled up in my eyes. There is nothing quite like the blindless devotion of a child to God to bring everything into perspective. I know he hears about God in pre-school and Sunday school. And I know he goes to chapel once a week and insists on being the one to say prayers at every meal. But to see him pray when he doesn't know anyone is watching or listening? It is a beautiful reassurance to me that there is a God.
One of my favorite episodes of The Cosby Show was when Rudy's goldfish dies. The entire family dresses up for the funeral, and, as they are all gathered solemnly around the toilet, Rudy asks if she can go watch TV instead of attending the funeral she insisted they have for the fish.
Today we had our own version of this episode. The events started last night when I did a double-take as I passed the fish tank if Robbie's room. It took a few minutes for me to realize what had happened. But, after my eyes recovered from the bright orange of the water, I realized what had happened. Robbie had fed his fish. The entire canister. Every last flake. The tank looked like it had been hit by an epic snow storm. Even the firetruck was up to its windshield in flakes.
Concerned about the welfare of the fish, I took the tank downstairs with me after Robbie went to bed. The fish went into a glass on the counter with as little of the sludge as I could manage. I painstakingly sorted through the gravel and decomposing fish flakes. Unfortunately, we didn't have any water treatment, so the fish had to spend the night in the glass.
As soon as we got home this afternoon, Robbie raced to the kitchen to check on his fish. He was no more. The sludge had claimed him. So, we opted for a burial at sea. Robbie and I headed to the bathroom, glass and fish in hand. Robbie helped me pour the glass into the toilet. He said a few words about what a good fish he had been and then looked to me. I asked Robbie if he wanted to say a quick prayer, and he nodded his head and folded his hands.
"God is great; God is…" Then it occurred to him that we weren't eating. He looked up a little confused and unsure of what to do. We managed to finish the prayer together, and then it was time to flush.
I once read an article in a magazine that claimed women were more likely to ask their children to do chores around the house than their husbands. And it's probably true. After all, what could our husbands possibly know about keeping a house clean and running? Probably best to let them do their "husband" thing while we become more and more frustrated at their lack of interest as we load the dishwasher, start another load of laundry, step on toys left all over the living room floor, and get dinner on the table. Right?
Turns out, no. Not at all. But, in most cases, it's really not the fault of our husbands. We don't want to turn the control over – or, at least, I don't. I like to know that the dishes are all in their right places and the laundry is sorted into three laundry baskets to make putting it away easier (or, to be totally honest, grabbing clothes in the morning because we didn't put them away). I don't know why I do; I just do.
Except I don't anymore. Slowly, over the past eighteen months, I've lost control of our house. Although he travels for work quite a bit, Justin actually runs most of our house now. It's a bizarre sensation, asking Justin where certain dishes are or finding the bed made almost every day – and not by me minutes before climbing into it again. It really hit home last Sunday when, as I finished breakfast, Justin undecorated our tree and packed up the ornaments. Then, he put all the rest of the Christmas decorations away, taking care to pack them all in the ginormous "Christmas box" he's had since he was little.
I'll admit; this was hard for me. I have certain boxes for my ornaments, certain boxes for Robbie's. But this year, they were all jumbled in together with Justin's childhood ornaments. In random boxes. And our Christmas decorations weren't put into the plastic bins I painstakingly labeled "CHRISTMAS" with green Sharpie years ago. It took every ounce of willpower in me to not correct the job Justin was doing. And you know what? It felt pretty good, especially because within an hour, all the decorations were put away and in the basement. And I didn't have to do any of it. I've also figured this will make decorating even more fun next year because I have no idea what I'll be unpacking.
A beautiful thing happened when I didn't correct the job Justin was doing; he continued doing other things around the house. And let's be clear – there was nothing to correct, which is the key here. Just because Justin's way isn't mine doesn't make it wrong. Hard to believe, I know. Yesterday, for instance, I got a text while I was at school. It simply said, "What can I do to help you today?" Funny thing: I did need help. The rabbit in my classroom desperately needed a new water bottle. And so Justin stopped (at two stores) to pick one up for me.
Today, he stopped to buy fabric we (read, I) were supposed to pick up over a month ago for a chair I want reupholstered. And while he was in the neighborhood? He stopped to buy new shoes for Robbie. Tonight, while I was putting Robbie to bed, Justin cleaned the kitchen. All of it. Later, when he could have been watching TV or puttering around on his iPad, Justin clipped coupons while I wrote a grocery list. Did I mention that I don't go to the grocery anymore?
Don't get me wrong; I still do a lot around the house (especially when Justin is traveling). However, as Justin would be more than happy to tell you, when he is in town, he does a lot more around the house than I do. So, give the husband a little more positive reinforcement when he makes the move to help out. For goodness sake, don't correct him. In the scheme of things, he's not doing anything wrong; it's just different. And, when you don't have to do it and can spend a little more time having fun together, different is a wonderful thing.
Remember those long, luxurious hours spend soaking in a hot bubble bath? Good book in hand? Maybe a little music playing? Yeah. Me, either. After a long run tonight, there was nothing my body wanted more than to soak in a near-scalding bathtub, letting my mind unwind after a long first day back at school. And then the bathroom door came flying open.
It was all I could do to not slide under the water. I knew it wasn't Barkley; the intruder had used the doorknob to enter. It could only be one person, since Justin was watching the Sugar Bowl. It was he-who-should-have-been-sleeping-but-wasn't. Robert Gaetano Manna, himself. Fortunately, between the incredible bath bomb from my sister and the powerful tub jets, I was completely covered with bubbles.
"I need to take a bath, too, Mom. I'm dirty," Robbie declared as he started to take off his shirt.
"Wait!' I screamed – we've reached the age where this whole situation was about to get extremely awkward. "You're not dirty! You don't need to take a bath! Go back to bed!" And, I'll admit, my voice got a little more high-pitched with each sentence, envisioning all of the therapy sessions that were sure to occur somewhere down the road when Robbie remembered taking a bath with his mother.
But then I remembered what I wrote yesterday – about cherishing every moment – and sent Robbie to go get Justin. Ten minutes later, Robbie and I were both in our bathing suits, enjoying bubbles that covered Robbie when he sat in the tub. We pretended Robbie was a snowman – a melting one, as he watched the bubbles fall from his shoulders down his body and back into the water. Robbie slipped all over the place, pretending to be a merman (after a brief lesson in gender, as he was declaring himself a mermaid initially).
All of a sudden it happened – the moment that makes giving up a moment of tranquility completely worthwhile. Robbie grabbed my face in both of his hands, stared into my eyes, and said, "Mom, we're having fun together, aren't we?" Before I could respond, before I could even fully hold onto the moment, it – and Robbie – had slipped away. I sure am glad I was paying attention.
As so many of you know, Justin and I had a difficult time adjusting to Robbie's sensory processing difficulties. It seemed like we were never going to have a "normal" child. We were never going to make it through a meal at a restaurant or a church service. We would always have to leave playgrounds because Robbie acted aggressively toward other children. Bedtime would involve epic meltdowns for all eternity. And Robbie would spend the rest of his life running around like his pants were on fire. And so, desperate for some sort of reprieve, we started Robbie in occupational therapy.
I was skeptical. How well could this really work? Robbie was going to go once a week. We were supposed to brush him and perform joint compressions every two hours. And this was really going to change our lives? Brushing was going to make Robbie calmer? No way. Not even possible. But we did it anyway. Justin and I saw some changes but then, as soon as we thought we were good, there would be a setback. A page-long letter from his pre-school teacher about three trips to the school office in one day for kicking kids, climbing cabinets, and leaping from cot to cot during nap time. Or the phone call asking us to come pick him up because he had slapped a child. Which is how we headed into Christmas vacation.
But, all of a sudden, something changed. I'm not sure if we were more consistent about brushing and incorporating a sensory diet into our day or if things just started to click with Robbie, but the past twelve days have been wonderful. Sure, we've had our moments. Like maybe when I told Robbie he couldn't get a gun as a prize at Gattitown and he threw a fit, so we left without a prize at all (one of my strongest parenting moments). On our way out, he punched me in the face twice. But, this gave us the opportunity to be firm in punishing him, and I think it's the first time he's realized that his actions have implications.
This break, we survived (barely) a Christmas Eve church service. Robbie went shopping with me and held my hand the entire time – not running around and making the entire experience a nightmare. He wants hugs and kisses and snuggling. He talks about how happy he is and how much he loves us – without prompting. And he seems so much freer to just be himself. Which I think is the most any parent can ask for.
All last year, Robbie waited for snow. It makes sense, right? We lived in Boston. There should have been lots of snow; the year before we had endured over 100 inches of the white stuff. But not last year – there was never enough to even make a good snowball. As soon as Thanksgiving was over this year, Robbie started asking for snow again. In his mind, it made sense. December means Christmas and Christmas means snow.
Finally, after waiting almost two years for a decent snow, Robbie's hopes and dreams came true. It wasn't much – maybe three inches – but it was more than enough to make a certain three-year-old happy. As soon as I walked through the door from the gym, Robbie ran to me, thrilled to tell me that it had snowed. We raced through the house, collecting one Spider-Man snow boot from my closet and another from his, gathering gloves and winter coats. And a carrot.
Then we were off, racing outside to see if the snow was right for snowmen. It was perfect. We rolled huge snowballs, stacking them on top of each other to create a snowman. Robbie proudly handed me the carrot he had remembered to bring for the snowman's nose and then we collected rocks for the eyes, mouth, and buttons. We stood back to examine our handiwork, and Robbie declared that we needed to make more snowmen. He ran back to the house two more times to get carrots and helped me find just the right rocks.