本文整理汇总了Java中java.awt.geom.AffineTransform.TYPE_GENERAL_SCALE属性的典型用法代码示例。如果您正苦于以下问题：Java AffineTransform.TYPE_GENERAL_SCALE属性的具体用法？Java AffineTransform.TYPE_GENERAL_SCALE怎么用？Java AffineTransform.TYPE_GENERAL_SCALE使用的例子？那么恭喜您, 这里精选的属性代码示例或许可以为您提供帮助。您也可以进一步了解该属性所在类java.awt.geom.AffineTransform的用法示例。
在下文中一共展示了AffineTransform.TYPE_GENERAL_SCALE属性的3个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Java代码示例。
示例1: userSpaceLineWidth
private float userSpaceLineWidth(AffineTransform at, float lw) {
double widthScale;
if ((at.getType() & (AffineTransform.TYPE_GENERAL_TRANSFORM |
AffineTransform.TYPE_GENERAL_SCALE)) != 0) {
widthScale = Math.sqrt(at.getDeterminant());
} else {
/* First calculate the "maximum scale" of this transform. */
double A = at.getScaleX(); // m00
double C = at.getShearX(); // m01
double B = at.getShearY(); // m10
double D = at.getScaleY(); // m11
/*
* Given a 2 x 2 affine matrix [ A B ] such that
* [ C D ]
* v' = [x' y'] = [Ax + Cy, Bx + Dy], we want to
* find the maximum magnitude (norm) of the vector v'
* with the constraint (x^2 + y^2 = 1).
* The equation to maximize is
* |v'| = sqrt((Ax+Cy)^2+(Bx+Dy)^2)
* or |v'| = sqrt((AA+BB)x^2 + 2(AC+BD)xy + (CC+DD)y^2).
* Since sqrt is monotonic we can maximize |v'|^2
* instead and plug in the substitution y = sqrt(1 - x^2).
* Trigonometric equalities can then be used to get
* rid of most of the sqrt terms.
*/
double EA = A*A + B*B; // x^2 coefficient
double EB = 2*(A*C + B*D); // xy coefficient
double EC = C*C + D*D; // y^2 coefficient
/*
* There is a lot of calculus omitted here.
*
* Conceptually, in the interests of understanding the
* terms that the calculus produced we can consider
* that EA and EC end up providing the lengths along
* the major axes and the hypot term ends up being an
* adjustment for the additional length along the off-axis
* angle of rotated or sheared ellipses as well as an
* adjustment for the fact that the equation below
* averages the two major axis lengths. (Notice that
* the hypot term contains a part which resolves to the
* difference of these two axis lengths in the absence
* of rotation.)
*
* In the calculus, the ratio of the EB and (EA-EC) terms
* ends up being the tangent of 2*theta where theta is
* the angle that the long axis of the ellipse makes
* with the horizontal axis. Thus, this equation is
* calculating the length of the hypotenuse of a triangle
* along that axis.
*/
double hypot = Math.sqrt(EB*EB + (EA-EC)*(EA-EC));
/* sqrt omitted, compare to squared limits below. */
double widthsquared = ((EA + EC + hypot)/2.0);
widthScale = Math.sqrt(widthsquared);
}
return (float) (lw / widthScale);
}
示例2: userSpaceLineWidth
private final double userSpaceLineWidth(AffineTransform at, double lw) {
double widthScale;
if (at == null) {
widthScale = 1.0d;
} else if ((at.getType() & (AffineTransform.TYPE_GENERAL_TRANSFORM |
AffineTransform.TYPE_GENERAL_SCALE)) != 0) {
widthScale = Math.sqrt(at.getDeterminant());
} else {
// First calculate the "maximum scale" of this transform.
double A = at.getScaleX(); // m00
double C = at.getShearX(); // m01
double B = at.getShearY(); // m10
double D = at.getScaleY(); // m11
/*
* Given a 2 x 2 affine matrix [ A B ] such that
* [ C D ]
* v' = [x' y'] = [Ax + Cy, Bx + Dy], we want to
* find the maximum magnitude (norm) of the vector v'
* with the constraint (x^2 + y^2 = 1).
* The equation to maximize is
* |v'| = sqrt((Ax+Cy)^2+(Bx+Dy)^2)
* or |v'| = sqrt((AA+BB)x^2 + 2(AC+BD)xy + (CC+DD)y^2).
* Since sqrt is monotonic we can maximize |v'|^2
* instead and plug in the substitution y = sqrt(1 - x^2).
* Trigonometric equalities can then be used to get
* rid of most of the sqrt terms.
*/
double EA = A*A + B*B; // x^2 coefficient
double EB = 2.0d * (A*C + B*D); // xy coefficient
double EC = C*C + D*D; // y^2 coefficient
/*
* There is a lot of calculus omitted here.
*
* Conceptually, in the interests of understanding the
* terms that the calculus produced we can consider
* that EA and EC end up providing the lengths along
* the major axes and the hypot term ends up being an
* adjustment for the additional length along the off-axis
* angle of rotated or sheared ellipses as well as an
* adjustment for the fact that the equation below
* averages the two major axis lengths. (Notice that
* the hypot term contains a part which resolves to the
* difference of these two axis lengths in the absence
* of rotation.)
*
* In the calculus, the ratio of the EB and (EA-EC) terms
* ends up being the tangent of 2*theta where theta is
* the angle that the long axis of the ellipse makes
* with the horizontal axis. Thus, this equation is
* calculating the length of the hypotenuse of a triangle
* along that axis.
*/
double hypot = Math.sqrt(EB*EB + (EA-EC)*(EA-EC));
// sqrt omitted, compare to squared limits below.
double widthsquared = ((EA + EC + hypot) / 2.0d);
widthScale = Math.sqrt(widthsquared);
}
return (lw / widthScale);
}
示例3: userSpaceLineWidth
private final float userSpaceLineWidth(AffineTransform at, float lw) {
float widthScale;
if (at == null) {
widthScale = 1.0f;
} else if ((at.getType() & (AffineTransform.TYPE_GENERAL_TRANSFORM |
AffineTransform.TYPE_GENERAL_SCALE)) != 0) {
widthScale = (float)Math.sqrt(at.getDeterminant());
} else {
// First calculate the "maximum scale" of this transform.
double A = at.getScaleX(); // m00
double C = at.getShearX(); // m01
double B = at.getShearY(); // m10
double D = at.getScaleY(); // m11
/*
* Given a 2 x 2 affine matrix [ A B ] such that
* [ C D ]
* v' = [x' y'] = [Ax + Cy, Bx + Dy], we want to
* find the maximum magnitude (norm) of the vector v'
* with the constraint (x^2 + y^2 = 1).
* The equation to maximize is
* |v'| = sqrt((Ax+Cy)^2+(Bx+Dy)^2)
* or |v'| = sqrt((AA+BB)x^2 + 2(AC+BD)xy + (CC+DD)y^2).
* Since sqrt is monotonic we can maximize |v'|^2
* instead and plug in the substitution y = sqrt(1 - x^2).
* Trigonometric equalities can then be used to get
* rid of most of the sqrt terms.
*/
double EA = A*A + B*B; // x^2 coefficient
double EB = 2.0d * (A*C + B*D); // xy coefficient
double EC = C*C + D*D; // y^2 coefficient
/*
* There is a lot of calculus omitted here.
*
* Conceptually, in the interests of understanding the
* terms that the calculus produced we can consider
* that EA and EC end up providing the lengths along
* the major axes and the hypot term ends up being an
* adjustment for the additional length along the off-axis
* angle of rotated or sheared ellipses as well as an
* adjustment for the fact that the equation below
* averages the two major axis lengths. (Notice that
* the hypot term contains a part which resolves to the
* difference of these two axis lengths in the absence
* of rotation.)
*
* In the calculus, the ratio of the EB and (EA-EC) terms
* ends up being the tangent of 2*theta where theta is
* the angle that the long axis of the ellipse makes
* with the horizontal axis. Thus, this equation is
* calculating the length of the hypotenuse of a triangle
* along that axis.
*/
double hypot = Math.sqrt(EB*EB + (EA-EC)*(EA-EC));
// sqrt omitted, compare to squared limits below.
double widthsquared = ((EA + EC + hypot) / 2.0d);
widthScale = (float)Math.sqrt(widthsquared);
}
return (lw / widthScale);
}