我有一个可以旋转的箭头。我想知道是否有可能在曲线上旋转箭头?我做了一些研究,我认为它被称为贝塞尔路径?是否有可能使用此代码在bezier路径上旋转精灵,如果可以,我该如何整合它?
UITouch *touch = [touches anyObject];
//acquire the previous touch location
CGPoint firstLocation = [touch previousLocationInView:[touch view]];
CGPoint location = [touch locationInView:[touch view]];
//preform all the same basic rig on both the current touch and previous touch
CGPoint touchingPoint = [[CCDirector sharedDirector] convertToGL:location];
CGPoint firstTouchingPoint = [[CCDirector sharedDirector] convertToGL:firstLocation];
CGPoint firstVector = ccpSub(firstTouchingPoint, _arrow.position);
CGFloat firstRotateAngle = -ccpToAngle(firstVector);
CGFloat previousTouch = CC_RADIANS_TO_DEGREES(firstRotateAngle);
CGPoint vector = ccpSub(touchingPoint, _arrow.position);
CGFloat rotateAngle = -ccpToAngle(vector);
CGFloat currentTouch = CC_RADIANS_TO_DEGREES(rotateAngle);
//keep adding the difference of the two angles to the dial rotation
arrowRotation += currentTouch - previousTouch;
例如,我有一个球放在地上,它的正上方有一个箭头。当你触摸屏幕并移动箭头时,箭头在半圆轴上移动。
曲线将看起来像这个半圆,箭头将绕轴旋转。
如果我需要更清楚,请让我知道。我真的需要一些帮助。
我几天前就遇到了同样的问题。这个答案中的大多数链接都是坏的,所以我在这里和这里找到了材料,并制作了这个代码。就像魔术一样。希望对大家有所帮助。
小描述:我有一个对象(self),它通过手指围绕另一个对象(self.target)旋转,我有一些动画精灵,比如自我运动指南,它围绕自我旋转。目标由bezier函数实现。算法是相当快的,我有100+指南的永久初始化,它的工作没有CPU过载。
/**
Each bezier curve is an array with 8 floats, x1, y1, x2, y2, x3, y3, x4, y4., where x1,y1 and x4,y4 are the arc's end points and x2,y2 and x3,y3 are the cubic bezier's control points.
@note adapted for xCode by Valentine Konov valentine@konov.su 2013
@return a array of objects that represent bezier curves which approximate the circular arc centered at the origin.
@param startAngle to endAngle (radians) with the specified radius.
*/
-(NSArray*)createArcWithRadius:(float)radius_ withStartAngle:(float)startAngle_ withEndAngle:(float)endAngle_;
{
// OMLog(@"radius:%.2f startAngle:%.2f endAngle:%.2f",radius_,startAngle_,endAngle_);
// normalize startAngle, endAngle to [-2PI, 2PI]
float twoPI = M_PI * 2;
float startAngle = startAngle_;
float endAngle = endAngle_;
// float startAngle = fmodf(startAngle_,twoPI);
// float endAngle = fmodf(endAngle_,twoPI);
// Compute the sequence of arc curves, up to PI/2 at a time. Total arc angle
// is less than 2PI.
NSMutableArray* curves = [NSMutableArray array];
float piOverTwo = M_PI / 2.0;
float sgn = (startAngle < endAngle) ? 1 : -1;
float a1 = startAngle;
for (float totalAngle = fminf(twoPI, fabsf(endAngle - startAngle)); totalAngle > 0.00001f /*FLT_EPSILON*/; nil) {
float a2 = a1 + sgn * min(totalAngle, piOverTwo);
[curves addObject: [self createSmallArc:radius_ a1:a1 a2:a2]];
totalAngle -= fabsf(a2 - a1);
a1 = a2;
}
return curves;
}
/**
Cubic bezier approximation of a circular arc centered at the origin,
This algorithm is based on the approach described in:
A. Riškus, "Approximation of a Cubic Bezier Curve by Circular Arcs and Vice Versa,"
Information Technology and Control, 35(4), 2006 pp. 371-378.
@note adapted for xCode by Valentine Konov valentine@konov.su 2013
@param from (radians) a1 to a2, where a2-a1 < pi/2
@return an array with 8 floats, x1, y1, x2, y2, x3, y3, x4, y4. where x1,y1 and x4,y4 are the arc's end points and x2,y2 and x3,y3 are the cubic bezier's control points.
*/
-(NSArray*)createSmallArc:(float)r a1:(float)a1 a2:(float)a2
{
// Compute all four points for an arc that subtends the same total angle
// but is centered on the X-axis
float a = (a2 - a1) / 2.0; //
float x4 = r * cosf(a);
float y4 = r * sinf(a);
float x1 = x4;
float y1 = -y4;
float k = 0.5522847498;
float f = k * tan(a);
float x2 = x1 + f * y4;
float y2 = y1 + f * x4;
float x3 = x2;
float y3 = -y2;
// Find the arc points actual locations by computing x1,y1 and x4,y4
// and rotating the control points by a + a1
float ar = a + a1;
float cos_ar = cosf(ar);
float sin_ar = sinf(ar);
return [NSArray arrayWithObjects: //
[NSNumber numberWithFloat:(r * cosf(a1))], //startPoint.x
[NSNumber numberWithFloat:(r * sinf(a1))], //startPoint.y
[NSNumber numberWithFloat:(x2 * cos_ar - y2 * sin_ar)], //ctrlPoint1.x
[NSNumber numberWithFloat:(x2 * sin_ar + y2 * cos_ar)], //ctrlPoint1.y
[NSNumber numberWithFloat:(x3 * cos_ar - y3 * sin_ar)], //ctrlPoint2.x
[NSNumber numberWithFloat:(x3 * sin_ar + y3 * cos_ar)], //ctrlPoint2.y
[NSNumber numberWithFloat:(r * cosf(a2))], //endPoint.x
[NSNumber numberWithFloat:(r * sinf(a2))], //endPoint.y
nil];
}
/**
Bezier approximation example
@note adapted for xCode by Valentine Konov valentine@konov.su 2013
@param inSprite_ is sprite, angle_ signed angle radiants
@return CCSequence of [CCSpawns of (CCBezierTo and CCRotateBy)]
*/
-(id)calcBezierCircle:(CCSprite*)inSprite_ withAngle:(float)angle_
{
double speed = 100; //points per second
CGPoint positionOffset = ccpSub(((CCNode*)self.target).position, self.position);
//((CCNode*)self.target).position is circle center
double startAngle = [self calcAngle:inSprite_.position ownerRelated:false];
while (startAngle<0) startAngle += 2*M_PI;
while (startAngle>=2*M_PI) startAngle -= 2*M_PI;
double endAngle = startAngle + angle_;
float radius = [self calcRadius];
NSArray* curves = [self createArcWithRadius:radius withStartAngle:startAngle withEndAngle:endAngle];
NSMutableArray* bezierActions = [NSMutableArray array];
for (NSArray* curve in curves) {
CGPoint startPoint = ccpAdd(ccp([[curve objectAtIndex:0] floatValue], [[curve objectAtIndex:1] floatValue]), positionOffset);
CGPoint controlPoint1 = ccpAdd(ccp([[curve objectAtIndex:2] floatValue], [[curve objectAtIndex:3] floatValue]), positionOffset);
CGPoint controlPoint2 = ccpAdd(ccp([[curve objectAtIndex:4] floatValue], [[curve objectAtIndex:5] floatValue]), positionOffset);
CGPoint endPoint = ccpAdd(ccp([[curve objectAtIndex:6] floatValue], [[curve objectAtIndex:7] floatValue]), positionOffset);
ccBezierConfig bezier;
bezier.controlPoint_1 = controlPoint1;
bezier.controlPoint_2 = controlPoint2;
bezier.endPosition =endPoint;
float bezierAngle = ccpAngleSigned(ccpSub(startPoint, positionOffset), ccpSub(endPoint, positionOffset));
float bezierDuration = radius*fabsf(bezierAngle)/speed;
id bezierTo = [CCBezierTo actionWithDuration:bezierDuration bezier:bezier];
id rotateBy = [CCRotateBy actionWithDuration:bezierDuration angle:CC_RADIANS_TO_DEGREES(-bezierAngle)];
CCAction * bezierToAndRotateBy = [CCSpawn actions:bezierTo, rotateBy, nil];
[bezierActions addObject:bezierToAndRotateBy];
}
if ([bezierActions count]<1) {
return nil;
}
return [CCSequence actionWithArray:bezierActions];
}
/**
Calculates angle
@param position_ current position of sprite on sircle, ownerRelated boolean, wich is startPoint is {1,0} or owner.position
@return angle (radiant)
*/
-(float)calcAngle:(CGPoint)position_ ownerRelated:(bool)ownerRelated {
if (ownerRelated) {
CGPoint v1 = ccpSub(((CCNode*)self.target).position, self.position);
CGPoint v2 = ccpSub(ccpSub(((CCNode*)self.target).position, self.position),position_);
return ccpAngleSigned(v1, v2);
}
else {
CGPoint v1 = ccp([self calcRadius], 0.0f);
CGPoint v2 = ccpSub(position_,ccpSub(((CCNode*)self.target).position, self.position));
return ccpAngleSigned(v1, v2);
}
}
/**
Calculates radius
@return radius
*/
-(float)calcRadius;
{
return sqrt(pow(self.position.x-((CCSprite*)self.target).position.x, 2)+pow(self.position.y-((CCSprite*)self.target).position.y, 2));
}
从未尝试过,但我认为这是可能的,基本上你想要做的是当用户触摸屏幕时箭头移动到曲线上的触摸位置并且箭头在曲线上旋转?
首先你需要让箭头旋转然后执行贝塞尔动作
//on touch
CGSize s = [[CCDirector sharedDirector] winSize];
//rotate action we make the arrow rotate forever
id actionBy = [CCRotateBy actionWithDuration:2 angle: 360];
[arrow runAction: [CCRepeatForever actionWithAction:actionBy]];
//bezier action
ccBezierConfig bezier;
bezier.controlPoint_1 = ccp(0, s.height/2);
bezier.controlPoint_2 = ccp(300, -s.height/2);
bezier.endPosition = ccp(300,100);
id bezierForward = [CCBezierBy actionWithDuration:3 bezier:bezier];
id action = [CCCallFunc actionWithTarget:self selector:@selector(endAction)];
[arrow runAction:[CCSequence actions:bezierForward, action]];
//write a method endAction
[arrow stopAllActions];