问题
I have an arrow that I rotate with touch. I was wondering if it was possible to rotate the arrow on a curved line? I've done some research and I think it is called a bezier path? Is it even possible to rotate a sprite on a bezier path using this code and if so how would I even incorporate it?
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;
For example,
I have a ball sitting on the ground and an arrow right above it. When you touch the screen and move the arrow, the arrow moves on a half circle axis.
The curve would look like this Half Circle and the arrow would be rotating on the axis.
Please let me know if I need to be more clear. I really need some help with this.
回答1:
Have faced with same problem couple of days. Most of links in this answer is broken, so I have found material here and here and made this code. Works like magic. Hope it will help someone.
Small description: I have object (self) witch rotates by finger around another object (self.target), and i have some animated sprites like guides of self movement, which rotates around self.target by bezier function. algoritm is quite fast, i have permanent initialization of 100+ guides and it works without CPU overload.
/**
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));
}
回答2:
Never tried but i think it's possible, basically what you want to do is when the user touch the screen the arrow move to the touch location on a curve line and the arrow rotate on the curve ?
first you need to make the arrow rotate and then perform the bezier action
//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];
来源：https://stackoverflow.com/questions/7494795/rotate-a-sprite-on-a-bezier-path-with-touch-cocos2d-box2d