I'm would like to achieve something like Object A will jump to position A in parabolic way. For example:
I am able to make the object jump but i do not know how to calculate to make the object jump to the precise location.
1 Answer
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SpriteKit uses SKAction objects to accomplish most of its functionality. What you're seeking is SKAction's followPath:duration: action.
The following accomplishes this:
UIBezierPath *path = [UIBezierPath bezierPath];
[path moveToPoint:CGPointZero];
[path addQuadCurveToPoint:CGPointMake(deltaX/2, desiredHeight) controlPoint:CGPointMake(0, desiredHeight)];
[path addQuadCurveToPoint:CGPointMake(deltaX, deltaY) controlPoint:CGPointMake(deltaX, desiredHeight)];
CGPathRef pathref = path.CGPath;
[jumpingNode runAction:[SKAction followPath:pathref duration:desiredDuration]];
This allows the Node to move along the specified path. It looks very uniform, but you can always add easing (related StackOverflow question) to give a more natural look
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\$\begingroup\$ Thank you for your answer, it's really work as I want, but still not so smooth like jumping. Maybe still need some value tweaking \$\endgroup\$ Commented Jul 31, 2014 at 5:31
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\$\begingroup\$ Anytime. Re: smoothness, use the easing link I provided. Or create your own. You can theoretically do the same with the physics engine in SpriteKit, but this solution is much simpler, and likely just needs some easing to look realistic. What you might want to do is set up two paths (one with the first quadCurve, one with the second quadCurve) and use two different easing methods with both. \$\endgroup\$ Commented Jul 31, 2014 at 5:36
f(x) = sin( pow(x,0.5f) * Pi);. Consider having point A and B and you want to jump ObjectA from A to B =>ObjectA.position = lerp(A,B,Time) + Vector.Up * f(Time)andTimeis from 0 to 1. No approach is better than the other, "fake" physics can look better or worse , it all depends on your imagination \$\endgroup\$