I am developing an app that animates a motion on a UIBezierPath (made of several curves).
In some use cases I need to place an item so it will start moving from some point on the route, and not from its beginning. E.g put item in the middle or 2/3 point of the path. How can I calculate the location of such point?
Thanks!
A Bezier curve is parametric curve a http://en.wikipedia.org/wiki/B%C3%A9zier_curve meaning you have two functions of a parameter T over a range. One function generates the X coordinate, the other generates the Y coordinate. If you know the two functions, just pick a value of T halfway or 2/3rds of the way between the endpoints of the range and plug that into the two functions to get the X & Y coordinates of the desired point.
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So I'm trying to find the location between point A & B. Point A will be my phone and point be will be a dot on a small object. Point A is a point in the middle of my phone with an X/Y axis. Using the front facing camera on an iPhone, point B attached to the small object, moves into the cameras view. Point B will never be more than a foot away from point A at any time.
Is it possible to determine the exact coordinates of point B in relation to point A's X/Y axis (Within a millimeter or so)? Would a signal need to be sent between the two points, or would the camera be able to pick up point B and determine its coordinates on point A's X/Y axis?
I've attached an image below to hopefully explain what I am trying to describe a little better.
The coordinates of two points are A(-6, 0) and B(10, 4). Find the coordinates of the point C on the y-axis such that AC=BC.
I have two moving objects with position and orientation data (Euler Angles, Quaternions) relative to ECI coordinate frame. I would like to calculate AZ/EL from what I'm guessing is the "body frame" of the first object. I have attempted to convert both objects into the body frame through rotation matrices (X-Y-Z and Z-Y-X rotation sequence) and calculate a target vector AZ/EL this way but have not had success. I've also tried to get body frame positions and calculate the body axis/angles and convert back to Eulers (relative to body frame). I'm never sure how the coordinate system axes I'm supposedly creating are aligned along my object.
I have found a couple other questions like this answered with Quaternion solutions so that may be the best path to take, however my Quaternion background is weak at best and I fail to see how the computations given result in a target vector.
Any advice would be much appreciated and I'm happy to share my progress/experiences going forward.
get the current NEH transform matrix for the moving object
you must know position and at least two directions from North,East,Height(Up or Altitude) of the moving object otherwise is your problem unsolvable no matter what. This matrix/frame is called NEH (X=North,Y=East,Z=Height) or sometimes also ENU (X=East,Y=North,Z=Up). Look here transform matrix anatomy and here Earth's NEH construction and change the position and radius to match your moving object.
convert point P0 from GCS (global coordinate system) to NEH
simply: P1=Inverse(NEH)*P0 where P1 is now in NEH LCS (Local coordinate system). Both P0,P1 are in homogenous coordinates { x,y,z,w=1 } to allow multiplications with 4x4 matrix so you can compute azimut and elevation directly from it:
Azimut=atanxy(P1.x,P1.y);
Elevation=atan(P1.z/sqrt((P1.x*P1.x)+(P1.y*P1.y)));
where atanxy is mine atan2 (4 quadrant atan) first is dx then dy. I think atan2 in matlab has it in reverse.
[Notes]
Always visually check all frames (especially NEH). Just draw the 3 axises as lines of some length to validate if the result is correct. It should look like on image, just different color for each axis. You can move to next point only if NEH is OK !!!
Check atan2/atanxy operands order and also check goniometric functions units (rad,deg) to avoid confusions.
In the following screen shot:
when you drag the tail of the word balloon (the thing that connects from the balloon to the persons mouth), the shape curves (as illustrated by the difference between the two balloon tails in the picture). I'm wondering, how is this done? I'm assuming you need to start with a CGPath and do something to it, does anyone happen to know what this is?
Update: So if I wanted to curve the following shape:
Would I use the following code:
CGPathAddCurveToPoint(mutablePath, NULL, x1, y1, x2, y2 + constant, x5, y5);
CGPathAddCurveToPoint(mutablePath, NULL, x3, y3, x4, y4 + constant, x5, y5);
Where the constant readjusts the y position of point 2 and point 4 to make the curve?
You need to exploit the fact that, mathematically, a straight-line segment is just a kind of curve segment.
(It's easier than it sounds, trust me.)
Bézier path segments have something called “order” that essentially determines how many points there are in the segment, not counting the point you're coming from.
A straight-line segment is a first-order curve, meaning that it only has the destination point. These “curves” are always straight lines because there are no control points to curve toward.
Quadratic curves are second-order curves (one control point plus the destination).
Cubic curves are third-order curves (two control points).
(The math doesn't put any limit on this, but Quartz stops here. No fourth-order curves for you without rolling your own rasterizer.)
This matters because any lower-order curve—including a straight line—can be expressed as a higher-order curve.
So, the secret?
For even a straight tail, use a curve.
(Namely, a cubic curve, since you want the curve going in two different directions: One, more or less into the tail, and the other, more or less along the edge of the balloon.)
From each of the two points at the base of the tail, you want one of the control points to be about halfway to the destination. This much is unconditional.
The direction of each of the control points gives you three options:
The straight-out tail
Notice the two control points along the blue line at the vertical center of the image.
Notice the direction of these two control points, relative to the base point it's connected to. They are angled inward, toward the tip—indeed, exactly on the straight line to the tip.
The oblique tail
Here, the tip point is no longer horizontally between the two base points. The control points have moved, but only to follow: each one is still halfway along the straight line between the corresponding base point and the tip.
The curved tail
For a curved tail, you move the tip, but you keep the control points at the same position as for a straight tail. Thus, the tail starts out straight out (following the control points), but as it gets farther from the base points, their influence wanes, and the tail begins curving toward the tip.
This is a lot easier to describe visually than to put into code, so you may want to consider using something like PaintCode or Opacity to draw each kind of tail using a pen tool and then see what the code they generate for it looks like.
You can use the CGContextAddCurveToPoint() functions:
CGContextMoveToPoint(ctx, x, y);
CGContextAddCurveToPoint(ctx, outTangentX, outTangentY, inTangentX, inTangentY, newX, newY);
... // more points or whatever you need here
CGContextFillPath(ctx); // Fill with white
CGContextStrokePath(ctx); // stroke the edges with black
The in/out tangents can be hardcoded to be something that looks good based on the point on the mouth of the picture and the point where it meets the balloon bubble. You might try something like making their angles half-way between perpendicular and the slope of the straight line between the 2 points or something like that as a starting place.
I need to create flight routes in Google Earth. Example from point A to point B, How do i get the equivalent middle point for both and along point A to B, there are also many different coordinates joining so that the line would be a curve.
Generally you should use the great-circle distance to compute the distance of two point on the surface of a sphere, like in this case the Earth. Even WolframAlpha uses this to compute direct travel times.
This would also define the midpoint for you uniquely.
Use e.g., Perl's Math::Trig module. It comes complete with the great circle functions you need.
I have an array of CGPoints (basic struct with two floats: x and y). I want to use OpenGL ES to draw a textured curve using these points. I can do this fine with just two points, but it gets harder when I need to make a line from several points.
Currently I draw a line horizontally, calculate its angle from the points given, and then rotate it. I don't think doing this for all lines in a curve is a good idea. There's probably a faster way.
I'm thinking that I can "enlarge" or "constrict" all the points at once to make a curve with some sort of width.
I'm not positive what you want to accomplish, but consider this:
Based on a ordered list of points, you can draw a polyline using those points. If you want to have a polyline with a 2D texture on it, you can draw a series of quadrilaterals (using two triangles each, of course). You can generate these quadrilaterals using an idea similar to catmul-rom spline generation.
Consider a series of points p[i-1], p[i], p[i+1]. Now, for each i, you can find two points each an epsilon distance away from p[i] along the line perpendicular to the line connecting p[i-1] and p[i+1]. You can determine the two points generated for the endpoints in various ways, like using the perpendicular to the line from p[0] to p[1].
I'm not sure if this will be faster than your method, but you should be caching the results. If you are planning on doing this every frame, another type of solution to your problem may be needed.