I am new to bullet physics and my usecase is like
I Know start position (x1 ,y1)and end position (x2,y2) .now by applying some force at the center of rigid body,it should move to end position (x2,y2) and then because of gravity it should fall (similar to kinematics reach maximum position by a projectile motion and falls down due to gravity).
I am using as below body->applyCentralForce(force);
But it was not reaching expected or end poistion.
Thanks in advance.
Related
I want to draw either the lower or the upper half of an elliptical arc using Win2D in a C++/winrt app. I can draw the curve, but it includes a diagonal line at its start that looks as if I had begun with a straight line segment - as if the figure didn't begin at the arc but rather some distance down and to the right. How I can restrict drawing to just the arc? Here is the code and an image of the result:
float2 arcSize(100, 6);
auto pathBuilder = CanvasPathBuilder(drawingSession.Device());
pathBuilder.BeginFigure(244, 175);
//Starting at 3 o"clock, sweep pi radians, that is, to 9:00
pathBuilder.AddArc(arcSize,50,6,0.0, 3.14);
pathBuilder.EndFigure(CanvasFigureLoop::Open); //Don't close path
auto geometry = CanvasGeometry::CreatePath(pathBuilder);
session.DrawGeometry(geometry,244, 175, Colors::Black(),1.5);
OK, thanks to that hint from Inspectable I have the solution: the problem lies in where the path begins. In my code I had mistakenly used the proposed arc size as the first argument to AddArc when really that should be the arc center coordinates. And the BeginFigure in the case of this arc must be the point lying at the right edge, at what would be 3:00 in a circle. With the ArcCenter correct in relation to the BeginFigure then AddArc doesn't draw the extra line.
[Update:] p.s. The x and y coordinates for the DrawGeometry call should be zero in this case; that draws it at its original coordinates from BeginFigure, not offset. Maybe this p.s can gain me another -1 for this question.
Say I have shapes represented by a set of points on the edges. A hexagon and a circle, with the points defining them in green
Some points may be collinear or continuous (as in the circle).
How can I write an algorithm detect the corners of the shapes?
In this case it should be that the corners of the hexagon are returned. However, if the shape has a smooth curve it should not return any corners
Thanks
If the points are ordered, you can compute the turning angle between vec{P(i-m),P(i)} and vec{P(i),P(i+m)} for each point P(i). If the turning angle is greater than a pre-defined threshold, then P(i) can be considered as a "corner point". You will have to experiment a bit to find out the proper value for m and the threshold.
I have 3 sprites that all have the same angle, so I'm just going to say arm sprite.
Arm sprite's angle, at the moment, is equal to one point1 (60,60 but this does not matter)
to another point2, the point where the player thumb pressed.
During the ccTime function I update everything, the angles and stuff. So whenever the user touches a spot on the screen, the angle is immediately changed and the arm's angle is equal to the vector from point1 to point2.
I don't want the angle change to take .016 seconds to complete (ccTime gets called every 1/60'th of a second). What I want is for the angle to increment/decrement faster/slower depending on how far away the new vector is from the current vector. Basically I want the arm to raise/lower at a certain speed, maybe accelerate a bit, depending on the vector.
I've tried many times to make it work, but I'm not getting anywhere. Please help me, rotation can go from 90 degrees straight up to almost 180 degrees straight down (the angles in cocos2d are changed, however, so I had to add 90 here and there).
If you need anymore information, just leave a comment and I'll give you the info asap.
You should set the new angle as a destinationAngle then on your update loop:
//Instead of checking for equality, you might want to check the angle is close enough, e.g. if they are withing 1 degree of each other e.g.(if (abs(destinationAngle - angle) < 1)
if (angle != destinationAngle)
{
//move towards destination
angle += ((destinationAngle - angle) / 10.0f);
}
Working with iPhone and Objective C.
I am working on a game and I need to correctly reflect a ball off a circle object. I am trying to do it as a line and circle intersection. I have my ball position outside the circle and I have the new ball position that would be inside the circle at the next draw update. I know the intersect point of the line (ball path) and the circle. Now I want to rotate the ending point of the ball path about the intersection point to get the correct angle of reflection off the tangent.
The following are known:
ball current x,y
ball end x,y
ball radius
circle center x,y
circle radius
intersection point of ball path and circle x and y
I know I need to find the angle of incidence between the tangent line and the incoming ball path which will also equal my angle of reflection. I think once I know those two angles I can subtract them from 180 to get my rotation angle then rotate my end point about the angle of intersection by that amount. I just don't know how.
First, you should note that the center of the ball doesn't have to be inside of the circle to indicate that there's a reflection or bounce. As long as the distance between ball center and circle is less than the radius of the ball, there will be a bounce.
If the radius of the circle is R and the radius of the ball is r, things are simplified if you convert to the case where the circle has radius R+r and the ball has radius 0. For the purposes of collision detection and reflection/bouncing, this is equivalent.
If you have the point of intersection between the (enlarged) circle and the ball's path, you can easily compute the normal N to the circle at that point (it is the unit vector in the direction from the center of the circle to the collision point).
For an incoming vector V the reflected vector is V-2(N⋅V) N, where (N⋅V) is the dot product. For this problem, the incoming vector V is the vector from the intersection point to the point inside the circle.
As for the reflection formula given above, it is relatively easy to derive using vector math, but you can also Google search terms like "calculate reflection vector". The signs in the formula will vary with the assumed directions of V and N. Mathworld has a derivation although, as noted, the signs are different.
I only know the solution to the geometry part.
Let:
r1 => Radius of ball
r2 => Radius of circle
You can calculate the distance between the two circles by using Pythagoras theorem.
If the distance is less than the r1+r2 then do the collision.
For the collision,I would refer you Here. It's in python but I think it should give you an idea of what to do. Hopefully, even implement it in Objective C. The Tutorial By PeterCollingRidge.
Let's say I have circle bouncing around inside a rectangular area. At some point this circle will collide with one of the surfaces of the rectangle and reflect back. The usual way I'd do this would be to let the circle overlap that boundary and then reflect the velocity vector. The fact that the circle actually overlaps the boundary isn't usually a problem, nor really noticeable at low velocity. At high velocity it becomes quite clear that the circle is doing something it shouldn't.
What I'd like to do is to programmatically take reflection into account and place the circle at it's proper position before displaying it on the screen. This means that I have to calculate the point where it hits the boundary between it's current position and it's future position -- rather than calculating it's new position and then checking if it has hit the boundary.
This is a little bit more complicated than the usual circle/rectangle collision problem. I have a vague idea of how I should do it -- basically create a bounding rectangle between the current position and the new position, which brings up a slew of problems of it's own (Since the rectangle is rotated according to the direction of the circle's velocity). However, I'm thinking that this is a common problem, and that a common solution already exists.
Is there a common solution to this kind of problem? Perhaps some basic theories which I should look into?
Since you just have a circle and a rectangle, it's actually pretty simple. A circle of radius r bouncing around inside a rectangle of dimensions w, h can be treated the same as a point p at the circle's center, inside a rectangle (w-r), (h-r).
Now position update becomes simple. Given your point at position x, y and a per-frame velocity of dx, dy, the updated position is x+dx, y+dy - except when you cross a boundary. If, say, you end up with x+dx > W (letting W = w-r), then you do the following:
crossover = (x+dx) - W // this is how far "past" the edge your ball went
x = W - crossover // so you bring it back the same amount on the correct side
dx = -dx // and flip the velocity to the opposite direction
And similarly for y. You'll have to set up a similar (reflected) check for the opposite boundaries in each dimension.
At each step, you can calculate the projected/expected position of the circle for the next frame.
If this lies outside the rectangle, then you can then use the distance from the old circle position to the rectangle's edge and the amount "past" the rectangle's edge that the next position lies at (the interpenetration) to linearly interpolate and determine the precise time when the circle "hits" the rectangle edge.
For example, if the circle is 10 pixels away from the rectangle's edge, then is predicted to move to 5 pixels beyond it, you know that for 2/3rds of the timestep (10/15ths) it moves on its orginal path, then is reflected and continues on its new path for the remaining 1/3rd of the timestep (5/15ths). By calculating these two parts of the motion and "adding" the translations together, you can find the correct new position.
(Of course, it gets more complicated if you hit near a corner, as there may be several collisions during the timestep, off different edges. And if you have more than one circle moving, things get a lot more complex. But that's where you can start for the case you've asked about)
Reflection across a rectangular boundary is incredibly simple. Just take the amount that the object passed the boundary and subtract it from the boundary position. If the position without reflecting would be (-0.8,-0.2) for example and the upper left corner is at (0,0), the reflected position would be (0.8,0.2).