As I wanted to animate an image in projectile motion,
My code is as follows, but it did not reach the target and give projectile animation, any help please?
-(void)timeLine
{
dTime += 0.1;
.................
double s_x = inVel * cos(angle1) ; // the X speed
double s_y = inVel * sin(angle1) ; // the Y speed
NSLog(#"sx = %i",s_x);
NSLog(#"sy = %i",s_y);
x = oX + (s_x * dTime);
y = oY + ( ( s_y * dTime) - (0.5 * 9.8 * dTime * dTime));
NSLog(#"x = %i",x);
NSLog(#"y = %i",y);
imageViewForAnimation.x += x;
imageViewForAnimation.y -= y;
}
imageViewForAnimation.x += x;
imageViewForAnimation.y -= y;
These lines don't seem right to me. You are calculating the actual x and y each time, not the difference moved since the last time. I'm also uncertain why one was being added and one was being subtracted, but that's beside the point. Basically, try changing the lines to
imageViewForAnimation.x = x;
imageViewForAnimation.y = y;
Also, you're doing some calculations over and over which only need to be done once. v_x == s_x (or it should within floating point error) as well as v_y == s_y. You only need to calculate v_x and V_y once beforehand rather than calculating them every time you update the coordinates.
Related
So I was trying to follow the code in this question to get a turret that can fire ballistic projectiles with a fixed starting velocity and no drag to a given point on a 3D surface.
Find an angle to launch the projectile at to reach a specific point
But It's not quite working. The turret ends up aiming too high when the target is close, and too low when the target is further away. There is of course a specific distance at which it does hit the target but that distance is arbitrary, so that's not at all helpful to me.
The way the error scales makes me think I have a multiplication mistake, or am missing some multiplication or division, but I can't for the life of me figure out where I am going wrong. Can anyone point me in the right direction?
Code Below:
float CalculateAngle(float velocity)
{
float gravity = -Physics.gravity.y;
Vector3 modPos = target.position;
if (modPos.x < 0) modPos.x -= 2 * modPos.x;
if (modPos.y < 0) modPos.y -= 2 * modPos.y;
if (modPos.z < 0) modPos.z -= 2 * modPos.z;
modPos.x /= 10;
modPos.y /= 10;
modPos.z /= 10;
float deltaX = modPos.x - FirePoint.position.x;
float deltaZ = modPos.z - FirePoint.position.z;
float deltaY = modPos.y - FirePoint.position.y;
float horzDelta = Mathf.Sqrt(deltaX * deltaX + deltaZ * deltaZ);
float RHSFirstPart = (velocity * velocity) / (gravity * horzDelta);
float RHSSecondPart = Mathf.Sqrt(((velocity * velocity) * ((velocity * velocity) - (2 * gravity * deltaY))/ (gravity * gravity * horzDelta * horzDelta)) - 1);
float tanθ = RHSFirstPart - RHSSecondPart;
float angle = Mathf.Atan2(tanθ, 1) * Mathf.Rad2Deg;
if (angle < 0) return angle;
return -angle;
}
Edit 1:
Still struggling heavily with this. I just can't get the math to work. I went back to the original root of the knowledge here https://physics.stackexchange.com/questions/56265/how-to-get-the-angle-needed-for-a-projectile-to-pass-through-a-given-point-for-t then wrote a function that did the exact equation given in the answers, copying the input values and everything. Except when I run it it fails, as one of the values that needs to be squared is negative which throws a NaN. I assume I am going wrong somewhere in my equation but I've gone over it a hundred times and I am not spotting the error. My code:
float CalculateAngle3(float velocity)
{
float deltaX = 500;
float deltaY = 20;
float v = 100;
float vSqr = v * v;
float g = 9.81f * 9.81f;
float a = vSqr * (vSqr - 2 * g * deltaY);
float b = (g * g) * (deltaX * deltaX);
float c = a / b - 1;
float d = Mathf.Sqrt(c); //c is negitive causing an NaN
float e = vSqr / g * deltaX;
float tanθ = e - d;
return tanθ;
}
Edit 2:
Gave up. This guy solved it so I am just going to use his logic instead
: P
https://www.forrestthewoods.com/blog/solving_ballistic_trajectories/
Using it like such:
Vector3 s0;
Vector3 s1;
if (fts.solve_ballistic_arc(FirePoint.position, bomb.StartingVelocity.z, target.position, -Physics.gravity.y, out s0, out s1) > 0)
{
targetPosition = transform.position + s1;
SafetyEnabled = false;
}
else
{
//Don't fire if we don't have a solution
SafetyEnabled = true;
}
I'm going to leave the question open for now since it's still technically not answered. I still don't know why the original implementation wasn't working.
It is possible your quadratic formula is incorrect (I do not know why you did not code a separate small function that solves the quadratic equation for any three given coefficients, to make your code more readable and less prone to errors)
float RHSFirstPart = velocity / (gravity * horzDelta);
float RHSSecondPart = Mathf.Sqrt(RHSFirstPart*RHSFirstPart - 2*RHSFirstPart*deltaY/horzDelta - 1);
float tanθ = RHSFirstPart - RHSSecondPart;
A comment: In most applications we do not really need the actual angle but the values of cos(angle) and sin(angle) because these are the components of the unit vector which usually is sought (just like in your case). So no need to use inverse trigonometry to find an actual number, which slows down calculations and is may introduce unnecessary round-off errors.
I am trying to learn how to animate dots and update their x and y positions from frame to frame. I have found a tutorial online, and I am really struggling with the formulas that are used in dx and dy:
dots.speed = 3; %degrees/second
dots.duration = 5; %seconds
dots.direction_deg = 30; %degrees (clockwise from straight up)
dots.direction_rad = dots.direction_deg * pi /180 %direction converted to radians
% Calculate the distance travelled by a dot (in degrees), which is the speed (degrees/second) divided by the frame rate (frames/second).
dx = dots.speed*sin(dots.direction_rad)/display.frameRate;
dy = -dots.speed*cos(dots.direction_rad)/display.frameRate;
With my knowledge in trigonometry, I end up with these formulas for dx and dy:
Cos(angle) = Adjacent/Hypotenuse --> cos(dots.direction_rad) = 1/dx (1 is the distance from the origin to a point on the circle,which is equal to 1 radian). --> dx = 1/cos(dots.direction_rad) --> the distance is then equal to :
dots.speed*1/cos(dots.direction_rad)*display.frameRate
Sin(angle) = Opposite/Hypotenuse --> sin(dots.direction_rad) = dy/1 --> the distance is then equal to :
dots.speed * sin(dots.direction_rad) / display.frameRate.
My formulas are not giving the expected result. So what I am doing wrong. I don't seem to understand how the dx and dy were computed in the tutorial. Please help! thanks
I have a video stream coming from a 180 degree fisheye camera. I want to do some image-processing to convert the fisheye view into a normal view.
After some research and lots of read articles I found this paper.
They describe an algorithm (and some formulas) to solve this problem.
I used tried to implement this method in a Matlab. Unfortunately it doesn't work, and I failed to make it work. The "corrected" image looks exactly like the original photograph and there's no any removal of distortion and secondly I am just receiving top left side of the image, not the complete image but changing the value of 'K' to 1.9 gives mw the whole image, but its exactly the same image.
Input image:
Result:
When the value of K is 1.15 as mentioned in the article
When the value of K is 1.9
Here is my code:
image = imread('image2.png');
[Cx, Cy, channel] = size(image);
k = 1.5;
f = (Cx * Cy)/3;
opw = fix(f * tan(asin(sin(atan((Cx/2)/f)) * k)));
oph = fix(f * tan(asin(sin(atan((Cy/2)/f)) * k)));
image_new = zeros(opw, oph,channel);
for i = 1: opw
for j = 1: oph
[theta,rho] = cart2pol(i,j);
R = f * tan(asin(sin(atan(rho/f)) * k));
r = f * tan(asin(sin(atan(R/f))/k));
X = ceil(r * cos(theta));
Y = ceil(r * sin(theta));
for k = 1: 3
image_new(i,j,k) = image(X,Y,k);
end
end
end
image_new = uint8(image_new);
warning('off', 'Images:initSize:adjustingMag');
imshow(image_new);
This is what solved my problem.
input:
strength as floating point >= 0. 0 = no change, high numbers equal stronger correction.
zoom as floating point >= 1. (1 = no change in zoom)
algorithm:
set halfWidth = imageWidth / 2
set halfHeight = imageHeight / 2
if strength = 0 then strength = 0.00001
set correctionRadius = squareroot(imageWidth ^ 2 + imageHeight ^ 2) / strength
for each pixel (x,y) in destinationImage
set newX = x - halfWidth
set newY = y - halfHeight
set distance = squareroot(newX ^ 2 + newY ^ 2)
set r = distance / correctionRadius
if r = 0 then
set theta = 1
else
set theta = arctangent(r) / r
set sourceX = halfWidth + theta * newX * zoom
set sourceY = halfHeight + theta * newY * zoom
set color of pixel (x, y) to color of source image pixel at (sourceX, sourceY)
I have tried several different solutions but no luck so far.
- (CGPoint)contractLineTemp:(CGPoint)point :(CGPoint)circle :(float)circleRadius {
CGFloat x,y;
x = point.x - circle.x;
y = point.y - circle.y;
CGFloat theta = atan2(x, y);
CGPoint newPoint;
newPoint.x = circle.x + circleRadius * sin(theta);
newPoint.y = circle.y + circleRadius * cos(theta);
return newPoint;
}
- (CGPoint)contractLineTemp:(CGPoint)startPoint :(CGPoint)endPoint :(float)scaleBy {
float dx = endPoint.x - startPoint.x;
float dy = endPoint.y - startPoint.y;
float scale = scaleBy * Q_rsqrt(dx * dx + dy * dy);
return CGPointMake (endPoint.x - dx * scale, endPoint.y - dy * scale);
}
Both of these solutions kind of work. If I draw the line to the center of the circle you can see that it intersects the circle exactly where it should.
http://www.freeimagehosting.net/le5pi
If I use either of the solutions above and draw to the circumference of the circle depending on the angle it is no longer going towards the center of the circle. In the second image the line should be in the middle of the right edge of the circle and going straight right.
http://www.freeimagehosting.net/53ovs
http://www.freeimagehosting.net/sb3b2
Sorry for the links. I am to new to currently post images.
Thanks for you help.
It's easier to treat this as a vector problem. Your second approach is close, but you don't correctly scale the vector between the two points. It's easier to work with a normalized vector in this case, although you have to assume that the distance between the two points on the line is non-zero.
Given:
double x0 = CIRC_X0; /* x-coord of center of circle */
double y0 = CIRC_Y0; /* y-coord of center of circle */
double x1 = LINE_X1; /* x-coord of other point on the line */
double y1 = LINE_Y1; /* y-coord of other point on the line */
Then the vector between the two points is (vx,vy):
double vx = x1 - x0;
double vy = y1 - y0;
It's easier to work with a unit vector, which we can get by normalizing (vx,vy):
double vmag = sqrt(vx*vx + vy*vy);
vx /= vmag; /* Assumption is vmag > 0 */
vy /= vmag;
Now, any point along the line can be described as:
x0 + dist * vx
y0 + dist * vy
where dist is the distance from the center. The intersection of the circle and the line must be a distance of CIRC_RADIUS from the center, so:
double x_intersect = x0 + CIRC_RADIUS * vx;
double y_intersect = y0 + CIRC_RADIUS * vy;
I think that there may be a convention conflict on what theta, x and y are. The atan2 function yields values in the range -pi..pi, by taking the convention of theta as the angle growing from the X axis towards Y. However you are considering theta as the angle from Y to X.
Try changing the code:
CGFloat theta = atan2(y, x);
CGPoint newPoint;
newPoint.x = circle.x + circleRadius * cos(theta);
newPoint.y = circle.y + circleRadius * sin(theta);
Although your formulae are consistent within a coordinate system, it may have conflict with the screen/display device coordinate system.
I have a problem regarding positioning an image according to the touches location, however limited to a circle.
It works for the most part, but if the angle (from the touches location to the desired location) is less than 0, it positions the image on the wrong side of the circle.
Perhaps it's some maths that I've done wrong.
Anyway, here's the code:
float newHeight, newWidth, centerPointX, centerPointY;
newHeight = -(invertedY.y - (view.frame.origin.y+view.frame.size.height/2));
newWidth = -(invertedY.x - (view.frame.origin.x+view.frame.size.width/2));
float tangent = newHeight/newWidth;
float calculatedAngle = atanf(tangent);
float s, c, d, fX, fY;
d = view.frame.size.width/2+30;
if (calculatedAngle < 0) {
s = sinf(calculatedAngle) * d;
c = cosf(calculatedAngle) * d;
} else {
s = -sinf(calculatedAngle) * d;
c = -cosf(calculatedAngle) * d;
}
fX = view.center.x + c;
fY = view.center.y + s;
[delegate setPoint:CGPointMake(fX, fY)];
NSLog(#"angle = %.2f", calculatedAngle);
Any help appreciated.
I think the best way to limit location to a circle is calculate vector from center to touch location. Calculate vector length then divide it by that length so it would be normalized. Then multiply normalized vector by radius of circle and finally add this vector to the center to compute new location.
CGPoint touch, center;
CGPoint vector = CGPointMake(touch.x-center.x, touch.y-center.y);
float length = sqrtf(vector.x*vector.x + vector.y*vector.y);
// Normalize and multiply by radius (r)
vector.x = r * vector.x / length;
vector.y = r * vector.y / length;
[delegate setPoint:CGPointMake(center.x + vector.x, center.y + vector.y)];