I've written iPhone - Mac, Client - Server app that allows to use mouse via touchpad.
Now on every packet sent I move cursor by pecific amount of pixels (now 10px).
It isn't accurate. When i change sensitivity to 1px it's to slow.
I am wondering how to enhance usability and simulate mouse acceleration.
Any ideas?
I suggest the following procedure:
ON THE IPHONE:
Determine the distance moved in x and y direction, let's name this dx and dy.
Calculate the total distance this corresponds to: dr = sqrt(dx^2+dy^2).
Determine how much time has passed, and calculate the speed of the movement: v = dr/dt.
Perform some non-linear transform on the velocity, e.g.: v_new = a * v + b * v^2 (start with a=1 and b=0 for no acceleration, and then experiment for optimal values)
Calculate a new distance: dr_new = v_new * dt.
Calculate new distances in x/y direction:
dx_new = dx * dr_new / dr and dy_new = dy * dr_new / dr.
Send dx_new and dy_new to the Mac.
ON THE MAC:
Move the mouse by dx_new and dy_new pixels in x/y direction.
NOTE: This might jitter a lot, you can try averaging the velocity after step (3) with the previous two or three measured velocities if it jitters to much.
Related
With only 1 dimension, you want to get to position X and stop there. you have a maximum acceleration A you can apply yourself; each frame you choose what direction to accelerate.
So if you have a velocity V, and want to stop at position X as fast as possible, how much of your maximum acceleration A do you apply yourself?
(If you are far away, you apply your maximum acceleration, but when you are close, you start braking. So based on your velocity and remaing distance, you need to decide when you begin to brake.)
The governing equation is x = v t + 0.5 a t^2
where x is distance, v is velocity, t is time, and a is acceleration in compatible units.
With no other constraints, in order to minimize your travel time, you will always apply maximum acceleration: Accelerate in the direction of the goal until you are halfway there, then accelerate in the direction of the origin until you stop.
If you have a maximum velocity, accelerate until you reach that maximum velocity, which will happen at some distance X away from the origin. When you are distance X away from the goal, accelerate back toward the origin.
Just dropping in with the answer to a more advanced version of this question I solved lately. You have v_0 at x_0, and want to reach position X, and have velocity V when reaching there. Acceleration is applied each small time step, e.g. 60 times per second.
So using the equation of motion x = v t + 0.5 a t^2, solve it for t with positive/negative acceleration applied, to see how fast one can reach x, regardless of end speed.
Also calculate how long it would take to reach V by applying the positive or negative acceleration. Whichever t is the highest, is the positive or negative acceleration to apply.
This causes objects to smoothly follow pahts such as sin(x), and if they are thrown off, they will elegantly slide back into it.
I'm trying to create a solar system simulation, and I'm having problems trying to figure out initial velocity vectors for random objects I've placed into the simulation.
Assume:
- I'm using Gaussian grav constant, so all my units are AU/Solar Masses/Day
- Using x,y,z for coordinates
- One star, which is fixed at 0,0,0. Quasi-random mass is determined for it
- I place a planet, at a random x,y,z coordinate, and its own quasi-random mass determined.
Before I start the nbody loop (using RK4), I would like the initial velocity of the planet to be such that it has a circular orbit around the star. Other placed planets will, of course, pull on it once the simulation starts, but I want to give it the chance to have a stable orbit...
So, in the end, I need to have an initial velocity vector (x,y,z) for the planet that means it would have a circular orbit around the star after 1 timestep.
Help? I've been beating my head against this for weeks and I don't believe I have any reasonable solution yet...
It is quite simple if you assume that the mass of the star M is much bigger than the total mass of all planets sum(m[i]). This simplifies the problem as it allows you to pin the star to the centre of the coordinate system. Also it is much easier to assume that the motion of all planets is coplanar, which further reduces the dimensionality of the problem to 2D.
First determine the magnitude of the circular orbit velocity given the magnitude of the radius vector r[i] (the radius of the orbit). It only depends on the mass of the star, because of the above mentioned assumption: v[i] = sqrt(mu / r[i]), where mu is the standard gravitational parameter of the star, mu = G * M.
Pick a random orbital phase parameter phi[i] by sampling uniformly from [0, 2*pi). Then the initial position of the planet in Cartesian coordinates is:x[i] = r[i] * cos(phi[i]) y[i] = r[i] * sin(phi[i])
With circular orbits the velocity vector is always perpendicular to the radial vector, i.e. its direction is phi[i] +/- pi/2 (+pi/2 for counter-clockwise (CCW) rotation and -pi/2 for clockwise rotation). Let's take CCW rotation as an example. The Cartesian coordinates of the planet's velocity are:vx[i] = v[i] * cos(phi[i] + pi/2) = -v[i] * sin(phi[i])vy[i] = v[i] * sin(phi[i] + pi/2) = v[i] * cos(phi[i])
This easily extends to coplanar 3D motion by adding z[i] = 0 and vz[i] = 0, but it makes no sense, since there are no forces in the Z direction and hence z[i] and vz[i] would forever stay equal to 0 (i.e. you will be solving for a 2D subspace problem of the full 3D space).
With full 3D simulation where each planet moves in a randomly inclined initial orbit, one can work that way:
This step is equal to step 1 from the 2D case.
You need to pick an initial position on the surface of the unit sphere. See here for examples on how to do that in a uniformly random fashion. Then scale the unit sphere coordinates by the magnitude of r[i].
In the 3D case, instead of two possible perpendicular vectors, there is a whole tangential plane where the planet velocity lies. The tangential plane has its normal vector collinear to the radius vector and dot(r[i], v[i]) = 0 = x[i]*vx[i] + y[i]*vy[i] + z[i]*vz[i]. One could pick any vector that is perpendicular to r[i], for example e1[i] = (-y[i], x[i], 0). This results in a null vector at the poles, so there one could pick e1[i] = (0, -z[i], y[i]) instead. Then another perpendicular vector can be found by taking the cross product of r[i] and e1[i]:e2[i] = r[i] x e1[i] = (r[2]*e1[3]-r[3]*e1[2], r[3]*e1[1]-r[1]*e1[3], r[1]*e1[2]-r[2]*e1[1]). Now e1[i] and e2[i] can be normalised by dividing them by their norms:n1[i] = e1[i] / ||e1[i]||n2[i] = e2[i] / ||e2[i]||where ||a|| = sqrt(dot(a, a)) = sqrt(a.x^2 + a.y^2 + a.z^2). Now that you have an orthogonal vector basis in the tangential plane, you can pick one random angle omega in [0, 2*pi) and compute the velocity vector as v[i] = cos(omega) * n1[i] + sin(omega) * n2[i], or as Cartesian components:vx[i] = cos(omega) * n1[i].x + sin(omega) * n2[i].xvy[i] = cos(omega) * n1[i].y + sin(omega) * n2[i].yvz[i] = cos(omega) * n1[i].z + sin(omega) * n2[i].z.
Note that by construction the basis in step 3 depends on the radius vector, but this does not matter since a random direction (omega) is added.
As to the choice of units, in simulation science we always tend to keep things in natural units, i.e. units where all computed quantities are dimensionless and kept in [0, 1] or at least within 1-2 orders of magnitude and so the full resolution of the limited floating-point representation could be used. If you take the star mass to be in units of Solar mass, distances to be in AUs and time to be in years, then for an Earth-like planet at 1 AU around a Sun-like star, the magnitude of the orbital velocity would be 2*pi (AU/yr) and the magnitude of the radius vector would be 1 (AU).
Just let centripetal acceleration equal gravitational acceleration.
m1v2 / r = G m1m2 / r2
v = sqrt( G m2 / r )
Of course the star mass m2 must be much greater than the planet mass m1 or you don't really have a one-body problem.
Units are a pain in the butt when setting up physics problems. I've spent days resolving errors in seconds vs timestep units. Your choice of AU/Solar Masses/Day is utterly insane. Fix that before anything else.
And, keep in mind that computers have inherently limited precision. An nbody simulation accumulates integration error, so after a million or a billion steps you will certainly not have a circle, regardless of the step duration. I don't know much about that math, but I think stable n-body systems keep themselves stable by resonances which absorb minor variations, whether introduced by nearby stars or by the FPU. So the setup might work fine for a stable, 5-body problem but still fail for a 1-body problem.
As Ed suggested, I would use the mks units, rather than some other set of units.
For the initial velocity, I would agree with part of what Ed said, but I would use the vector form of the centripetal acceleration:
m1v2/r r(hat) = G m1 m2 / r2 r(hat)
Set z to 0, and convert from polar coordinates to cartesian coordinates (x,y). Then, you can assign either y or x an initial velocity, and compute what the other variable is to satisfy the circular orbit criteria. This should give you an initial (Vx,Vy) that you can start your nbody problem from. There should also be quite a bit of literature out there on numerical recipes for nbody central force problems.
I am trying to calculate the forces that will act on circular objects in the event of a collision. Unfortunately, my mechanics is slightly rusty so i'm having a bit of trouble.
I have an agent class with members
vector position // (x,y)
vector velocity // (x,y)
vector forward // (x,y)
float radius // radius of the agent (all circles)
float mass
So if we have A,B:Agent, and in the next time step the velocity is going to change the position. If a collision is going to occur I want to work out the force that will act on the objects.
I know Line1 = (B.position-A.position) is needed to work out the angle of the resultant force but how to calculate it is baffling me when I have to take into account current velocity of the vehicle along with the angle of collision.
arctan(L1.y,L1.x) is am angle for the force (direction can be determined)
sin/cos are height/width of the components
Also I know to calculate the rotated axis I need to use
x = cos(T)*vel.x + sin(T)*vel.y
y = cos(T)*vel.y + sin(T)*vel.x
This is where my brain can't cope anymore.. Any help would be appreciated.
As I say, the aim is to work out the vector force applied to the objects as I have already taken into account basic physics.
Added a little psudocode to show where I was starting to go with it..
A,B:Agent
Agent {
vector position, velocity, front;
float radius,mass;
}
vector dist = B.position - A.position;
float distMag = dist.magnitude();
if (distMag < A.radius + B.radius) { // collision
float theta = arctan(dist.y,dist.x);
flost sine = sin(theta);
float cosine = cos(theta);
vector newAxis = new vector;
newAxis.x = cosine * dist .x + sine * dist .y;
newAxis.y = cosine * dist .y - sine * dist .x;
// Converted velocities
vector[] vTemp = {
new vector(), new vector() };
vTemp[0].x = cosine * agent.velocity.x + sine * agent.velocity.y;
vTemp[0].y = cosine * agent.velocity.y - sine * agent.velocity.x;
vTemp[1].x = cosine * current.velocity.x + sine * current.velocity.y;
vTemp[1].y = cosine * current.velocity.y - sine * current.velocity.x;
Here's to hoping there's a curious maths geek on stack..
Let us assume, without loss of generality, that we are in the second object's reference frame before the collision.
Conservation of momentum:
m1*vx1 = m1*vx1' + m2*vx2'
m1*vy1 = m1*vy1' + m2*vy2'
Solving for vx1', vy1':
vx1' = vx1 - (m2/m1)*vx2'
vy1' = vy1 - (m2/m1)*vy2'
Secretly, I will remember the fact that vx1'*vx1' + vy1'*vy1' = v1'*v1'.
Conservation of energy (one of the things elastic collisions give us is that angle of incidence is angle of reflection):
m1*v1*v1 = m1*v1'*v1' + m2*v2'+v2'
Solving for v1' squared:
v1'*v1' = v1*v1 - (m2/m1)v2'*v2'
Combine to eliminate v1':
(1-m2/m1)*v2'*v2' = 2*(vx2'*vx1+vy2'*vy1)
Now, if you've ever seen a stationary poolball hit, you know that it flies off in the direction of the contact normal (this is the same as your theta).
v2x' = v2'cos(theta)
v2y' = v2'sin(theta)
Therefore:
v2' = 2/(1-m2/m1)*(vx1*sin(theta)+vy1*cos(theta))
Now you can solve for v1' (either use v1'=sqrt(v1*v1-(m2/m1)*v2'*v2') or solve the whole thing in terms of the input variables).
Let's call phi = arctan(vy1/vx1). The angle of incidence relative to the tangent line to the circle at the point of intersection is 90-phi-theta (pi/2-phi-theta if you prefer). Add that again for the reflection, then convert back to an angle relative to the horizontal. Let's call the angle of incidence psi = 180-phi-2*theta (pi-phi-2*theta). Or,
psi = (180 or pi) - (arctan(vy1/vx1))-2*(arctan(dy/dx))
So:
vx1' = v1'sin(psi)
vy1' = v1'cos(psi)
Consider: if these circles are supposed to be solid 3D spheres, then use a mass proportional to radius-cubed for each one (note that the proportionality constant cancels out). If they are supposed to be disklike, use mass proportional to radius-squared. If they are rings, just use radius.
Next point to consider: Since the computer updates at discrete time events, you actually have overlapping objects. You should back out the objects so that they don't overlap before computing the new location of each object. For extra credit, figure out the time that they should have intersected, then move them in the new direction for that amount of time. Note that this time is just the overlap / old velocity. The reason that this is important is that you might imagine a collision that is computed that causes the objects to still overlap (causing them to collide again).
Next point to consider: to translate the original problem into this problem, just subtract object 2's velocity from object 1 (component-wise). After the computation, remember to add it back.
Final point to consider: I probably made an algebra error somewhere along the line. You should seriously consider checking my work.
I want to simulate a free fall and a collision with the ground (for example a bouncing ball). The object will fall in a vacuum - an air resistance can be omitted. A collision with the ground should causes some energy loss so finally the object will stop moving. I use JOGL to render a point which is my falling object. A gravity is constant (-9.8 m/s^2).
I found an euler method to calculate a new position of the point:
deltaTime = currentTime - previousTime;
vel += acc * deltaTime;
pos += vel * deltaTime;
but I'm doing something wrong. The point bounces a few times and then it's moving down (very slow).
Here is a pseudocode (initial pos = (0.0f, 2.0f, 0.0f), initial vel(0.0f, 0.0f, 0.0f), gravity = -9.8f):
display()
{
calculateDeltaTime();
velocity.y += gravity * deltaTime;
pos.y += velocity.y * deltaTime;
if(pos.y < -2.0f) //a collision with the ground
{
velocity.y = velocity.y * energyLoss * -1.0f;
}
}
What is the best way to achieve a realistic effect ? How the euler method refer to the constant acceleration equations ?
Because floating points dont round-up nicely, you'll never get at a velocity that's actually 0. You'd probably get something like -0.00000000000001 or something.
you need to to make it 0.0 when it's close enough. (define some delta.)
To expand upon my comment above, and to answer Tobias, I'll add a complete answer here.
Upon initial inspection, I determined that you were bleeding off velocity to fast. Simply put, the relationship between kinetic energy and velocity is E = m v^2 /2, so after taking the derivative with respect to velocity you get
delta_E = m v delta_v
Then, depending on how energyloss is defined, you can establish the relationship between delta_E and energyloss. For instance, in most cases energyloss = delta_E/E_initial, then the above relationship can be simplified as
delta_v = energyloss*v_initial / 2
This is assuming that the time interval is small allowing you to replace v in the first equation with v_initial, so you should be able to get away with it for what your doing. To be clear, delta_v is subtracted from velocity.y inside your collision block instead of what you have.
As to the question of adding air-resistance or not, the answer is it depends. For small initial drop heights, it won't matter, but it can start to matter with smaller energy losses due to bounce and higher drop points. For a 1 gram, 1 inch (2.54 cm) diameter, smooth sphere, I plotted time difference between with and without air friction vs. drop height:
For low energy loss materials (80 - 90+ % energy retained), I'd consider adding it in for 10 meter, and higher, drop heights. But, if the drops are under 2 - 3 meters, I wouldn't bother.
If anyone wants the calculations, I'll share them.
I wanted to simulate the following through animation :
A ball starts with a certain velocity at the bottom most point of
a vertical circular loop and keeps rolling in it until its velocity permits.
For this, I wanted to find velocity/x/y vs. time equation.
For e.g. if the ball had mass : 5Kg, radius of the circular loop = 10m,
and initial velocity of the ball is 200 m/s, what will its velocity and (x,y) position
be after 5 seconds?
thanks.
Sliding, frictionless case with a point-particle ball
In this case we aren't worrying about rotational energy and are assuming that the ball is actually a point particle. Then, in order for the ball to stay on at the top, the centripetal force condition has to be satisfied:
m * v_top^2 / r = m * g
so
v_top = sqrt(r * g)
So the minimum initial velocity is determined by:
1 / 2 * m * v0^2 >= 1 / 2 * m * v_top^2 + m * g * 2 * r
v0 >= sqrt(5 * r * g)
This is similar to what Pete said, except that he forgot the centripetal force condition to stay on at the top.
Next, the acceleration tangential to the track is given by:
a = - g * sin(theta)
but a = r * alpha = r * d^2(theta)/dt^2 where alpha is the rotational acceleration. Thus, we get
r * d^2(theta)/dt^2 = g * sin(theta)
However, I don't know of an analytical solution to this differential equation and Mathematica was stumbling with finding one too. You can't just move the dts to the other side and integrate because theta is a function of t. I would recommend solving it by numerical means such as a Runga-Kutte or maybe the Verlet method. I solved it using Mathematica for the parameters you gave, but with the ball moving so quickly, it doesn't really slow down much in going around. When I lowered the initial velocity though, I was able to see the speeding up and slowing down by plotting theta as a function of time.
Adding in other things like a finite ball radius, rotational energy and friction are certainly doable, but I would worry about being able to solve this first case before moving on because it only gets more complicated from here. By the way, with the friction you will have to choose some kinetic coefficient of friction for your given materials which will of course be proportional to the normal force exerted on the ball by the track which can be solved for by summing the force components along the radius of the circle and don't forget to include the centripetal force condition.
If you haven't done this sort of physics before, I definitely recommend getting a introductory good book on physics (with calculus) and working through it. You only need to bother with the sections that apply to mechanics though that is a very large section of the book probably. There might be better routes to pursue though like some of the resources in this question.
If there are no acceleration (x,y) =(xstart+ vx*time ,ystart + vy*time) and speed remain the same, and it is not related to the radius
Since the velocity is constant you will have an angular velocity of omega = vel / radius. You will obtain how many radians you ball will move per second over its circular path.
To get the position at time t you just have to exploit polar coordinates:
x = x_center + sin( 3/2*PI + omega*t)*radius
y = y_center + cos( 3/2*PI + omega*t)*radius
This because you start from bottom point of the circle (so its 3/2*PI) plus how many radiants you move every second (we obtained it from tangential velocity). All multiplied for the radius, otherwise you will consider a unity circle.
EDIT: Since you wonder how to find a position of an object that is subject to many different forces I can tell you that usually a physical engine doesn't care about finding equations of moving objects. It just applies forces to objects considering their intended motions (like your circular one) or environmental factors (like gravity or friction) and calculates coordinates step by step by applying forces and using an integrator to see the results.
Ignoring friction, the forces on the ball are gravity and the track.
First, there are two main cases - is the velocity enough for the ball to loop-the-loop or not:
initial energy = 1/2 m v² = 0.5 * 5 * 200 * 200
potential energy = m g h = 5 * 9.8 * 20
so it will go round the whole loop.
Initially the ball is at the bottom of the loop, theta = 0
The acceleration on the ball is the component of g along the track
a = g⋅sin theta
The distance travelled is theta * radius. It is also the double integral of acceleration against time.
theta ⋅ radius = double integral of acceleration against time
Integrating acceleration once gives velocity, integrating velocity gives distance.
so solve this for t:
theta ⋅ r = ∫(∫ g⋅sin theta.dt).dt
then your x and y are trivial functions of theta.
Whether you solve it analytically or numerically is up to you.
With dynamic friction, friction is usually proportional to the normal force on the bodies. So this will equal the centripetal force - proportional to the square of the angular velocity, and the component of gravity normal to the track (g sin theta)
You didn't tell anything about how you want your velocity to change. Do you have any friction model? If there is no friction, then the formulas are simple:
length = velocity*t
x = sin(length)*radius
y = -cos(length)*radius
If the velocity is changing, then you have to change length to something like
length = integral over dt[0..t] (velocity dt)
The only thing I wanted to add is the if this is real ball (sphere) with mass 5kg then it must have a diameter dia=(6*m/(PI*rho))^(1/3) where rho is the density of the material. For steel (rho=7680) the diameter is dia=0.1075 meters. Therefore the pitch radius (radius at which the center of gravity of the ball rides on) is equal to R=10-(dia/2) or R=9.9466 meters.
The problem gets a little more complex when friction is included. For one you have to consider the direction of friction (assuming dry friction theory). That depends on the amount the ball rotates in its axis and that depends on moment of inertia of the ball.
When you do the simulation you might want to monitor the total kinetic energy + the total potential energy and make sure your are not adding energy to the system (or taking away). [Don't forget to include the rotational component for the kinetic energy]
Get a standard book on dynamics, and I am sure a similar problem is already described in the book.I would recommend "Vector Mechanic for Engineers - Dynamics".