I want an object to smoothly stop at a specified point. I have the initial velocity, V0, of the object, the end velocity (which is set to 0), V1, and the distance, d, required to reach the destination. V1 and V0 are measured in radians per frame, and d is also measured in radians.
I've tried using the following formula:
a = (V0*V0 - V1*V1) / (2.0 * d);
But it always seems to overshoot its target.
Edit:
Essentially, I have a wheel that starts spinning at an initial velocity of V0. The wheel has to do a certain number of spins before stopping at a specified location. The distance, d is the amount of radians required to do the specified amount of spins and stop at a specified location. The velocity is number of radians per frame.
We have enough information now for an educated guess.
The formula is correct, but only in the ideal case, where the length of an iteration time is very small. My guess is that in your loop you are updating position before velocity, so that for that time period the wheel can advance with undiminished velocity, and you overshoot. (If you updated velocity before position, you would undershoot.)
You can either make your frames shorter, which will make the overshoot less severe, or you can modify the formula to eliminate it:
a = (V0*V0) / (2.0 * d - V0*tdelta);
where tdelta is the length of time of a single frame. (I've assumed V1=0.)
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.
To make animation independent of frame rate, is it necessary to multiply the delta value by both velocity and acceleration?
// Multiply both acceleration and velocity by delta?
vVelocity.x += vAcceleration.x * delta;
vVelocity.y += vAcceleration.y * delta;
position.x += vVelocity.x * delta;
position.y += vVelocity.y * delta;
Should I apply delta to the velocity only and not acceleration?
Assuming your "delta" is the amount of time passed since last update:
Short answer: yes.
Long answer:
One way to check this sort of thing is to see if the units work out. It's not guaranteed, but usually if your units work out, then you've figured things correctly.
Velocity measures distance per unit time, and delta is time. So velocity times delta is (picking arbitrary units meters and seconds) (m/s) * s = m. So you can see that velocity times delta does create a distance, so that appears reasonable for position.
Acceleration measures velocity per unit time, that is, with the same units (m/s)/s. So, acceleration times delta is ((m/s)/s) * s = m/s. Looks like a velocity to me. We're good!
Yes, it is necessary to involve delta with both the velocity and the acceleration. They're both properties that are defined with respect to time (m/s for one, m/s/s for the other - units may vary), so delta should be used whenever they have to change non-instantaneously.
Does anyone know of a tutorial that would deal with gravitational pull of two objects? Eg. a satellite being drawn to the moon (and possibly sling shot past it).
I have a small Java game that I am working on and I would like to implement his feature in it.
I have the formula for gravitational attraction between two bodies, but when I try to use it in my game, nothing happens?
There are two object on the screen, one of which will always be stationary while the other one moves in a straight line at a constant speed until it comes within the detection range of the stationary object. At which point it should be drawn to the stationary object.
First I calculate the distance between the two objects, and depending on their mass and this distance, I update the x and y coordinates.
But like I said, nothing happens. Am I not implementing the formula correctly?
I have included some code to show what I have so far.
This is the instance when the particle collides with the gates detection range, and should start being pulled towards it
for (int i = 0; i < particle.length; i++)
{
// **************************************************************************************************
// GATE COLLISION
// **************************************************************************************************
// Getting the instance when a Particle collides with a Gate
if (getDistanceBetweenObjects(gate.getX(), particle[i].getX(), gate.getY(), particle[i].getY()) <=
sumOfRadii(particle[i].getRadius(), barrier.getRadius()))
{
particle[i].calcGravPull(particle[i].getMass(), barrier.getMass(),
getDistanceBetweenObjects(gate.getX(), particle[i].getX(), gate.getY(), particle[i].getY()));
}
And the method in my Particle class to do the movement
// Calculate the gravitational pull between objects
public void calcGravPull(int mass1, int mass2, double distBetweenObjects)
{
double gravityPull;
gravityPull = GRAV_CONSTANT * ((mass1 * mass2) / (distBetweenObjects * distBetweenObjects));
x += gravityPull;
y += gravityPull;
}
Your formula has problems. You're calculating the gravitational force, and then applying it as if it were an acceleration. Acceleration is force divided by mass, so you need to divide the force by the small object's mass. Therefore, GRAV_CONSTANT * ((mass1) / (distBetweenObjects * distBetweenObjects)) is the formula for acceleration of mass2.
Then you're using it as if it were a positional adjustment, not a velocity adjustment (which an acceleration is). Keep track of the velocity of the moving mass, use that to adjust its position, and use the acceleration to change that velocity.
Finally, you're using acceleration as a scalar when it's really a vector. Calculate the angle from the moving mass to the stationary mass, and if you're representing it as angle from the positive x-axis multiply the x acceleration by the cosine of the angle, and the y acceleration by the sine of the angle.
That will give you a correct representation of gravity.
If it does nothing, check the coordinates to see what is happening. Make sure the stationary mass is large enough to have an effect. Gravity is a very weak force, and you'll have no significant effect with much smaller than a planetary mass.
Also, make sure you're using the correct gravitational constant for the units you're using. The constant you find in the books is for the MKS system - meters, kilograms, and seconds. If you're using kilometers as units of length, you need to multiply the constant by a million, or alternately multiply the length by a thousand before plugging it into the formula.
Your algorithm is correct. Probably the gravitational pull you compute is too small to be seen. I'd remove GRAV_CONSTANT and try again.
BTW if you can gain a bit of speed moving the result of getDistanceBetweenObjects() in a temporary variable.
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".
I've tried the typical physics equations for this but none of them really work because the equations deal with constant acceleration and mine will need to change to work correctly. Basically I have a car that can be going at a large range of speeds and needs to slow down and stop over a given distance and time as it reaches the end of its path.
So, I have:
V0, or the current speed
Vf, or the speed I want to reach (typically 0)
t, or the amount of time I want to take to reach the end of my path
d, or the distance I want to go as I change from V0 to Vf
I want to calculate
a, or the acceleration needed to go from V0 to Vf
The reason this becomes a programming-specific question is because a needs to be recalculated every single timestep as the car keeps stopping. So, V0 constantly is changed to be V0 from last timestep plus the a that was calculated last timestep. So essentially it will start stopping slowly then will eventually stop more abruptly, sort of like a car in real life.
EDITS:
All right, thanks for the great responses. A lot of what I needed was just some help thinking about this. Let me be more specific now that I've got some more ideas from you all:
I have a car c that is 64 pixels from its destination, so d=64. It is driving at 2 pixels per timestep, where a timestep is 1/60 of a second. I want to find the acceleration a that will bring it to a speed of 0.2 pixels per timestep by the time it has traveled d.
d = 64 //distance
V0 = 2 //initial velocity (in ppt)
Vf = 0.2 //final velocity (in ppt)
Also because this happens in a game loop, a variable delta is passed through to each action, which is the multiple of 1/60s that the last timestep took. In other words, if it took 1/60s, then delta is 1.0, if it took 1/30s, then delta is 0.5. Before acceleration is actually applied, it is multiplied by this delta value. Similarly, before the car moves again its velocity is multiplied by the delta value. This is pretty standard stuff, but it might be what is causing problems with my calculations.
Linear acceleration a for a distance d going from a starting speed Vi to a final speed Vf:
a = (Vf*Vf - Vi*Vi)/(2 * d)
EDIT:
After your edit, let me try and gauge what you need...
If you take this formula and insert your numbers, you get a constant acceleration of -0,0309375. Now, let's keep calling this result 'a'.
What you need between timestamps (frames?) is not actually the acceleration, but new location of the vehicle, right? So you use the following formula:
Sd = Vi * t + 0.5 * t * t * a
where Sd is the current distance from the start position at current frame/moment/sum_of_deltas, Vi is the starting speed, and t is the time since the start.
With this, your decceleration is constant, but even if it is linear, your speed will accomodate to your constraints.
If you want a non-linear decceleration, you could find some non-linear interpolation method, and interpolate not acceleration, but simply position between two points.
location = non_linear_function(time);
The four constraints you give are one too many for a linear system (one with constant acceleration), where any three of the variables would suffice to compute the acceleration and thereby determine the fourth variables. However, the system is way under-specified for a completely general nonlinear system -- there may be uncountably infinite ways to change acceleration over time while satisfying all the constraints as given. Can you perhaps specify better along what kind of curve acceleration should change over time?
Using 0 index to mean "at the start", 1 to mean "at the end", and D for Delta to mean "variation", given a linearly changing acceleration
a(t) = a0 + t * (a1-a0)/Dt
where a0 and a1 are the two parameters we want to compute to satisfy all the various constraints, I compute (if there's been no misstep, as I did it all by hand):
DV = Dt * (a0+a1)/2
Ds = Dt * (V0 + ((a1-a0)/6 + a0/2) * Dt)
Given DV, Dt and Ds are all given, this leaves 2 linear equations in the unknowns a0 and a1 so you can solve for these (but I'm leaving things in this form to make it easier to double check on my derivations!!!).
If you're applying the proper formulas at every step to compute changes in space and velocity, it should make no difference whether you compute a0 and a1 once and for all or recompute them at every step based on the remaining Dt, Ds and DV.
If you're trying to simulate a time-dependent acceleration in your equations, it just means that you should assume that. You have to integrate F = ma along with the acceleration equations, that's all. If acceleration isn't constant, you just have to solve a system of equations instead of just one.
So now it's really three vector equations that you have to integrate simultaneously: one for each component of displacement, velocity, and acceleration, or nine equations in total. The force as a function of time will be an input for your problem.
If you're assuming 1D motion you're down to three simultaneous equations. The ones for velocity and displacement are both pretty easy.
In real life, a car's stopping ability depends on the pressure on the brake pedal, any engine braking that's going on, surface conditions, and such: also, there's that "grab" at the end when the car really stops. Modeling that is complicated, and you're unlikely to find good answers on a programming website. Find some automotive engineers.
Aside from that, I don't know what you're asking for. Are you trying to determine a braking schedule? As in there's a certain amount of deceleration while coasting, and then applying the brake? In real driving, the time is not usually considered in these maneuvers, but rather the distance.
As far as I can tell, your problem is that you aren't asking for anything specific, which suggests that you really haven't figured out what you actually want. If you'd provide a sample use for this, we could probably help you. As it is, you've provided the bare bones of a problem that is either overdetermined or way underconstrained, and there's really nothing we can do with that.
if you need to go from 10m/s to 0m/s in 1m with linear acceleration you need 2 equations.
first find the time (t) it takes to stop.
v0 = initial velocity
vf = final velocity
x0 = initial displacement
xf = final displacement
a = constant linear acceleration
(xf-x0)=.5*(v0-vf)*t
t=2*(xf-x0)/(v0-vf)
t=2*(1m-0m)/(10m/s-0m/s)
t=.2seconds
next to calculate the linear acceleration between x0 & xf
(xf-x0)=(v0-vf)*t+.5*a*t^2
(1m-0m)=(10m/s-0m/s)*(.2s)+.5*a*((.2s)^2)
1m=(10m/s)*(.2s)+.5*a*(.04s^2)
1m=2m+a*(.02s^2)
-1m=a*(.02s^2)
a=-1m/(.02s^2)
a=-50m/s^2
in terms of gravity (g's)
a=(-50m/s^2)/(9.8m/s^2)
a=5.1g over the .2 seconds from 0m to 10m
Problem is either overconstrained or underconstrained (a is not constant? is there a maximum a?) or ambiguous.
Simplest formula would be a=(Vf-V0)/t
Edit: if time is not constrained, and distance s is constrained, and acceleration is constant, then the relevant formulae are s = (Vf+V0)/2 * t, t=(Vf-V0)/a which simplifies to a = (Vf2 - V02) / (2s).