Trajectory Calculator (W/ Drag)

[gnunes]

Yes, drag is complicated, but also important. If your launcher spins the ball, then lift is important too.

If you don't want this complication, then your best approach is empirical. Get a lot of data on distance vs. your launch parameters and make a table. Then, the robot can use interpolation to work the other way: given a distance, it can get values for the launch parameters. If you are varying two things (launch angle and ball velocity), then this could get complicated. You will have several tables, and will need an algorithm to determine which one to use.

Even though I wrote the trajectory calculator, our team is going to use the interpolation approach!

[engunneer]

Originally Posted by gnunes

Get a lot of data on distance vs. your launch parameters and make a table. Then, the robot can use interpolation to work the other way: given a distance, it can get values for the launch parameters.

This is pretty much our plan.

[Hinfoiltat]

I would appreciate a video of this in action.

[Brandon_L]

I see a formula for finding the angle to launch at with a constant velocity, but I can't find the opposite (Velocity to launch at with a constant angle) anywhere for the life of me. Does anyone know/have it? I've been going through my kinematic equations for the past 6 hours (I kid you not) and have nothing

[Ether]

Originally Posted by Brandon_L

I see a formula for finding the angle to launch at with a constant velocity, but I can't find the opposite (Velocity to launch at with a constant angle) anywhere for the life of me. Does anyone know/have it? I've been going through my kinematic equations for the past 6 hours (I kid you not) and have nothing

Let

d = horizontal distance (in feet) from launch point to hoop

h = vertical height (in feet) of hoop above the launch point

a = angle of launch (radians) above the horizontal

v = initial speed of ball (in ft/sec) as it leaves the launcher

Then the no-drag, no magnus formula for speed is:

v = 4*d/(cos(a)*sqrt(d*tan(a)-h))

[Lalaland1125]

Originally Posted by Brandon_L

I see a formula for finding the angle to launch at with a constant velocity, but I can't find the opposite (Velocity to launch at with a constant angle) anywhere for the life of me. Does anyone know/have it? I've been going through my kinematic equations for the past 6 hours (I kid you not) and have nothing

Our plan is to simply do the simple drag-free formula for an initial guess, and then slowly increase the velocity till the non-reversible drag including formula "hits" the target.

The computers are definitely fast enough for this, but who knows if the ball actually acts like its supposed to.

[gnunes]

Originally Posted by Lalaland1125

Our plan is to...slowly increase the velocity till the non-reversible drag including formula "hits" the target.

The problem is that there is no "formula" for the case where drag is included. If you are using LabView, then you could take my trajectory calculator (see post #11) and turn it into a sub-VI that would find the distance for you (just cut out all the graphing stuff which you won't need).

I would recommend using the trajectory calculator "off-line" to calculate the distance for, say 100 or 200 launch velocities between your fastest and slowest launch speeds. Then store that table of distance vs launch speed in your competition code, and use interpolation to find the velocity you want in a single pass. Much, much faster.

[Ether]

Originally Posted by Lalaland1125

Our plan is to simply do the simple drag-free formula for an initial guess, and then slowly increase the velocity till the non-reversible drag including formula "hits" the target.

Originally Posted by gnunes

The problem is that there is no "formula" for the case where drag is included.

For a drag model where drag is either linear or quadratic in velocity, I believe there is a closed form solution* for height, given distance, launch angle, drag coefficient, and launch velocity. So you could vary the launch velocity and calculate the height until the height is the desired value.

I'm not saying that's the best way to do it, but it could be done.


* I haven't checked it. It's rather ugly.

[Ether]

Originally Posted by Ether

* I haven't checked it. It's rather ugly.

Here it is, if anyone wants to try it or check my math. I'm sure this could be simplified with a bit of work. Maybe even solved explicitly for Vo?

h=log(1-Vo^2*K/g)/(2*K)-log(sec(sqrt(-g)*sqrt(K)*((%e^(d*K)-1)/(cos(theta)*Vo*K)-atan((Vo*sqrt(K))/sqrt(-g))/(sqrt(-g)*sqrt(K)))))/K;

See attached PDF for derivation.

[Ether]

@gnunes: can you run your trajectory program with the following parameters, and post height vs distance data (say, every foot) and a PNG of a plot of height (above the launch point, in feet) vs distance (horizontal distance from the launch point) with the following parameters: Vo=23 feet/sec, theta=45 degrees, g=-32 ft/sec², drag = -0.03*V² ft/sec², and no spin?

[Lalaland1125]

I ran your numbers through my dead simple approximator(code at bottom) and got a different result(image + source code is attached)

I am fairly sure that your result cannot be correct due to running it through the standard projectile motion equations.

Assuming no loss of energy, the ball will follow the equation: .5*Vy^2 = g * h.
(mgh = 1/2 * m * v^2)

Plugging this into wolfram alpha(http://www.wolframalpha.com/input/?i=%28%2823+feet+per+second+in+meters+*+sin%2845+d egrees%29%29+**+2++*+.5+%29%2F9.8+meters+per+secon d+squared) ( using meters because I am more used to them) gets me 4.113 ft as the max possible height. Your graph shows the ball approaching 7 m.

(When I set the drag to 0 on my code the height reaches a max of about 4ft and change)

Oh, and the code attachment shows what I mean by trial and error. Look at findBestCurveWithin.

void calcY(double& x,double& y, double &xVel, double& yVel)
{
   double oldxVel = xVel;
   double oldyVel = yVel;


   xVel += drag * xVel * xVel * timeIncrement;
   yVel += (g + drag * yVel * yVel) * timeIncrement;

   x += oldxVel * timeIncrement + .5 * timeIncrement * (xVel - oldxVel);
   y += oldyVel * timeIncrement + .5 * timeIncrement * (yVel - oldyVel);
}

QVector<QPointF> findCurve(double velocity, double angle, double targetX)
{
   QVector<QPointF> result;

   double rads = angle * M_PI/180;
   double xVel = velocity * cos(rads);
   double yVel = velocity * sin(rads);
   double x = 0;
   double y = 0;

   do
   {
      calcY(x,y,xVel,yVel);
      result.push_back(QPointF(x,y));
   }
   while (fabs(x - targetX) > .01);


   return result;
}

[Ether]

Originally Posted by Lalaland1125

I ran your numbers through my dead simple approximator(code at bottom) and got a different result(image + source code is attached)

Thank you. I have should have caught that.

x3: Vo =subst(0,t,rhs(x2));  <= a small typo on Page 2
x3: Vyo=subst(0,t,rhs(x2));  <= should have been

I'll see if a closed-form solution is still possible with this correction.