Why is my simulation of projectile trajectory with air resistance wrong?

  • Thread starter fisicist
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  • #1
fisicist
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Hey!

This started very harmless... A friend and I were throwing stones in a lake. Mine didn't get very far, he was teasing me "What was the ideal angle again?". Of course, I know it should be 45°. I replied in jest: "That's because I'm considering air resistance!" Then we had a discussion what the ideal angle should be with air resistance considered.
So I wrote a simulation in Python. It was just for fun... But I encountered an issue that I don't quite understand: The simulated trajectory is wrong. I hope someone can help me understand where this is coming from, and perhaps give me some hint how to fix it.

My code is:
Python:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp

c = 7
v = 5

# f gives phase space velocity
# z = [x, y, v_x, v_y]
# f(t, z) = [v_x, v_y, a_x, a_y]
def f(t, z):
    global c
    vel = z[2:]
    friction = -c * np.absolute(vel) * vel
    gravity = [0, -0.5]
    accel = friction + gravity
    return np.concatenate((vel, accel))

def hit_ground(t, z): return z[1]
hit_ground.terminal = True
hit_ground.direction = -1

t_eval = np.linspace(0, 100, 10000)

def v_init(theta):
    global v
    return v * np.array([np.cos(theta), np.sin(theta)])

thetas = np.linspace(0.0, np.pi/2, 181)
dists = np.array([])
plt.subplot('211')
for counter,theta in enumerate(thetas):
    init = np.concatenate((np.array([0., 0.]), v_init(theta)))
    sol = solve_ivp(f, [0, 100], init, t_eval = t_eval, events=hit_ground)
    if counter % 40 == 0:
        plt.plot(sol.y[0,:], sol.y[1,:], label='{:4.0f} °'.format(theta*180/np.pi))
        plt.gca().quiver(sol.y[0,::70], sol.y[1,::70], 1000*sol.y[2,::70], 1000*sol.y[3,::70])
    dists = np.append(dists, [sol.y[0,-1]])
#plt.gca().set_aspect('equal', adjustable='box')
plt.legend(loc='upper left')
plt.xlabel('$x$')
plt.ylabel('$y$')

plt.subplot('212')
plt.plot(thetas*180/np.pi, dists, 'r-')
best_angle = thetas[np.argmax(dists)]
plt.gca().axvline(x = best_angle*180/ np.pi)
print('{:3.1f}'.format(best_angle*180/ np.pi))
plt.xlabel('angle [°]')
plt.ylabel('distance')

plt.show()

Result:
angles.png


This shouldn't happen... As you can see, the curves bend upward, i.e. are not concave. Even without knowing the exact solution [itex]y(x)[/itex], it is easy to prove it should be a concave function.
The model is [itex]\ddot{x} = - c \dot{x}\sqrt{\dot{x}^2+\dot{y}^2}, \ddot{y} = - c \dot{y} \sqrt{\dot{x}^2+\dot{y}^2} - \frac 1 2 [/itex]
By the chain rule, [itex]\frac{d^2 y}{d^2 x} (x(t)) = \frac{\ddot{y}}{\dot{x}^2} - \dot{y}\frac{\ddot{x}}{\dot{x}^3} = \frac{1}{\dot{x}^2} \left( -2 c \dot y \sqrt{\dot{x}^2+\dot{y}^2} - \frac 1 2 \right)[/itex]
Therefore, at least as long as [itex]\dot{y} \geq 0[/itex], this should be a concave function. If I through it downwards with high speed, it might get convex there, but not as long as the motion is upward.

CORRECTION: By the chain rule: [itex]\frac{d^2 y}{d^2 x} (x(t)) = -\frac{1}{2 \dot{x}^2} \leq 0[/itex]

So, what is wrong with the simulation? I know the time subdivision is not very high. But that shouldn't matter, because also in the discretized model, the velocity vector should be shortened and then get a downward contribution in every step. That's not what I see in the plot.

Any ideas, anyone?
 
Last edited:

Answers and Replies

  • #2
33,711
11,305
Therefore, at least as long as [itex]\dot{y} \geq 0[/itex], this should be a concave function.
Hmm, I am not seeing that, where do you get that condition from?
 
  • #3
fisicist
46
7
Hmm, I am not seeing that, where do you get that condition from?
[itex]1/\dot{x}^2 \geq 0[/itex] manifestly, so it remains to check that [itex]-2 c \dot{y} \sqrt{\dot{x}^2+\dot{y}^2}-\frac 1 2 \leq 0 [/itex] in order for [itex]d^2 y/ d x^2 \leq 0[/itex] (i.e. concavity). But [itex]-2 c \sqrt{\dot{x}^2 +\dot{y}^2}[/itex] is always negative, so if [itex]\dot{y} \geq 0[/itex], everything is negative.
 
  • #5
A.T.
Science Advisor
11,670
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Hmm, I am not seeing that, where do you get that condition from?
An informal argument for concavity: The drag is always tangential so it doesn't change the direction, while gravity is always bending the path down.

Looks like some lift sneaked in somewhere.
 
Last edited:
  • #6
fisicist
46
7
Hey sorry, I've made a sign error. The two friction terms compensate each other, not add up. Actually not surprising: Without gravity, motion is along a straight line.
I will correct it. Correct is
[itex] d^2 y / dx^2 = - \frac{1}{2 \dot{x}^2} [/itex]
Still, it's concave. In fact, it is *always* concave, even if I throw the stone downwards.
 
  • #7
A.T.
Science Advisor
11,670
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As you can see, the curves bend upward, i.e. are not concave.
I would call bending upward and then downward "concave". Bending only downward would be "convex", and what you would expect here.

Have you tried coding a simple Euler integration yourself?
 
  • #8
fisicist
46
7
I would call bending upward and then downward "concave". Bending only downward would be "convex", and what you would expect here.

Have you tried coding a simple Euler integration yourself?

The standard definition used in mathematics (in fields such as calculus of variations, 'convex analysis' or 'convex optimization') is that a function is convex if its epigraph [itex]\operatorname{epi}f := \lbrace (x, y) : y \geq f(x) \rbrace[/itex] is a convex set. For a twice differentiable function, this is equivalent to [itex]f'' \geq 0[/itex]. A function [itex]f[/itex] is concave if [itex]-f[/itex] is convex.
 
  • #9
A.T.
Science Advisor
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The standard definition used in mathematics (in fields such as calculus of variations, 'convex analysis' or 'convex optimization') is that a function is convex if its epigraph [itex]\operatorname{epi}f := \lbrace (x, y) : y \geq f(x) \rbrace[/itex] is a convex set. For a twice differentiable function, this is equivalent to [itex]f'' \geq 0[/itex]. A function [itex]f[/itex] is concave if [itex]-f[/itex] is convex.
Ok, thanks. I would suggest coding a simple Euler integration. It should never bend upwards.
 
  • #10
fisicist
46
7
Thank you to everyone who replied here!
I found the error. It's not a mathematical or numerical curiosity, but a simple coding error. Namely,
Python:
np.absolute(vel)
does not give [itex]\vert \dot{\mathbf x} = \sqrt{\dot{x}^2+\dot{y}^2}[/itex], as I thought it would, but [itex](\vert \dot x \vert, \vert \dot y \vert )^T[/itex], and
Python:
np.absolute(vel) * vel
does not give [itex]\dot{\mathbf{x}} \vert \dot{\mathbf{x}} \vert[/itex], as I thought it would, but [itex](\dot{x} \vert \dot{x} \vert, \dot{y} \vert \dot{y} \vert )^T[/itex].

The correct way to do it is
Python:
np.sqrt(vel.dot(vel)) * vel
 
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  • #11
William Crawford
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Thank you to everyone who replied here!
I found the error. It's not a mathematical or numerical curiosity, but a simple coding error. Namely,
Python:
np.absolute(vel)
does not give [itex]\vert \dot{\mathbf x} = \sqrt{\dot{x}^2+\dot{y}^2}[/itex], as I thought it would, but [itex](\vert \dot x \vert, \vert \dot y \vert )^T[/itex], and
Python:
np.absolute(vel) * vel
does not give [itex]\dot{\mathbf{x}} \vert \dot{\mathbf{x}} \vert[/itex], as I thought it would, but [itex](\dot{x} \vert \dot{x} \vert, \dot{y} \vert \dot{y} \vert )^T[/itex].

The correct way to do it is
Python:
np.sqrt(vel.dot(vel)) * vel

Congratulation! So what is the ideal angle when you include air resistance?
 
  • #12
fisicist
46
7
For anyone interested in the physics: Qualitatively, the result is similar.
angles-corr.png

Qualitatively, the best angle decreases rapidly the higher the friction is:
bestangles.png
 
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  • #13
Mister T
Science Advisor
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Run your sim on a baseball and see if 40 deg gives the maximum range.
 
  • #14
A.T.
Science Advisor
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The correct way to do it is
Python:
np.sqrt(vel.dot(vel)) * vel
Or:
Python:
np.linalg.norm(vel) * vel
https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.norm.html

For anyone interested in the physics: Qualitatively, the result is similar.
View attachment 247485
Qualitatively, the best angle decreases rapidly the higher the friction is:
View attachment 247487
Regarding your stone throwing contest:

You could plot the optimal angle as function of the object's density for a given realistic object shape/size and throw speed. It could be that for stones the angle is pretty close to 45° at the speed you can throw, while for light balls (like tennis) it makes sense to aim lower.

Bio-mechanics complicates matters because you won't achieve the same throw speeds for any angle.

Finally, your friend might simply throw much faster, and thus further, even at sub optimal angles.
 
Last edited:

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