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

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SUMMARY

The forum discussion centers on a Python simulation of projectile motion with air resistance, specifically addressing the incorrect trajectory results produced by the code. The user implemented the simulation using libraries such as NumPy, Matplotlib, and SciPy's solve_ivp function. The primary issue identified was a coding error in calculating the velocity vector, which led to incorrect drag force computations. The correct approach involves using np.linalg.norm(vel) to accurately compute the magnitude of the velocity vector.

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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:
[CODE lang="python" highlight="11-17, 33,34, 23, 25-27"]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()
[/CODE]

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 y(x), it is easy to prove it should be a concave function.
The model is \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
By the chain rule, \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)
Therefore, at least as long as \dot{y} \geq 0, 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: \frac{d^2 y}{d^2 x} (x(t)) = -\frac{1}{2 \dot{x}^2} \leq 0

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:
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fisicist said:
Therefore, at least as long as \dot{y} \geq 0, this should be a concave function.
Hmm, I am not seeing that, where do you get that condition from?
 
Dale said:
Hmm, I am not seeing that, where do you get that condition from?
1/\dot{x}^2 \geq 0 manifestly, so it remains to check that -2 c \dot{y} \sqrt{\dot{x}^2+\dot{y}^2}-\frac 1 2 \leq 0 in order for d^2 y/ d x^2 \leq 0 (i.e. concavity). But -2 c \sqrt{\dot{x}^2 +\dot{y}^2} is always negative, so if \dot{y} \geq 0, everything is negative.
 
What about the 1/2?
 
Dale said:
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:
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
d^2 y / dx^2 = - \frac{1}{2 \dot{x}^2}
Still, it's concave. In fact, it is *always* concave, even if I throw the stone downwards.
 
fisicist said:
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?
 
A.T. said:
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 \operatorname{epi}f := \lbrace (x, y) : y \geq f(x) \rbrace is a convex set. For a twice differentiable function, this is equivalent to f'' \geq 0. A function f is concave if -f is convex.
 
fisicist said:
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 \operatorname{epi}f := \lbrace (x, y) : y \geq f(x) \rbrace is a convex set. For a twice differentiable function, this is equivalent to f'' \geq 0. A function f is concave if -f is convex.
Ok, thanks. I would suggest coding a simple Euler integration. It should never bend upwards.
 
  • #10
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 \vert \dot{\mathbf x} = \sqrt{\dot{x}^2+\dot{y}^2}, as I thought it would, but (\vert \dot x \vert, \vert \dot y \vert )^T, and
Python:
np.absolute(vel) * vel
does not give \dot{\mathbf{x}} \vert \dot{\mathbf{x}} \vert, as I thought it would, but (\dot{x} \vert \dot{x} \vert, \dot{y} \vert \dot{y} \vert )^T.

The correct way to do it is
Python:
np.sqrt(vel.dot(vel)) * vel
 
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Likes jim mcnamara and Dale
  • #11
fisicist said:
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 \vert \dot{\mathbf x} = \sqrt{\dot{x}^2+\dot{y}^2}, as I thought it would, but (\vert \dot x \vert, \vert \dot y \vert )^T, and
Python:
np.absolute(vel) * vel
does not give \dot{\mathbf{x}} \vert \dot{\mathbf{x}} \vert, as I thought it would, but (\dot{x} \vert \dot{x} \vert, \dot{y} \vert \dot{y} \vert )^T.

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
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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Likes William Crawford
  • #13
Run your sim on a baseball and see if 40 deg gives the maximum range.
 
  • #14
fisicist said:
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
fisicist said:
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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