Differential Equation for Projectile Motion with Air Drag

[ChanYoung Park]

Hi! I was wondering how I could come up with a differential equation for projectile motion on a 2D plane when air resistance is not negligible. I'm trying to guess the position of a projected ball at a certain time period by approximating the coordinates using the Euler's method.
Here, I would assume that I know the angle, mass, and initial force; θ, m and F are constants. What I modeled so far are the equations below, but I'm pretty sure that they're completely wrong. And I do not know how I could decompose x(t) and y(t) in terms of the time.$$F_x= Fcosθ - mg - kv_{x}^2$$
$$F_y= Fsinθ - mg - kv_{y}^2$$
$$\frac{dx}{dt} = v_x$$
$$\frac{dy}{dt} = v_y$$
$$\frac{d^{2}x}{dt^2} = a_x$$
$$\frac{d^{2}y}{dt^2} = a_y$$Please help!

 

[willem2]

I don't think the initial force is relevant here. This kind of problems starts after the projectile has an initial velocity, and the only forces present will be gravity and air resistance. Only the initial velocity (and position) matter. I have no idea at all what you mean with decomposing x(t) and y(t) in terms of the time.
With eulers method you start out with a position x(t), y(t) and velocity v_x(t), v_y(t), and you compute:

x(t+dt) = x(t) + dt * v(t)
y(t+dt) = x(t) + dt * v(t)
v_x(t+dt) = v_x(t) + a_x
v_y(t+dt) = v_y(t) + a_y
with the timestep dt something like 0.01s. repeat as often as needed.
To find a, you have to use the forces in x and y direction.

 

[phyzguy]

I think there are several things wrong here. First, gravity only acts downward, so it should not be present in both equations. Assuming the y-axis is up-down, then the mg term should only be in the Fy equation. Second, the air resistance depends on the square of the total velocity, and acts in the direction of the current velocity, so it should be decomposed as follows:
[tex]
F_x = k (v_x^2 + v_y^2)\frac{v_x}{|v|} = k (v_x^2 + v_y^2)\frac{v_x}{\sqrt{v_x^2 + v_y^2}} = k v_x\sqrt{v_x^2 + v_y^2} [/tex]

and similarly for Fy. This is the same as what you wrote if vx or vy is zero, but otherwise not.

 

[ChanYoung Park]

I don't think the initial force is relevant here. This kind of problems starts after the projectile has an initial velocity, and the only forces present will be gravity and air resistance. Only the initial velocity (and position) matter. I have no idea at all what you mean with decomposing x(t) and y(t) in terms of the time.
With eulers method you start out with a position x(t), y(t) and velocity v_x(t), v_y(t), and you compute:

x(t+dt) = x(t) + dt * v(t)
y(t+dt) = x(t) + dt * v(t)
v_x(t+dt) = v_x(t) + a_x
v_y(t+dt) = v_y(t) + a_y
with the timestep dt something like 0.01s. repeat as often as needed.
To find a, you have to use the forces in x and y direction.

Thanks for your reply! So the way I can get the x coordinate of the ball at t given the intial velocity v_x is
#1. to find the velocity at time t+dt using:
$$v_{x}(t+dt)=v_{x}(t)+a_{x}\cdot{dt}$$
#2. then to get the x coordinate at time t by using:
$$x(t+dt)=x(t)+v_{x}(t)\cdot{dt}$$
and to find a_x, I would need to use force equation which is...
$$F_{x}=ma_{x}=kv_{x}\sqrt{v_{x}^{2}+v_{y}^{2}}$$

...but as I think... is #1 necessary??

 

[ChanYoung Park]

I think there are several things wrong here. First, gravity only acts downward, so it should not be present in both equations. Assuming the y-axis is up-down, then the mg term should only be in the Fy equation. Second, the air resistance depends on the square of the total velocity, and acts in the direction of the current velocity, so it should be decomposed as follows:
[tex]
F_x = k (v_x^2 + v_y^2)\frac{v_x}{|v|} = k (v_x^2 + v_y^2)\frac{v_x}{\sqrt{v_x^2 + v_y^2}} = k v_x\sqrt{v_x^2 + v_y^2} [/tex]

and similarly for Fy. This is the same as what you wrote if vx or vy is zero, but otherwise not.

Thanks for corrections! But I don't quite get what's happening with v_x and v_y.
Could you please explain me how it's being done?

 

[phyzguy]

Do you know how to decompose a vector into its components? What is the length of the vector v in terms of vx and vy, and how much of v points along x and how much along y?

 

[vanhees71]

I'd also say the EoM is
$$m \ddot{\vec{x}}=-m \vec{g} - C \vec{v} |\vec{v}|,$$
where ##m## is the mass of the particle, ##\vec{g}## the (constant) gravitational acceleration close to the surface of the Earth, and ##C## a constant related to the ##C_w## drag coefficient by
$$C=\frac{1}{2} C_w \varrho A.$$

 

[ChanYoung Park]

Do you know how to decompose a vector into its components? What is the length of the vector v in terms of vx and vy, and how much of v points along x and how much along y?

Though what is the reason for decomposing the vector? Can't it be just F_x=kv_x²?

 

[phyzguy]

Though what is the reason for decomposing the vector? Can't it be just F_x=kv_x²?

The force due to air resistance is always in the (negative) direction of the velocity vector. Quoting from Wikipedia, "In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid." With your proposed equation, this is not the case, and the air resistance would cause the projectile to change direction, not just slow down. This can happen if the projectile has a non-spherical shape, but I doubt if this is the case you want to analyze. As vanhees71 said, another way to write the drag force which ensures that it is in the same direction as the velocity vector, is [itex], F_D = -C \vec{v} |\vec{v}| [/itex]