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In summary: You can solve it by using a couple of techniques, but I'm not really an expert in that area. So, you might want to consult a more experienced person for that. In summary, integrating both sides of the equation won't help you solve the problem, and you might want to consult someone more experienced in solving differential equations.

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Note this post only describes the math for the y-axis. (The x-axis movement would not involve g).

This can be integrated directly to determine velocity versus time, but apparently not position versus time. Wiki article with the answers, but not the derivation:

http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object

http://en.wikipedia.org/wiki/Terminal_velocity#Derivation_for_terminal_velocity

derivation in this pdf file (see section 3, falling object with Newton drag):

http://math.kennesaw.edu/~sellerme/sfehtml/classes/math2202/fallingbodies.pdf

You can use numerical integration using the equation for acceleration versus velocity, to calculate velocity and position over a series of small time steps, or use the equation linked to above to calculate positions using the equation for velocity versus time. Example using acceleration versus velocity:

Acceleration = F(v) = g - ( Cd ρ A v^{2}) / (2 m)

For each time step, you calculate

new_velocity = old_velocity + time_step x average_acceleration

One way to do this is to use trapezoidal method

average_accleration = 1/2 (old_acceleration + new_acceleration)

new_velocity = old_velocity + time_step x 1/2 (old_acceleration + new_acceleration)

where old_acceleration is the acceleration at the start of a time step and new_acceleration is the acceleration at the end of a time step. This requires you be able to calculate new_acceleration directly or to be able to predict it via an iterative method (see below)

You can then calculate position once you calculate the new velocity, using the same method:

new_position = old_position + time_step x 1/2 x (old_velocity + new_velocity)

An iterative corrector method will improve the results. In the algorithm shown below, a_{n}, v_{n}, and p_{n}, are successive "guesses" that should converge quickly. F(...) calculates the acceleration based on v_{n}(t), and Δt is the elapsed time per step. You may want to do 6 to 8 interations instead of the 4 shown in this example. The first step is essentially Euler (since a_{1}(t) is set = F(v_{0}(t)) (= a(t-1)), the remaining steps are trapezoidal. Even though each step of this algorithm will converge to a specific set of values, the algorithm is based on trapezoidal rule, a linear approximation, so you need to use small time steps (Δt).

F(v) = g - ( Cd ρ A v^{2}) / (2 m)

v_{0}(t) = v(t-1)

p_{0}(t) = p(t-1)

a_{1}(t) = F(v_{0}(t))

v_{1}(t) = v(t-1) + 1/2 (a(t-1) + a_{1}(t)) Δt

p_{1}(t) = p(t-1) + 1/2 (v(t-1) + v_{1}(t)) Δt

a_{2} = F(v_{1}(t))

v_{2}(t) = v(t-1) + 1/2 (a(t-1) + a_{2}(t)) Δt

p_{2}(t) = p(t-1) + 1/2 (v(t-1) + v_{2}(t)) Δt

a_{3} = F(v_{2}(t))

v_{3}(t) = v(t-1) + 1/2 (a(t-1) + a_{3}(t)) Δt

p_{3}(t) = p(t-1) + 1/2 (v(t-1) + v_{3}(t)) Δt

a_{4} = F(v_{3}(t))

v_{4}(t) = v(t-1) + 1/2 (a(t-1) + a_{4}(t)) Δt

p_{4}(t) = p(t-1) + 1/2 (v(t-1) + v_{4}(t)) Δt

...

v(t) = v_{n}(t)

p(t) = p_{n}(t)

a(t) = F(p_{n}(t))

time += Δt

t += 1

This is a predictor-corrector type algorithm:

http://en.wikipedia.org/wiki/Predictor-corrector_method#Euler_trapezoidal_example

Since you have an equation for velocity versus time v = V(t), you can use it directly and just use trapezoidal rule (time steps still need to be small):

t = 0

time = 0

v(t) = v(time)

p(t) = p(time)

loop:

t += 1

time += Δt

v(t) = V(time)

p(t) = p(t-1) + 1/2 (v(t-1) + v(t)) Δt

This can be integrated directly to determine velocity versus time, but apparently not position versus time. Wiki article with the answers, but not the derivation:

http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object

http://en.wikipedia.org/wiki/Terminal_velocity#Derivation_for_terminal_velocity

derivation in this pdf file (see section 3, falling object with Newton drag):

http://math.kennesaw.edu/~sellerme/sfehtml/classes/math2202/fallingbodies.pdf

You can use numerical integration using the equation for acceleration versus velocity, to calculate velocity and position over a series of small time steps, or use the equation linked to above to calculate positions using the equation for velocity versus time. Example using acceleration versus velocity:

Acceleration = F(v) = g - ( Cd ρ A v

For each time step, you calculate

new_velocity = old_velocity + time_step x average_acceleration

One way to do this is to use trapezoidal method

average_accleration = 1/2 (old_acceleration + new_acceleration)

new_velocity = old_velocity + time_step x 1/2 (old_acceleration + new_acceleration)

where old_acceleration is the acceleration at the start of a time step and new_acceleration is the acceleration at the end of a time step. This requires you be able to calculate new_acceleration directly or to be able to predict it via an iterative method (see below)

You can then calculate position once you calculate the new velocity, using the same method:

new_position = old_position + time_step x 1/2 x (old_velocity + new_velocity)

An iterative corrector method will improve the results. In the algorithm shown below, a

F(v) = g - ( Cd ρ A v

v

p

a

v

p

a

v

p

a

v

p

a

v

p

...

v(t) = v

p(t) = p

a(t) = F(p

time += Δt

t += 1

This is a predictor-corrector type algorithm:

http://en.wikipedia.org/wiki/Predictor-corrector_method#Euler_trapezoidal_example

Since you have an equation for velocity versus time v = V(t), you can use it directly and just use trapezoidal rule (time steps still need to be small):

t = 0

time = 0

v(t) = v(time)

p(t) = p(time)

loop:

t += 1

time += Δt

v(t) = V(time)

p(t) = p(t-1) + 1/2 (v(t-1) + v(t)) Δt

Last edited:

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I agree that it looks nasty: right now you've got what's called an integral equation. The velocity depends on its own integral in some way. My advice would be NOT to take that step of integrating both sides of the equation. Instead, leave it as it is. Then you have a differential equation -- the velocity is related to its derivative. That's something that I, at least, have a better idea of how one might go about solving. For instance, in the x-direction where drag is the only force acting, you have:

[tex] F_x = -\frac{1}{2}\rho A C_d v_x^2 [/tex]

[tex] ma_x = -\frac{1}{2}\rho A C_d v_x^2 [/tex]

You can write this as

[tex] m\frac{dv_x}{dt} = -\frac{1}{2}\rho A C_d v_x^2 [/tex]

This differential equation is separable.

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rcgldr said:v(t) = v_{n}(t)

p(t) = p_{n}(t)

a(t) = F(p_{n}(t))

time += Δt

t += 1

a(t) is a function of velocity, so that third line should be:

a(t) = F(v

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I would like to commend you for your curiosity and initiative in creating a program that deals with 2D projectile motion with wind resistance. It is a fascinating and complex topic to explore.

Regarding your question about integrating velocity for velocity in the acceleration equation, I would like to offer some insights. The equation you have mentioned, a = CdPAV^2/2*m, is a simplified version of the drag force equation. In reality, drag force is not only dependent on velocity, but also on other factors such as the shape and size of the object, the density of the fluid, and the viscosity of the fluid.

In order to accurately model the motion of an object with wind resistance, you would need to take into account the changing drag force as the object moves through the fluid. This can be done by using differential equations and numerical methods to calculate the velocity and position of the object at different time intervals.

In addition, the integration of velocity for velocity is not a straightforward process and requires advanced mathematical techniques. Therefore, it may be more practical to use numerical methods to solve for the position and velocity of the object at different time intervals, rather than trying to derive a general position v. position equation.

I hope this helps to clarify your doubts. Keep exploring and learning about projectile motion and wind resistance – it is a fascinating and important topic in science and engineering.

Velocity-dependent acceleration within drag force is a phenomenon in which the acceleration of an object is affected by the velocity at which it is moving through a fluid. This is due to the presence of a drag force, which is a resistance force that opposes the motion of the object.

Drag force acts in the opposite direction of the object's motion and increases as the object's velocity increases. This means that as the object moves faster, the drag force increases and therefore, the acceleration decreases. This is known as velocity-dependent acceleration within drag force.

Drag force is caused by the interaction between the object and the fluid it is moving through. As the object's velocity increases, it pushes more fluid out of its way, resulting in a larger force in the opposite direction. This is why drag force increases with velocity.

The formula for velocity-dependent acceleration within drag force is a = (F - mg)/m, where F is the drag force, m is the mass of the object, and g is the acceleration due to gravity. This formula takes into account the decrease in acceleration due to the opposing drag force.

Air resistance, which is a type of drag force, can significantly affect velocity-dependent acceleration. In cases where an object is moving through a fluid with a high air resistance, the drag force can be significant enough to cancel out the object's acceleration, resulting in a constant velocity or even a decrease in velocity over time.

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