Excel Projectile Motion: Predicting Trajectory with Wind Calculation

[mysticjbyrd]

I am attempting to create an excel worksheet that accurately predicts the trajectory of a projectile.

Anyways, I am having some trouble with calculating wind along the Z axis, left or right of the target.
Y axis is up in the air
X axis is towards the target
Z axis is left or right of target

Fd = force of drag
Fw = force of wind
Fb = force applied on golf ball
p = Rho, density of air
Cd = Drag coefficient
V = velocity
A = Area, which for a sphere I am using A = PI(Diameter)^2 /4
m= mass of golf ball
a = acceleration of ball
t = time
z = distance off from z-axis

Force of wind(Fw) ->
<- Force of drag(Fd)I am using,
Fd = 1/2*p*Cd*V^2*A

So,
Fw = 1/2*p*Cd*Vw^2*A
Vw = simply wind speed, Fw is constant

and

Fd = 1/2*p*Cd*Vd^2*A
Vd = Fw / mass * time

then
Fw - Fd = Fb
Fb = force applied on ball

Fb/mass = acceleration of ball

then
z = Vot + 1/2at^2

I am getting an acceleration that is roughly constant since I am dealing with low wind speeds and short time of flight. However, I am getting extremely low #'s for the distance the ball strays off on Z axis. I get the same answer if I use a constant acceleration or time stamps.

Information with just variables is fine with me, but if you want the data for some reason...
mass of golf ball = 0.04593
diameter of golf ball = 0.04267
Wind velocity = .44704 m/s
Rho = 1.1735 kg/m^3
Cd = .3878
time of flight = 1 sec

I am getting 0.0007221 meters off on z-axis.

Not sure where I am going wrong... Any help would be much appreciated.

 

[fatra2]

Look here: https://www.physicsforums.com/showthread.php?t=351066

I just answered the same question to this dude.

Cheers

 

[willem2]

Information with just variables is fine with me, but if you want the data for some reason...
mass of golf ball = 0.04593
diameter of golf ball = 0.04267
Wind velocity = .44704 m/s
Rho = 1.1735 kg/m^3
Cd = .3878
time of flight = 1 sec

I am getting 0.0007221 meters off on z-axis.

I'm getting 10 times that when I calculate it.

The way you calculate the drag and the force of the wind is wrong, you can't separate them
You need to add the vectors of the wind speed and the apparent wind because of the velocity of the ball, calculate the magnitude of the result, and calculate the total force from that, and then separate the force again in x, y and z components.
You'll get a much bigger force in the z direction that way. This is the same effect that makes cyclists feel more drag if there is a wind from the side.

Treating the drag force or the wind force as constant will only work if the velocity of the ball
changes little. Even in 1 second you get 10 m/s velocity change in the downward direction. This will have an influence on the windforce too. You really need to compute the solution of a set of differential equations for any realistic golfball trajectory. I suppose it can be done in excel with one row for each timestep.

 

[willem2]

Look here: https://www.physicsforums.com/showthread.php?t=351066

I just answered the same question to this dude.

Cheers

You were too late. Also your method won't work if the drag force is proportional to the square of the velocity, because you no longer can compute the x and y components separately.

 

[fatra2]

You were too late. Also your method won't work if the drag force is proportional to the square of the velocity, because you no longer can compute the x and y components separately.

In the case of a projectile thrown in the air, the drag force is linearly proportional to the velocity. Of course, you must consider that the air is homogenous.

Cheers

 

[willem2]

In the case of a projectile thrown in the air, the drag force is linearly proportional to the velocity. Of course, you must consider that the air is homogenous.

It's only linear for small velocities. While mysticjbyrd doesn't give the ball velocity, he is using quadratic friction. Everything I can find about friction with golfball sized objects uses a
quadratic friction force, as did mysticjbyrd so I have to assume a quadratic friction force here.

 

[mysticjbyrd]

I'm getting 10 times that when I calculate it.

The way you calculate the drag and the force of the wind is wrong, you can't separate them
You need to add the vectors of the wind speed and the apparent wind because of the velocity of the ball, calculate the magnitude of the result, and calculate the total force from that, and then separate the force again in x, y and z components.
You'll get a much bigger force in the z direction that way. This is the same effect that makes cyclists feel more drag if there is a wind from the side.

Treating the drag force or the wind force as constant will only work if the velocity of the ball
changes little. Even in 1 second you get 10 m/s velocity change in the downward direction. This will have an influence on the windforce too. You really need to compute the solution of a set of differential equations for any realistic golfball trajectory. I suppose it can be done in excel with one row for each timestep.

Could you give me an example of what you did? I am still a little confused.

 

[willem2]

suppose the ball is moving at 10 m/s, and you have 1 m/s wind from the side.
The speed of the ball relative to the air is $$v = \sqrt{v_{ball}^2+{v_wind}^2}$$ = 10.05 m/s
the friction force is (1/2)*C*A*v^2. The sideways component of this is (1/2)*C*A*v*v_wind
this more than 10 times as big as the force of the wind on a stationary ball

 

[mysticjbyrd]

Ok so what you are saying is that you have to consider the velocity of the ball from both the X and Z.

But what about the velocity of the Y axis as the ball falls?

Sorry to be so annoying, I have only had 1 semester of physics thus far, and I appreciate the help.

 

[mysticjbyrd]

can anyone else give me some more input? Thanks

 

[mysticjbyrd]

suppose the ball is moving at 10 m/s, and you have 1 m/s wind from the side.
The speed of the ball relative to the air is $$v = \sqrt{v_{ball}^2+{v_wind}^2}$$ = 10.05 m/s
the friction force is (1/2)*C*A*v^2. The sideways component of this is (1/2)*C*A*v*v_wind
this more than 10 times as big as the force of the wind on a stationary ball

You are using wind relative to the ground and not the ball... The ball won't be traveling anywhere near 1 m/s on z axis

 

[mysticjbyrd]

can anyone else give me some more input? Thanks

 

[ideasrule]

If you're using small timesteps, the most naive possible algorithm works well; you don't need to use y=v0t+1/2gt^2. In each timestep, calculate:

1) acceleration (in x, y, z)
2) velocity
3) position
4) total drag force (remember to use the speed of the air as the ball sees it, not as the ground sees it)
5) components of drag force (you'll have to do some trigonometry for this one)
6) in the next time step acceleration will be calculated again

Also, golf balls usually have a time of flight much longer than one second.