# Homework Help: Wind & Quadratic drag

1. Nov 7, 2009

### mysticjbyrd

1. The problem statement, all variables and given/known data

This is a real world problem and not a problem in a book, so if you feel I am missing some information just let me know.

A ball is flying at 25 m/s
The wind is blowing perpendicular to the trajectory at .44704 m/s

$$\rho = \frac{1.17 kg}{{m}^3}$$

$$Cd = 1/2$$

$$Diameter_{ball} = 0.04593 m$$

$$Mass_{ball} = 0.04267 kg$$

$$gravity = 9.8 \frac{m}{s}$$

X-Y-Z plane
X-axis is straight towards the target
Y-axis is up in the air
Z-axis is left or right to the target

Using time steps of .1 seconds,
What is the force the wind applies on the ball?
How far will the ball fly off the trajectory in 2 seconds?

2. Relevant equations

$$F_{d} = .5*rho*A*{v}^2$$

$$F_{d} = drag force$$

$$\rho = air density$$

$$v = velocity$$

A = frontal area_{ball}
I am using,
$$A=\pi/4*{d}^2$$

3. The attempt at a solution
Force of wind on the ball,

Using time steps of .1 s, Find all velocities to find total velocity.
$$v_{x}=25\frac{m}{s}$$

$$v_{y} = a*t => v_{y}=9.8*.1 = 0.98\frac{m}{s}$$

Force applied on ball from wind
$$F_{ball}=F_{wind}-F_{drag}$$

$$F_{w} = .5*\rho*Cd*{V_{w}^2*A => F_{w}= .5*1.17*.5*{.44704}^2*\frac{\pi}{4}*{0.04593}^2 => F_{w}=0.0000968504 N$$

$$F_{w}=m_{ball}*a_{ball} => \frac{F}{m}=a => \frac{0.0000968504}{0.04267} = 0.00226975\frac{m}{s^2} => v=at => {0.00226975}*.1 = 0.000226975\frac{m}{s}$$

Velocity of ball no drag
$$v_{bnd} = 0.000226975\frac{m}{s}$$

$$F_{d}= .5*\rho*Cd*{V_{b}^2*A => F_{d}= .5*1.17*.5*{0.000226975}^2*\frac{\pi}{4}*{0.04593}^2 => F_{d}= {{2.49669}^{-11}}N$$

$$F_{ball}=F_{wind}-F_{drag} = 0.0000968504-{2.49669^{-11}}= {0.0000968504} N$$

$$F_{b}=m_{ball}*a_{ball} => \frac{F}{m}=a => \frac{{0.0000968504}}{0.04267} = 0.00226975\frac{m}{s^2} => v=at => {{0.00226975}}*.1 = {0.000226975}\frac{m}{s}$$

$$v_{z}= 0.000226975\frac{m}{s}$$

Force of Drag on ball X-axis
$$F_{d}= .5*1.17*.5*{25}^2*\frac{\pi}{4}*{0.04593}^2 = 0.302893N => a=\frac{F}{m} => \frac{0.302893}{0.04267} = 7.09849\frac{m}{s^2} => v=at => 7.09849 * .1 = .70949\frac{m}{s}$$

Total Velocity
$$v_{xy} ={({v_{ball_x}^2+v_{ball_y}^2)}^\frac{1}{2} => v_{xy}=({{25}^2+{.98}^2)}^\frac{1}{2} = 25.0192\frac{m}{s}$$

$$V_{total}= {({v_{ball_xy}^2+v_{ball_z}^2)}^\frac{1}{2} => v_{total}=({{25.0192}^2+({8.9776*{10}^{-6})^2)}^\frac{1}{2} = 25.0192\frac{m}{s}$$

If you would prefer to just use variables that would be fine with me... Since I am new to this style of formatting, it took me a long time to post this, and I do not expect someone else to do the same. However, it would be amazing if someone did.

Last edited: Nov 7, 2009
2. Nov 7, 2009

### Andrew Mason

The drag force would be based on the velocity of the air relative to the ball, not the velocity of the air relative to the ground. It appears from your question that the ball relative to the air is 25 m/s. You also do not seem to be factoring in the area of the ball.

AM

3. Nov 7, 2009

### mysticjbyrd

Yah I was still in the process of editing it Andrew. I am new to posting things here and it took me a lot longer than I thought to post the information.

I could really use some more input though, thanks.

4. Nov 10, 2009

### mysticjbyrd

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5. Nov 20, 2009

### mysticjbyrd

can anyone else give me some more input? Thanks

6. Nov 20, 2009

### ideasrule

Wow. Do you know how to program, or use Excel, to simulate this for you? Even learning a programming language (or to use Excel) from scratch is going to be quicker than manually time-stepping. It will be more accurate, too, since rounding errors are tiny and you can make each timestep one nanosecond.

That said, depending on what this projectile really is, there might be other problems with the solution. The drag equation F=Cd*A*rho*v^2 doesn't work for small Reynolds numbers. The Reynolds number for the object you gave is about 60 000, which is good enough if the object is rough, but if it's smooth Cd varies significantly with speed up until a Reynold number of 10 million. If this object is a ball used for sports, the ball is likely designed to maximize lift. This makes a HUGE difference, and is probably even more significant than drag. Mythbusters once tested dimpled vs. non-dimpled golf balls and found the dimpled ones travelled nearly 40% farther. If the ball is rotating, aerodynamic effects are going to curve the ball's trajectory. This is the physics behind curve balls, and as you probably know, balls curve significantly when rotating.

7. Nov 21, 2009

### mysticjbyrd

I am actually using excel to perform the time steps, as you may know from my previous post that was in another thread. I would never attempt to do this by hand, that would be insane and worthless. I am rather new to excel, no experience with programming algorithms or excel macros, and 1 semester of physics.

Maple can easily solve this problem for me, but I was really wanting to use excel because It would make for a better presentation.

Thank you for the information on the Reynolds number... that really explains my inability to get anything that made sense.

I am launching a golf ball that is not dimpled in order to try to eliminate MOST of the effects of lift on the ball. This ball is not designed to maximize lift, but to provide for a straight accurate golf shot. The ball will be pushed along inside a barrel, slightly larger than the ball, and launched out the other end, thus it should have little to no spinning but freely rotating. The balls will travel btn 20-50 yards, at roughly 90 ft/sec, and will be in the air for less than 4 seconds.

I have as of yet to try to figure out the loss of force from the rotation as well.

PS: Mythbusters is my favorite show and I did catch that episode.

Last edited: Nov 21, 2009