1. Concept of Drag Viscous Drag and 1D motion

2. Concept of Drag Drag is the retarding force exerted on a moving body in a fluid medium It does not attempt to turn the object, simply to slow it down It is a function of the speed of the body, the size (and shape) of the body, and the fluid through which it is moving

3. Drag Force Due to Air The drag force due to wind (air) acting on an object can be found by: F D = ½ ρ C D V 2 A where: F D = drag force (N) C D = drag coefficient (no units) V = velocity of object (m/s) A = projected area (m 2 ) ρ = density of air (kg/m 3 ) {1.2 kg/m 3 }

4. Drag Coefficient: C D The drag coefficient is a function of the shape of the object and the Reynolds Number (Re) Re = (velocity * length-scale) / (kinematic viscosity) For a spherical shape the drag coefficient ranges from 0.1 to 300, as shown on the next slide.

5. Drag Coefficient for Spheres

6. Dropping a Ping Pong Ball If you dropped a ping pong ball down the stairwell in this building (height 50 feet), and the stairwell had a vacuum in it, how long would it take for the ping pong ball to hit the floor? If you left the air in the stairwell would it take longer, shorter, or the same time to hit the bottom?

7. Looking at the ball in detail Drawing a Free Body Diagram (FBD) of the ball is shown to the right Since all the drag force is doing is slowing the ball down, it is directly vertical and upwards F D =f(v) F G =mg

8. Numerical Analysis If you have two data points (time, position), then you can approximate the velocity of the body. Given the points (2 s, -15m) and (2.1 s, -17m), what is the approximate velocity at 2.1 seconds? If the next data point is (2.2 s, - 19.05m), what is the velocity at 2.2 seconds?

9. Solution

10. Now Find Acceleration Given the velocities at 2.1s and 2.2s, what is the acceleration at 2.2s? Data points are (time, velocity): (2.1s, -20 m/s) (2.2s, -20.5 m/s)

11. Acceleration Solution

12. Continuing the process The ultimate goal of this numerical analysis is to find the drag force on the body Now that we have the acceleration, we can find the total force acting on the body (F=ma), the force of gravity (F g =mg), and Drag Force (F D )

13. Implementing this in Matlab (Section 8.3) Suppose you have two arrays defined, one with the elevation of the ping pong ball and the other with the time intervals You can use numerical methods to determine the drag coefficient by estimating the acceleration of the ball at each time step

14. Central Differencing To find the average velocity over a time interval, you divide the change in position by the change in time. This is called forward differencing. A better estimate is central differencing, where you estimate the velocity at a point by considering the points on either side of it.

15. Sample Data Type the sample data into Matlab and plot Height versus time Use central differencing to find the velocity at the points 0.1 through 0.9

16. The plotted data

17. Central Differencing for Velocity To find the velocity at time 0.1 seconds, use the formula: >> t1=[0:.1:1]; >> h1=[0 -.05 -.2 -.44 -.76 -1.17 -1.65 -2.19 -2.78 -3.41 -4.08]; >> v1=(h1(3:10)-h1(1:8))./(t1(3:10)-t1(1:8)) v1 = -1.0000 -1.9500 -2.8000 -3.6500 -4.4500 -5.1000 -5.6500 -6.1000

18. Central Differencing for Acceleration To find the acceleration at time 0.2 seconds, use the formula: Notice you must be careful with the indices >> a1=(v1(3:8)-v1(1:6))./(t1(4:9)-t1(2:7)) a1 = -9.0000 -8.5000 -8.2500 -7.2500 -6.0000 -5.0000

19. Finding the Drag Force With the net acceleration known at each time step, the acceleration due to drag can be found be subtracting the acceleration due to gravity. Then the drag force, at each time step, can be found from Newton’s second law (for a mass of 0.0025 kg). >> a2=a1-(-9.8) a2 = 0.8000 1.3000 1.5500 2.5500 3.8000 4.8000 >> f2=a2*.0025 f2 = 0.0020 0.0032 0.0039 0.0064 0.0095 0.0120

20. Finding the Drag Coefficient If you know the velocity at each time step and the drag force at each time step, you can plot F versus v 2 and fit a straight line to the data. Care must be taken to make sure you line up the correct force (a 1x6 array) with the correct velocity (a 1x8 array). Force 1 matches velocity 2, etc.

21. Using Matlab >> v2=v1(2:7) v2 = -1.9500 -2.8000 -3.6500 -4.4500 -5.1000 -5.6500 >> v3=v2.^2 v3 = 3.8025 7.8400 13.3225 19.8025 26.0100 31.9225 >> plot(v3,f2,v3,f2,'o')

22. Using Least Squares Cast the data points into array form using the base equation and then solve using matrix math. Note that currently V 2 and F D are row vectors and need to be transposed. >> k=v3'\f2' k = 3.5928e-004 For this problem, \ is the least squares operator!

23. Final Form of Drag Equation The final form of the drag equation can be written, which will allow the drag force to be calculated at any velocity. where: F D is in Newtons V is in meters per second