Concept of Drag Viscous Drag and 1D motion. Concept of Drag Drag is the retarding force exerted on a moving body in a fluid medium It does not attempt.

Concept of Drag

Viscous Drag and 1D motion

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

Drag Force Due to Air The drag force due to wind (air) acting on

an object can be found by:

FD = ½ ρ CDV2A

where: FD = drag force (N)

CD = drag coefficient (no units)

V = velocity of object (m/s)A = projected area (m2)ρ = density of air (kg/m3) {1.2

kg/m3}

Drag Coefficient: CD

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.

Drag Coefficient for Spheres

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?

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

FD=f(v)

FG=mg

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?

Solution

s

m

ss

mmV 20

)1.22(

1715]1.2[@

s

m

ss

mmV 5.20

)2.21.2(

05.1917]2.2[@

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)

Acceleration Solution

25

2.21.2

5.2020

]2.2[@s

m

ss

sm

sm

A

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 (Fg=mg), and Drag Force (FD)

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

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.

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

T (s) h (m)0 0

0.1 -0.050.2 -0.20.3 -0.440.4 -0.760.5 -1.170.6 -1.650.7 -2.190.8 -2.780.9 -3.411 -4.08

The plotted data

Central Differencing for Velocity

To find the velocity at time 0.1 seconds, use the formula:

s

m

ss

mm

tt

hhv 0.1

02.0

020.0

02.0

02.01.0

>> 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

Central Differencing for Acceleration

To find the acceleration at time 0.2 seconds, use the formula:

Notice you must be careful with the indices

21.03.0

1.03.02.0 0.9

1.03.0

0.180.2

s

m

sssm

sm

tt

vva

>> 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

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

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 v2 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.

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')

Using Least Squares Cast the data points

into array form using the base equation and then solve using matrix math.

Note that currently V2 and FD are row vectors and need to be transposed.

0120.0

0095.0

0064.00039.0

0032.0

0020.0

*

9225.31

0100.26

8025.193225.13

8400.7

8025.3

*2

k

Fkv D

>> k=v3'\f2'k = 3.5928e-004 For this problem, \ is

the least squares operator!

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: FD is in Newtons V is in meters per second

2000359.0 vFD