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Author Topic: Projectile Motion with Air Drag  (Read 140758 times)
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Fu-Kwun Hwang
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on: January 29, 2004, 02:44:09 pm » posted from:,,Satellite Provider

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Normally you find projectile in ideal case, i.e. without any air drag.
The following applet add airdrag to simulate more realistic case.
Drag Force is proportional to F = -kV2




Mass of the object is assumed to be 1kg.
If the mass of the object is different, you should divide k by real mass k->k/m .

Normally, the drag force is very small for small projectile motion in air.
So the initial velocity was set to 626 m/s (very large value) and the number display for distance was always in unit of km, so that the air drag effect can be visible more easily.
The right most text field (initial value=100) is used for scaling the X-Y coordinate.
For example: You want to have a smaller initial velocity v=62.6 m/s.
You might want to change the sclae from 100. to 1. (The same trajectory will be displayed for v=626m/s and scale=100.)
If you want to have smaller velocity, i.e. v=20 m/s , you might want to change scale to 0.1


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There are 16 translations,
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*** There are 1 more attached files. You need to login to acces it!
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Reply #1 on: January 30, 2004, 12:43:30 pm » posted from:,,Satellite Provider

Subject:
Date: Wed, 14 Jul 1999 23:23:57 +1000
From: "Antony Tribbick" <tribbick@lavalink.com.au>
To: <hwang@phy03.phy.ntnu.edu.tw>
Nice java applet..
Demonstrates very nicely what I was experimenting on with golf balls
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Reply #2 on: February 04, 2004, 03:36:28 pm » posted from:Vilnius,Vilniaus Apskritis,Lithuania

Tongue Cool :lol: :wink:
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janesf
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Reply #3 on: January 05, 2007, 04:15:48 pm »

why is it that the projectile stays on the ground when I set k > 0? of course I clicked start after setting k to some nonzero value...
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Reply #4 on: February 19, 2007, 06:57:56 pm »

Hello!
My name is Marko Podpecan from Slovenia, Europe. My diploma is Physical and computer modelling with EJS. I'm doing simulations with ideal conditions (without air).

Now i have found a simulations from you, where you include air.
In which program did you do this simulations ?
And which program you use to open and see *.class files.

You have very interesting and great simulations. I'm amazed.

Thank you, for answer.

Marko Podpecan, Slovenia
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Fu-Kwun Hwang
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Reply #5 on: February 19, 2007, 10:23:05 pm »

All the program in this category are written by JDK1.0.2 between 1996-2001.
Java simulations created by EJS can be found at Easy java simulation category.

It is very easy to include air drag into ejs.
Just add air drag terms into evolution equations in ejs.

For this simulation : The drag force is propotional to V2
For projectile motion you should have the following equations in EJS evolution page:

dx/dt=vx;
dy/dt=vy;
dvy/dt=-g;

If you want to add drag force like this one. In this model, the drag force is proportional to F=-b*V2, where b is the drag force coefficient.
Fx=-b*V2*(vx/|V|)=-b*vx*|V|;
Fx=-b*V2*(vy/|V|)=-b*vy*|V|;

You should modify the evolution equations to the following:
dx/dt=vx;
dy/dt=vy;
dvx/dt=-b*Math.sqrt(vx*vx+vy*vy)/m*vx;
dvy/dt=-b*Math.sqrt(vx*vx+vy*vy)*vy/m-g;

The following is EJS version of air drag simulation.

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
  • Please feel free to post your ideas about how to use the simulation for better teaching and learning.
  • Post questions to be asked to help students to think, to explore.
  • Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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Reply #6 on: September 24, 2007, 07:33:36 pm »

Dr. Hwang,
 
A question concerning your projectile simulation with air drag.  I am trying to plug in parameters to simulate a car traveling through the air.  Assuming that your drag force of the form F=-kv^2 is equivalent to the drag force of the form D=.5(rho)S(Cd)*v^2, I was assuming your use of SI units and calculating your k to be equal to .5*1.25*1.1m^2*1.2; therefore, coming up with a k=.825.  Placing -0.825 into the k= box and making the velocity = 20 (assuming this to be 20m/s), I am getting no x,y coordinate pairs and a time value that is very long.  Assuming that you are using a Runge-Kutta 4th order approximation, I am very interested in the results of this simulation.  Any insight into what I am doing wrong would be much appreciated.

Thanks in advance.
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Fu-Kwun Hwang
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Reply #7 on: September 24, 2007, 11:10:07 pm »

I just added some more description for the simulation.

Normally, the drag force is very small for small projectile motion in air.
So the initial velocity was set to 626 m/s (very large value) and the number display for distance was always in unit of km, so that the air drag effect can be visible more easily.
The right most text field (initial value=100) is used for scaling the X-Y coordinate.
For example: You want to have a smaller initial velocity v=62.6 m/s.
You might want to change the sclae from 100. to 1. (The same trajectory will be displayed for v=626m/s and scale=100.)
If you want to have smaller velocity, i.e. v=20 m/s , you might want to change scale to 0.1

The k value for the simulation should always be positive. (The negative sign already entered in the code)
Because you enter a negative k value, it was simulating some push force was added(instead of drag force).
Please check the unit of your formula.  Why the density is 1.25 ? This is not the density of air (also check the unit for the density in your formula)!
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kaadkins
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Reply #8 on: September 25, 2007, 12:08:39 am »

Forgive me if I am not understanding something but using D=1/2*rho*S*Cd*v^2, I would take k to be equivalent to 1/2*rho*S*Cd.  Plugging in values and units I get .5*1.229 kg/m^3*2.6m^2*.45 (dimensionless drag coefficient) = .72 kg/m^3 * m^2 =.72kg/m.  When multiplied with the v^2 term, I get units of kg/m * m^2/s^2 = kg * m/s^2 which is of course equal to a Newton (appropriate for a force).

I am plugging in values that I would expect for something like an automobile.  I believe 1.229kg/m^3 is an appropriate value associated unit for density of the atmosphere at sea level.

Plugging in v=20m/s, k=.72 and an x,y scale factor of .01, I get an xmax=2.91e-3.  I believe the value not acounting for air resistance to be just over 41m.

Thanks.  Sorry for any misunderstanding.
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Fu-Kwun Hwang
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Reply #9 on: September 25, 2007, 10:45:10 am » posted from:Taipei,T\'ai-pei,Taiwan

The force F=-kV2  I need to know the acceleration for the calculation.
So I assume the mass is equal to 1kg in the above calculation.
For you case, you should divide k by the mass( in unit of kg for your car).
I am sorry. It is something I did not write out clearly.

The air-drag is not a very strong effect for small velocity and large mass.
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Reply #10 on: February 03, 2008, 11:09:08 pm » posted from:Surabaya,Jawa Timur,Indonesia

My name is falah. I had been studying in Institute of Technology Sepuluh November (ITS) Surabaya Indonesia majoring in Physics. I have an assignment related to a movement of tennis ball in the air (at parabolic motion). However, there are some questions I would like to ask, would you mind to answer?
By the way, these are my questions list:
1.  How is the projectile motion for the sphere thing at 3 dimension by seeing air resistance, spin, and buoyant force ?How are the equation for simulating it?
2.   What forces are worked on the tennis ball moving in the air, with and without spin?
3.   How is the equation for the movement of tennis ball as it spins and doesn’t spin?
Thanks
my email: hanunfalah@gmail.com
« Last Edit: February 19, 2008, 12:48:57 pm by falah » Logged
fadiga
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Reply #11 on: April 28, 2008, 09:50:26 pm » posted from:Birkirkara,,Malta

Dear Dr. Hwang,

  I am Daniel Baldacchino, a second year mechanical engineering student at the University of Malta. I have an assignment to build a simple catapult to launch a 100g water balloon, as far as possible but also to try and predict as accurately as possible the range it will achieve with kinetic equations. Since this has to be an accurate calculation, I don't think we can afford to simply neglect drag on the balloon.

I now refer to your equations posted on the "Projectile Motion with Air Drag" forum.
  Fx= - b* V2 * (vx/|V|) = - b * vx * |V|;
  Fy= - b* V2 * (vy/|V|) = - b *vy * |V|;

What exactly are the terms (vx/|V|) ? initially i thought these were a position vector , but the fact you use vx has confused me.



Also,  when you plug this force into the acceleration equations ,

  dvx/dt = - b * ( Math.sqrt(vx*vx+vy*vy) / m ) * vx;
  dvy/dt = - b * ( Math.sqrt(vx*vx+vy*vy) * vy / m-g ;

how would you integrate to solve for these values? are vx and vy intial values or would you have to integrate them from some intial value to 0 ?

I was considering, for the sake of simplifying the calculation, to eliminate effect of drag in the vertical direction, and consider only a force in the horizontal direction, say Fx = -bv  ( or v^2, im not sure which to use, ive found both versions!)

therefore d(vx)/dt = bv/m

and using (ax)dx =  v(dv), i would integrate x from 0 to x, and v from some initial velocity to a final value. 
I would obtain another equation in the y direction, assuming a constant acceleration -g and try and find a way to solve...

Any insight into what your thoughts are would be much appreciated.
Thank you for your time Smiley

Best Regards,
Daniel Baldacchino
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Fu-Kwun Hwang
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Reply #12 on: April 28, 2008, 10:13:38 pm »

Because one of the model for air drag is the drag force is proportional to v2.
So if the velocity is v. the magnitude of the drag force is b*v2 (where b is a constant).
Since the force F is a vector, there are two components Fx and Fy in my simulation.
So I have to calculate the component of the force in the x and y direction.
Fx= - b* V2 * (vx/|V|);  // vx/|V| will give me the component of the force in the x component.

Once I have dvx/dt =Fx/m =  - b/m * V2 * (vx/|V|);
I can integrate the above equation with numerical method. The method I used is Runge-Kutta 4-th order method, which is a standard method for numerical calculation. You can find related information from any numerical calculation textbook. If you use the EJS tool (which is available on our server), it will solve the differential equation for you. You just need to provide the equations and initial values. :-)
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Reply #13 on: October 24, 2008, 10:39:55 am » posted from:Singapore,,Singapore

Hi Sir,

I don't quite understand how you get Fx and Fy from the equation F = - bV^2.

Is it possible for you to show the steps to get Fx and Fy from F = -bV^2?

Sorry for any inconveniences caused.

Thank You.

Regards.
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Reply #14 on: October 24, 2008, 04:49:50 pm » posted from:Taipei,T'ai-pei,Taiwan

It was explained in the previous message.
The magnitude of the drag force is F=-b*V2 and the direction of the force is the same as velocity.
So we can calculate the force component in x direction as Fx=-b*V2*Vx/|V|;  (i.e. calculate inner product between F and univ vector in the direction of velocity vector.)

You can download EJS source code and load it into ejs
Or Click  load ejs as signed applet button to load ejs into your browser and find out how it was calculated in the program!

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Reply #15 on: May 08, 2012, 04:26:57 am » posted from:,,Satellite Provider

Quote from: link=topic=145.msg729#msg729 date=1075883788
Tongue Cool :lol: :wink:

Thanks to share these informative rules. I don't know anything about this issue before.
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Reply #16 on: May 31, 2012, 07:03:09 pm » posted from:Dhaka,Dhaka,Bangladesh

It is a nice site.-*-
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Reply #17 on: July 07, 2012, 12:23:40 pm » posted from:,,Satellite Provider

 Shocked Wink
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Reply #18 on: October 01, 2012, 04:06:35 pm » posted from:Bangkok,Krung Thep,Thailand

 Tongue Tongue Tongue
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Reply #19 on: December 29, 2012, 01:17:56 pm » posted from:,,Satellite Provider

 Cheesy Cheesy Cheesy
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