Experiment to Investigate Air Resistance

In summary, the conversation discussed an experiment involving air resistance where paper disks with different radii were timed as they fell over a fixed distance. It was found that as the radius increased, the falling time also increased. The individual was trying to explain these results but was having difficulty due to the observed weight increase of the disk not matching up with the expected increase in drag force. The conversation then delved into the use of equations for air resistance and the coefficient of drag, as well as the effects of size and weight on parachutes. Ultimately, it was concluded that the weight of the object being carried by the parachute has a larger impact on the overall weight and drag force than the weight of the parachute itself.
  • #1
Jimmy87
686
17

Homework Statement


I collected data for an experiment involving air resistance. We made paper disks and timed how long they took to fall over a fixed distance. We changed the radius of the disk. We found that as you increase the radius they take longer to fall. I am trying to explain my results but with some difficulty.

Homework Equations

The Attempt at a Solution


We have been taught about forces expeienced by falling object e.g. weight and drag. I was thinking that if you double the radius of the circle then the area will increase by a factor of 4 which means the weight force will increase by a factor of 4. Since we found that it took longer for a bigger radius, the air resistance must not be proportional to the area. We have not been given any equations for air resistance but I did some research and found this one (https://en.wikipedia.org/wiki/Drag_equation). You can see that the drag force is in fact proportional to the surface area. However, if you double the radius the area should increase by a factor of 4 which means the weight will increase by a factor of 4 but according to this equation so should the drag force which means a paper circle of a bigger radius should take the same time to fall. What am I missing here? I know I am missing somehting because if this were true then for a given material e.g. a parachute it wouldn't matter how big you make the surface area and we all know that it does.
 
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  • #2
Usually paper circles doesn't fall in a smooth and repeatable way. How did you perform your experiment?
 
  • #3
What about the coefficient of drag? Which is part of the drag equation.
 
  • #4
mfb said:
Usually paper circles doesn't fall in a smooth and repeatable way. How did you perform your experiment?
Hi. We actucally used a pair of scissor and made a single cut to the centre of the circle and gently twisted it into a cone. We then dropped the cone with the tip pointing downwards so that it fell nicely. Surely I have something wrong with the physics though because if you had some material to make a parachute then a parachute with a bigger surface area would be more effective? However, since the parachute is roughly circular then doubling the surface area should double the weight?
 
  • #5
Jimmy87 said:
Hi. We actucally used a pair of scissor and made a single cut to the centre of the circle and gently twisted it into a cone. We then dropped the cone with the tip pointing downwards so that it fell nicely. Surely I have something wrong with the physics though because if you had some material to make a parachute then a parachute with a bigger surface area would be more effective? However, since the parachute is roughly circular then doubling the surface area should double the weight?

It doesn't double the weight of the parachutist, though!
 
  • #6
PeroK said:
What about the coefficient of drag? Which is part of the drag equation.
Thanks. So would a paper circle with a bigger surface area have a different coefficient?
 
  • #7
PeroK said:
It doesn't double the weight of the parachutist, though!
Why not? If it has roughly the same thickness then wouldn't a parachute with twice the surface area effectively mean you have twice as much parachute (so twice as much weight)?
 
  • #8
Jimmy87 said:
Thanks. So would a paper circle with a bigger surface area have a different coefficient?

What do you think? Does size matter?
 
  • #9
Jimmy87 said:
Why not? If it has roughly the same thickness then wouldn't a parachute with twice the surface area effectively mean you have twice as much parachute (so twice as much weight)?

The parachutist is the person hanging on the parachute. Their weight doesn't change.

7467702-parachutist--Stock-Vector-parachute-skydiving.jpg
 
  • #10
PeroK said:
The parachutist is the person hanging on the parachute. Their weight doesn't change.
No, I meant that if you designed two separate parachutes - the second one with twice the surface area.
 
  • #11
Jimmy87 said:
No, I meant that if you designed two separate parachutes - the second one with twice the surface area.

I'd quite happily throw a toy soldier out of an aeroplane with a pocket hankerchief as a makeshift parachute, but I wouldn't bet on it holding my weight as I fell to Earth! Would you?
 
  • #12
You should realize that generally the weight of a parachute itself is only a portion of the weight it is carrying; and, as a result, the small percentage of added material weight has much less effect than the increased area drag. For example, if the parachutist weighs 200 lbs and the parachute weighs 15 lbs., then, a 10% increase in the parachute weight is 1.5 lbs or 16.5 lbs total. As a result. the total % increase in the parachutist + parachute = 216.5 / 215 = 1. 007% as apposed to a 10% increase in drag.

Actually, the total packed weight of a parachute including an aerobatic harness with a 240 lb rated load is 15 lbs.
 
  • #13
PeroK said:
I'd quite happily throw a toy soldier out of an aeroplane with a pocket hankerchief as a makeshift parachute, but I wouldn't bet on it holding my weight as I fell to Earth! Would you?
No indeed not. I understand that though because with much smaller parachute you would have a lot less air interacting with it as it fell so it would not work with someone who has a much larger weight force. I am more interested with object itself falling and not with a parachute with someone on the end. If a paper circle has four times the surface area then why does it fall slower because it should also have increased its weight by four times?
 
  • #14
Jimmy87 said:
No indeed not. I understand that though because with much smaller parachute you would have a lot less air interacting with it as it fell so it would not work with someone who has a much larger weight force. I am more interested with object itself falling and not with a parachute with someone on the end. If a paper circle has four times the surface area then why does it fall slower because it should also have increased its weight by four times?

Well, then we're back to coefficient of drag. Can you think of any reason why the coefficient might depend on size, not just shape?
 
  • #15
No not really. The way I see it is that an object with four times the surface area will experience four times more resistive forces as it interacts with the air molecules. Could you kindly explain?
 
  • #16
Jimmy87 said:
No not really. The way I see it is that an object with four times the surface area will experience four times more resistive forces as it interacts with the air molecules. Could you kindly explain?

Where do the air molecules go? Perhaps think about something flat falling.
 
  • #17
PeroK said:
Where do the air molecules go? Perhaps think about something flat falling.
PeroK said:
Where do the air molecules go? Perhaps think about something flat falling.
They would get deflected downwards as they collide with the surface?
 
  • #18
Jimmy87 said:
They would get deflected downwards as they collide with the surface?

And perhaps sideways as well?
 
  • #19
PeroK said:
And perhaps sideways as well?
Yeh I guess some of them would get deflected sideways.
 
  • #20
Jimmy87 said:
Yeh I guess some of them would get deflected sideways.

How far sideways?
 
  • #21
PeroK said:
How far sideways?
To the extent of the surface? So are you trying to say that a bigger surface area would have greater friction with air molecules being deflected sideways along the surface?
 
  • #22
You should repeat your experiment using a shallow cone. Any flat paper disc will oscillate as it falls and this results in both effective area and form Cd changes that are both too unstable to predict or calculate. In addition, the oscillations allow lateral flow that nothing to do with Cd.
 
  • #23
JBA said:
You should repeat your experiment using a shallow cone. Any flat paper disc will oscillate as it falls and this results in both effective area and form Cd changes that are both too unstable to predict or calculate. In addition, the oscillations allow lateral flow that nothing to do with Cd.
I'm very confused now. Ignoring these oscillations should doubling the radius of a paper disk cause a slower falling time or should it remain the same for the reasons outlined in my first post?
 
  • #24
Jimmy87 said:
To the extent of the surface? So are you trying to say that a bigger surface area would have greater friction with air molecules being deflected sideways along the surface?

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.
 
  • #25
I suspect you will see close to an equal falling time; but you must be careful to insure that the cone angle is exactly same on every cone size because any variations in that angle will definitely affect the Cd of test piece.
 
  • #26
PeroK said:
For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.
I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?
 
  • #27
PeroK said:
For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.
So for a larger surface area the air molecules being deflected sideways along the surface will travel further and fas
PeroK said:
For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.
If I understand you correctly are you saying that for a larger surface area the air molecules being deflected sideways along the surface will travel further and faster resulting in more friction at that surface and this effectively increases the coefficient? So accoding to the physics as you described then should a paper circle with twice the radius fall slower despite having a four times increase in weight? This is what I get from what you have said but not the same as what JB has said as he said that the falling time should be the same for a larger object?
 
  • #28
haruspex said:
I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?
Sorry didn't see this before my last thread. Just to understand everyone correctly are we saying that a paper disk with double the radius has four times the weight but it would have more air resistance due to a different drag coefficient? I will think about what you just said and post a reply after I have thought it through.
 
  • #29
haruspex said:
I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?
The volume of air displaced would be volume of the cylinder of air so: pi r^2 h = pi r^2 (vt). For this cylinder to move out of the way in time 't' would need a sideways velocity of 2r/t. So the kinetic energy to move it out of the way would be 1/2 m (2r/t)^2. I think I see your point. If we double the radius then we get four times the area so four times the weight. However, in terms of knietic energy required to move the air out of the way we get 8 times more energy lost to air resistance which is why it would fall slower?
 
  • #30
PeroK said:
For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.
No, the overall scale and therefore the time the air has increases as well, the same speed will still work. You scale everything up by the same factor.
 
  • #31
With regard to PeroK's statement, he may have a point, in that as the air flows outward along the face of the cone it could add a skin drag factor to the total drag effect. I think we are getting into an area that would require a detailed aerodynamic analysis that is beyond your scope for any hope of trying to match your test results with any predicted performance. See if you can convince whomever may be required that the testing and presentation of its results is of sufficient to stand on its own merits.

Even in the early space program, in spite of all of the design calculations and wind tunnel testing, the ultimate determination of their capsules' ability to perform as required was only established by multiple aircraft drop and unmanned ballistic re-entry tests.
 
  • #32
Can we have a look at the raw data?
 
  • #33
mfb said:
Can we have a look at the raw data?
I don't have the raw data to hand now but I will get it tomorrow and upload it onto this post.
 
  • #35
Thanks for everyone's insights. I was hoping for a very simple explanation but I guess the situation is actually quite complex. I would very much like to see if my reply to Haruspex's post is along the right lines? Thinking about it further you could actually get an expression for the mass term in the kinetic energy formula as: mass = density x volume so m = p (pir^2 vt) which gives the sideways kinetic energy term as: 1/2 p (pi r^2 vt) (2r/t) ^2

That doesn't look right to me though as that would mean a doubling of the radius would increase the sideways kinetic energy by a factor of 32?
 
<h2>1. What is air resistance?</h2><p>Air resistance is a type of frictional force that acts on objects as they move through the air. It is caused by the collision of air molecules with the surface of the object, which creates a resistance force that opposes the motion of the object.</p><h2>2. Why is it important to investigate air resistance?</h2><p>Understanding air resistance is important because it affects the motion of objects in the air, such as airplanes, birds, and even falling objects. It can also impact the efficiency and design of various devices, such as parachutes and wind turbines.</p><h2>3. How can air resistance be measured?</h2><p>Air resistance can be measured using various methods, such as by measuring the force required to move an object through the air at a constant speed, or by measuring the acceleration of an object as it falls through the air. Additionally, specialized equipment such as wind tunnels can also be used to measure air resistance.</p><h2>4. What factors affect air resistance?</h2><p>The amount of air resistance experienced by an object depends on several factors, including the speed and surface area of the object, as well as the density and viscosity of the air. Other factors such as the shape and texture of the object's surface can also impact air resistance.</p><h2>5. How can we reduce air resistance?</h2><p>There are various ways to reduce air resistance, such as by streamlining the shape of an object to minimize its surface area and creating a smooth surface to reduce drag. Additionally, using materials with lower air resistance, such as aerodynamic materials, can also help to reduce air resistance.</p>

1. What is air resistance?

Air resistance is a type of frictional force that acts on objects as they move through the air. It is caused by the collision of air molecules with the surface of the object, which creates a resistance force that opposes the motion of the object.

2. Why is it important to investigate air resistance?

Understanding air resistance is important because it affects the motion of objects in the air, such as airplanes, birds, and even falling objects. It can also impact the efficiency and design of various devices, such as parachutes and wind turbines.

3. How can air resistance be measured?

Air resistance can be measured using various methods, such as by measuring the force required to move an object through the air at a constant speed, or by measuring the acceleration of an object as it falls through the air. Additionally, specialized equipment such as wind tunnels can also be used to measure air resistance.

4. What factors affect air resistance?

The amount of air resistance experienced by an object depends on several factors, including the speed and surface area of the object, as well as the density and viscosity of the air. Other factors such as the shape and texture of the object's surface can also impact air resistance.

5. How can we reduce air resistance?

There are various ways to reduce air resistance, such as by streamlining the shape of an object to minimize its surface area and creating a smooth surface to reduce drag. Additionally, using materials with lower air resistance, such as aerodynamic materials, can also help to reduce air resistance.

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