Investigating Drag with Paper Cones

In summary: If you are constructing a cone by taking a circular piece of paper, carving out a wedge and taping the cut edges then the size of the wedge...
  • #1
Jimmy87
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17
Hi,

I have been doing an experiment at school looking into how the angle of a cone affects the time it takes for it to fall 2m. The data consistently shows an inverse relationship but it is unclear whether or not they are inversely proportional i.e. if the angle doubles, the time taken to fall halves. I wanted to see if I could relate the angle of the cone to the force of drag in some way and wanted to know if someone could help me?

The drag equation we are given is:

FD = CDApv2 (FD = force of drag, CD = drag coefficient, A = cross-sectional area, p = density of fluid, v = velocity)

From looking at the equation I am thinking that the angle of the cone will affect the cross sectional area. So I need some way of expressing the cross-sectional area in terms of the angle of the cone so I see what the relationship would be between the angle of cone and the force of drag. Any ideas?

Many thanks!
 
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  • #2
Jimmy87 said:
Any ideas?
What are you holding constant as you change the angle of the cone?
 
  • #3
jbriggs444 said:
What are you holding constant as you change the angle of the cone?

Yeh good point, I didn't really consider that. I'm holding velocity constant as I am timing it for the last 2m of a 3m fall so should have reached terminal velocity. The density of air is constant. I know that changing the angle is like an indirect measure of cross sectional area. So I'm actually unsure if changing the angle changed the drag force or the drag coefficient or both?
 
  • #4
What else are you holding constant as you change the angle of the cone? Is it the same cone canted at an angle? Or is it the same paper rolled up into a pointier cone? Or what?
 
  • #5
Is the area of the base of the cone constant?
 
  • #6
CWatters said:
Is the area of the base of the cone constant?

No it isn't and the cone is actually dropped upside down (i.e. the pointed tip facing down) otherwise it tumbles and the data is awful. The radius of the flat circle I originally cut out is the same. I basically cut 6 circles each with a radius of 6cm. I then made a straight cut along the radius. I then twisted the paper to form a cone. I measured the height of the cone and new radius once twisted to find the angle. I have attached a diagram to show what I mean.
 

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  • #7
jbriggs444 said:
What else are you holding constant as you change the angle of the cone? Is it the same cone canted at an angle? Or is it the same paper rolled up into a pointier cone? Or what?

The last one - the same bit of paper rolled into a pointier cone as shown by the attached diagram on post #6.
 
  • #8
jbriggs444 said:
What are you holding constant as you change the angle of the cone?

Also just realized that if the velocity is constant after falling 1m that drag will equal the weight and therefore be constant! So the only variables changing are the angle (which affect cross-sectional area) and the drag coefficient I guess?
 
  • #9
Ok so in addition to the angle, the frontal area is also changing. I would try and make cones that have the same frontal area (eg same base diameter) and same weight. May need to add ballast to them.
 
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  • #10
CWatters said:
Ok so in addition to the angle, the frontal area is also changing. I would try and make cones that have the same frontal area (eg same base diameter) and same weight. May need to add ballast to them.

Thanks. But then I don't see how the angle could be changed. The angle changes as a result of changing the height and base diameter. How can you make cones which have the same base diameter but a different angle?
 
  • #11
Jimmy87 said:
Thanks. But then I don't see how the angle could be changed. The angle changes as a result of changing the height and base diameter. How can you make cones which have the same base diameter but a different angle?
Modify the height.

If you are constructing a cone by taking a circular piece of paper, carving out a wedge and taping the cut edges then the size of the wedge affects the height and circumference of the cone. Increase the size of the wedge and you increase the height while reducing the circumference. The diameter of the circle from which you cut affects the height and the circumference. Increase the diamter and you increase both height and circumference. For any desired height and circumference, there is a way to construct a cone to match.
 
  • #12
jbriggs444 said:
Modify the height.

If you are constructing a cone by taking a circular piece of paper, carving out a wedge and taping the cut edges then the size of the wedge affects the height and circumference of the cone. Increase the size of the wedge and you increase the height while reducing the circumference. The diameter of the circle from which you cut affects the height and the circumference. Increase the diamter and you increase both height and circumference. For any desired height and circumference, there is a way to construct a cone to match.

Thanks. So should I change my method or not? Seeming as though all variables are constant except for angle (which surely is a function of cross-sectional area) then the only thing to change would be drag coefficient. I didn't actually cut out a wedge, I simply made a cut along the radius (so no paper was taken away). I then moved one side over the other to twist into a cone. This will decrease the original radius and the angle as shown in the diagram. Does it matter if they both change? Because I was thinking I could get an equation for the cross-sectional area - which is just one variable - or is this hard to do? So are you saying you would keep the base area the same but just change the height by cutting out different sized wedges? Because I can't see how you can get the base area and angle the same by cutting out the same wedge? If you did cut out different sized wedges would this not change the mass though?
 
  • #13
Jimmy87 said:
Thanks. So should I change my method or not? Seeming as though all variables are constant except for angle (which surely is a function of cross-sectional area) then the only thing to change would be drag coefficient. I didn't actually cut out a wedge, I simply made a cut along the radius (so no paper was taken away). I then moved one side over the other to twist into a cone. This will decrease the original radius and the angle as shown in the diagram. Does it matter if they both change? Because I was thinking I could get an equation for the cross-sectional area - which is just one variable - or is this hard to do? So are you saying you would keep the base area the same but just change the height by cutting out different sized wedges? Because I can't see how you can get the base area and angle the same by cutting out the same wedge? If you did cut out different sized wedges would this not change the mass though?
If you want to keep the area of the base of the cone fixed but vary the angle then you will have to vary both the diameter of the circle you use to construct the cone and the overlap when you roll it up.

If you want to then keep the mass of the cone fixed then you will need to add a enough ballast (sand? chewing gum?) so that each cone masses the same.
 
  • #14
I'm going to first try to tackle your biggest problem which is the controlling of the variables. I must say at this point that I think you are following the wrong variable, since changing the angle of the cone will probably not effect the velocities, since this is not what the drag force depends on. It is the area of the base of the cone you should be trying to change instead of the angle. You could reason in your school experiment that it is not required to control the angle, while you change the base area. In simple words, change the base area, forget the angle. The angle has little to no effect on the drag force and this should have been obvious.

Secondly, your issue with relating the drag force and velocity should be easier to solve if you simply use the base area. It should be easy to calculate the drag force knowing the area, and therefore the terminal velocity.

I have studied in the IB where school experiments make up the majority of the final grades you receive. I have three years worth of creating experiments and labs, try to keep it simple.
 
  • #15
lekh2003 said:
I'm going to first try to tackle your biggest problem which is the controlling of the variables. I must say at this point that I think you are following the wrong variable, since changing the angle of the cone will probably not effect the velocities, since this is not what the drag force depends on. It is the area of the base of the cone you should be trying to change instead of the angle. You could reason in your school experiment that it is not required to control the angle, while you change the base area. In simple words, change the base area, forget the angle. The angle has little to no effect on the drag force and this should have been obvious.

Secondly, your issue with relating the drag force and velocity should be easier to solve if you simply use the base area. It should be easy to calculate the drag force knowing the area, and therefore the terminal velocity.

I have studied in the IB where school experiments make up the majority of the final grades you receive. I have three years worth of creating experiments and labs, try to keep it simple.

I don't quite understand your reasoning. None of the variables affect the drag force as the drag force is constant isn't it? This experiment where t=0s is terminal velocity so drag force is constant? If I keep the base area the same wouldn't the only variable that can change be the drag coefficient?
 
  • #16
Jimmy87 said:
I don't quite understand your reasoning. None of the variables affect the drag force as the drag force is constant isn't it? This experiment where t=0s is terminal velocity so drag force is constant? If I keep the base area the same wouldn't the only variable that can change be the drag coefficient?

I don't understand what you mean. If you change the surface area of the cone (the base area effects this most), the drag increases, therefore the terminal velocity decreases. If you simply change base area, the terminal velocities will change. You will have a working experiment and be able to identify a trend. If you change the cone angle, the change in terminal velocity will be close to 0. Try and work with the base area instead. Even your calculations for drag will be easier since you have your cross-sectional area right in front of you. Angles will complicate things.
 
  • #17
Jimmy87 said:
I have been doing an experiment at school looking into how the angle of a cone affects the time it takes for it to fall 2m.

Jimmy87 said:
So should I change my method or not?

If you want to know how the angle affects the time it takes to fall 2m then you should try and keep all other variables constant. Otherwise you can't tell if the angle or something else is responsible for the change/different performance.

So I would make a set of cones that only differ by the cone angle and have the same base area and the same mass. That way you know they aren't effecting the results.

Edit: Likewise if you wanted to know how the frontal area affects the time you should vary that and keep everything else (mass and cone angle) constant.
 
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  • #18
lekh2003 said:
I don't understand what you mean. If you change the surface area of the cone (the base area effects this most), the drag increases, therefore the terminal velocity decreases. If you simply change base area, the terminal velocities will change. You will have a working experiment and be able to identify a trend. If you change the cone angle, the change in terminal velocity will be close to 0. Try and work with the base area instead. Even your calculations for drag will be easier since you have your cross-sectional area right in front of you. Angles will complicate things.

Ah yes I agree the velocity will change so this will be the dependent variable but the drag force won't change at terminal velocity. If the cones have the same mass then they have the same weight and terminal velocity means drag = weight so drag will be a constant in my experiment won't it?
 
  • #19
Jimmy87 said:
Ah yes I agree the velocity will change so this will be the dependent variable but the drag force won't change at terminal velocity. If the cones have the same mass then they have the same weight and terminal velocity means drag = weight so drag will be a constant in my experiment won't it?

The drag is exactly what you hope to see changing. Only if the drag changes will you see a result. Separate your independent and dependent and controlled. Your independent will be the base area. The dependent will be the velocity which is determined by drag ( drag is determined by base area). The controls will be mass, height, etc. These are controlled and should not be affected by base area (ideally).

I think I understand what you mean by "drag=weight". Even though drag could be increasing, the weight definitely is not. You are confusing their relationship. You can change the weight which could affect drag, but it doesn't work the other way around. Drag could increase since you are increasing cross-sectional area, but that doesn't mean you are effecting the weight. You can still keep that constant. Drag is not controlled, weight is.
 
  • #20
lekh2003 said:
Your independent will be the base area.
If you are investigating how the point angle affects things, the independent will be the point angle. That's pretty basic.
 
  • #21
jbriggs444 said:
If you are investigating how the point angle affects things, the independent will be the point angle. That's pretty basic.

Initially, the OP wanted to use the point angle. I would just suggest that the base area would prove easier, and give more solid results. I assume in school experiments that independent variables can be chosen. It would be easier to use base area. You can predict results in hypotheses as well as back up your claim with math.
 
  • #22
lekh2003 said:
Initially, the OP wanted to use the point angle. I would just suggest that the base area would prove easier, and give more solid results. I assume in school experiments that independent variables can be chosen. It would be easier to use base area. You can predict results in hypotheses as well as back up your claim with math.

Right guys, thanks for the help so far. I think I am going to change the base area and forget the angle. However, our teacher said to hold drag constant by letting it reach terminal velocity first and then measuring the velocity for the last 1m of its fall. Therefore, the drag MUST be the same surely? I agree that the terminal velocity will change because of the larger drag during the initial fall but I am measuring the terminal velocity so drag is surely a constant?
 
  • #23
Jimmy87 said:
Right guys, thanks for the help so far. I think I am going to change the base area and forget the angle. However, our teacher said to hold drag constant by letting it reach terminal velocity first and then measuring the velocity for the last 1m of its fall. Therefore, the drag MUST be the same surely? I agree that the terminal velocity will change because of the larger drag during the initial fall but I am measuring the terminal velocity so drag is surely a constant?

I think you're confusing something. When your teacher probably said drag is "constant", that means during the fall. When we are saying drag is "changing", we mean between experiments. Your teacher is simply saying that in a calculation drag is constant during the fall, but drag is not constant when going between different experiments with different base areas.
 
  • #24
lekh2003 said:
I think you're confusing something. When your teacher probably said drag is "constant", that means during the fall. When we are saying drag is "changing", we mean between experiments. Your teacher is simply saying that in a calculation drag is constant during the fall, but drag is not constant when going between different experiments with different base areas.

But we are only looking at the part of the fall when it has reached terminal velocity. Terminal velocity is when drag is equal to the weight. It is impossible for drag to change between experiments because the weight of the paper is the same. I agree drag changes during the part of the fall BEFORE terminal velocity but all our measurements of time and distance are taken when it has reached terminal velocity. I remember we had to do an exam question about a skydiver which was similar to this. The skydiver has a terminal velocity before h opens his parachute and another (lower) terminal velocity after he opens his parachute. The force of drag changes all the time EXCEPT at both terminal velocities where it is equal to the skydivers weight.
 
  • #25
Jimmy87 said:
But we are only looking at the part of the fall when it has reached terminal velocity. Terminal velocity is when drag is equal to the weight. It is impossible for drag to change between experiments because the weight of the paper is the same. I agree drag changes during the part of the fall BEFORE terminal velocity but all our measurements of time and distance are taken when it has reached terminal velocity. I remember we had to do an exam question about a skydiver which was similar to this. The skydiver has a terminal velocity before h opens his parachute and another (lower) terminal velocity after he opens his parachute. The force of drag changes all the time EXCEPT at both terminal velocities where it is equal to the skydivers weight.
Ohhh, I think I get it now. All of the accelerations and velocity are completely useless, since in the end we are looking at terminal velocity regardless of time and which reaches the ground first. Am i right?

I think you should carry ahead with any independent variable, execute the experiment and hope that your independent variable had an effect. If not, then try and look at your equations and see why. Go ahead with the experiment, if anything doesn't go perfectly, you can just find the problem and explain this in whatever you might submit to your teacher. There's no harm in being wrong or not getting the results you expected.
 
  • #26
lekh2003 said:
Ohhh, I think I get it now. All of the accelerations and velocity are completely useless, since in the end we are looking at terminal velocity regardless of time and which reaches the ground first. Am i right?

I think you should carry ahead with any independent variable, execute the experiment and hope that your independent variable had an effect. If not, then try and look at your equations and see why. Go ahead with the experiment, if anything doesn't go perfectly, you can just find the problem and explain this in whatever you might submit to your teacher. There's no harm in being wrong or not getting the results you expected.

Ok, well thanks for all peoples help. I think I will assume terminal velocity has been reached so drag is constant. However, the terminal velocity will change and this will simply be the last 2m of the fall divided by the time it takes to fall these last 2m. I will now change my independent variable to the radius of the CONE but will keep the original radius of the flat CIRCLE I cut out the same. Going back to the OP now. I have two questions:

1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down. Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?
2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)

Thanks again!
 
  • #27
Jimmy87 said:
Ok, well thanks for all peoples help. I think I will assume terminal velocity has been reached so drag is constant. However, the terminal velocity will change and this will simply be the last 2m of the fall divided by the time it takes to fall these last 2m. I will now change my independent variable to the radius of the CONE but will keep the original radius of the flat CIRCLE I cut out the same. Going back to the OP now. I have two questions:

1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down. Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?
2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)

Thanks again!
You're welcome.
 
  • #28
Jimmy87 said:
1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down.

You are on the right lines except the "A" in the drag equation is the maximum cross sectional area (sometimes referred to as the frontal area) which is the area of the base of a cone when it's pointing downwards.

Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?

The drag equation is..

FD = ½ρ CDAV2

At terminal velocity FD = mg

so
mg = ½ρCDAV2

Rearrange to give

1/V2 = ½mgρCDA

The drag coefficient typically depends on the shape of the object so I would expect it to vary with the cone angle. If you keep the mass and cone angle constant then I would simplify the equation to..

1/V2 = KA

where K is a constant = ½mgρCD

If you plot y = 1/V2 and x = A you should get a straight line with slope = K.

You could also consider square rooting both sides and ploting 1/V against R (the radius of the base of the cone). That works because the area is proportional to R2 .

2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)

Over the narrow speed and size range of this experiment I would hope the drag coefficient Cd is reasonably constant. Over a wider speed range or a large range of sizes Cd isn't constant. eg sub sonic vs super sonic, or model vs full size aircraft.
 
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  • #29
CWatters said:
You are on the right lines except the "A" in the drag equation is the maximum cross sectional area (sometimes referred to as the frontal area) which is the area of the base of a cone when it's pointing downwards.

Thank you - this is extremely useful! Just to make sure I understand what 'A' represents - let's say I have two cones, each with an equal base diameter (and therefore area) but one is simply a flat piece of paper (a circle) and the other is a cone. Would I enter the same 'A' into the drag equation? Would I also enter the same 'A' even though the cone is falling with its tip facing downward? In this example, the paper circle will fall very slowly compared to the cone so what would explain the greater drag - would it be the drag coefficient?
 
  • #30
Jimmy87 said:
Thank you - this is extremely useful! Just to make sure I understand what 'A' represents - let's say I have two cones, each with an equal base diameter (and therefore area) but one is simply a flat piece of paper (a circle) and the other is a cone. Would I enter the same 'A' into the drag equation? Would I also enter the same 'A' even though the cone is falling with its tip facing downward?

Yes (provided you can get the flat piece of paper to fall in the horizontal position and not swoop about like a falling leaf.).

In this example, the paper circle will fall very slowly compared to the cone so what would explain the greater drag - would it be the drag coefficient?

Yes I would expect a cone with a wider angle to fall more slowly than a cone with a narrow angle because Cd is higher.
 
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  • #31
There are some drag coefficients here..

https://en.wikipedia.org/wiki/Drag_coefficient
The Cd of a cone point first is given in the top right hand table as 0.5.
The Cd of a flat plate is given in the lower table as = 1.28

Note these are only a rough guide. The angle of the cone isn't specified.
 
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  • #32
CWatters said:
Yes I would expect a cone with a wider angle to fall more slowly than a cone with a narrow angle because Cd is higher.

Would a reasonable scientific explanation of this be to discuss how the rate in change of momentum of air particles changes between the flat circle and cone. Looking at the flat circle, air particles will undergo a change in momentum of 2mv when they strike the surface. The force would be the rate of change of momentum which would be dictated by the velocity of the object. If we look at the cone then the air particles strike the surface of the cone which is now at some angle compared to the normal of the cone. Therefore the momentum will go down by the sine of the angle between the air particle striking the surface and the normal. Is this ok to explain why a cone falls faster through the air?
 
  • #33
Perhaps but its not that simple. The shape behind the point of maximum cross sectional area also matters, perhaps not in this case (the back of a disk and the back of a cone are similar) but certainly for wing sections.
 
  • #34
You might also consider adding a short string ( ~ 10 CM?) with a little weight on one end, and the other pulled through the tip of the cone, and attached on the inside. This would help keep the cone pointed down, and the weight would be more consistent cone-to-cone, as small variations in cone weight would not affect total weight very much (%-wise).
 
  • #35
There seems to be a lot more floundering around than necessary in this discussion. You want to know how drag depends on the angle of your cone. It sounds like all your cones are constructed of the same amount of paper so their masses are equal. The area A is the orthographic projection which, simply put, is the area presented to the onrushing wind and that is the area of the base. The area of the base is A=πR2 where R=rsinθ and r is the radius of the paper circle (6" for you) and θ is the angle of the cone. Therefore A=π(rsinθ )2. You should therefore expect the measured drag to be proportional to the square of the sine of θ, FD∝sin2θ.
 
Last edited:

1. What is drag and how does it affect objects?

Drag is a force that opposes the motion of an object through a fluid, such as air or water. It is caused by the friction between the fluid and the surface of the object, and it can slow down or even stop the object's motion.

2. How can paper cones be used to investigate drag?

Paper cones can be used as a simple and inexpensive tool to investigate drag. By dropping the cones from a height and measuring their fall time, we can compare the drag forces acting on the cones and gather data on how different factors, such as cone shape or weight, affect drag.

3. What are the variables that can affect drag in this experiment?

The variables that can affect drag in this experiment include the shape and weight of the paper cones, the height from which they are dropped, and the air resistance in the environment. Other factors such as wind or air temperature can also affect drag.

4. How can we measure the drag force acting on the paper cones?

The drag force acting on the paper cones can be measured by using the equation Fd = ½ρAv², where ρ is the density of the fluid (air), A is the cross-sectional area of the cone, and v is the velocity of the cone. By measuring the fall time and using the equation, we can calculate the drag force acting on the cone.

5. What are some potential applications of understanding drag?

Understanding drag is important in many fields, such as aerodynamics, engineering, and sports. It can help in designing more efficient vehicles, improving the performance of athletes, and even predicting weather patterns. By investigating drag with paper cones, we can gain a better understanding of this force and its effects on various objects.

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