Experiment to Investigate Air Resistance

[haruspex]

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?

Yes, there's something not making sense to me either. Need to think about it some more...

 

[Jimmy87]

Yes, there's something not making sense to me either. Need to think about it some more...

is my working correct?

 

[haruspex]

is my working correct?

Yes, but I see what is wrong with my model. It treats the cylinder of air height h under the disc as the only air able to move. In practice, air will start to move further ahead of the disc the broader the disc. This suggests a model in which a cone of air in front of the disc is accelerated (the cone angle being independent of v and r). However, that leads us to the usual kAv² formula, so no further forward.

Of course, the cone model breaks down in the final stages of landing, though at what height I'm not sure - maybe only a few centimetres. In the final millimetres, the cylinder model looks right. That would explain slower descent with large r, but only if those final stages are a significant fraction of the total descent.

I found this link describing descent rate of a parachute: http://www.pcprg.com/rounddes.htm, but it sheds no light on your problem. It implies the descent rate would be independent of radius when there's no load.
One thing: you mentioned cutting a small hole and forming a cone in the centre of the disc to prevent oscillation. Were the holes identical across different radii, or were they in proportion to the disc area?

 

[PeroK]

The rate at which something falls depends on its density (for a given shape). Here we have essentially a hollow object, and the mass (unless you use thicker paper) increases only 4 times if you double the size. So, the effective density of the larger cone decreases. Is it as simple as that?

A solid cone should fall at the same rate regardless of its size. But, a hollow cone must fall more slowly the larger it gets (for material of a constant thickness).

 

[Nidum]

http://www.diracdelta.co.uk/science/source/d/r/drag coefficient/source.html#.VjHwIitgS70

http://www.roymech.co.uk/Related/Fluids/Fluids_Drag.html

 

[256bits]

The rate at which something falls depends on its density (for a given shape). Here we have essentially a hollow object, and the mass (unless you use thicker paper) increases only 4 times if you double the size. So, the effective density of the larger cone decreases. Is it as simple as that?

A solid cone should fall at the same rate regardless of its size. But, a hollow cone must fall more slowly the larger it gets (for material of a constant thickness).

The density part seems to be reasonable.

solid cone though - the mass increase is cubic, and so is the increase of the downward force due to gravity. A larger solid object has a higher terminal velocity than one of similar shape, but smaller. Dropping 2 similar shaped solid objects of the same density from the same height, the larger of the two will reach the ground earlier.

 

[haruspex]

The rate at which something falls depends on its density (for a given shape). Here we have essentially a hollow object, and the mass (unless you use thicker paper) increases only 4 times if you double the size. So, the effective density of the larger cone decreases. Is it as simple as that?

No. All the standard theory and equations say the drag, for a given shape and speed, is proportional to the area. Thus two discs of the same thickness and material should fall at the same speed regardless of radius. If you scale up all three dimaensions it should fall faster.

 

[Jimmy87]

Yes, but I see what is wrong with my model. It treats the cylinder of air height h under the disc as the only air able to move. In practice, air will start to move further ahead of the disc the broader the disc. This suggests a model in which a cone of air in front of the disc is accelerated (the cone angle being independent of v and r). However, that leads us to the usual kAv² formula, so no further forward.

Of course, the cone model breaks down in the final stages of landing, though at what height I'm not sure - maybe only a few centimetres. In the final millimetres, the cylinder model looks right. That would explain slower descent with large r, but only if those final stages are a significant fraction of the total descent.

I found this link describing descent rate of a parachute: http://www.pcprg.com/rounddes.htm, but it sheds no light on your problem. It implies the descent rate would be independent of radius when there's no load.
One thing: you mentioned cutting a small hole and forming a cone in the centre of the disc to prevent oscillation. Were the holes identical across different radii, or were they in proportion to the disc area?

Thanks for your help - this is very insightful (and thanks to everyone else). No - the holes were not cut to prevent oscillations. A slit was cut along the radius to be able to form a cone. The reason for forming a cone was to stop the oscillations as the falling motion is very smooth after being formed into a cone.

 

[JBA]

Jimmy87,

If you go to the below wesite that Nidum previously posted and do a little plug and play on thd disc drag calculator you will find that, based upon that calculator, your test disc will fall at the same rate regardless of its diameter, i.e keeping all other parameters equal, if you double the disc area then the darg force doubles.

http://www.diracdelta.co.uk/science/source/d/r/drag coefficient/source.html#.VjHwIitgS70

 

[haruspex]

Jimmy87,

If you go to the below wesite that Nidum previously posted and do a little plug and play on thd disc drag calculator you will find that, based upon that calculator, your test disc will fall at the same rate regardless of its diameter, i.e keeping all other parameters equal, if you double the disc area then the darg force doubles.

http://www.diracdelta.co.uk/science/source/d/r/drag coefficient/source.html#.VjHwIitgS70

Yes, but the whole point of this thread was to come up with some explanation of why that was not matched by observation.

 

[mfb]

Yes, but the whole point of this thread was to come up with some explanation of why that was not matched by observation.

Right, and so far we don't have a good explanation. Noting that we don't expect different times is a good first step, and certainly better than wrong explanations.

Raw data, a picture of the setup and so on could help.

 

[Nidum]

(1) It may be that with test objects of relatively large area and weighing almost nothing chance events distort the results of any experiments in uncontrolled conditions so much that no reliable results can be obtained .

Room draughts , thermals and small differences in geometry may have large effect .

(2) There is an experiment which might give some useful indications of what is going on .

Suspend the test objects on fine threads and then raise a low power fan slowly up from an initially large distance below . Note distance between fan and test objects at which (a) disturbance is first observed , (b) levitation occurs and (c) strong upward movement occurs .

Weigh each test object . Repeat each experiment . Plot results .

A way to suspend test objects would be to have radial threads arranged trampoline like out to a big hoop .