[FliegenderZirkus]

Coefficient of drag (CD) of a smooth sphere is not constant, but depends on the Reynolds number. At around Re=3e5 there's an abrupt drop of the CD which is known as the drag crisis. For convenience, let's assume the following closed-form formula as a fit to the Schlichting CD vs Re data (first attached plot)
https://pages.mtu.edu/~fmorriso/Data...reDrag2016.pdf
Next, let's evaluate the drag force for a solid aluminium golf-ball-sized smooth sphere falling in earth's atmosphere (second attached plot). My understanding is that the terminal velocity (neglecting buoyancy) is the intercept between the blue and red curves. But in this plot there are three such intercepts! So as the ball is falling, it accelerates until it reaches ~60 m/s. But if we somehow gave it a nudge to accelerate it further, it could reach another equilibrium speed at ~130 m/s? I'm attaching a python snippet to reproduce the plot. Maybe it's just a mistake in the code?

import numpy as np
import matplotlib.pyplot as plt
Re_vals = np.logspace(4, 5.8, num=100)
Cd_vals = 24/Re_vals + (2.6*Re_vals/5)/(1+(Re_vals/5)**1.52) + (0.4111*(Re_vals/2.63e5)**-7.94)/(1+(Re_vals/2.63e5)**-8) + (0.25*Re_vals/1e6)/(1+Re_vals/1e6)

diameter = 0.045 # m
density_solid = 3000 # kg/m3
density_air = 1.22 # kg/m3
viscosity_air = 1.84e-5 # kg/(m.s)
g = 9.81 # m/s2
volume_sphere = 4/3*np.pi*(diameter/2)**3 # m3
mass_sphere = density_solid * volume_sphere # kg
gravity_force_sphere = mass_sphere*g # N
projected_area_sphere = np.pi * (diameter/2)**2

velocity_vals = Re_vals*viscosity_air/(diameter*density_air)
drag_force_vals = 0.5*Cd_vals*velocity_vals**2*projected_area_sphere

## Drag force vs. Velocity
fig, ax = plt.subplots()
ax.set_title('Drag Force, Smooth Sphere')
ax.set_xlabel('Velocity (m/s)')
ax.set_ylabel('Force (N)')
ax.plot(velocity_vals, drag_force_vals, label='Drag Force')
ax.axhline(y=gravity_force_sphere, linestyle='--', linewidth=1, color='r',label = 'Gravity Force')
ax.grid()
ax.legend()

[LuckyTran]

I didn't check that the code is 10000% correct but yes there is 3 intercepts for this unstable problem.

Criticality phenomenon are actually quite common. If you heat up water in a microwave it can immediately boil or stay in the superheated liquid state. Give the superheated liquid a little nudge and it liquid flashes immediately into vapor.

Furthermore, this is exactly why you put dimples on a golf ball to force it to pick a regime. Similarly, you shape bullets and add stabilizers to avoid the unstable region to prevent your bullet from tumbling.

[FliegenderZirkus]

But isn't only the middle intercept unstable? In the sense that any tiny perturbation breaks the force equilibrium and the falling ball switches to one of the other two equilibrium speeds, depending on the direction of the perturbation. 60 m/s and 130 m/s seem to be so far apart, that it's hard to see how could there be any switching between the two. Or perhaps this example is too simplistic and a real ball will always spin, so that more complex physics would need to be modelled?
In any case, my main point doesn't concern the stability, but rather the simple observation that a smooth sphere with diameter 0.045m in air experiences the same drag force at 60 m/s and 130 m/s. Say, in a controlled wind tunnel measurement. If that's correct, then it's totally counterintuitive. It makes me wonder if a similar drag crisis phenomenon can be found in more complex shapes as well. Wouldn't this have some real-world engineering applications?

[sbaffini]

Your drag force seems to miss the density... which, again, is not correct to assume constant in a free fall under gravity, at least not if it is meant to represent an actual fall on earth.

But, besides this, it might or not happen that you get another equilibrium, what is important is that it is not anymore for a true free fall. Also, in general, for anything falling, you typically want to have the slowest fall and not the opposite.

If you need to get faster, and you need propulsion to reach the higher velocity equilibrium (assumed it really exists), you are probably going to use propulsion all along the path or not having propulsion at all (I'm thinking about bombs and missiles here)

[LuckyTran]

You have to draw the work diagram to find the equilibria. For these exact numbers, the only equilibrium is 58 m/s. However, if you tinker with and increase the mass of the ball, you could potentially encounter an unstable problem so your question is still relevant. If you do this work diagram, you can show that 60 m/s and 130 m/s are both stable locations, the ball does not flip-flop between 60 and 130. Once at either of those velocities, it stays there. So in a freefall, the ball accelerates from 0 to 60 m/s and stops accelerating. There is no freefall paradox.

[FliegenderZirkus]

Originally Posted by sbaffini

Your drag force seems to miss the density

Thanks for pointing out the missing air density. Given that in this example it's indeed assumed constant, I'm not going to edit the first post because this typo doesn't affect the conclusions.

You make good points about the free fall. In retrospect I shouldn't have put that in the thread title as I'm mainly interested in the general result from wind tunnel measurement, which I still find quite surprising. I'm not trying to model atmospheric reentry of an aluminum golf-ball-sized sphere.

[FliegenderZirkus]

Originally Posted by LuckyTran

You have to draw the work diagram to find the equilibria.

Thanks for the hint about stability, I'll look up work diagrams and see if I get the same results as you. The 130 m/s seemed stable by intuition (increasing speed leads to higher drag force, so the ball wants to stay at that speed). But intuition is of course not to be trusted.

Still, my main point was that many engineers are unaware of the fact that drag force can decrease with speed. Take for example the Wikipedia article on drag force, which says

A smooth sphere, for example, has a Cd that varies from high values for laminar flow to 0.47 for turbulent flow. Although the drag coefficient decreases with increasing Re, the drag force increases.

I would say that this statement is inaccurate, do you agree? Drag force can decrease with speed, provided that the drag coefficient decreases faster than with square of speed.

[LuckyTran]

I forgot to mention, to get the work rate you just multiply the gravity/drag force by the velocity. And then you can plot this versus velocity. Whatever work rate is higher tells you whether the kinetic energy of the ball increases or decreases. You are just plotting a work-energy diagram instead of force-momentum diagram like you already have done for the drag force.

I would not split hairs to a sentence meant to explain why drag force can go up even though drag coefficient tends to go down.

Originally Posted by FliegenderZirkus

I would say that this statement is inaccurate, do you agree? Drag force can decrease with speed, provided that the drag coefficient decreases faster than with square of speed.

I would say this statement is also inaccurate because you are assuming that the density of the gas is a constant. You also have not considered that there might be an elephant in the way. I'm kidding of course. The explicit formula is given, all consequences of the formula are implicitly implied. For pedagogical purposes, don't make semantic arguments that can be semantically challenged. All the information is contained in the formula. If the formula is wrong, only then should we bring out the pitchforks.

However, I do agree that engineers do lack awareness of a lot of things that have been well known for a very very long time. One example I always come to is the pressure ratio for a choked nozzle. 1D compresible flow equations says the choked pressure ratio is 1.8. Anyone that has actually used a sonic nozzle (usually technicians with no higher level education) knows it is choked as low as 1.2.

[sbaffini]

Originally Posted by FliegenderZirkus

Thanks for pointing out the missing air density. Given that in this example it's indeed assumed constant, I'm not going to edit the first post because this typo doesn't affect the conclusions.

You make good points about the free fall. In retrospect I shouldn't have put that in the thread title as I'm mainly interested in the general result from wind tunnel measurement, which I still find quite surprising. I'm not trying to model atmospheric reentry of an aluminum golf-ball-sized sphere.

Well, it actually does, because the main conclusion is that your gravity curve will now intersect the drag curve only once then (check yourself).

But you could always pick up another material for the sphere. And I could claim that the Cd formula is not accurate enough. Etc.

I don't think the general result here, that drag can be absolutely lower for a faster sphere, is actually surprising. Again, that's how golf balls came out that way.

[LuckyTran]

Maybe not engineers but baseball pitchers certainly know how to abuse laminar-to-turbulent transition to throw Knuckleballs that are very hard to respond to. They understand that applying very little spin to the ball makes it extremely unstable. Although it is unlikely that they are interested in the not-simple, non-linear, non-harmonic oscillator problem (but be careful because several professional athletes have gone on to pursue advanced degrees), there is clear intent that the outcome they are trying to achieve is to make this ball moving as close to the critical region as humanly possible.