Need help with deriving air drag equation

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SUMMARY

The discussion centers on deriving the air drag equation, specifically the relationship between force and velocity squared (F = ½DCAv²). Participants explore the principles of work, kinetic energy (KE), potential energy (PE), and the empirical nature of the drag coefficient. Key concepts include momentum conservation and the proportionality of collision frequency and mean velocity to speed. The conversation highlights the limitations of simplistic models in accurately representing drag forces in fluid dynamics.

PREREQUISITES
  • Understanding of basic physics concepts such as work, kinetic energy, and potential energy.
  • Familiarity with fluid dynamics principles, particularly drag forces and Reynolds number.
  • Knowledge of empirical modeling and its application in physics.
  • Basic grasp of momentum conservation and its relevance to collision dynamics.
NEXT STEPS
  • Research the derivation of the drag equation in fluid dynamics.
  • Study the implications of Reynolds number on drag force behavior.
  • Explore advanced models for drag that account for varying flow conditions.
  • Learn about the empirical methods used to determine drag coefficients in real-world applications.
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Physics students, engineers, and researchers interested in fluid dynamics, particularly those focusing on drag force calculations and empirical modeling techniques.

Johnkrkr
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△Work = △KE + △PE
Work = Force x Distance
W=F△x Pressure = Force/Area
W=PA△x Volume=Area x length
W=PV Work = Pressure x Volume

KE=(1/2)mv^2
Density = mass/volume
mass = density x volume m=DV
KE=(1/2)DVv^2 (work done by air pressure?)

PE=mgy
PE=DVgy (work done by gravity and air pressure??)

PV=(1/2)DVv^2 + DVgy
P=(1/2)Dv^2 + Dgy
F=PA
F=(1/2)Dv^2A + DgyA (final equation?)   

Final equation I was expecting was F= ½DCAv2
I'm so confused right now.
can you guys help me fix the equation?
 
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You cannot strictly "derive" the V^2 drag law from something like F = m*a. It is largely an empirical result. You can show that it works by dimensional analysis, but the drag coefficient is still an empirical value.
 
Dr.D said:
You cannot strictly "derive" the V^2 drag law from something like F = m*a.
Not strictly, but the ~v2 relationship is based on momentum conservation.
 
A.T. said:
Not strictly, but the ~v2 relationship is based on on momentum conservation.

Really? That's news to me. I'd like to know how you think that is done.
 
Dr.D said:
Really? That's news to me. I'd like to know how you think that is done.
- Number of collisions per second is proportional to speed
- Mean velocity of collisions is proportional to speed
- The force (momentum transfer) is proportional to the product of both above, so it's proportional to speed squared.
 
A.T. said:
- Number of collisions per second is proportional to speed
- Mean velocity of collisions is proportional to speed
- The force (momentum transfer) is proportional to the product of both above, so it's proportional to speed squared.

That sounds good, but I'd like to examine these statements more closely.

You say that "the number of collisions per second is proportional to speed" where we presume that this refers to the speed of the object subject to drag. So, does this mean that a moving car hits more gas molecules per unit of time than a stationary car does, remembering that the gas molecules are wandering around in random directions all the time? That seems questionable. Can you justify this, please?

What is the "mean velocity of collisions"? In air, with a mix of molecules of various sorts, all whizzing around in random directions, this term does not seem to me to have much meaning. Can you define it more precisely, please?
 
Dr.D said:
That sounds good, but I'd like to examine these statements more closely.
If you look more closely, you will find that this simple drag formula doesn't really work. But if you want a simple formula you use a simplistic model.

Dr.D said:
...gas molecules are wandering around in random directions ...all whizzing around in random directions...
Since it's random in all directions, it doesn't produce any net force and can be ignored in a simplistic model.
 
A.T. said:
If you look more closely, you will find that this simple drag formula doesn't really work. But if you want a simple formula you use a simplistic model.

I'll bite: How closely do I have to look? This formula is certainly extremely widely used for all cases where the Reynold's Number is too high for a viscous model to apply. There are countless folks who think it works well enough to use it everyday, so a vast number of us need your wisdom and insight. Where do I look for a better formula?

A.T. said:
Since it's random in all directions, it doesn't produce any net force and can be ignored in a simplistic model.

But, but, but ... wasn't this the basis for your first two statements at the beginning of our discussion?

Nice dodge to avoid addressing both of my points!
 
Dr.D said:
But, but, but ... wasn't this the basis for your first two statements at the beginning of our discussion?
Nope. I was talking about a model which ignores Brownian motion, because it doesn't produce any net force.
 
  • #10
Looks to me like your original assertion, "Not strictly, but the ~v2 relationship is based on momentum conservation" has no basis in fact at all. You have steadfastly refused to answer any of my questions for clarification. Sounds like hot air to me!
 
  • #11
Dr.D said:
You have steadfastly refused to answer any of my questions for clarification.
What is still unclear?
 

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