Need help with deriving air drag equation

In summary, there are various equations related to work, force, and energy that involve terms such as mass, velocity, area, length, pressure, and volume. The final equation that was expected is F=½DCAv2, but it cannot strictly be derived from F=m*a and is largely an empirical result. The relationship between drag and velocity is based on the number of collisions per second, the mean velocity of collisions, and the force (momentum transfer), but these statements may not hold true in all cases and can be ignored in a simplistic model.
  • #1
Johnkrkr
1
0
△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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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!
 
  • #9
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?
 

Related to Need help with deriving air drag equation

1. What is the air drag equation and how is it derived?

The air drag equation, also known as the drag force equation, is a mathematical formula used to calculate the force of air resistance on an object moving through a fluid, such as air. It is derived from the fundamental principles of fluid mechanics, specifically the concept of drag force being proportional to the square of an object's velocity. The equation takes into account factors such as the object's shape, density, and the fluid's viscosity to calculate the force of air resistance.

2. How is the air drag equation used in scientific research?

The air drag equation is used in a variety of scientific research fields, including aerodynamics, fluid mechanics, and physics. It is often used in experiments and simulations to study the effects of air resistance on different objects, such as airplanes, cars, and projectiles. The equation is also used in engineering and design processes to optimize the performance and efficiency of various objects moving through fluids.

3. Can the air drag equation be applied to all objects moving through air?

The air drag equation is applicable to most objects moving through air, but it may not accurately predict the drag force for objects with complex shapes or those moving at extremely high velocities. In these cases, additional factors may need to be considered, such as turbulence and compressibility effects. Additionally, the air drag equation may not be accurate in extreme environmental conditions, such as high altitudes or low air densities.

4. Are there any limitations or assumptions associated with the air drag equation?

Like any mathematical model, the air drag equation has some limitations and assumptions. It assumes that the fluid flow is laminar and that the object is moving through the fluid at a constant velocity. It also assumes that the object's shape and orientation do not change during its motion. These assumptions may not hold true in all real-life scenarios, but the equation is still a useful tool for estimating air resistance in many situations.

5. Is there a simplified version of the air drag equation?

Yes, there are simplified versions of the air drag equation that are commonly used in introductory physics and engineering courses. These simplified equations may assume a constant drag coefficient or ignore certain factors, such as the object's size or fluid viscosity. While these simplified equations may not provide the most accurate results, they can still give a general understanding of the concept of air resistance and how it affects moving objects.

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