Drag Force Equation Useless in Finding Terminal Velocity

In summary: I appreciate your patience and assistance.In summary, the drag force equation Fd=1/2*p*Cd*V^2*Cd*Ac can be used to find the weight (W) of an object at terminal velocity. However, if the values of all variables except Cd and V are known, this equation will not provide a solution for Cd. The terminal velocity equation Vt= sqrt 2weight/p*Cd*Ac is derived from Fd= W and can be used to find the terminal velocity (Vt) of an object with a known drag coefficient. To find the drag coefficient, one can guess a value and calculate the Reynolds Number, then use a chart to find the corresponding Cd. Iterations may be
  • #1
picklefeet
16
0
As most of you know, the drag force equation is Fd=1/2*p*Cd*V^2*Cd*Ac.
At terminal velocity, Fd equals weight. If you find the value of all the variables except Cd and V, you still won't have anything. But the terminal velocity equation is Vt= sqrrt of 2weight/p*Cd*Ac. So if V is terminal velocity, it is reasonable to conclude that V^2 is equivalent to 2W/p*Cd*Ac. So if you replace V^2 with what I've stated above, you'd get W= 1/2p*Cd*2W/p*Cd*Ac.
This equation is true, but only because it cancels out everything but W. You are left with W=W. This is useless for finding Cd. You could also do the opposite. Replace Cd with (1/V^2*2weight)
pAcCd

In this case everything cancels out too. In conclusion, this formula is useless in finding terminal velocity.
 
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  • #2
Well, no. Clearly the forumla is useful for finding terminal velocity of an object with a known drag coefficient.
 
  • #3
Since I am assuming that you are carrying this over from the previous thread about the quarter...

I'm going to neglect the buoyant force...
[tex]F_d = W[/tex]

[tex]\frac{1}{2} \rho V^2 A C_d = m*g[/tex]

For air you can use STP value of :
[tex]\rho = 1.201 \frac{kg}{m^3}[/tex]

[tex]\mu = 1.79 x 10^{-5} \frac{N*s}{m^2}[/tex]

The quarter has values of:
[tex]d = 24.15 mm[/tex]
[tex]A = 4.522 x 10^{-4} m^2[/tex]
[tex]m=5.67 x 10^{-3} kg[/tex]

So now we have:
[tex]\frac{1}{2}\left(1.201 \frac{kg}{m^3}\right)\left(V^2\right)\left(4.522 x 10^{-4} m^2\right)\left(C_d\right) = \left(5.67 x 10^{-3} kg\right)\left(9.81 \frac{m}{s^2}\right)[/tex]

[tex]V^2 C_d = 206.22 \frac{m^2}{s^2}[/tex]

Now, it just so happens, that if you look up the drag coefficient for a circular flat plate, perpendicular to flow,
it is a relatively constant value of 1.1 which makes this a trivial solution.

[tex]V = \sqrt{\frac{206.22 \frac{m^2}{s^2}}{1.1}}[/tex]

[tex]V = 13.69 \frac{m}{s}[/tex]

However, if you did not have a constant [tex]C_d[/tex] you would guess at the [tex]C_d[/tex] and calculate the resultant [tex]V[/tex].

From that [tex]V[/tex] you would then calculate the Reynolds Number for that flow using

[tex]Re = \frac{\rho*V*D}{\mu}[/tex]

You would then use the calculated Reynolds Number and look up on a chart the corresponding [tex]C_d[/tex]. If that value you
just looked up comes very close to your guess, you are done. If it isn't, use the new value of [tex]C_d[/tex] you just looked up
and recalculate the velocity. Repeat the Reynolds Number calculation to get another [tex]C_d[/tex] value.

Keep doing that until you converge on a solution. It shouldn't take that many iterations to solve it.

The fact that you got W=W in your result means that you had circular reasoning in your analysis. You did some algebra on the original equation and then plugged it right back into the original equation. If you do that I should hope you get W=W.
 
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  • #4
russ_watters said:
Well, no. Clearly the forumla is useful for finding terminal velocity of an object with a known drag coefficient.

The fact is, if I already knew the Cd, I wouldn't have to use the Fd equation.
I'd just use the terminal velocity equation of Vt= sqrt 2weight
pAcCd
And I did find the terminal velocity ofa quarter. It's 83.136 miles per hour (rounded to the nearest thousandth.) on a different not, How do I use the mathematical symbols? i can't find them.
 
  • #5
picklefeet said:
The fact is, if I already knew the Cd, I wouldn't have to use the Fd equation.
I'd just use the terminal velocity equation of Vt= sqrt 2weight
pAcCd
But the terminal velocity equation is derived from Fd = Weight. You're going in a circle again.
And I did find the terminal velocity ofa quarter. It's 83.136 miles per hour (rounded to the nearest thousandth.)
You're way off, you slipped a digit or two somewhere. Per Fred's response, the terminal velocity of a quarter falling belly up is about 14m/s, or a mere 30 mph.
 
  • #6
I'm realize now that I placed the decimal point one space to the left while figuring out the cross-sectional area (oops). But for some reason I don't see where you got 30 m.p.h. If I plug in 1.17 for Cd, 11.34 g. for weight, .46353465 m^2 for Ac and 1.184 for air density, I get about 10.9914 miles per hour. My problem might be the units I'm using. If so, please correct me. thanks.
 
  • #7
You need to use Newtons for the weight, not grams. Also, your area is off. Take a look at my post and you'll see that the area is [tex]A = 4.522 x 10^{-4} m^2[/tex]
 
  • #8
Thanks for the info. For the cross-sectional area, do I use centimeters or meters? And I still don't know where to find the mathematical symbols to put in my replies. Can you tell me where they are?
 
  • #10
Thank you to all who helped me with this problem. And thanks for the link to the mathematical symbols.
 

1. What is the drag force equation and how is it used to find terminal velocity?

The drag force equation is a mathematical formula that calculates the force of air resistance acting on an object moving through a fluid, such as air or water. Terminal velocity is the maximum speed that an object can reach when falling through a fluid. The drag force equation is used to estimate the terminal velocity of an object by equating the drag force with the weight of the object.

2. Why is the drag force equation considered useless in finding terminal velocity?

The drag force equation is considered useless in finding terminal velocity because it assumes that the drag force remains constant throughout an object's acceleration. However, in reality, the drag force increases as an object's speed increases, and eventually, the drag force will equal the weight of the object, resulting in a constant terminal velocity. Therefore, the drag force equation cannot accurately predict the true terminal velocity of an object.

3. Are there any situations where the drag force equation can be used to find terminal velocity?

The drag force equation can be used to estimate the terminal velocity of a small, spherical object falling through a fluid at low speeds. In this scenario, the drag force remains relatively constant, and the equation can provide a rough estimate of the terminal velocity. However, for larger or more complex objects, the drag force equation is not reliable.

4. How does air resistance affect an object's terminal velocity?

Air resistance, also known as drag force, acts in the opposite direction of an object's motion and increases as the object's speed increases. As a result, the drag force eventually equals the weight of the object, causing it to reach a constant terminal velocity where the forces are balanced.

5. What other factors besides air resistance affect an object's terminal velocity?

In addition to air resistance, an object's weight, density, shape, and surface area can also affect its terminal velocity. Objects with a greater weight or cross-sectional area experience a greater drag force, resulting in a lower terminal velocity. Objects with a streamlined shape experience less drag force and can reach higher terminal velocities. The density of the fluid also plays a role, with denser fluids resulting in higher drag forces and lower terminal velocities.

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