# A Constant Ratio of Drag coefficient and terminal velocity

1. Oct 2, 2016

### jhan

Guys,, please check my own way to get Drag coefficient and Terminal Velocity by given Drag force,.. okay,,
lets for example: we know drag force is Fd= Cd*pf*v^2*A/2,, ill give little example (Cd=.02)*(pf=1027kg/m^3)*(v^2=.0025m/s)*(A=24m)/2 =( Fd .6162N) so drag force is .6162N,, okay..according to my analysis drag coefficient and terminal velocity are have constant ratio.. where???how??
okay,, lets do a recheck a given example above,, (.6162Nx2 = 1.2324N) next (1.2334N divide 24m of A= .05135) next (.05135 divide 1027 of fluid density= .00005)next ( .00005* of my own ratio 8= .0004)next (square root of .0004 is = .02 this your drag coefficient) next back to .05135 divide 1027 = .00005) next ( .00005 divide drag coefficient .02 = .0025) next ( square root of .0025 = .05 m/s so this your terminal velocity) :)

Last edited: Oct 2, 2016
2. Oct 2, 2016

### Simon Bridge

Welcome to PF;
I guess ... is this a personal theory or some homework?

... so your model for the drag force is that it is proportional to the square of the speed.
Unless some of the extra factors depend on speed.

... that would be a specific example of the drag force at a particular velocity in a articular fluid.

Are you taking $C_d$ as the drag coefficient?
Are you asserting that there must be some situation where $v/C_d=$ constant??
Well since the drag coefficient is a constant, the ratio is constant at any time that the speed is constant.
ie. when I travel down the street in my car at a constant speed of 50kmph, the ratio of the speed of my car to the car's drag coefficient is a constant.

Here's what I got:
You seem to have calculated first: $$C_d=\sqrt{8\frac{2F_d}{\rho_fA}}$$ ... which seems to have this factor of 8 in it for no reason, why did you do this?
...then you did: $$v_t = \sqrt{\frac{2F_d}{\rho_fAC_d}}$$ ... which is just the usual way to find the terminal velocity.
Considering you were given the drag coefficient in the first place, why did you need to do the first calculation?

But if I sub the expression for the drag coefficient into the expression for terminal velocity I get:
$$v_t = \sqrt{\frac{2F_d}{\rho_fA} \frac{1}{8}\frac{\rho_fA}{2F_d} } = \sqrt{\frac{1}{8}} \neq 0.05$$ ... so you have not actually used the first calculation.

Basically - calculating something is physics is not a sequence of steps to be followed blindly, that just happens to give you the right numbers: it is a consequence of a physical model which depends on the rules for how Nature works. The numbers for specific situation do not demonstrate anything, it is the relationships between the physical quantities that are important.

3. Oct 2, 2016

### jhan

what i mean.. is if the given is drag force, area, and fluid density,, then 8 is the ratio to find the Cd and velocity.. means (your Drag force is .6162N ) (your area is 24m^2) (then fluid density is 1027kg/m^3) then to find the drag coefficient and velocity is used to 8 ratio,,

4. Oct 2, 2016

### Simon Bridge

... how do you figure that?
How did you arrive at the number 8 for that purpose?
What part of the physics did you use to determine that "8" was the correct number to put in there to find the drag coefficient?

Here's the trouble:
The value for v that you got was not for the terminal velocity, it was just the velocity that went into the 1st equation (remember you wrote that $v^2=0.025$(m/s)2, well that means that $v=0.05$m/s ...) when you worked out what the drag force was for the object moving through the fluid at speed v. That is only the terminal velocity for the special case that the object has mass exactly 0.0629kg. Any other mass and it is wrong.

[edit - I got some of the algebra wrong in post #2]
Note: the formula you ended up with was: $$v_t = \sqrt{\frac{2F_d}{\rho_fA}\frac{1}{\sqrt{8}}\sqrt{\frac{\rho_fA}{2F_d}}} =\sqrt{\sqrt{\frac{F_d}{4\rho_fA}}}$$ ... if I follow you correctly.

Where are you up to in your education - particularly algebra and physics?

Last edited: Oct 2, 2016
5. Oct 2, 2016

### jhan

Fd=.6162N
Area=24m^2
Fluid=1027kg/m^3
Cd=?
√V=?

(Fd .6162N)= (Cd=?)×(pf=1027kg/m^3)×(V^2=?)×(A=24m)÷2

remember Cd is unknown, V is unknown

what i did is

(Cd)= √(Fd×2)÷(A)÷(pf)×8
√(.6162N×2)÷(24m^2)÷(1027kg/m^3)×8
equals to .02 so i think this is the Cd

Then

V= √(Fd×2)÷(A)÷(pf)÷(Cd)
√(.6162N×2)÷(24m^2)÷(1027kg/m^3)÷(.02)
equals to .05 m/s

So Drag force

Fd= (Cd×pf×V^2×A)÷2

= (.02)×(1027kg/m^3)×((.05m/s)^2)×(24m^2)÷2 is equals to .6162N

am i right???or no??

6. Oct 2, 2016

You still haven't shown where you are coming up with that factor of 8 that is in your calculations, and even then, it is hard to follow them because you insist on using numbers instead of symbols. If you can manipulate symbols such that the ratios have the same formulation, then you may be onto something.

As it stands, the drag force is defined using the drag coefficient by
$$F_D = \dfrac{1}{2}C_D\rho v^2 A.$$
For some object falling through a fluid of constant properties, a force balance yields
$$m\dfrac{dv}{dt} = F_D - mg,$$i
and in the case of terminal velocity, where $\frac{dv}{dt} = 0$, that is
$$0 = F_D - mg.$$
If you want to solve for terminal velocity, then
$$\dfrac{1}{2}C_D\rho v_t^2 A = mg,$$
and after some manipulation,
$$v_t = \sqrt{\dfrac{2mg}{\rho A C_D}}.$$

So, if the drag coefficient is
$$C_D = \dfrac{2 F_D}{\rho v^2 A},$$
and the above is true of the terminal velocity, then
$$\dfrac{C_D}{v_t} = \dfrac{\dfrac{2 F_D}{\rho v^2 A}}{\sqrt{\dfrac{2mg}{\rho A C_D}}}.$$
You can simplify that a little bit to
$$\dfrac{C_D}{v_t} = \dfrac{F_D}{v^2}\sqrt{\dfrac{2C_D}{\rho A mg}}.$$

So, the only way that $C_D/v_t$ is constant is if all of the above parameters are, in conjunction, constant. $C_D$ is always essentially constant over a wide range of parameters, and assuming the properties of the ball and fluid don't change, then $\rho$, $A$, $m$, and $g$ are constants. That means for your theory to hold, then $F_D/v^2$ has to be a constant. So,
$$\dfrac{F_D}{v^2} = \dfrac{1}{2}C_D\rho v^2 A\dfrac{1}{v^2} = \dfrac{1}{2}C_D\rho A.$$
In point of fact, that value is a constant. I still don't see where you are getting a factor of 8, nor do I see why this is in any way useful. In fact, this ratio depends on one of its own parameters:
$$\dfrac{C_D}{v_t} = \sqrt{\dfrac{\rho A C_D^3}{2 mg}}.$$
Perhaps the better relationship would then be
$$v_t\sqrt{C_D} = \sqrt{\dfrac{2 mg}{\rho A}}.$$
I am still not sure exactly why this is useful. It just says that two constants multiplied by one another is a constant.

I guess my question, then, is where are you going with this?