Terminal velocity of a rocket using dimensional analysis

[Enthu]

I started taking MIT's online physics course a couple of days ago, and since I have no physics background at all, I'm getting a bit confused with dimensional analysis.
I'm trying to find a formula for the terminal velocity of a rocket, using air density ρ, gravity g, area of rocket affected by air A, mass m and the drag coefficient Cd. I have reached the following equation of dimensional analysis:
LT^-1 = [M]^V$\cdot$[LT^-2]^W$\cdot$[ML^-3]^X$\cdot$[L²]^Y

I'm pretty sure that is completely wrong. I tried solving for the coefficients and ended up with V = 1 X = 1/2 W = 1 Y = 1/2, and I'm certain that is wrong as well.

Can somebody help me with this? I just can't seem to get a correct formula.
Also, as far as I understood it, from the equation above it is implied that mass should have no effect, but the actual formula for the terminal velocity of a rocket does have mass in it.

--this isn't a homework/coursework question, I'm doing this out of my own curiosity--

 

[mfb]

Well, you have 4 variables and just 3 equations. Possible fix: Use the gravitational force of the rocket, m*g, as single variable.

Anyway:
Solving for M gives V=-X (note the sign).
Solving for T gives W=1/2
The equation for L is now 1=W-3X+2Y, which can be simplified to 1/2 = -3X+2Y.
If m and g can appear as m*g only, this implies V=W and therefore V=1/2, X=-1/2. This allows to determine Y=-1/2.

As final formula:

Spoiler

$$v \propto \sqrt{\frac{mg}{\rho A}}$$

 

[Enthu]

Ah, thanks. I get it now. Any idea how the actual formula got to $v = \sqrt{\frac{mg}{\frac{1}{2}ρC_dA}}$ (i.e. where's that 0.5 from)?

 

[Infinitum]

Ah, thanks. I get it now. Any idea how the actual formula got to $v = \sqrt{\frac{mg}{\frac{1}{2}ρC_dA}}$ (i.e. where's that 0.5 from)?

That is the constant of proportionality.

If, for example,

$$v \propto \sqrt{\frac{mg}{C_d\rho A}}$$

then to introduce an equality sign, we have to introduce a constant,

$$v = k \sqrt{\frac{mg}{C_d\rho A}}$$

Usually, this constant is evaluated by experiments. In this particular case k came out to be $\sqrt{2}$ You cannot get it by dimensional analysis because the number has no dimension by itself.

 

[Enthu]

That is the constant of proportionality.

If, for example,

$$v \propto \sqrt{\frac{mg}{C_d\rho A}}$$

then to introduce an equality sign, we have to introduce a constant,

$$v = k \sqrt{\frac{mg}{C_d\rho A}}$$

Usually, this constant is evaluated by experiments. In this particular case k came out to be $\sqrt{2}$ You cannot get it by dimensional analysis because the number has no dimension by itself.

Oh, that's something I must have missed going through my papers. Thanks.

 

[Rap]

I know this is not a "standard" analysis, but you can say that the x, y, and z distances have different dimensions. In other words, you have 5 fundamental variables m, t, Lx, Ly, and Lz. Then, assuming the z direction is the direction of motion (up):
$$v\propto L_z/t$$
$$\rho\propto\frac{m}{L_x L_y L_z}$$
$$g\propto L_z/t^2$$
$$A\propto L_x L_y$$
and you get
$$\frac{L_z}{t}\propto\left(m\right)^a\left(\frac{L_z}{t^2}\right)^b\left(\frac{m}{L_xL_yL_z}\right)^c\left(L_xL_y\right)^d$$

You can solve to get a=b=1/2 and c=d=-1/2, and get $$v\propto\sqrt{\frac{m g}{A \rho}}$$ same as above. Check out http://en.wikipedia.org/wiki/Dimensional_analysis#Huntley.27s_extension:_directed_dimensions for an explanation of this refinement of dimensional analysis

(Note - the area is Lx Ly because the plane of the area is perpendicular to the direction of motion)