Terminal velocity equation for falling skydiver doesn't work for empirically determined values Try as I might, using the terminal velocity equation with empirically-determined $v_{T}$ and $C_{d}$ requires a skydiver to have a very small projected surface area.  I'm not sure what I am doing wrong, but I can't make the numbers work for the published assumptions.  (Here's a calculator so you can try it out yourself.)
The equation I am using is
$v_{T} = \sqrt{\frac{2mg}{\rho AC_{d}}}$
with the following constants:
$v_{T}$: Terminal velocity, in meters per second, assumed to be 54 m/s for a skydiver falling chest-down and 90 m/s for a skydiver falling head-first 
$m$: mass, assumed to be 70 kilograms 
$g$: Acceleration due to gravity, equal to 9.81 m/s${^2}$ 
$\rho$: Density of air, assumed to be 1.225 kg/m$^3$ 
$A$: Projected surface area, the parameter that seems to be going all wrong 
$C_{d}$: Coefficient of drag, assumed to be 1.1 for a chest-first skydiver and 0.6 for a head-first skydiver based on Wikipedia and this chart on human-powered vehicles 
The Mosteller formula gives a simple means of finding the total surface area of a human being, and for a 1.78 m tall 70 kg skydiver, it gives a value of 1.86 m$^2$.  I assume the belly-first skydiver's projected surface area to be about 40% of this, and the head-first skydiver's projected surface area to be about 12%.  (The Human Powered Vehicle chart lists an upright bike rider's frontal area as 5.5 square feet (0.5 of which is the bike), or 0.465 m$^2$ for the human alone.  That seems low to me, but the person is a little hunched over, even on an upright commuter bike.)  
With these assumptions, the head-first skydiver falls at about 91.5 m/s.  That's pretty close to the empirically-derived value, so I'm happy.  But it predicts the chest-first skydiver should be falling at a mere 37.0 m/s for a $C_{d}$ of 1.1 (already generously low.)
I thought, maybe I got the projected surface area wrong.  Maybe it's less than 40% of the total surface area.  But it needs to be as low as 18% of the total human surface area to get to 54 m/s, which is seems well out of bounds.  That, or the $C_{d}$ needs to be as low as 0.5 when it's listed at 1.1. Even if I use the suggested 0.465 m$^2$ frontal surface area, the terminal velocity is still 46.8 m/s, which is still out of bounds according to this aggregation of sources.
Because all the parameters are under that square root sign, I need to make a big adjustment in my assumptions to change $v_{T}$ even slightly.
So what is going on here? Do I have a number wrong, or do I need to take Reynolds number into account in this regime?  (It seems strange it'd show up for a lower $v$ and not for a higher one.)  Or is the $C_{d}$ of a belly-first skydiver really less than that of someone falling head-first?
 A: 
for a 1.78 m tall 70 kg skydiver, it gives a value of 1.86 m2

That doesn’t seem right. That area requires the person to be more than a meter wide! 
This page shows that the Mosteller formula is for the total surface area, I.e. amount of skin. Instead you want the projected a.k.a. frontal area. That’s more than a factor two smaller. 
A: You have not stated the conditions in which your empirically-derived values were obtained.
Air density decreases with height. While it is 1.225 kg/cu.m at ground level, it is 0.85 kg/cu.m at 12,000 ft which is the standard jump height for skydiving (see what is the maximum altitude a skydiving plane can fly).
A: Let's estimate error margins for your numbers, apply error propagation and see what happens.
Let the error estimate for the terminal velocity equation be:
$$\delta v_t = \sqrt{({\partial v_t \over \partial m} \delta m)^2 + ({\partial v_t \over \partial g} \delta g)^2 + ({\partial v_t \over \partial \rho} \delta \rho)^2 + ({\partial v_t \over \partial A} \delta A)^2 + ({\partial v_t \over \partial C_d} \delta C_d)^2} $$
This is a common technique for simple error propagation, where it is assumed the error of the variables are independent. The partial derivatives all follow a similar formula:
$$|{\partial v_t \over \partial x} \delta x| = \frac{v_t \delta x}{2 x} $$
Here is a table of the values I used:
$$\begin{array}{c}
\text{Variable} & \text{Min} & \text{Max} & \text{Value} & \text{Error} & |{\partial v_t \over \partial x} \delta x|  \\
g & 9.7903 & 9.81 & 9.80019 & 0.00981 & 0.02387 \\
m & - & - & 70 & 0.5 & 0.17035 \\
\rho & 1.269 & 1.225 & 1.247 & 0.022 & 0.42075 \\
A & 0.465 & 0.744 & 0.6045 & 0.1395 & \color{red}{5.5036} \\
C_d & 0.5 & 1.1 &  0.8 & 0.3 & \color{red}{8.9434} \\
\hline \\
v_t & - & - & 47.698 & 10.511 & - \\
\end{array} $$
Thus $v_t = 47.698 \pm 10.511$m/s is the result of error propagation. That does place $v_t$ within error margins of 54 m/s.
For each variable (except mass) I took the minimum value and maximum value you provided. The value column is then the mid-point and the error is half the range. Mass was given an error of simply half the last digit. Gravitational acceleration was assumed maximum at ground and about 0.2% less at 5km height (arbitrary). Air density was assumed the minimum value  at 15C and the maximum value at 5C.
I represented you having uncertain values for $C_d$ and $A$ by giving them wide errors; clearly these were the contributing factors. It should be illustrative now where your major sources of uncertainty are. 
