Why do flat-plate friction studies only consider half the plate?

I'm trying to learn friction (only - nothing else - just skin friction) on a surface as it moves through water. I'm hoping to find a working formula, including all necessary units, in terms of wetted area, viscosity, and velocity (while, along the way, absorbing the subject well).

More than twice now, I've encountered almost-perfect papers, except for this critical weirdness:-

3.5. Experimental condition The experiments were conducted as follows. The towing speeds were set at U0 = 0.25, 0.5, 0.75, 1.0, 1.125, 1.25, 1.375, 1.5, 1.625, 1.75, 1.875, 2.0 m s−1 For each speed, the draft was varied among h = 0.590, 0.635, 0.700, 0.785 m, where h is defined as the distance from the bottom of the plate to the water surface. The total drag was measured under the same conditions for plate A (L = 3.3 m) and plate B (L = 4.3 m). Wetted surface area S is defined as L ×h. The Reynolds number ReL varied from 8×10 to 10 .

A flat plate has two sides - so "Wetted surface area" should be 2* L * h.

The paper goes on using this mistake, complete with comparisons to other tables and so on... thus, either I'm missing something important in my comprehension, or, someone somewhere made that mistake and now it's become the standard, or more than a few papers are both wrong, and fudging their results to match reference tables ?

For those curious, I'm working on examining the difference between rotating a propeller inside a fixed shroud, as opposed to rotating a propeller that is physically attached to the shroud (i.e. rotating the shroud too)

I am aware that skin friction formulas are abundant, and that there's two (one for slow, one for fast), however, they're typically all missing units, making them not practically useful.

In the test situation the drag force is obviously the same on both sides of the plate. But in real life applications there is often only flow over one side of the plate.

For example, the flow over the inside and the outside of your propeller shroud will be different, so a "two sided" formula for the drag that assumed they are the same would be useless.

The fact that the formulas do not "contain units" but are presented in terms of non-dimensional parameters is what does make them practically useful.

Reynolds number and Mach number in aerodynamics are probably the most common non-dimensional parameters of "numbers", but there are many others. There is a list of more than 50 here.

• Your answer makes no sense. The area always has units, so does viscosity and velocity etc. It is impossible to practically use any formula that includes any of those things, if it doesn't mention what they're measured in (meters? foot? miles? etc). I also very clearly (and twice) mentioned I'm interested in FRICTION. Drag is something else - I only want the friction component here.
– cnd
Jul 8, 2020 at 13:06
• @cnd Units are always omitted in useful formulae. For example Lift $= \kappa \rho V^2$ (Area), where $\kappa$ is a dimensionless number depending on the shape an angle of attack of the wing, gives the same physical answer in any units because the two sides of the equation have dimensions MLT$^{-2}$. Jul 8, 2020 at 13:22
• For convenience let $\kappa=1$. In SI so $\rho$ is in kg per cubic meter and $v$ is in ms$^{-1}$ then lift is in kg m s$^{-2}$ which is the same Newtons. if we use fps so that $\rho$ is pounds per cubic foot and $v$ is feet per second then lift is in pound feet per second squared = poundals. If we use cgs so that that $\rho$ is in gm per cc and $v$ is in cm per second then Lift is in gram-centimeter per second squared =dynes. In each case the output number is in the approprate units for force without conversion factors. That's what being dimensionally correct means. Jul 10, 2020 at 0:21
• My comments relate to your misunderstanding of the need to specify units in formulae. I used the lift as an example because it is simple. Jul 15, 2020 at 11:35

3. The formulas are useful because they do not have units prescribed. You can decide whatever you want to get out and out in the appropriate units into the formula. i.e. $$F=ma$$. Does this have units? No. Is it useful? Extremely. If you out in $$kg$$ and $$\mathrm{m/s^2}$$ you get N of force, if you put in $$\mathrm{lbm}$$ and $$\mathrm{ft/s^2}$$ you get $$\mathrm{lbm \cdot ft/s^2}$$ out. The exact same reasoning applies to your skin friction problem