# Can surface roughing effects on an airfoil Reynolds number be quantified?

I am building balsa free flight gliders with a chord of 0.10 meters and a velocity of around 3 meters per second. This works out to a Reynolds number of around 15,000.

The wings are around 1 meter in span, and are made with the wood grain running lengthwise. They are thin undercambered (no bottom covering) with maximum thickness of 7 mm at 30% of chord from leading edge.

Because of the roughness of the unfinished grain, is my Reynolds more like 50,000, or am I still down in the range where a flat plate might be better?

Roughness has no influence on the Reynolds number. It will, however, help to trigger a laminar-turbulent transition. That is the reason for dimples in golf balls. Also, roughness will increase the friction drag in turbulent boundary layers.

Flat plate friction diagram (picture source). Note the horizontal lines: They show that drag does not drop any more with increasing Reynolds number once the surface roughness exceeds a certain value. Roughness is given here relative to the chord length as $$\frac{\epsilon}{l}$$, with $$\epsilon$$ being the mean height of the roughness and $$l$$ the chord length.

The diagram only starts at Re = 100,000, so it is clearly outside the range of your application. If you sand your wooden surface with fine grit sandpaper, I don't expect that the remaining roughness will increase drag. At those Reynolds numbers it might be advantageous to add a trip wire in order to induce a transition to turbulent flow – see the golf ball example.

• Frisbee grooves may be a bit closer. Roughness to chord length ratio interesting way of trying it, as well as location on the airfoil. Commented Oct 18, 2019 at 16:08
• What's causing the increasing in drag with Re for the low surface roughness lines?
– JZYL
Commented Oct 18, 2019 at 23:08
• @Jimmy Don't confuse "increasing drag" and "increasing the drag coefficient". For a given structure, lower Re means lower velocity and the drag coefficient is multiplied by $v^2$ to get the drag force. But for laminar flow the drag force is proportional to $v$ not $v^2$, so the drag coefficient used in the conventional formula appears to be bigger at low Re even though the drag force is less. Commented Oct 19, 2019 at 0:01
• @alephzero I didn't confuse. 1e6 Re is a standard wind tunnel Re for engineering design. No one will say the drag is linear to V.
– JZYL
Commented Oct 19, 2019 at 0:39
• What I'm getting is that larger roughness to chord ratio is "completely turbulent" at lower Reynolds, therefor "transitional turbulent" at a lower airspeed, where drag reduction occurs. This may have been what Kline-Fogelman were after, but their "disrupter" steps may have been too large. Commented Oct 19, 2019 at 11:53