Calculating Force of Drag Using the Reynolds Number

[luckis11]

http://www.grc.nasa.gov/WWW/BGH/reynolds.html

(Reynolds Number)=(Inertial Forces)/(Viscous Forces)

My question is: Force of drag Fd=?

Wild guess: Fd=Inertial Forces?

 

[stewartcs]

http://www.grc.nasa.gov/WWW/BGH/reynolds.html

(Reynolds Number)=(Inertial Forces)/(Viscous Forces)

My question is: Force of drag Fd=?

Wild guess: Fd=Inertial Forces?

Drag is a mechanical force:

http://www.grc.nasa.gov/WWW/K-12/airplane/drag1.html

CS

 

[Bob S]

See

http://en.wikipedia.org/wiki/Drag_(physics )

There are two regimes, turbulent (high Reynold's number) and linear (Stokes) drag.

Bob S

 

[luckis11]

I meant how Fd is related to:

(Reynolds Number)=(Inertial Forces)/(Viscous Forces)

 

[stewartcs]

I meant how Fd is related to:

(Reynolds Number)=(Inertial Forces)/(Viscous Forces)

Did you not read the link I gave you?

...For the gas, the magnitude depends on the viscosity of the air and the relative magnitude of the viscous forces to the motion of the flow, expressed as the Reynolds number...

CS

 

[Bob S]

See Navier Stokes equation in

http://en.wikipedia.org/wiki/Reynolds_number

Bob S

 

[luckis11]

"...For the gas, the magnitude depends on the viscosity of the air and the relative magnitude of the viscous forces to the motion of the flow, expressed as the Reynolds number..."

So? We have:
Re=(inertial forces)/(viscous forces)

At wiki it says
" Note that this (Re) is equal to the ratio between...which is the drag (up to a numerical factor, half the drag coefficient)..."

So, Fd=(inertial forces)?
Should't it be Fd=(inertial forces)+(viscous forces)?

Both wrong? Any site making all this a bit clear?