Force on dog's head out of car sunfoof

[oneshortplanck]

(i have made this thread as i previously had it in the aero section and it was probably not getting the attention such an important question deserves.)

lets use the sunroof as an example for simplicity.
what force is the wind acting on the dog at 50mph/22.3m/s?
dog head diameter = 0.2m
therefore frontal area of dog head is 0.0314m^2
Drag coefficient = 2
air density = 1.22

here is my attempt (#1) based on wall loading calculations:
0.00256 * 50 * 50 = 6.4 lb/ sq feet = 306.4Pa

0.0314 * 306.4 * 2=19.25N

here is my attempt (2) based on drag equation
2 x ((1.22 x 50^2)/2) * 0.0314 = 19N


any thoughts

 

[SteamKing]

Yayyy! They're approximately equal.

 

[billy_joule]

But both equally wrong unfortunately! You need to use the correct units in the drag equation, mph won't do. Also, 2 for Cd seems high for a dogs head, where did you get that from?

 

[HuskyNamedNala]

The tongue flapping contributes to an anamalous CD value.

 

[oneshortplanck]

But both equally wrong unfortunately! You need to use the correct units in the drag equation, mph won't do. Also, 2 for Cd seems high for a dogs head, where did you get that from?

the first is correct units, but the second should read
2 x ((1.22 x 22.352^2)/2) * 0.0314 = 19N

yes, i agree on the 2 cd, perhaps its the engineering factor of safety.