Automotive Force on dog's head out of car sunfoof

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The discussion focuses on calculating the wind force acting on a dog's head at 50 mph through a car sunroof, using the drag equation and wall loading calculations. The frontal area of the dog's head is noted as 0.0314 m², with a drag coefficient initially assumed to be 2. Participants identify that the drag coefficient may be too high and emphasize the need for correct unit conversions, specifically converting speed from mph to m/s. The calculations yield similar results, but both methods are deemed incorrect due to unit issues and assumptions about the drag coefficient. Accurate calculations are essential for understanding the forces involved in this scenario.
oneshortplanck
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(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 = 19Nany thoughts
 
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Yayyy! They're approximately equal.
 
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 tongue flapping contributes to an anamalous CD value.
 
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billy_joule said:
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.
 

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