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Old 09-08-2012, 11:18 PM   #25 (permalink)
Southern Squidbillie
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Yes, your clever formula is quite correct...

Originally Posted by aerohead View Post
I re-visited my books.And it looks like the 375 value is valid.
Working with the 550 lb'ft/sec value for horsepower calculation requires mph to be multiplied by 5,280 to get feet,then division by 3,600 sec/hr to get it into feet/sec..
If you already have a drag force at a given mph value,to get power,you can use this shortcut,multiplying the force by mph,then dividing by 375.
550/375 yields a constant percentage = to( feet/sec)/(mph) at any given velocity.It's always 1.466X.
Hope that helps!
Okay yes that helps and now i see where my confusion arose, and i agree that if you already have the Aero drag Force in unit of lbs, then you can get the Power in Horsepower by multiplying by speed in mph and dividing by 375:

1 mph x 5280 / 3600 = 1.466 ft/sec [the conversion from mph to ft/sec is 1.466].

power (in units of lb-ft/sec) / 550 = power (in units of Horsepower, HP).

So 1.466 / 550 = 1 / 375 does indeed convert mph to ft/sec and power to HP when multiplying Force in lbs times Speed in mph.

My confusion with your HP formula was three-fold: there was a factor of 1/2 missing inside the square brackets in your equation for the Aero drag "Force", and there was no conversion factor inside the brackets to convert speed in mph to ft/sec, and the constant term inside the brackets was so close to the value for the density of air.

Originally Posted by aerohead View Post
...The formulas,once posted,will allow everyone to calculate aerodynamic loads ...
The horsepower it takes at the drivewheel of your vehicle to overcome aerodynamic drag can be estimated by the formula

HP =V/375 [ 0.00256 X Cd X A X (V squared)]

where V= speed in miles per hour, Cd is your drag coefficient,A= frontal area of your vehicle,and (V squared) is your speed times itself.
So now i see how your formula is correct, and please allow me to expand on the derivation of this formula:

Aero drag Force = 1/2 x (air density) x V^2 x Cd x A ; air density = 0.00237 lbm/ft^3

Power = Force x Speed , so the fully expanded version of the formula for the Aero drag in Horsepower would be

HP = (V x 1.466/550) x [ 0.5 x 0.00237 x Cd x A x (V squared) x (1.466 squared)]

So the constant term inside the square brackets of your formula is really NOT the air density, but is the product of 0.5 x 0.00237 x (1.466^2) = 0.00256 , and this all simplifies down to the clever formula that you provided:

HP = (V/375) x [ 0.00256 x Cd x A x (V squared)]
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