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 Remember 02-17-2010, 10:13 PM   #21 (permalink)
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Magic number

Quote:
 Originally Posted by MetroMPG Frontal area can be approximated by multiplying your vehicles width times its height,times 0.84.
I missed the basis for 16% reduction. Is height corrected for ground clearance, sides corrected for tapers, or what?

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Popular topics Other popular topics in this forum... >> See all the most popular projects & topics in our Aerodynamics subforum. 02-18-2010, 12:26 AM #22 (permalink) (:   Join Date: Jan 2008 Location: up north Posts: 12,636 Blue - '93 Ford Tempo Last 3: 27.29 mpg (US) F150 - '94 Ford F150 XLT 4x4 90 day: 18.5 mpg (US) Sport Coupe - '92 Ford Tempo GL Last 3: 69.62 mpg (US) ShWing! - '82 honda gold wing Interstate 90 day: 33.65 mpg (US) Moon Unit - '98 Mercury Sable LX Wagon 90 day: 21.24 mpg (US) Thanks: 1,550 Thanked 3,426 Times in 2,151 Posts I think it was statistically derived from comparing w x h and known frontal area values. __________________     The Following User Says Thank You to Frank Lee For This Useful Post: 08-08-2012, 07:12 PM   #23 (permalink)
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375

Quote:
 Originally Posted by kennybobby The 375 in the denominator should be 349. (348.66) That term comes from 0.5 times the conversion factor for mph^3 to (ft/sec)^3 divided by the definition factor for horsepower = 550 lbs-ft/sec. conversion factor: mph * 5280 ft/mile / 3600 seconds/hr = ft/sec. The 375 value assumes a mile is only 5153 ft.
Three of my reference texts use the 375 value.I made a leap of faith that the Ph.Ds had checked their own work.Apologize for the trouble.  08-09-2012, 06:41 PM   #24 (permalink)
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375 seems okay

Quote:
 Originally Posted by kennybobby The 375 in the denominator should be 349. (348.66) That term comes from 0.5 times the conversion factor for mph^3 to (ft/sec)^3 divided by the definition factor for horsepower = 550 lbs-ft/sec. conversion factor: mph * 5280 ft/mile / 3600 seconds/hr = ft/sec. The 375 value assumes a mile is only 5153 ft.
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!  09-09-2012, 12:18 AM   #25 (permalink)
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Yes, your clever formula is quite correct...

Quote:
 Originally Posted by aerohead 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.

Quote:
 Originally Posted by aerohead ...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)] Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MetroMPG Aerodynamics 2 05-10-2013 06:34 PM MetroMPG Aerodynamics 7 08-08-2012 07:00 PM MetroMPG Aerodynamics 4 02-16-2009 08:15 PM MetroMPG Aerodynamics 4 01-29-2008 02:41 PM MetroMPG Aerodynamics 0 01-28-2008 09:36 PM

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