CD for fit? .34?
#21
05-13-2009 | 12:39 AM
Looks like I never responded in this thread, my apologies.
This thread contains this post by me and others discussing the underpanel. Short version: its a winner.
Likely to be JDM only because the USDM cars are longer on the front and rear for 5mph crash nonsense reasons. Looks like it would be easy enough to section in extra length in the center though. I've never actually seen a picture of a car with that piece on it.
This thread contains this post by me and others discussing the underpanel. Short version: its a winner.
Anybody found a place to buy the Mugen unit stateside? Or is it JDM only?
#24
03-26-2010 | 02:58 PM
Member
Joined: Sep 2008
Posts: 622
From: Québec city
Marko!
#25
04-08-2010 | 04:46 PM
New Member
Joined: Apr 2010
Posts: 14
From: FL
With some Physics, you could find out the drag yourself! =)
KE(i) + PE(i) = KE(f) + PE(f) + E(lost)
The sum of the initial kinetic and potential energies is equal to the sum of the final kinetic and potential energies and energy lost, which in the case of cars, is the rolling resistance and drag
We get:
½mv(i)² + mgh(i) = ½mv(f)² + mgh(f) + E(r.r.) + E(drag)
m = mass
v = velocity
g = gravitational acceleration
h = height
And we can change this from an energy problem to a work problem!
↓
½mΔv² + mgΔh = F(r.r.)d + F(C)d
d = distance traveled
d can be found by d = ½(v(i) + v(f))Δt
↓
½mΔv² + mgΔh = C(r.r.)Nd + ½ρv(f)²C(d)Ad
ρ = air density
A = cross-sectional area (largest)
C(d) = drag coefficient
C(r.r.) = coefficient of rolling friction of the tire(CRF)
↓
½mΔv² + mgΔh = d(C(r.r.)mg + ½ρv(f)²C(d)A)
ρ ≈ 1.2kg/m³
m = 1103.14kg
mg = 1103.14kg * 9.81m/s² = 10821.80N
↓
551.57Δv² + 10821.80Δh = C(r.r.)10821.80d + 0.6v(f)²C(d)Ad
↓
[551.57Δv² + 10821.80Δh - C(r.r.)10821.80d] \ [0.6v(f)²Ad] = C(d)
Ta-da! I ♥ Physics. And I hope I got all of that right!
What you'll need is the rolling resistance rating of your tires, and the cross-sectional area of the fit. Everything else can be found! =)
KE(i) + PE(i) = KE(f) + PE(f) + E(lost)
The sum of the initial kinetic and potential energies is equal to the sum of the final kinetic and potential energies and energy lost, which in the case of cars, is the rolling resistance and drag
We get:
½mv(i)² + mgh(i) = ½mv(f)² + mgh(f) + E(r.r.) + E(drag)
m = mass
v = velocity
g = gravitational acceleration
h = height
And we can change this from an energy problem to a work problem!
↓
½mΔv² + mgΔh = F(r.r.)d + F(C)d
d = distance traveled
d can be found by d = ½(v(i) + v(f))Δt
↓
½mΔv² + mgΔh = C(r.r.)Nd + ½ρv(f)²C(d)Ad
ρ = air density
A = cross-sectional area (largest)
C(d) = drag coefficient
C(r.r.) = coefficient of rolling friction of the tire(CRF)
↓
½mΔv² + mgΔh = d(C(r.r.)mg + ½ρv(f)²C(d)A)
ρ ≈ 1.2kg/m³
m = 1103.14kg
mg = 1103.14kg * 9.81m/s² = 10821.80N
↓
551.57Δv² + 10821.80Δh = C(r.r.)10821.80d + 0.6v(f)²C(d)Ad
↓
[551.57Δv² + 10821.80Δh - C(r.r.)10821.80d] \ [0.6v(f)²Ad] = C(d)
Ta-da! I ♥ Physics. And I hope I got all of that right!
What you'll need is the rolling resistance rating of your tires, and the cross-sectional area of the fit. Everything else can be found! =)
Also, what does Δh represent? Change in height I realize, but what height is changing? Distance above sea level?
#26
04-21-2010 | 09:31 AM
Member
Joined: Jul 2007
Posts: 1,584
From: Memphis, TN
It's been a while since I've taken physics, so I may very well be missing something, but it looks to me like you're suggesting that C(d) will be the same for any vehicle with identical cross-sectional area and tires.
Also, what does Δh represent? Change in height I realize, but what height is changing? Distance above sea level?
Also, what does Δh represent? Change in height I realize, but what height is changing? Distance above sea level?
I've since made a gigantic spreadsheet and figured the CdA is about 7.6ft^2, which would make for a Cd around 0.34, roughly.
I think it'd be safe to assume the 07-08 Fits have a Cd of about 0.34 +/-.01, and a CdA of about 7.6 +/-.1ft^2. That'll do for the average joe; an average Cd! The Fit is a breadbox, after all :P
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