CD for fit? .34?
Cd is the drag coefficient compared to a barn door of same area being = 1.0. Kinetic and potential energies have nothing to do with those. And tire rolling resistances aren't involved either.
Cd is meaured in a wind tunnel with load cells measuring the push to the rear for various wind speeds, compared to that of a flat surface of the same cross-sectional area.
And only a few muascle cars got over .4; I measured some of those.
Last edited by mahout; Jul 13, 2008 at 07:52 PM.
Interesting ... I can't seem to find a published cd figure for the Fit anywhere. Apparently Honda never published an official number.
One interesting think I did find is that Mugen makes a "Dynamite" underbody skirt kit for the Fit to reduce drag (shown below). Has anybody ever heard of this before or know about availability?
One interesting think I did find is that Mugen makes a "Dynamite" underbody skirt kit for the Fit to reduce drag (shown below). Has anybody ever heard of this before or know about availability?
I found the Beatrush unit:
BeatRush Aluminum Under Panel *SPECIAL ORDER* - S542020
It's only a partial underpanel -- looks like it covers back to about the firewall. It looks to be made of pretty stout aluminum so it would also provide some extra protection for the engine/trans etc., especially for anybody who takes their Fit off pavement (like me, occassionally). But it would also be a PITA to change the oil as you'd have to remove the whole panel first. Price ain't bad tho: $171.
Here's a couple threads about it:
https://www.fitfreak.net/forums/fit-...oys-fit-2.html
https://www.fitfreak.net/forums/fit-...nderpanel.html
BeatRush Aluminum Under Panel *SPECIAL ORDER* - S542020
It's only a partial underpanel -- looks like it covers back to about the firewall. It looks to be made of pretty stout aluminum so it would also provide some extra protection for the engine/trans etc., especially for anybody who takes their Fit off pavement (like me, occassionally). But it would also be a PITA to change the oil as you'd have to remove the whole panel first. Price ain't bad tho: $171.
Here's a couple threads about it:
https://www.fitfreak.net/forums/fit-...oys-fit-2.html
https://www.fitfreak.net/forums/fit-...nderpanel.html
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! =)
Last edited by cojaro; Jul 13, 2008 at 08:14 PM.
That's some crazy sh!t, which is why I was an English major and not a math/physics major. 
And then you hit us with this:
All that, and you still can't give us the answer? Just kidding. 
Come on now, somebody come through with the two "unknowns" and we'll have this baby solved!

And then you hit us with this:

Come on now, somebody come through with the two "unknowns" and we'll have this baby solved!
And someone can guess the cross-sectional area. Nothing completely wrong with an educated guess =P just measure it as if the Fit were a trapezoid. Measure the bottom width, top width, and height between those two. Voila!
Last edited by cojaro; Jul 13, 2008 at 07:01 PM.
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 + ½ρv²C(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 + ½ρΔv²C(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 + 0.6Δv²C(d)Ad
↓
[551.57Δv² + 10821.80Δh - C(r.r.)10821.80d] \ [0.6Δv²Ad] = 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 + ½ρv²C(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 + ½ρΔv²C(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 + 0.6Δv²C(d)Ad
↓
[551.57Δv² + 10821.80Δh - C(r.r.)10821.80d] \ [0.6Δv²Ad] = 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! =)
Design cars? haaaaa. =) I don't have that sort of artistic skill.
And bridges? haaaaa again. That's Civil Engineering territory. Mechanical Engineering is where it's at =P
But if any of that is wrong, please correct me! I don't like being wrong. I kinda have to know this Physics stuff for the next few years, if not for the rest of my working life.
And bridges? haaaaa again. That's Civil Engineering territory. Mechanical Engineering is where it's at =P
But if any of that is wrong, please correct me! I don't like being wrong. I kinda have to know this Physics stuff for the next few years, if not for the rest of my working life.
Last edited by cojaro; Jul 13, 2008 at 08:15 PM.
I found the Beatrush unit:
BeatRush Aluminum Under Panel *SPECIAL ORDER* - S542020
It's only a partial underpanel -- looks like it covers back to about the firewall. It looks to be made of pretty stout aluminum so it would also provide some extra protection for the engine/trans etc., especially for anybody who takes their Fit off pavement (like me, occassionally). But it would also be a PITA to change the oil as you'd have to remove the whole panel first.
BeatRush Aluminum Under Panel *SPECIAL ORDER* - S542020
It's only a partial underpanel -- looks like it covers back to about the firewall. It looks to be made of pretty stout aluminum so it would also provide some extra protection for the engine/trans etc., especially for anybody who takes their Fit off pavement (like me, occassionally). But it would also be a PITA to change the oil as you'd have to remove the whole panel first.
(Wish I'd seen the Mugen full-car underpanel first...)
I had one in my hands about 5 hours ago; it survived being extremely poorly packaged, then shipped 1200 miles by UPS, without harm. So, yeah, it's a stout piece. And lighter than the box it was shipped in. After I get under the car and compare it to the the stocker I'll let you know if it's worth consideration from an aerodynamic standpoint.
(Wish I'd seen the Mugen full-car underpanel first...)
(Wish I'd seen the Mugen full-car underpanel first...)

Anybody found a place to buy the Mugen unit stateside? Or is it JDM only?



