Calculating Cd and Crr coefficient experimentally

[jumpjack]

If I let pass a line through lower-speed points, I see that 4 points out of 8 match the line:

http://imageshack.com/a/img819/5869/qsub.jpg

A line passing through hi-speed points just matches with 2 point; i reverse the graph for better readibility of next one: now speed is on the X axis and seconds are on Y axis:

http://img850.imageshack.us/img850/469/exym.jpg


Overlapping the known graph of air drag and rolling friction forces, it appears clear why data get non-linear as speed increases:
http://imageshack.com/a/img546/1932/fmhx.jpg

Indeed, at higher speeds the v^2 term prevail.

So probably the low speed data should be used to determine Crr, then the high-speed data and the value of the now-known Crr would allow a precise calculation of Cd.

To be verified.

 

[voko]

I am not sure where you are going - or want to be going with that. #46 is a compete solution of the problem. I do not see why you need to invent anything else.

 

[jumpjack]

Charts: trying to understand/explain the phenomenon
New data: test the model

 

[jumpjack]

Motion equation:

$$ \dot v = a(1 + \alpha^2 v^2) $$


Integrated from initial speed $v_i$ to any speed v:

$$ arctan(\alpha v ) - \arctan (\alpha v_i) = a \alpha \Delta t $$
$\Delta t =t(v)-t(v_i)$


Reordering:

$$ arctan(\alpha v ) = a \alpha \Delta t + \arctan (\alpha v_i)$$

Applying tan():
$$ \alpha v = tan(a \alpha \Delta t + \arctan (\alpha v_i))$$
$$ v = \frac 1 \alpha tan(a \alpha \Delta t + \arctan (\alpha v_i))$$

Using new notation, more readable and easily usable in Excel and other math SW:
$$ v = \frac 1 B tan(A B T + \arctan (B v_i))$$
$a = A$
$\alpha = B$
$\Delta t = T$

v = (1/B) tan(A*B*T + arctan(B*I))
$I=v_i$

Considering initial time t=0 , and using x in place of t for final time, and y for v, we can write in the usal form y=f(x):

(4) y = (1/B) tan(A*B*x + arctan(B*I))


Where:

(2) $B = \alpha = \sqrt {\frac \rho {2g}} \sqrt {\frac S m} \sqrt {\frac { C_d }{C_r}} = 0.25 \sqrt {\frac S m} \sqrt {\frac { C_d }{C_r}}$

(3) $A = a = -gC_r$

S= frontal area (Surface)


We could play with equation (4) in a math SW by trimming A and B, but so we would not be able to directly know how Cd and Crr change as curve chages, so let's expand the constants to obtain a y=f(x) function depending on Cd and Crr rather than A and B:


Once S and m are known we can set:

$K = 0.25 \sqrt {\frac S m}$

so:
(2b) $B=K \sqrt {\frac { C_d }{C_r}}$

And:

$AB = (-gC_r) K \sqrt {\frac { C_d }{C_r}} = -g K \sqrt {\frac { C_d C_r^2}{C_r}} = -g K \sqrt {C_d C_r}$


So we can finally express v as a function of just Cd and Crr rather than A and B:

$$ v = \frac 1 B tan(A B T + \arctan (B v_i))$$
$$ v = \frac 1 {K \sqrt {\frac { C_d }{C_r}}} tan(-g T K \sqrt {C_d C_r} + \arctan ({K \sqrt {\frac { C_d }{C_r}}}v_i)) $$ =
$$\frac 1 {\frac 1 4 \sqrt {\frac S m} \sqrt {\frac { C_d }{C_r}}} tan(-g T \frac 1 4 \sqrt {\frac S m} \sqrt {C_d C_r} + \arctan ({\frac 1 4 \sqrt {\frac S m} \sqrt {\frac { C_d }{C_r}}}v_i)) $$=

$$4 \sqrt \frac m S \sqrt \frac {C_r}{C_d} tan(-g T \frac 1 4 \sqrt {\frac S m} \sqrt {C_d C_r} + \arctan ({\frac 1 4 \sqrt {\frac S m} \sqrt {\frac { C_d }{C_r}}}v_i)) $$

$$4 \sqrt \frac {mC_r}{SC_d}tan(-2.45 T \sqrt \frac{SC_dC_r}{m} + arctan(\frac {v_i}4 \sqrt \frac{SC_d}{mC_r}))$$


We can now express this equation in "planar" form for math SW:

(4d) y =4*sqrt((m*R)/(S*D)) * tan(-2.45 * x * sqrt((S*D*R)/m) + arctan ((I/4)*sqrt((S*D)/(m*R))))

A2=Cd=D
B2=Cr=R
C2=K=0.25*sqrt(S/m)
D2=S
E2=m
F2=Vi

B4...Bn: x values

(4e) y = 4*sqrt(($E$2*$B$2)/($D$2*$A$2)) * tan(-2.45 * $B4 * sqrt(($D$2*$A$2*$B$2)/$E$2) + arctan (($F$2/4)*sqrt(($D$2*$A$2)/($E$2*$B$2))))

So now we can plot it in WinPlot and play with D and R (Cd and Crr) to see how curve changes and fits to experimental data.

B4...Bn (x values) are time instants in seconds.
y is the speed after x seconds starting from Vi speed (Vi =I in Winplot, Vi=F2 in Excel).

 

[jumpjack]

Damn, it does not work both in excel and winplot... :-(

 

[jumpjack]

Damn, it does not work both in excel and winplot... :-(

Ok, now fixed.

 

[jumpjack]

Ok here it is:
http://jumpjack.altervista.org/Cx/CdCrr_EN.html

Thanks fotr your help.