Calculating Cd and Crr coefficient experimentally

[voko]

That is probably all correct, but I do not understand where that gets you. What is the purpose of those calculations?

Here is how you could employ the method of #14. I will use the data from the original excel sheet, namely $v_1 = 70 \ \text{km/h} = 19.4 \ \text{m/s}, v_2 = 60.3 \ \text{km/h} = 16.8 \ \text{m/s}, v_3 = 51.5 \ \text{km/h} = 14.3 \ \text{m/s}$, uniform $10 \ \text{s}$ time interval.

The basic equation in #14 was $\arctan \alpha v_{1f} - \arctan \alpha v_{1i} = \arctan \alpha v_{2f} - \arctan \alpha v_{2i}$, which in this case simplifies to $$ 2 \arctan \alpha v_2 - \arctan \alpha v_1 - \arctan \alpha v_3 = 0 $$ Cheating, I get $\alpha = 0.022931$. To determine $a$: $$ a = \frac 1 {\alpha T} (\arctan \alpha v_2 - \arctan \alpha v_1) = \frac 1 {\alpha T} (\arctan \alpha v_3 - \arctan \alpha v_2) $$ and I get $$ \frac 1 {\alpha T} (\arctan \alpha v_2 - \arctan \alpha v_1) = -0.221769 $$ and $$ \frac 1 {\alpha T} (\arctan \alpha v_3 - \arctan \alpha v_2) = -0.221769 $$ which I call pretty good agreement :)

Given $$ a = - \frac R m = - \frac {mgC_r}{m} = -gC_r $$ and $$ \alpha = \sqrt {\frac { \rho A }{2mg}} \sqrt {\frac { C_d }{C_r}}$$ we should have $$ C_r = -{a \over g}$$ and $$ C_d = {2 mg \over \rho A} \alpha^2 C_r $$ From the constants in the original sheet, I obtain $C_r = - 0.023$ and $C_d = 0.083$, which differ quite significantly from the results in the spreadsheet.

 

[jumpjack]

That is probably all correct, but I do not understand where that gets you. What is the purpose of those calculations?

I was leaving mathematical analysis in favour of graphical analysis to determine Cd and Crr.
I was also trying to figure out how the suggested equation ( 1/a tan(...) ) fits to real data, but apparently it does not fit as good as said.
How do I plot single points together with a function in http://www.wolframalpha.com/ ?

I obtain $C_r = - 0.023$ and $C_d = 0.083$, which differ quite significantly from the results in the spreadsheet.

They do not "differ quite significantly": that Cd value is just impossible (typical value for cars is around 0.35; 0.083 is Cd for an airplane!)

As the curve y=19.51343152 e^(-0.01576* x) fits very very well with data, now I'd like to figure out how I can obtain it algebrically, i.e. how 19.51343152 and -0.01576 can be correlated to Cd and Crr; any tip?

 

[voko]

They do not "differ quite significantly": that Cd value is just impossible (typical value for cars is around 0.35; 0.083 is Cd for an airplane!)

That is evident from the data in the sheet. During the first 20 seconds, which were used for the calculation, deceleration is almost constant. The explanation for that is that either the data are either completely fake, or taken for some other vehicle. The mass of the vehicle (1000 kg) and the frontal area (2.2 sq. m) combo is completely bogus.

As the curve y=19.51343152 e^(-0.01576* x) fits very very well with data, now I'd like to figure out how I can obtain it algebrically, i.e. how 19.51343152 and -0.01576 can be correlated to Cd and Crr; any tip?

You get those coefficients by fitting the data with the tan/arctan curve.

 

[jumpjack]

Wait.
For A = -0.07654 and B = 0.10394 a very very good approximation of experimental data!

This means, if my math is correct:

Crr = 0.008
Cd = 0.86

Math:
Crr = -A/9.81
$C_r$ =~ $- \frac A {10}$
B =~ $0.25 \sqrt{\frac Sm} * \sqrt{\frac {C_d} {C_r} } =~ 0.01 \sqrt{\frac {C_d} {C_r} }$
$B^2 = 0.0001 \frac {C_d} {C_r}$
$C_d = 10000 C_r * B^2 = 10000 * - \frac A {9.81} * B^2$ =~ $10000 * - \frac A {10} * B^2$
$C_d = -1000 AB^2$Cd is still strange, but at least it is... less impossible! Renault Twizy has Cd = 0.64

There's still something to be tuned in the tan() model...

 

[jumpjack]

That is evident from the data in the sheet. During the first 20 seconds, which were used for the calculation, deceleration is almost constant.

How can we take into account all samples?

You get those coefficients by fitting the data with the tan/arctan curve.

I don't get it, this is an exponential curve, how do its coefficients relate to tan/arctan curve?

 

[voko]

I also found, using this site, this curve which appears to approximate experimental data very well as long Y is "far" from 0:
y=19.51343152 e^(-0.01576* x)

This is actually completely wrong. The real curve crosses the X axis in some finite time, while this approximation never does.

 

[voko]

Wait.
For A = -0.07654 and B = 0.10394 a very very good approximation of experimental data!

I do not know what experimental data you are talking about. I can see that at least the mass you have is different from the one given in the original Excel sheet.

 

[voko]

How can we take into account all samples?

Using the minimisation method explained earlier.

I don't get it, this is an exponential curve, how do its coefficients relate to tan/arctan curve?

You are mistaken. It is not an exponential curve. It is not a polynomial curve. In the model with constant rolling resistance and speed-squared drag, it is a tangent function curve. Using this model AND trying to fit the data with some other curve makes no sense whatsoever. Don't do that.

 

[jumpjack]

This is actually completely wrong. The real curve crosses the X axis in some finite time, while this approximation never does.

That's why I said "as long as Y is far from 0"!
We have no experimental data down to Y=0 to put in the model.

I do not know what experimental data you are talking about. I can see that at least the mass you have is different from the one given in the original Excel sheet.

I played a lot with the excel sheet, probably too much...
Indeed I'm designing a WinPlot graph with 5 variables: A, B, V, S and m. By now I'm just playing with A,B and V, with S=2,2 and m=1500 , but you're right, S and m MUST match the original experimental data!

I don't get it, this is an exponential curve, how do its coefficients relate to tan/arctan curve?

You are mistaken. It is not an exponential curve.

You didn't understand... The subject of that sentence was curve y=19.51343152 e^(-0.01576* x) (see #38)

http://img844.imageshack.us/img844/1411/qqcl.jpg

Experimental values:
0 19.44
10 16.76
20 14.31
30 12.08
40 10.35
50 8.92
60 7.55


Equations in text form:
y = 0.001505952381*x^2 - 0.2878928571* x + 19.45190476
y = 19.51343152 e^(-0.01576382642 x)
y = (1/B) tan(A*B*X+arctan(V*B))

WinPlot file:
http://jumpjack.altervista.org/CdCrr.wp2

 

[jumpjack]

How to obtain Cd and Crr from A, B and generic car data:

$$C_r = -\frac{A}{9,81}$$
$$C_d = -\frac{1.63 * m}{S} AB^2 $$

Math:
$B = \frac 1 4 \sqrt{\frac S m} \sqrt {{C_d}{C_r}}$
$$B^2 = \frac 1 {16} \frac S m \frac{C_d}{C_r} = \frac 1 {16} \frac S m \frac{C_d}{-\frac A {9.81} } = - \frac {9.81} {16} \frac S m \frac{C_d}{A}$$
$$C_d = -\frac{16mAB^2} {9.81 * S} = \frac{1.63 * m}{S} AB^2 $$

For original data (m=1000, S=2.3):

$$C_r = -\frac{A}{9,81}$$
$$C_d = -708.7 * AB^2 $$

For A and B values of my WinPlot chart (A=-0.07654, B=0.09911) (best approximation for initial values)
Crr= 0.0078
Cd = 0.53

Best approximation for last value:
Crr=0.008
Cd=0.79

Other tests:

0.0088
0.69

0.0089
0.699

Original Excel solution (which uses Excel solver, unknow minimisation algorithm and unknown best fit curve):
Crr=0.011
Cd=0.37

 

[voko]

Using Wolfram Alpha, I get the following: $A = -0.0985426, B = 0.0773005$, where $A = a, B = \alpha$ from the original notation.

Looking at its plot, I find that it fits rather well in the 0 - 60 s range.

 

[jumpjack]

Found some literature coastdown data:
LINK

V Vehicle 1 Vehicle 1b Vehicle 2 Vehicle 2b Vehicle 3 Vehicle 4 Vehicle 5 Vehicle 6 Vehicle 6
120 5,88 6,28 5,64 4,75 4,88 7,06 5,91 6,08 6,47 6,68
100 7,55 8,03 7,25 6,21 6,66 8,88 7,73 7,6 8,15 8,38
80 9,4 10,34 9,46 8,07 8,98 10,91 10,02 9,5 10,31 10,2
60 12,1 13,52 13,28 11,01 10,93 13,6 12,97 12,25 12,75 12,93
40 15,19 15,08 18,03 14,6 15,92 15,07 15,48 15,76 16,38 16,24
20 18,88 19,42 24,43 19,92 21,58 18,68 20,18 20,86 20,34 20,57

Mass kg 1492 1566 1316 1277 1124 1962 1652 1523 1634 1724
Frontal area: n/a n/a n/a n/a n/a n/a n/a n/a n/a n/a


Nissan Leaf coastdown test:
http://img191.imageshack.us/img191/4828/dragfast1.jpg
http://www.mynissanleaf.com/viewtopic.php?f=31&t=5224

M = mass = 1605 = mass of LEAF 1525kg + mass of driver 80kg
g = gravity = 9.81 m/s^2
A = frontal area of LEAF = 2.27 m^2
Declared Cd = 0.29

a = { Crr * g } + { Cd * A * 1/2 * rho V^2 } / M
High speed dv/dt: 0.37904 = Crr * 9.81 + Cd * 0.8916 (using V = 32.7m/s)
Low speed dv/dt: 0.1328 = Crr * 9.81 + Cd * 0.044613 (using V = 7.32 m/s)

From these we get Crr = 0.01221 and Cd = 0.2907

 

[voko]

I do not see how all that is relevant for your problem. They say:

it gets reported about 50-100 times /sec

I said awhile ago:

You need subsecond time resolution to get anywhere near decent approximation. And you cannot get that, plain and simple.

 

[jumpjack]

Literature data are useful to verify if data in our excel file are "typical", to see how usually the v/t coastdown graph appears for different vehicles, to have additional data to experiment with...
I don't understand your doubts.
And I don't think subsecond resolution is necessary at all, instead you can see coastdown data are always taken several seconds apart. Look at the Leaf chart: although weird because an electric vehicle can't properly coast down due to lack of clutch, you can see a lot of noise in the plot: it's due to high sample rate. You'll never be able to find a function which exactly approximates such a noisy curve; but if you pick 10 values several seconds apart, you'll already have an approximated, but still good, curve, easier to model.

 

[voko]

And I don't think subsecond resolution is necessary at all, instead you can see coastdown data are always taken several seconds apart. Look at the Leaf chart: although weird because an electric vehicle can't properly coast down due to lack of clutch, you can see a lot of noise in the plot: it's due to high sample rate. You'll never be able to find a function which exactly approximates such a noisy curve; but if you pick 10 values several seconds apart, you'll already have an approximated, but still good, curve, easier to model.

This is completely wrong. It is not easier to model when you have less data. When you have more data, even if noisy, you can use simpler models. For example, you can trivially obtain good estimations for acceleration from the Leaf data and then you can use a linear model. You cannot do that from the Excel data, and you have to use a non-linear model.

 

[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.