How can I calculate the resistance force for an electric Go Kart?

[Pablo87]

Hello to all,

I am trying to design an electric Go kart as part of a personal project. I am now in the preliminary design phase, trying to find some literature regarding to the resistance force (Rolling resistance + aerodynamic drag).

I was able to find a document in which some guys did some testing, obtaining a graph speed/time (I understand that the controller was plotting a graph of values of speed and time). The graph shows how they disengaged the motor with a clutch and left the kart to decelerate by its "own" (in reality due to all the resistances = Rolling + aerodynamic...considering the bearing friction and rolling resistance negligible). I also consider the lifting down force negligible (measured to be around less than 1% the kart weight).

The steps I was following to find out the resistance force are the following:

I check the graph and wrote few +/- accurate points (speed in km/h, time s) contained in the graph. I got (115, 0) (80, 4) (60, 10) (40, 20).

I am going to consider the resistance force Fr= Frolling resistance +Drag = P^α*Z^β*(a+b*v+c*v²) + 1/2*ρ*C_D*A*v². I used the SAE J2452 for the rolling resistance (where v is in km/h). I am considering P, α, Z, β, a, b, c constant. Therefore, I ended up with a Fr = c₁+c₂*v+c₃*v² (v in km/h, and time in s)

My objective then is to find the constants c₁, c₂, and c₃, using the graph points I showed before.

since the kart is decelerating only by the resistance force Fr, then I know that the deceleration -a = Fr/m

and a=dv/dt, therefore dv/dt = -(c1+c2*b+c3*v²)/m → dv/(c1+c2*b+c3*v²) = -dt/m →
∫dv/(c1+c2*b+c3*v²) = -∫dt / m

the ∫dv/(c1+c2*b+c3*v²) would be between v₀ = 115 and v

the ∫dt would be between t₀ = 0 and t

The integration of ∫dv/(c₁+c₂*v+c₃*v²) is a very annoying one :

if 4ac-b² >0 I obtain the solution

-t/m = 2/√(4*c₃*c₁-c₂²) * arctg [(2*c₃*v + c₂) /√(4*c₃*c₁-c2²)] - 2/√(4*c₃*c₁-c₂²) * arctgh [(2*c₃*115 + c₂) /√(4*c₃*c₁-c₂²)]

if 4ac-b² <0

Similar but with the Ln.

When I try to resolve the function v (t) using excel solver...it cannot converge into any solution.

Could somebody help me finding a way to solve the problem?

I have other ideas to calculate the Resistance torque/Force, but it would required a torque transducer that would complicate the design of the vehicle.

Excuse me if I had any mistake or if the formulas are not very well explained. it isn't easy to write those formulas in the forum.

 

[anorlunda]

It's a good question. This is not a full answer, but I think you made an error or a typo in

(c1+c2*b+c3*v2)

Shouldn't it be (c1+c2*v+c3*v2) ?

 

[Pablo87]

Hi Anorlunda,

Thank you, yes it was a typo but only here at the forum, the (c₁+c₂*v+c₃*v²) is the correct polynomial I used in the problem.

Regards,

Pablo

 

[jrmichler]

Here's how I measured the drag and rolling resistance of my truck: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html. I used a simpler model, where total drag = fixed rolling resistance plus a speed squared term for aerodynamic drag. It solved easily because I was fitting only two coefficients to a much larger number of data points, and the data points went down to zero speed.

When fitting an equation to data, several conditions must be met to get good results:
1) Enough data points. More is better.
2) Data must cover a wide enough speed range. If you are fitting a curve with a quadratic term, you should have speed data down to zero speed.
3) Noise level must be low enough.

 

[jack action]

Here's an empirical method from Bosch Automotive Handbook 4th ed. (p.331):

Empirical determination of coefficients for aerodynamic drag and rolling resistance

Allow vehicle to coast down in neutral under windless conditions on a level road surface. The time that elapses while the vehicle coasts down by a specific increment of speed is measured from two initial velocities, $v_1$ (high speed) and $v_2$ (low speed). This information is used to calculate the mean deceleration rates $a_1$ and $a_2$. See the following example.

The example is based on a vehicle weighing $m$ = 1450 kg with a cross section $A$ = 2.2 m².

The method is suitable for application at vehicle speeds of less than 100 km/h.

1st trial (high speed)

Initial velocity: $v_{a1}$ = 60 km/h
Terminal velocity: $v_{b1}$ = 55 km/h
Interval between $v_a$ and $v_b$: $t_1$ = 6.5 s

Mean velocity: $v_1 = \frac{v_{a1} + v_{b1}}{2}$ = 57.5 km/h

Mean deceleration: $a_1 = \frac{v_{a1} - v_{b1}}{t_1}$ = 0.77 km/h/s

2nd trial (low speed)

Initial velocity: $v_{a2}$ = 15 km/h
Terminal velocity: $v_{b2}$ = 10 km/h
Interval between $v_a$ and $v_b$: $t_2$ = 10.5 s

Mean velocity: $v_2 = \frac{v_{a2} + v_{b2}}{2}$ = 12.5 km/h

Mean deceleration: $a_2 = \frac{v_{a2} - v_{b2}}{t_2}$ = 0.48 km/h/s

Drag Coefficient
$$C_d = \frac{6m(a_1-a_2)}{A(v_1^2 - v_2^2)} = 0.36$$

Coefficient of rolling resistance
$$f = \frac{28.2(a_2 v_1^2 - a_1 v_2^2)}{1000(v_1^2 - v_2^2)} = 0.013$$

 

[Pablo87]

Thank you for the responses!

jrmichler, Not sure why I didn't think about a polynomial regression, you just solved my issue right away!

jack action, Thank you! I didn't know about this empirical solution. It will be handy to compare it with a polynomial regression. Maybe I should check the bosh handbook, since its quite inexpensive