Jtdylktuy

Starting from this question I am trying figuring out the equation to calculate the needed time to get from 0 to 65 mph for a car, without using multiple excel formulas across cells as I've done till now.

I know that the forces acting on the car are:

1) $F_e = frac {P_e}{v}$

being Pe = Power of engine in W and v of course speed in m/s ; let's assume this power is constant at any speed.

2) $F_d = 0.5rho C_dAv^2-mgC_{rr} $

• rho = air denisty = 1.2 g/m3


• Cd = air drag coefficient = 0.3


• A = frontal area = 2.2 m2


• m = mass 1000 kg


• g = 9.18 m/s2


• Crr = wheels drag = 0.01

So the total force is:

Ft = Fe - Fd

This is of course = m * dv/dt :

$F_t = mfrac{dv}{dt}$

$F_e - F_d = mfrac{dv}{dt}$

$frac{P}{v} - 0.5 rho C_d A v^2 - mgC_{rr} = mfrac{dv}{dt}$

grouping constants and sorting by power:

$- K_1 v^2 + frac{P}{v} - K_2 = mfrac{dv}{dt}$

Now I don't know where to go from here, I can't get how to separate t and v with so many terms, powers and constants...

Start condition is v=0; end condition is v = 27.8; what I'm trying to get is t.

From

$$
- K_1 v^2 + frac{P}{v} - K_2 = mfrac{dv}{dt}
$$

we get

$$
frac{m v dv}{P- K_1 v^3 - K_2 v} = dt
$$

now

being $v_1, v_2 ,v_3$ the three roots of $P-K_1v^3-K_2 v = 0$ we have

$$
frac{m v dv}{(v-v_1)(v-v_2)(v-v_3)} = dt
$$

giving after integration

$$
frac{m (v_1 (v_2-v_3) ln (v-v_1)+v_2 (v_3-v_1) ln (v-v_2)+v_3 (v_1-v_2) ln(v-v_3))}{(v_1-v_2) (v_1-v_3) (v_2-v_3)} = t + C_0
$$

or

$$
(v_1 (v_2-v_3) ln (v-v_1)+v_2 (v_3-v_1) ln (v-v_2)+v_3 (v_1-v_2) ln
(v-v_3)) = frac{(v_1-v_2) (v_1-v_3) (v_2-v_3)}{m}(t+C_0)
$$

or

$$
(v-v_1)^{v_1(v_2-v_3)}(v-v_2)^{v_2(v_3-v_1)}(v-v_3)^{v_3(v_1-v_2)} = C_1e^{frac{(v_1-v_2) (v_1-v_3) (v_2-v_3)}{m}t}
$$

Adding constants for readability:

$$
(v-v_1)^{alpha}(v-v_2)^{beta}(v-v_3)^{gamma} = C_1e^{frac{delta}{m}t}
$$

now for $t = 0$ assuming $v = 0$ and at $t = t_f$ with $v_f = 27.8$ we have

$$
(-v_1)^{alpha}(-v_2)^{beta}(-v_3)^{gamma} = C_1\
(v_f-v_1)^{alpha}(v_f-v_2)^{beta}(v_f-v_3)^{gamma} = C_1e^{frac{delta}{m}t_f}
$$

and thus

$$
t_f = frac{m}{delta}lnleft(frac{(v_f-v_1)^{alpha}(v_f-v_2)^{beta}(v_f-v_3)^{gamma}}{(-v_1)^{alpha}(-v_2)^{beta}(-v_3)^{gamma}}right)
$$

NOTE

At this point supposing we have the inversion $v = f(t, C_1)$ we could follow with

$$
v = frac{ds}{dt} = f(t,C_1)
$$

obtaining after integration

$$
s(t) = g(t,C_1,C_2)
$$

anyway I would suggest the numerical approach.

Short answer

$$
t_{65mph} = frac{m}{delta}lnleft(frac{(27.8-v_1)^{alpha} * (27.8-v_2)^{beta} * (27.8-v_3)^{gamma}}{(-v_1)^alpha * (-v_2)^beta * (-v_3)^gamma }right)
$$

• $alpha = v_1(v_2-v_3)$


• $beta = v_2(v_3-v_1) $


• $gamma = v_3(v_1-v_2)$


• $delta = (v_1-v_2)(v_1-v_3)(v_2-v_3)$

v1, v2 and v3 are the roots of:

$ - K_1 v^3 - K_2 v + P = (v-v_1)(v-v_2)(v-v_3) $

• 
  

  $K_1 = 0.5 * 1.225 * Cd * A $

  
  
  
  
  

  • 1.225 = air density g/m3

  
  

  • $C_d$ air drag coefficient (~0.3 for cars, ~0.8 for cycles)

  
  

  • A = Frontal Area in $m^2$ (~2.2 for cars, ~0.8 for cycles)

  
  
  
  


• 
  

  $K_2 = m * 9.81 * C_{rr} $

  
  
  
  
  

  • m = mass of vehicle in kg

  
  

  • 9.81 = gravitational acceleration m/s2

  
  

  • $C_{rr}$ = rolling drag coefficient (~0.01 in cars, ~0.005 for cycles)

  
  
  
  

$ v_{1,2,3} = sqrt[3]{u + sqrt{u^2+v^3}} + sqrt[3]{u - sqrt{u^2+v^3}} $

$ u = frac{P}{rho C_dA} $

$v= frac{2mgC_{rr}}{3 rho C_dA}$

Generic final speed:

$$
t_f = frac{m}{delta}lnleft(frac{(v_f-v_1)^{alpha} * (v_f-v_2)^{beta} * (v_f-v_3)^{gamma}}{(-v_1)^alpha * (-v_2)^beta * (-v_3)^gamma }right)
$$

Start time and start speed are assumed to be t=0, end time is $t_f$ and final speed is $v_f$.

Demonstration

Last equation in the question, $- K_1 v^2 + frac{P}{v} - K_2 = mfrac{dv}{dt}$ , can be also written as:

$ - K_1 v^3 +P- K_2 v= mvfrac{dv}{dt} $

and sorting by power:

$ - K_1 v^3 - K_2 v + P= mvfrac{dv}{dt} $

Separating t and v:

$ dt= frac{mvdv}{- K_1 v^3 - K_2 v + P} $

Mirroring:

$ frac{mvdv}{- K_1 v^3 - K_2 v + P} = dt$

So we can now integrate on both sides

$int {frac{mv}{- K_1 v^3 - K_2 v + P}dv} = int dt$

$ mint {frac{v}{- K_1 v^3 - K_2 v + P}dv} = int dt$

Lower factor can be written as:

$ - K_1 v^3 - K_2 v + P = (v-v_1)(v-v_2)(v-v_3) $

being v1, v2 and v3 the roots of $ - K_1 v^3 - K_2 v + P$

Hence we can write:

$ mint {frac{v}{(v-v_1)(v-v_2)(v-v_3)}dv} = int dt $

getting (source):

$$ m frac{v_1 (v_2 - v_3) ln(v - v_1) + v_2 (v_3 - v_1) ln(v - v_2) + v_3 (v_1 - v_2) ln(v - v_3)}{(v_1 - v_2) (v_1 - v_3) (v_2 - v_3)} = t + C_0$$

Replacing with some constants for readability:

• $ alpha = v_1 (v_2 - v_3) $




• $ beta = v_2 (v_3 - v_1)$




• $ gamma = v_3 (v_1 - v_2)$




• $ delta = (v_1 - v_2) (v_1 - v_3) (v_2 - v_3) $

$$ m frac{alpha ln(v - v_1) + beta ln(v - v_2) + gamma ln(v - v_3)}{delta} = t + C_0 $$

or

$$ frac{m}{delta} (alpha ln(v - v_1) + beta ln(v - v_2) + gamma ln(v - v_3)) = t + C_0 $$

or

$$ alpha ln(v - v_1) + beta ln(v - v_2) + gamma ln(v - v_3) = frac{delta}{m} (t + C_0) $$

Being:

$a*ln(b) = ln (b^a) $

we can then rewrite in this form:

$$ ln(v - v_1)^alpha + ln(v - v_2)^beta + ln(v - v_3)^gamma = frac{delta}{m} (t + C_0) $$

Then we can transform into:

$$ e^{ln(v - v_1)^alpha + ln(v - v_2)^beta + ln(v - v_3)^gamma} = e^{frac{delta}{m} (t + C_0)} $$

But this can be split into:

$$ e^{ln(v - v_1)^alpha}* e^{ln(v - v_2)^beta} * e^{ln(v - v_3)^gamma} = e^{frac{delta}{m} t} * e^{C_0} $$

which means:

$$ (v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma = e^{frac{delta}{m} t} * C_1 $$

Bringing C1 to left:

$$ frac{(v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma}{C_1} = e^{frac{delta}{m} t} $$

Applying logarithm again:

$$ ln{frac{(v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma}{C_1}} = ln{e^{frac{delta}{m} t}} $$

$$ ln{frac{(v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma}{C_1}} = frac{delta}{m} t $$

and finally:

$$ t = frac{m}{delta} ln{frac{(v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma}{C_1}} $$

We need to calculate C1 value. This can be done considering initial conditions:

t=0, v=0

Putting these values in previous equation:

$$ frac{(v - v_1)^alpha * (v - v_2)^beta * (v - v_3)^gamma}{C_1} = e^{frac{delta}{m} t} $$

we get:

$$ frac{(0 - v_1)^alpha * (0 - v_2)^beta * (0 - v_3)^gamma}{C_1} = e^{frac{delta}{m} 0}$$

$$ frac{(-v_1)^alpha * (-v_2)^beta * (-v_3)^gamma}{C_1} = e^0 $$

$$ (-v_1)^alpha * (-v_2)^beta * (-v_3)^gamma = C1 $$

This means that, for final conditions tf = unknown and vf = known we have:

(1) $
t_f = frac{m}{delta}lnleft(frac{(v_f-v_1)^{alpha}(v_f-v_2)^{beta}(v_f-v_3)^{gamma}}{(-v_1)^alpha * (-v_2)^beta * (-v_3)^gamma }right)
$

V1, V2, V3 calculation

Now we only need v1, v2 and v3,which are the solutions of the equation:

$-K_1v^3 - K_2 v + P= 0$

Adding "missing" coefficient:

$-K_1v^3 + 0 v^2- K_2 v + P= 0$

This is a 3rd grade equation, whose solutions can be determined as follows:

It can be expressed in the form of

$ax^3 + bx^2 + cx + d = 0$

being:

• a = $-K_1 = - 0.5 rho C_dA$
  


• b = 0


• c = $-K_2 = -mgC_{rr}$


• 
  

  d = P

  
  
  

  Solutions are (source):

  
  

$ x = sqrt[3]{q + sqrt{q^2 + (r-p^2)^3}} + sqrt[3]{q - sqrt{q^2 + (r-p^2)^3}} + p = sqrt[3]{q + s} + sqrt[3]{q - s} + p $

where (considering b=0)

$p = frac{-b}{3a} = 0 $

$q = p^3 + frac{bc-3ad}{6a^2} = - frac{d}{2a} = - frac{P_{ower}}{2 (- 0.5 rho C_dA)} = frac{P_{ower}}{rho C_dA}$

$r = frac{c}{3a} = frac{-mgC_{rr}}{3(- 0.5 rho C_dA)} = frac{2mgC_{rr}}{3 rho C_dA}$

$s = sqrt{q^2+(r-p^2)^3} = sqrt{q^2+r^3} = sqrt{(frac{P_{ower}}{rho C_dA})^2+(frac{2mgC_{rr}}{3 rho C_dA})^3} $

Shortly:

$q = frac{P_{ower}}{rho C_dA}$

$s = sqrt{(frac{P_{ower}}{rho C_dA})^2+(frac{2mgC_{rr}}{3 rho C_dA})^3} $

Due to p = 0 (as b=0):

$ x = sqrt[3]{q + s} + sqrt[3]{q - s} $

replacing:

$ x_{1,2,3} = sqrt[3]{frac{P_{ower}}{rho C_dA} + sqrt{(frac{P_{ower}}{rho C_dA})^2+(frac{2mgC_{rr}}{3 rho C_dA})^3}} + sqrt[3]{frac{P_{ower}}{rho C_dA} - sqrt{(frac{P_{ower}}{rho C_dA})^2+(frac{2mgC_{rr}}{3 rho C_dA})^3}} $

This gives 3 values x1, x2 and x3 which must replace v1, v2 and v3 in the final equation (1) found before.