Calculating time-to-65mph for a car considering air drag

Question

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 = 0.5\rho 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 :

$F_t = m\frac{dv}{dt}$

$F_e - F_d = m\frac{dv}{dt}$

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

grouping constants and sorting by power:

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

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.

Answer 1

From

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

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

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

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.

Answer 2

Short answer

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

• $\alpha = v_1(v_2-v_3)$
• $\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:

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

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

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Demonstration

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

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

and sorting by power:

$ - K_1 v^3 - K_2 v + P= mv\frac{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

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

$ m\int {\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:

$ m\int {\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}\ln\left(\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.