# Calculating 0-60 mph time for a vehicle

1. Mar 24, 2014

### jumpjack

After figuring out how to calculate Cd and Crr coefficients, I'd like to continue my "vehicle physics exploration": I'd like to determine how much it takes to go from 0 to 60 mph depending on engine power. Or does it depends on torque? By sure it also depends on air drag, which depends on speed, which makes things very complex.

2. Mar 24, 2014

### TumblingDice

This sounds like a cool way to add some fun to physics. The 0-60 vehicle calculation could become a never-ending quest without some limitations on what you expect to calculate. Setting a reasonable target keeps it fun, and you can add new pieces in the future from your list.

At the track there are a bunch of factors to complicate further. Transmisson gearing and ratios and clutch performance, for example, if you want to gear for best torque. Pavement, temperature, tire compound, size, pressure, and more decide how much of your power is lost to friction/heat/slippage.

Maybe you'd consider doing this for an electric vehicle like Tesla, which would remove multiple gear/transmission/shifting from the equation?

3. Mar 24, 2014

### Staff: Mentor

Power alone gives you estimate of the shortest required time. Real time will be always longer.

4. Mar 24, 2014

### jumpjack

@Tumblingdice,
We can take into account just airdrag in a first instance.

@borek , your assumption is already included in my question...

5. Mar 24, 2014

### Staff: Mentor

It is not an assumption, it is the energy conservation.

6. Mar 24, 2014

### jumpjack

I don't need results which do not take into account airdrag.

7. Mar 24, 2014

### Staff: Mentor

I think you missed the point of my post. You stated

and all I said is that it is impossible - you need to know more than just power. Power alone can give an estimate of the lower boundary, but the real time will be always longer.

8. Mar 24, 2014

### jumpjack

Ok, it "looks like" $v(t)=v_f * tanh(\sqrt {cF}* \frac t m)$ ,where $v_f = 27.777 m/s$ (100 kph) and $c= \frac 1 2 \rho C_d A$.
Now how do I relate vehicle power to F? Maybe I should instead use vehicle torque... but wouldn't F then depend on wheel radius, which I do not know?

Last edited: Mar 24, 2014
9. Mar 24, 2014

### jumpjack

Oh, and let's suppose both power and torque constant, I'm interested on electric vehicles mostly.

10. Mar 27, 2014

### jumpjack

No ideas?

11. Mar 27, 2014

### AlephZero

You can't make both assumptions at once, unless the speed is also constant.

Given your apparent lack of knowledge about basic mechanics, it's hard to give any constructive advice, except "learn some basic mechanics first".

12. Mar 27, 2014

### jumpjack

Just help if you can, don't argument.
Thanks.

I reword my question:
I have a vehicle with given power and torque, say 100 kW / 100 Nm motor. Known it's Cd,frontal area , Crr (rolling friction coefficient) and mass, which is the minimum time it can achieve to go from 0 to 60 mph?

13. Mar 27, 2014

### BiGyElLoWhAt

well this is what I would do:
∑F=ma;
So
$F_{propulsion}-F_{drag}-F_{rolling friction}-F_{whatever else} = m_{car}a_{car}$

$F_{propulsion}$ is due to the tires spinning, which stems from the torque, right? we don't really care about the "direction" of the torque (it's a pseudo-vector), so we don't necessarily need R x F, we can just use RFsin(Θ), but Θ=pi/2, so sinΘ=1,
From using T = RF, we get $F_{propulsion}= \frac{T}{R}$.
so...
$\frac{T}{R} - F_{drag} -...=m_{car}a_{car}.$

$\frac{∑F}{m_{car}}=a_{car}$
integral(from 0 to t) a dt = $a_{SomeConstantFromNonDragTerms}t - inegral(from 0 to t)\frac{1}{2}ρC_{d}v^2A_{crossection}$= velocity change from time 0 to t. The tricky part is that v is a funtction of t, so you can't just treat it like a constant, I'm not entirely sure how to write v correctly to give you a good answer... soooo... lets act like physicists and forget drag? lol.
Actually I am curious to see what people have to say about this, any ideas to help finish this off?

Last edited: Mar 27, 2014
14. Mar 28, 2014

### jumpjack

Did you mean this?
$\int_0^t a dt = a_{SomeConstantFromNonDragTerms}t - \int_0^t\frac{1}{2}ρC_{d}v^2A_{crossection}$= velocity change from time 0 to t.

15. Mar 28, 2014

### jumpjack

I don't know if it can help, anyway in this thread we determined the solution of equation motion when no traction force is acting on the vehicle (coasting down).

$ma = -mgC_r - \frac 1 2 \rho C_d S v^2$
$m \dot v = -R - D$

$R -mgC_r$
$D = \frac 1 2 \rho C_d S v^2$

If we put $D = d v^2$, we have $m \dot v = -R - dv^2$ and so we have $$\dot v = -{R \over m} - {d \over m}v^2$$ So let $$A = - \frac R m$$ and $$B = \sqrt {\frac d R }$$ then $$\dot v = A(1 + B^2 v^2)$$

This post explains how to solve it to:

$$v = \frac 1 B tan(A B T + \arctan (B v_i))$$

$ma = -mgC_r - \frac 1 2 \rho C_d S v^2 +F_t$
$m \dot v = -R - D +F_t$
$\dot v = -{R \over m} - {d \over m}v^2 + {F_t \over m}$

New differential equation to solve:

$\dot v = A(1 + B^2 v^2) + C$

16. Mar 28, 2014

### jumpjack

$\dot v = -{R \over m} - {d \over m}v^2 + {F_t \over m}$

I could put the equation into form $A(1+B^2v^2)$ to be able to solve it like previous one:

$\dot v = (-{R \over m} + {F_t \over m}) - {d \over m}v^2$

"Extracting" $-{R \over m} + {F_t \over m}$ :

$$\dot v = (-{R \over m} + {F_t \over m}) (\frac {-{R \over m} + {F_t \over m}} {-{R \over m} + {F_t \over m} } - \frac {d \over m} {-{R \over m} + {F_t \over m}} v^2)$$

$$\dot v = (-{R \over m} + {F_t \over m}) (1 - \frac {d \over m} {-{R \over m} + {F_t \over m}} v^2)$$

$$\dot v = (-{R \over m} + {F_t \over m}) (1 + \frac {d \over m} {{R \over m} - {F_t \over m}} v^2)$$

$$\dot v = (-{R \over m} + {F_t \over m}) (1 + \frac {d } {{R } - {F_t }} v^2)$$

So if we put:

$A = -{R \over m} + {F_t \over m}$
$$B^2 = \frac {d } {{R } - {F_t }}$$
$$B= \sqrt {\frac {d } {{R } - {F_t }}}$$

we get our familiar equation:

$\dot v = A(1 + B^2 v^2)$

which as said solves to:

$$v = \frac 1 B tan(ABT+arctan(Bv_i))$$

but with different A and B:
$A = -{R \over m} + {F_t \over m}$ .................. vs ..........$A = -{R \over m}$

$B= \sqrt {\frac {d } {{R } - {F_t }}}$ .............. vs ......... $\sqrt {\frac d R}$

But this must be wrong, because speed should be increasing in case of applied traction force!

So where is the error?!?

17. Mar 28, 2014

### craigi

Not necessairily. You can apply traction and still lose speed, when going up a steep slippy hill for instance.

I didn't read the full thread, but you might want to check that you picked the right B when you squarerooted it.

18. Mar 28, 2014

### jumpjack

I also found the solution for free fall motion eqauation:
$F = ma = -cv^2 + mg$

where $c = \frac 1 2 \rho C_d S$

solves to:

(1) $v=\sqrt{\frac{mg}{c}}tanh(\sqrt{\frac{gc}{m}}t)$

In place of mg I have traction force and rolling friction, which sum to $F_t-mgC_r$:
$F = ma = -cv^2 + F_t-mgC_r$

Grouping weirdly :-)

$F = ma = -cv^2 + m(\frac {F_t} m -gC_r)$

If this is correct, than I have just to replace $g$ by $(\frac {F_t} m -gC_r)$ in (1) to get my solution:

$v=\sqrt{\frac{m(\frac {F_t} m -gC_r)}{c}}tanh(\sqrt{\frac{(\frac {F_t} m -gC_r)c}{m}}t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\sqrt{\frac{(\frac {F_t} m -gC_r)c}{m}}t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\sqrt{(\frac {F_t} {m^2} - \frac {gC_r} m ) c}t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\sqrt{(\frac {F_t} {m^2} - \frac {mgC_r} {m^2} ) c }t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\frac 1 m \sqrt{(F_t -mgC_r ) c }t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\frac t m \sqrt{(F_t -mgC_r ) c })$

$v=\sqrt{\frac{(F_t -mgC_r)}{\frac 1 2 \rho C_d S}}tanh(\frac t m \sqrt{(F_t -mgC_r ) \frac 1 2 \rho C_d S })$

I think for $F_t$ I should use $\frac T r$, with T = engine torque and r = wheel radius.

$$v=\sqrt{\frac{(\frac T r -mgC_r)}{\frac 1 2 \rho C_d S}}tanh(\frac t m \sqrt{(\frac T r -mgC_r ) \frac 1 2 \rho C_d S })$$

Now it's "just" a matter of extracting t from this mess to be able to calculate 0-60 time given Cd, Cr, S and T!

(but of course only if my math is right!!)

19. Mar 28, 2014

### jumpjack

I simplify again to try to extract t:

$$v=\sqrt{\frac{Q}{c}}tanh(\frac t m \sqrt{Q c})$$
$Q=\frac T r -mgC_r$
$c=\frac 1 2 \rho C_d S$

In linear form:

v = sqrt(Q/c) * tanh( (t/m) sqrt(Qc))

y = sqrt(Q/c) * tanh( (x/m) sqrt(Qc))

This should solve to:

$$t = \frac {m * atanh \frac {v} {\sqrt{\frac Q c}}} {\sqrt {cQ}}$$

Now back to impossible form :-)

$$t = \frac {m * atanh \frac {v} {\sqrt{\frac {\frac T r -mgC_r} {\frac 1 2 \rho C_d S}}}} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

$$t = \frac {m * atanh \frac {v * \sqrt {\frac 1 2 \rho C_d S}} {\sqrt{\frac T r -mgC_r }}} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

$$t = \frac {m * atanh (v * \sqrt \frac {\frac 1 2 \rho C_d S} {\frac T r -mgC_r } )} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

20. Mar 28, 2014

### jumpjack

In planar form:

t=M*atanh( v*sqrt((0.5*1.22*X*S)/((T/(d/2))-MGR)) ) / sqrt((0.5*1.22*X*S)*((T/(d/2))-MGR))
M=mass (kg)
v = speed (m/s)
X = Cd = Cx = Air Drag coefficient
S = Frontal Area (m2)
T = Torque (Nm)
d = wheel diameter (m)
G = 9.81 m/s2
R = Cr = Rolling Friction coefficient