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Calculating 0-60 mph time for a vehicle

  1. Mar 24, 2014 #1
    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. jcsd
  3. Mar 24, 2014 #2


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    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? :wink:
  4. Mar 24, 2014 #3


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    Power alone gives you estimate of the shortest required time. Real time will be always longer.
  5. Mar 24, 2014 #4
    We can take into account just airdrag in a first instance.

    @borek , your assumption is already included in my question...
  6. Mar 24, 2014 #5


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    It is not an assumption, it is the energy conservation.
  7. Mar 24, 2014 #6
    I don't need results which do not take into account airdrag.
  8. Mar 24, 2014 #7


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    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.
  9. Mar 24, 2014 #8
    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
  10. Mar 24, 2014 #9
    Oh, and let's suppose both power and torque constant, I'm interested on electric vehicles mostly.
  11. Mar 27, 2014 #10
    No ideas?
  12. Mar 27, 2014 #11


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    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".
  13. Mar 27, 2014 #12
    Just help if you can, don't argument.

    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?
  14. Mar 27, 2014 #13


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    well this is what I would do:
    [itex]F_{propulsion}-F_{drag}-F_{rolling friction}-F_{whatever else} = m_{car}a_{car}[/itex]

    [itex]F_{propulsion}[/itex] 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 [itex]F_{propulsion}= \frac{T}{R}[/itex].
    [itex]\frac{T}{R} - F_{drag} -...=m_{car}a_{car}.[/itex]

    integral(from 0 to t) a dt = [itex]a_{SomeConstantFromNonDragTerms}t - inegral(from 0 to t)\frac{1}{2}ρC_{d}v^2A_{crossection} [/itex]= 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
  15. Mar 28, 2014 #14
    Did you mean this?
    [itex]\int_0^t a dt = a_{SomeConstantFromNonDragTerms}t - \int_0^t\frac{1}{2}ρC_{d}v^2A_{crossection} [/itex]= velocity change from time 0 to t.
  16. Mar 28, 2014 #15
    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:

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

    Adding traction force:

    ## 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##
  17. Mar 28, 2014 #16
    ##\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?!?
  18. Mar 28, 2014 #17
    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.
  19. Mar 28, 2014 #18
    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!!)
  20. Mar 28, 2014 #19
    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 :-)

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

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

    $$t = \frac {m * atanh (v * \sqrt \frac {\frac 1 2 \rho C_d S} {\frac T r -mgC_r } )} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$
  21. Mar 28, 2014 #20
    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
  22. Apr 2, 2014 #21
    ##v=\sqrt{\frac{m(\frac {F_t} m -gC_r)}{c}}tanh(\sqrt{\frac{(\frac {F_t} m -gC_r)c}{m}}t)##

    I found confirmation to my formula in this book;
    §, formula 2.11:
    ##v(t) = \sqrt \frac {K_1}{ K_2} tanh (\sqrt {K_1 K_2} t)##

    It matches with mine considering that book defines (p.30):
    ##K_1 = \frac F m - gC_r##
    ##K_2 = \frac 1 2 \rho C_d A * \frac 1 m ##

    So in my equation I have:
    ##K_1 = \frac Q m, Q= mK_1##
    ##K_2 = \frac c m ## , ##c = m K_2##
    ##Qc = m^2 K_1 K_2 ##
    ##\frac Q c = \frac {K_1}{K_2}##

    My formula for 0-60 time:

    $$t_{60} = \frac {m * atanh \frac {27.7} {\sqrt{\frac Qc}}} {\sqrt {Qc}} = \frac {m * atanh \frac {27.7} {\sqrt{\frac {K_1}{K_2}}}} {m\sqrt {K_1K_2}} = \frac {m * atanh (\sqrt{\frac {K_2}{K_1}}27.7 ) } {m\sqrt {K_1K_2}} = \frac {atanh (\sqrt{\frac {K_2}{K_1}}27.7 ) } {\sqrt {K_1K_2}} $$

    Book's formula:

    $$t_{60} = \frac 1 {\sqrt{K_1 K_2}} atanh(\sqrt{\frac {K_2} {K_1} } v_f) $$

    But both using my formula and looking at book's example I get impossible times to get to 60 mph: 60-90 seconds rather than 10-20.

    What is it going wrong in this math?
  23. Apr 2, 2014 #22
    What values are you using with units. It is hard to check where you might be going wrong, if I don't know the details of the object you are trying to make calculations for.
  24. Apr 2, 2014 #23
    I think your biggest mistake is probably where you started using T / r with T = engine torque, and r = wheel radius. The wheel radius alone is not enough, because the torque on the wheel is dependent both on the gear ratio and the velocity of the car. Unless you are assuming that power is on a never ending climb to the heavens, torque needs to be an equation in relationship to velocity. You can't keep torque as a constant value unless you increase power.

    Power = Work / Time

    The torque needed to produce work over time at a specific RPM changes as the RPMs change:

    Power = Torque * RPM / (Unit conversion constant)

    You can't keep torque as a constant unless you are assuming your engine is constantly increasing power no matter what speed you are at. If your engine is constantly changing power then it has a limit. Gasoline engines can get better as their crank speed increases, but even gasoline engines eventually hit a power peak, and then you are going to have to decrease torque at the wheel by shifting gears or watch the torque drop in the same gear as has to happen if power doesn't increase. You might get away with using an average torque, but then it is going to have to be the average torque at the wheel, not the engine. Torque at the engine needs gear ratios to determine how much force is applied in relationship to the wheel radius. Torque at the wheel is directly related to the wheel radius, but it is based on both gear ratio and power loss moving power from the engine to the wheel. Keep in mind to figure out average torque at the wheels you are going to need to know the power being sent to the wheels in relationship to time as well as the vehicle velocity in relationship to time. You still end up needing an equation for torque.

    What might be easier is going back and determining how exactly you want to define force at specific times. That F in itself needs to become F(t), and my personal opinion is that power is easier to work with in that equation, because it doesn't rely on the gears as much. A power in first gear is going to apply very similar power compared to second gear at the wheel. A gear ratio 2:1 vs. 1:1 applies different torques at the wheel by a factor of 2. Torque is more important if you are trying to figure out whether or not the car has enough torque to overcome the high friction of the vehicle at rest, but working it back into power is easier when dealing with the transfer of force from the engine to the wheel.
    Last edited: Apr 2, 2014
  25. Apr 2, 2014 #24


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    Power = Torque / Time [taken to rotate through one radian]

    This assumes you are using a system of units where torque and energy share the same units, e.g. both in foot-pounds or both in kg meter/sec2

    I expect that RedRook is perfectly well aware of the above.
  26. Apr 2, 2014 #25


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    Torque is computed as a vector cross product. Force times perpendicular distance.
    Work is computed as a vector dot product. Force times parallel distance.

    They share the same units (more or less) but they measure different quantities. They are connected by the fact that if you exert a given torque through a rotation angle of one radian, the torque and the resulting work are numerically equal.
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