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Test Driving a Car Virtually

  1. Apr 3, 2014 #1
    I played around in another thread with mathematically determining where a car should be time wise on 1/4 mile run. I had a chance to drive this car myself giving me more information than you normally have on a car you have yet to buy. As I thought more deeply about my calculations, I became interested in estimating characteristics of a car modified to a new engine that I can not drive based on data commonly available about all vehicles. I'll make a new post for each step I take on this endeavor. I have a feeling it will take days.
  2. jcsd
  3. Apr 3, 2014 #2
    So my starting point is the equation for acceleration based on power.

    A(t) = P(V(t))/ (M * V(t))

    A = Acceleration
    t = Time
    P = Power
    V = Velocity
    M = Mass

    My first goal is to come up with an equation for power based on velocity as velocity changes over time. The first component of power that I want to look at is the power curve. Now normally, I will only have two power readings at the peak torque RPM and peak power RPM. I need to somehow translate this into an equation that can approximate the power at any RPM.

    W = P * C / RPM

    P = Power
    C = Unit conversion constant
    W = Unknown Work given a specific number of rotations
    RPM = rotations per minute

    As an example I'm going to use an engine with these characteristics

    Power: 245 HP
    Power RPM: 4300 RPM
    Torque: 345 ft-lb
    Torque RPM: 3200 RPM

    With this we can see that

    W(4300 RPM) = 245 HP * 5252 / (4300 RPM) = 299 ft-lbs
    W(3200 RPM) = 345 ft. lbs

    Two things are happening here. The first is that we are getting a linear increase in energy due to every increase in revolutions leading to an increase in fuel placed into the engine. At the same time there is a loss in efficiency due to a wide variety of factors including friction and heat transfer. Unfortunately, 2 points do not give us enough information to determine the characteristics of these many factors. All we can hope to deduce is an average constant deceleration of linearly increasing energy. Think of it like a ball being thrown in the air. Our energy starts out increasing linearly as the RPM's increase, just like a ball starts out going into the air with a linear vertical velocity as inertia attempts to keep the velocity constant. Just like the inverse parabola of our ball facing near constant gravity, we are going to assume our torque forms an inverse parabola as it faces an average constant deceleration. This curve should peak where the torque peaks, which is why people watching fuel consumption try to straddle the max torque peak rather than the power peak when shifting. To make this easier we will recreate our axis, so that the torque peak is sitting on the y-axis.

    W(z) = (Tp - Tt) * z2 + Tp

    W = Work
    z = (RPM - RPMt) / (RPMp - RPMt)
    RPM = Current RPM we want to know work for
    RPMt = RPM at peak torque
    RPMp = RPM at peak power
    Tp = Torque at peak power
    Tt = Torque at peak torque

    Now that we have a rough equation to estimate our torque curve the best we can with limited information, we can estimate the power curve by simply using the equation

    P(z) = W(z) * RPM / C

    P = Power
    z = (RPM - RPMt) / (RPMp - RPMt)
    RPM = Current RPM we want to know work for
    RPMt = RPM at peak torque
    RPMp = RPM at peak power
    C = Unit Conversion Constant

    To change this into P(V(T)) we are going to need to start examining how that power gets to the wheels. So far this estimated power curve is a good start.
    Last edited: Apr 3, 2014
  4. Apr 5, 2014 #3
    In an ideal world the power from the engine would go directly toward moving the vehicle. We already saw this was not happening in the engine. Gasoline has a given potential energy per mass, but the power of our engine did not increase purely based on the amount of gasoline burned. A four stroke engine brings in gas every two cycles, so a constant pressure on the gas pedal should release a linear increase of potential energy from the gas. It does do this for all of the gasoline burned, but most of that energy is transformed into things like heat, and heat doesn't move cars.

    http://www.rookeye.com//images/efficiency.jpg [Broken]

    This MIT report gives a good breakdown of where the rest of the energy goes in a 2.5L 2005 Camry. Details on the efficiencies of specific cars are hard to come by, but this benchmark can help us get an idea of what to expect on any car. Loss in the driveline is not a flat line over RPM's, but it is unrealistic to expect a person to find exact numbers on those details. The only force we see here that should dramatically change based on velocity is drag.
    [tex]P_d = 1/2 \rho V^3 C_d A[/tex]
    Fd = Force of drag
    [itex]\rho[/itex] = density of air
    V = velocity
    Cd = Drag coefficient
    A = cross-sectional area of object

    Ex: cd = .34, A = 21.23 ft2, V = 60MPH
    .0807 lbs(mass)/ft3 / 2 * 1 lbs(force) / 32.2 lbs(mass) * .34 *
    21.23 ft2 * ((60 MPH) * (5280 ft / 3600 s))3 *
    1 HP / 550 (ft-lbs(force)/s) = 11.2 HP

    Aerodynamic drag has a staggering cubic increase as speed increases. We can't simply take the numbers from the picture above for this number. There are several Wiki Lists that keep up to date drag coefficients and frontal areas. Even if you can't find the exact car in question, you should be able to make an educated guess from a similar car.

    This still leaves the question of how the factors other than drag effect our vehicle when examining power loss from the engine to the road. I don't expect to have the details to be very precise on this measurement. I would like to find a rough average that I can apply at all times. For cars rated for sale in America we have an amazing tool. The end of the EPA regulations governing MPG tests gives us the exact speeds a vehicle was tested under for a given amount of gas and distance. It also gives us the precise energy per pound in the gasoline burned, and it measures the gas burned by looking at the gasses leaving the exhaust, so we know that we are looking at the energy actually released rather than the energy in the gas leaving the tank. We still have to factor in road load and drag, because the EPA forces them to simulate these forces. We can look at the energy that should have been produced and then separate it out into the energy lost in the engine and overcoming inertia verses the energy lost in everything else.

    Let's start with figuring out the energy that should have been produced. Your life is easy if you are looking at a car made before 2008. The test in the earlier link is all that is used with the EPA knocking off an arbitrary 22% from the number. If you have a vehicle with a model after 2008 you are going to have to deal with the much less laboratory accurate MPG that the EPA uses today. The highway efficiency for these later models is a mixture of the old test run once, and the highway segment of the new US06 test run about 6 times with an extremely small adjustment from other tests that would effect the value by less than a 1/10 of a percent. They then give it an arbitrary 7% bonus followed by the old arbitrary 22% docking. On the mixed test, we don't have an actual test time, so the easiest way to work this is to multiply the test figures by their weight. The old test is 765 seconds long. The new segment is 360 seconds long. The old test weighted by 21% and the new one weighted by 79% makes the combined test's time to be 445.04 seconds long. The old test was 10.25 miles long. The new test was 6.25 miles long. This makes the comined test 7.09 miles long

    Model Year < 2008:
    10.25 Miles / (Fe / .78) * 6.22 lbs(mass) / 1 Gallon * 18478 BTU / 1 lbs(mass) /
    765 seconds / 1 hour * 3600 seconds * 1 HP / 2544 (BTU/hr)

    Model Year >= 2007:
    7.09 Miles / (Fe / .78 / 1.07) / 6.22 lbs(mass) * 1 Gallon *
    18478 BTU / 1 lbs(mass) / 765 seconds / 1 hour * 3600 seconds * 1 HP / 2544 (BTU/hr)

    Fe = fuel efficiency in miles per gallon Ex. 1990 model 24MPG:
    10.25 Miles / (24MPG / .78) * 6.22 lbs(mass) / 1 Gallon * 18478 BTU / 1 lbs(mass) /
    765 seconds / 1 hour * 3600 seconds * 1 HP / 2544 (BTU/hr) = 70.8 HP

    This obviously seems very low when you compare it to the horsepower of the engine, but you have to remember that this test was not done with the engine running at full throttle the whole time. For a good comparison, let's take a look at the power required to move this vehicle over the distance of the test. For this example we'll say my car's weight with driver is 3627 lbs(mass).

    W = F * D = 3627 lbs(mass) * 10.25 miles * 5280 ft / 1 mile * 1 lbs(force) / 32.17 lbs(mass) = 6101748 (ft-lbs)
    P = W / t = 6101748 (lbs-ft) / 765 seconds * 1 HP / 550 (ft-lbs/s) = 14.5 HP = 20.4% of gasoline HP

    My next job is going to be determining how this power is used. This is already quite a bit, so I'll start on that in the next post.
    Last edited by a moderator: May 6, 2017
  5. Apr 7, 2014 #4
    It would be really nice if you could use metric units rather than imperial ones: Great Britain World Empire (="imperium" in latin) is currently just a small United Kingdom limited to a couple of islands... ;-)
  6. Apr 8, 2014 #5
    My post will be short today as time has just disappeared somehow. I'll go back and try to do some metric unit examples later. SI are what I'm used to, so it is easier for me. Today I think I'll just give some idea of where I'm heading with all of this in case someone wants to work it out faster than I am. The law of the conservation of energy also let's us know that power is conserved within a time frame.

    W = ΔE
    P = W/T

    So we know that power is a way of describing the energy changed over a given time. My end equation for the power is going to be this:

    PG = PE + PF + PD + PM + PB
    PG = the power we determined comes from the gas in the last post.
    PE = the power we lose in the engine.
    PF = the loss from rolling friction and driveline loss
    PD = the cubic loss from aerodynamic drag
    PM = power going into forward motion of the car
    PB = power lost in the brakes

    Most of the loss in the driveline and rolling of the tires will be from friction and slippage. If we want to get more detailed, we need to figure out down force, because the equation for friction is based on the force perpendicular to motion, and the tires are moving against the ground. To make things more complicated, you are getting loss through the deformation of the tires. Overall though, you should end up with a roughly linear component to the overall power equation. I plan to determine a rough coefficient of friction for all the components combined between the engine and the road.

    [itex]F_f = \mu F_n[/itex]
    [itex]F_f = [/itex]force of friction
    [itex]\mu = [/itex]coefficient of friction
    [itex]F_n = [/itex]normal force, which is the vector of forces perpendicular to the plane of friction

    Pf = Ff * V(t)

    The engine loss is going to be the difference in potential torque from an increase in RPM and the actual torque curve we saw earlier. We can figure this out by looking at the displacement of the engine. The engine displacement is not actually the volume of the cylinders. It is the amount of volume covered by the motion of the pistons. This is roughly the amount of air that could be sucked into the engine in an ideal world, and all cylinders should be filled once for every two revolutions of the standard four stroke engine. Air is the limiting component in the reaction, so we can use it to determine the ideal energy output per revolution. The brakes are going to be what is left over once we hit a certain deceleration threshold that seems impossible for the other power losses to account for.

    The end power equation for full throttle between the engine and the wheels should look something like this:

    PWheels = PEgnine - AV(t)3 - CV(t)

    At that point the sky is the limit. You can even find the top speed through a special case of the cubic formula:

    Top Speed = [itex]\sqrt[3]{P_M / 2A+\sqrt[2]{(P_M / 2A)^2+(C/3A)^3}}+\sqrt[3]{P_M / 2A-\sqrt[2]{(P_M / 2A)^2+(C/3A)^3}}[/itex]

    PM = Peak power fighting losses in the drive train and drag

    Once I get to that point I can crank out the maximum turn speed by using the friction equation above, and at that point we can drive our little car all around the track. We'll know the power of acceleration, the speeds of turns by radius, and the loss in resistance. So that is where I'm going. Feel free to take this further. It may be a few days until I have the free time to sit down and crank out more of the calculations on a sample car.
    Last edited: Apr 8, 2014
  7. Apr 8, 2014 #6
    As far as I know:

    $$A = \frac 1 2 \rho C_d A$$
    $$C = m g C_{rr}$$

    ##C_d## = air drag coefficient (~ 0.3 for cars, 0.8 for motorbikes)
    ##\rho## = air density = 1.225 kg/m3
    A = frontal area (~ 2.2 m^2 for cars, 1.0 m^2 for motorbikes)
    m = mass
    g = 9.81 m/s2
    ##C_{rr}## = rolling friction coefficient (~0.01 for cars, 0.008 for motorbikes, almost constant wrt speed)
  8. Apr 8, 2014 #7
    Some figures for losses:
    http://auto.opinionzine.com/wp-content/uploads/2014/03/energia-auto-benzina-citta1-333x580.png [Broken]
    http://auto.opinionzine.com/come-ridurre-i-consumi-della-propria-auto-elettrica-e-non/6781 [Broken]
    Last edited by a moderator: May 6, 2017
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