3
$\begingroup$

How can I calculate the current (instantaneous) mpg of my car if I know the speed and acceleration of the car? From reading various answers for the "car going level or up/down hill" question asked earlier by someone, I understand there are 4-5 factors to consider:

  1. Rolling resistance (how do I get or calculate this value for my make/model?)

  2. Engine resistance (how do I get or calculate this value for my make/model?)

  3. Air drag coefficient (assume I know this)

  4. Mass of car (assume I know this)

  5. "Power curve" of car? What is power curve? Do I need to know this? How do I get or calculate this?

  6. Slope of road (assume level for simplification)

Improve this question
$\endgroup$
5
  • $\begingroup$ This qustion seems a little rediculous to me; are you (for example) merely looking for a way to see how much it is costing you to push harder on the accelerator when the light goes green?? I.e. a more instantaneous mpg reading than modern cars currently offer. If so, the simlest answer would be to design a more accurate fuel level guage that took temperature and gaseous pressure in the fuel tank, as well as the actual fuel level, into account. $\endgroup$
    – qftme
    Commented Apr 28, 2011 at 23:55
  • $\begingroup$ My car is not a newer model with instantaneous mpg readout. Also, there are some onboard diagonostic meters that may do this for $200-300. I wanted to know of it could be done using physics and engineering. $\endgroup$
    – user3344
    Commented Apr 29, 2011 at 16:49
  • $\begingroup$ How do you think the $200 gadgets work? I don't feel this answers the question. $\endgroup$
    – qftme
    Commented Apr 30, 2011 at 2:50
  • $\begingroup$ Some approximate calculation is possible in principle, but it is not clear what would constitute and "answer" to this battery of questions. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented May 30, 2011 at 19:00
  • $\begingroup$ Sender resistance is missing in the 6 factors given. $\endgroup$
    – Inquisitive
    Commented Dec 29, 2012 at 14:04

5 Answers 5

Reset to default
2
$\begingroup$

You can't. At least, not in any accurate way. The reason being that you have too many variables. Each variable adds a significant amount of error due to inaccuracies in measurement to your calculation. This might be a nice math problem, but as a physics problem, it has little application to the real world. My WAG is that the error in your calculation would probably exceed 50%.

If you want to determine fuel consumption, try just accurately measuring how much fuel is being consumed at any given moment in time. Your engine's fuel pump will "know" this information based on how hard it needs to work. It's a pretty trivial task to convert this information into something usable if you're in the process of designing a car's computer, which is why this feature is included in many newer cars.

As for the math that would give you a "miles per gallon" number, it would go like this:

For argument's sake, consumption rate R is measured by the fuel pump in ml/second. For the time being, let's measure speed (V) in metric as well, since that too is being measured accurately. Speed needs to be reduced to meters/second to match the fuel pump's consumption rate figures, allowing them to cancel out. The amount of fuel used over any distance would be:

Consumption = R / V

and Consumption would be measured in millilitres/meter. Multiply by 1000 for litres/km.

Improve this answer
$\endgroup$
1
$\begingroup$

there is an online calculator which gives answer to your question:

https://www.gribble.org/cycling/power_v_speed.html

It presents the result for the given parameters. The only thing you need to do is convert power to mpg. This part is a bit tricky but there are many websites with ready answers: https://www.aqua-calc.com/convert/electric-car-energy-economy/miles-per-gallon-gasoline-equivalent-to-watt-hour-per-mile

This is hardly a scientific approach but is fairly quick:))

Improve this answer
$\endgroup$
1
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Vincent Thacker
    Commented Apr 16 at 8:47
0
$\begingroup$

You can, using kinematics and there are some third party libraries to do this using factors about the vehicle within 10% of the engine controller it also uses the gradient of terrain. I saw someone also refine this approach to also use average wind speed. I have some python code to do this.

Improve this answer
$\endgroup$
0
$\begingroup$

I was asking a similar question about general energy consumption. The approach I took is presented below. I hope some of you could verify this in terms of physics, specifically in regard to the Law of conservation of energy.

**

The mass of a car equals 1000 kg. Its frontal area is $2x1,5 = 3 m^2$ and the aerodynamic coefficient is 0,3. The average rolling coefficient is 0,01. Let’s assume the drive-chain is 100% efficient and there is no AC nor heating. The brakes use no recuperation unfortunately, for ease of calculation.

This took me quite a while to find the right formula to calculate this. I hope it is convincing:

$$E_{el} = E + E_l$$

Where: $E_el$ – is total energy consumed by the motor, $E$ – energy consumption of the motion, $E_l$ – drive train and idling losses [ $E_l = 0$ in our case ]

$$E = E_{rr} + E_{air} + E_{gr} + E_k$$

Where: $E_{rr}$ – energy lost to rolling resistance, $E_{air}$ – energy lost to air resistance, $E_{gr}$ – enrgy lost to climb grades = 0 as assumed, $E_k$ – energy used to increase kinetic energy.

$$E = ½ mv^2$$ - during acceleration only

$$E_{rr} = mgC_{rr}(L_a + L_d)$$ and $$E_{rr} = m g C_{rr} L_{avg}$$

Where $m$ is mass, $g$ – gravity, $C_{rr}$ – rolling resistance coefficient, $L$ - distance covered, $a$ – acceleration, $d$ – deceleration, $avg$ – average velocity for the steady case.

$$E_{air} = AC_d(V_{avg}^2L_a + V_{avg}^2L_d)$$ and $$E_{air} = AC_d V_{avg}^2L_{avg}$$

Where $A$ - frontal area, $C_d$ - air friction coefficient [ $\rho = 1$, negligible ]

Improve this answer
$\endgroup$
-1
$\begingroup$

No, you cant. Because it is simple to know mpg we have to calculate fuel flow. Only measuring speed and acceleration you know variation distance time and need to calculate fuel flow also.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.