Math: Calculate force from fuel throughput

Hi, I hope this is the right newsgroup for this question, otherwise please direct me to the correct one. My math and my knowledge of car mechanics range from rusty to inexistent, I hope you can still bear with me. Having said that: I am trying to set up a very simplicistic car simulation. My first step looks basically like this: // time frame in s, applied thrust in N public void step(double time,double thrust) { double force=thrust-getResistance(); // rolling and air resistance double acceleration=force/weight; velocity+=acceleration*time; position+=velocity*time; } That seems to work. Now I would like to specify the amount of fuel burnt in the given time frame instead of applied thrust. (I assume that I get the whole fuel energy without any loss for the moment.) But I have problems figuring out how to calculate the resulting thrust force from the fuel energy. I think I need the distance traveled in this time frame, however, this value is not available at the moment I'm calculating thrust. Am I totally on the wrong track? Could somebody please enlighten me or point me to a decent web page on this topic? Regards, Patrick

As long as you don't mind using simplistic calculations you could use brake specific fuel consumption to convert from fuel burned to power, and then divide the power by the velocity to get thrust. Any way you slice it up though you are going to need to refer to either a length or a velocity to get from thrust to fuel use, due to the dimensionality of the problem. If you don't have velocity or distance yet then perhaps you can extrapolate them from the previous step, and then use a predictor-corrector method to refine the calculation. "Patrick Roemer" <[email hidden]> wrote in message news:chsoo2$kko$00$[email hidden]...

Quoted text (30 lines)

> Hi, > > I hope this is the right newsgroup for this question, otherwise please > direct me to the correct one. My math and my knowledge of car mechanics > range from rusty to inexistent, I hope you can still bear with me. > > Having said that: I am trying to set up a very simplicistic car > simulation. My first step looks basically like this: > > // time frame in s, applied thrust in N > public void step(double time,double thrust) { > double force=thrust-getResistance(); // rolling and air resistance > double acceleration=force/weight; > velocity+=acceleration*time; > position+=velocity*time; > } > > That seems to work. Now I would like to specify the amount of fuel burnt > in the given time frame instead of applied thrust. (I assume that I get > the whole fuel energy without any loss for the moment.) But I have > problems figuring out how to calculate the resulting thrust force from > the fuel energy. > > I think I need the distance traveled in this time frame, however, this > value is not available at the moment I'm calculating thrust. Am I > totally on the wrong track? Could somebody please enlighten me or point > me to a decent web page on this topic? > > Regards, > Patrick

You can't go exactly from rate of fuel consumed (g/s or kW or similar) to thrust, but you can go from fuel to power if you know the efficiency of the engine. Efficiency (or brake specific fuel consumption) is often plotted as a fcn of torque and rpm, but you can start with a number like 30% if you wish (varies from about 20-35 or 40% depending on load, rpm, engine technology). Anyway, in power units: fuel consumed (power)*efficiency = engine power = torque (engine)* rpm torque (wheel) = gear ratio * torque (engine) * drivetrain efficiency road force = torque(wheel)*wheel efficiency/radius

Haqsau wrote: > As long as you don't mind using simplistic calculations you could use brake > specific fuel consumption to convert from fuel burned to power, and then > divide the power by the velocity to get thrust. I don't mind simplicity at all, quite the contrary. But my initial velocity is zero. How do I get the car started? Regards, Patrick

Dave wrote: > You can't go exactly from rate of fuel consumed (g/s or kW or > similar) to thrust, but you can go from fuel to power if you know > the efficiency of the engine. I'd like to assume an efficiency of 1.0 for the sake of simplicity. How do I calculate the force resulting from expending a certain amount of fuel within a given time frame? Regards, Patrick

In article <chtkm0$hma$05$[email hidden]>, Patrick Roemer <[email hidden]> wrote: >I don't mind simplicity at all, quite the contrary. But my initial >velocity is zero. How do I get the car started? That's where the power model (force = power/velocity) breaks down. Gotta use the dynamical approach (force = torque/radius). The power at launch is actually quite low. Well, essentially nil right at the wheel, but probably a fair chunk going to heat!

Dave wrote:

Quoted text (7 lines)

>>I don't mind simplicity at all, quite the contrary. But my initial >>velocity is zero. How do I get the car started? > > That's where the power model (force = power/velocity) breaks down. > Gotta use the dynamical approach (force = torque/radius). The > power at launch is actually quite low. Well, essentially nil > right at the wheel, but probably a fair chunk going to heat!

Thanks. I'll have a closer look at the torque model then. Best regards, Patrick

Sorry guys, I ready every post and reply, and tried to follow what was happening, but got a nose bleeed from the mental extertion. Hope you find a simple solution - but I'd hate to see the complex version! Keith "Dave" <[email hidden]> wrote in message news:09q0d.221947$[email hidden]...

Quoted text (14 lines)

> > You can't go exactly from rate of fuel consumed (g/s or kW or > similar) to thrust, but you can go from fuel to power if you know > the efficiency of the engine. Efficiency (or brake specific fuel > consumption) is often plotted as a fcn of torque and rpm, but you > can start with a number like 30% if you wish (varies from about > 20-35 or 40% depending on load, rpm, engine technology). > > Anyway, in power units: > fuel consumed (power)*efficiency = engine power = torque (engine)* > rpm > torque (wheel) = gear ratio * torque (engine) * drivetrain > efficiency > road force = torque(wheel)*wheel efficiency/radius

He can combine both approaches. Use throttle position and rpm to come up with fuel usage. Convert to power using some sort of BSFC curve. From there, torque = power/rpm. Etc. "Dave" <[email hidden]> wrote in message news:u7u0d.21901$[email hidden]... > In article <chtkm0$hma$05$[email hidden]>, Patrick Roemer<[email hidden]> wrote: > > That's where the power model (force = power/velocity) breaks down. > Gotta use the dynamical approach (force = torque/radius). The > power at launch is actually quite low. Well, essentially nil > right at the wheel, but probably a fair chunk going to heat!

Patrick Roemer wrote: > I'd like to assume an efficiency of 1.0 for the sake of simplicity. How > do I calculate the force resulting from expending a certain amount of > fuel within a given time frame? To do what you ask in your question, you _must_ include efficiency. Engine power is proportional to stored energy in the fuel times the efficiency of the engine in converting that chemical energy into mechanical energy times the rate at which fuel is consumed. Power is proportional to torque times speed. Torque is the number you need to determine thrust at the contact patch. If you want to convert from fuel rate to thrust, you will need to crunch the numbers in the equations to determine a simple factor such that Thrust=FuelRate * Factor. I believe the theoretically perfect thermodynamic engine operating between peak cumbustion temperature and ambient temperature is capable of about 20% efficiency. Real engines are less efficient than that. Later, Jason C. Wells

Jason C. Wells wrote: > To do what you ask in your question, you _must_ include efficiency. > Engine power is proportional to stored energy in the fuel times the > efficiency of the engine in converting that chemical energy into > mechanical energy times the rate at which fuel is consumed. That's only a constant factor AFAICS? So I can assume that engine power is about 44MJ * fuel throughput in kg per time frame * 0.2 for 'perfect' efficiency? Regards, Patrick

In article <[email hidden]>, "Jason C. Wells" <[email hidden]> wrote: >I believe the theoretically perfect thermodynamic engine operating >between peak cumbustion temperature and ambient temperature is capable >of about 20% efficiency. Real engines are less efficient than that. Nah, gasoline ICE's peak efficiencies are in the 35% range. That's usually at mid rpm and high (but not full) throttle. Overall average efficiency over a typical drive cycle (city/highway) is about 20-25%. That's defined as engine net crank power over fuel energy (lower heating value). Diesels peak out at about 40%, sometimes a couple of points higher. And I believe they don't drop off as much as gasoline engines do at lower throttle settings.

In article <ci1gna$m71$03$[email hidden]>, Patrick Roemer <[email hidden]> wrote: >That's only a constant factor AFAICS? So I can assume that engine power >is about 44MJ * fuel throughput in kg per time frame * 0.2 for 'perfect' >efficiency? See my previous. 35% near peak, 20-25% typical. But again, what are you trying to do??

Dave wrote:

Quoted text (8 lines)

> Nah, gasoline ICE's peak efficiencies are in the 35% range. > That's usually at mid rpm and high (but not full) throttle. > Overall average efficiency over a typical drive cycle > (city/highway) is about 20-25%. That's defined as engine net > crank power over fuel energy (lower heating value). Diesels peak > out at about 40%, sometimes a couple of points higher. And I > believe they don't drop off as much as gasoline engines do at > lower throttle settings.

Going back to my books, I see that you are correct, if not a little low. A 10:1 compression engine running at fuel/air lamba of 1.0 and a specific fuel is about %47 efficient. The Otto cycle effiency is around 60% for 10:1. My big thick books are a bit dusty. Real would be less, which means you are closer than I am. Patrick, use Dave's number. Later, Jason C. Wells

Dave wrote: > See my previous. 35% near peak, 20-25% typical. But again, what > are you trying to do?? Nothing serious in fact, just being curious. I just wanted to try to get a car moving the simplest way possible and then see where that leads me. Regards, Patrick

Diesels generally don't have a throttle. "Dave" <[email hidden]> wrote in message news:rTY0d.64094$[email hidden]... > > Diesels peak > out at about 40%, sometimes a couple of points higher. And I > believe they don't drop off as much as gasoline engines do at > lower throttle settings.

In article <[email hidden]>, "Haqsau" <[email hidden]> wrote: >Diesels generally don't have a throttle. I meant throttle pedal, which in this case I admit would be a misnomer. Lack of an actual throttle is in part why they get better efficiency in turndown.