Car Engine Problem: Acceleration Function of RPMs

[whozum]

I made this problem up as I was driving to school.
Suppose you press the gas pedal on your car and your rpm's rev from 800 to 6200. You also know that your car's max horsepower is 150 (at 6200rpm). Find avg acceleration and acceleration function of the car as a function of RPMs.

Time is defined in seconds as the time it takes to get from 800 to 6200rpms. For simplicity we will assume a gear ratio of 1, and of course neglect friction.

I know this can be done using kinematic equations and a few functions but I wanted to do it this way and this is what I got:

P = W/t
For every infinitesimally small increment of RPM the engine produces a power P in an infinitesimally small time dt or,

dW = P(t) dt

Assuming a quadratic relationship between power and time, we'll say P = c*t^2 where c is some proportionality constant.

dW = c*t^2 dt and
W = c/3*t^3

Work is F*d and F = m*a and the equation becomes
3mad = ct^3 and solved for a is:
a(t) = (ct^3)/(3md)

I don't have maple to graph the function to see if it makes sense, but a cubic function isn't really what I was expecting. So with a linear relationship between power and time we get:
a(t) = ct^2/(2md) which still doesn't sound right.


Something tells me I should be integrating with respect to Power, I am not sure though. All I can get for certain in my head is that at a certain RPM constant power is produced. Constant power means constant jerk (skipping some kinematics here). So variable power should mean constant jounce (whatever the 4th derivative of position is) and quadratic acceleration.

 

[whozum]

bump? Someones got to have some input.

 

[Andrew Mason]

I made this problem up as I was driving to school.
Suppose you press the gas pedal on your car and your rpm's rev from 800 to 6200. You also know that your car's max horsepower is 150 (at 6200rpm). Find avg acceleration and acceleration function of the car as a function of RPMs.

...

Something tells me I should be integrating with respect to Power, I am not sure though. All I can get for certain in my head is that at a certain RPM constant power is produced. Constant power means constant jerk (skipping some kinematics here). So variable power should mean constant jounce (whatever the 4th derivative of position is) and quadratic acceleration.

I am not sure anyone can understand your question. I am not sure why you assume there is constant power. If there is no gear change and maximum power occurs at max rpm, then the car will not have maximum power as it accelerates. You don't accelerate with constant rpm (unless you are driving a hybrid).

AM

 

[whozum]

Well at 0 rpms no power is being produced
at 6200 rpms there is 150hp being produced.
There has to be some relationship between horsepower and RPM.
Does it not make sense that the car will produce most power when it is completing the most cycles per second?

A car typically doesn't have max power as it accelerates, that's why you accelerate much faster in the 4000+ rpm range than in the 2000rpm range.

If Power = \frac{Mass * Acceleration * Distance}{Time} there is a relationship between power and time, here it looks like it isn't linear because as acceleration increases so will distance for each increment time.

P dt = m \frac{d^2x}{dt^2} \frac{dx}{dt}

At a given RPM, a certain amount of power is produced, that's the goal of the engine, to produce power per RPMs. You don't accelerate at an rpm once the power produced by the engine has been displaced on to the wheels, but for example if you put your car in 1st and at 6000rpms and drop your clutch, you will certainly accelerate.

Work = Mass*Acceleration*Time

Given a changing acceleration during an increase in RPM:

\frac{d^2x}{dt^2} = a(t) = c*rpm(t)

and a changing power during increase in RPM:

\frac{dP}{dt} = k*\frac{drpm(t)}{dt}

Is it not reasonable to say that there is a relationship between power increase and changing acceleration (jerk)? :

dP = k*drpm(t) and therefore P = k\int{rpm(t)} {dt}

 

[Shockwave]

Is it not reasonable to say that there is a relationship between power increase and changing acceleration (jerk)? :

I agree that there is a relationship.

 

[whozum]

So if there is a relationship between power and RPMs, and there is definitely a relationship between Power and acceleration, that there is a relationship between acceleration and RPMs. This relationship is the one I am trying to model.

rpm(t) \Rightarrow r(t)

m\frac{d^2x}{dt}\frac{dx}{dt} = k\int{r(t)}{dt}

Lets assume that P(t) \mbox{ has a linear relationship to } r(t):

P(t) = k\int{dt}

P(t) = k*r(t)

Then:

m\frac{d^2x}{dt}\frac{dx}{dt} = k*r(t)

and \frac{d^2x}{dt} = \frac{k*r(t)}{m\frac{dx}{dt}}

Or algebraically a(t) = \frac{k}{m}\frac{r(t)}{v(t)}

 

[Andrew Mason]

Well at 0 rpms no power is being produced
at 6200 rpms there is 150hp being produced.

Engines are measured in terms of brake horsepower. - ie with a load. If you just race the engine to 6200 rpms with a small load (eg. first gear) you aren't going to get much power or acceleration. That is why you have a transmission. You seem to be omitting the load factor in your calculation of power.

Electric motors are quite different. They draw more current at slow speeds so they produce more horsepower at slow speed and less horsepower at higher speeds, which is exactly what you want to have in a car. Cars with electric drives don't need transmissions.

AM

 

[whozum]

ok you've successfully confused me.

Each gear has a moment of inertia I. Gear 1 has the lowest I and Gear 5 has the highest. The engine provides a torque T that it applies to Gear X with moment I and the result is an angular acceleration alpha. The relationship between power and force would be proportional to the relationship between power and torque, but with that why would the moment of inertia affect the power produced?

Torque is a function of power, not the other way around. (and inherently angular acceleration)

Im pretty sure hte power produced by an engine isn't dependant on where that power is going..

 

[Cliff_J]

Few more things you may or may not want to include in your calculations.

The power produced by the engine is a function of the torque and the RPM its produced at. HP = TQ * RPM / 5252 when TQ is ft-lbs, you'd need to convert for Nm.

The torque is a function of the volumetric efficiency. Typically the volumetric efficiency is ok and improves until 1/2 to 2/3 the RPM range where it declines again. So until that point where it declines again, the increasing efficiency means an increase in torque, and as it declines the torque falls off again. The increasing RPM means the power increases even with this decline but it makes for a somewhat complex parabolic curve.

On some of the newer high tech engines even the marketing boasts a flat torque curve over a specified RPM range, meaning the volumetric efficiency has been very well managed over that range. In that case, the power curve would indeed be a linear straight line and much easier to calculate.

Here's a graph of the dyno results on a big Chevrolet motor (referred to as a rat engine because it is the larger design, the smaller design is referred to as the mouse engine) to demonstrate the typical effects of unmanaged volumentric efficiency changes.
http://superchevy.com/tech/0411sc_rat_15_z.jpg

Electric motors do not have constant power either. They produce roughly constant torque from 0 RPM (max current) to about 1/2 maximum freewheel RPM (roughly max efficienty) where the inductance takes over and the torque declines to zero at the max freewheel RPM. So they have a parabolic power curve as well and it changes depending on the loading and available electrical power and these are only rough rule-of-thumb gross approximations.

Here's a few calcs on modeling an electric scooter acceleration.
http://www.ent.ohiou.edu/~urieli/scooter/scooter.html

Hope this is useful!

 

[whozum]

Thats pretty cool. Thanks a lot for that enlightment. I realize I was wrong in assuming torque as a function of power, as obviously the power depends on how quick the torque is created. I am trying to keep this as simplified as possible, so efficiencies arent a concern for now.

A quick question to AM though, if power is a function of torque, and the torque applied is always the same for a given RPM then woudlnt the power output for that RPM be the same regardless of what gear it is connected to? Given 100% efficiency and 100% torque transfer to a gear with the same moment of inertia I, the angular acceleration of the engine (RPMs) will be directly proportional to angular acceleration of the transmission.

Cliff: You are saying Power vs Torque is linear for the new engines, or power vs RPMs?

 

[krab]

if power is a function of torque, and the torque applied is always the same for a given RPM then woudlnt the power output for that RPM be the same regardless of what gear it is connected to? Given 100% efficiency and 100% torque transfer to a gear with the same moment of inertia I, the angular acceleration of the engine (RPMs) will be directly proportional to angular acceleration of the transmission.

Power is a function of engine rpm. This same power appears at the wheels (neglecting friction losses in transmission). This is because energy is conserved. Torque is NOT conserved. So a 2:1 gear reduction results in twice the torque and half the rotation speed. You see that the product (rotation speed x torque) is unchanged. That's power. That's why power is the useful quantity. I think you will also see that "100% torque transfer" makes no sense.

 

[whozum]

100% torque transfer meant that the torque applied by the engine is the same torque onto the wheels, aka there was no gear reduction.

Power is a function of torque because torque is a function of RPM? I am getting lost here. I just want a simple relationship between power and RPM because from power i can derive acceleration. Were involving all these middle steps now.

 

[krab]

THere is no simple relationship. That's why all those car mags publish graphs; every engine has its own unique graph. That said, modern engines have overcome effects of resonances and so they are not "peaky", in fact most torque curves within something like redline/4 to redline, are flat. That means that in this rpm range, the power rises linearly with rpm.

 

[whozum]

the power rises linearly with rpm.

Thats all I need.

 

[krab]

Back to yr original problem: If you adopt this simple model, then power is proportional to speed, and force at the rear wheels, being power divided by speed, is constant. So you can model it as F=constant. To find F, you need what is the speed at 6200rpm. Let's say it is 60mph, or 88 ft/s. 150hp = 82,500 ft-lb/s. Divide this by speed and you get 937.5 lb. Now assume the car is 2800 pounds = mg. Then acceleration is F/m and number of g's is F/mg = 0.33, or 11 ft/s/s = 7.3 mph per second. Zero to 60 is 8.2 seconds.
Of course this neglects air resistance. If you are willing to go to that detail, let me know; I have the equation for it.

 

[whozum]

How is power proportional to just speed?

P = m\frac{d^2x}{dt}\frac{dx}{dt}

\frac{6200 rev}{60 sec} = \frac{103 rev}{sec}

(16 inch rim diameter = 50.7" circumference)

103*2*50.7*\pi = \frac{32811 in}{sec}
32811 * \frac{2.54cm}{1in} * \frac{1m}{100cm} = 833.4 \frac{m}{sec}

Yeah that seems a bit fast for a car... Gear ratios in a car are up to about 5, right? Factoring that in would knock it down to 166m/s.. still really high.

 

[krab]

You've mistakenly put a 2 pi in there: you only need that if you are given the radius. But you already know the circumference. This brings it to 132 m/s. And you are right, the overall ratio is about 5, bringing it to just the 60mph I originally guessed.

 

[whozum]

Back to yr original problem: If you adopt this simple model, then power is proportional to speed, and force at the rear wheels, being power divided by speed, is constant. So you can model it as F=constant. To find F, you need what is the speed at 6200rpm. Let's say it is 60mph, or 88 ft/s. 150hp = 82,500 ft-lb/s. Divide this by speed and you get 937.5 lb. Now assume the car is 2800 pounds = mg.

I'm not following what's going on here. Does this have anything to do with the work I showed earlier? It looks like your using something completely different.

 

[krab]

Ignore the work you did earlier; it looks mostly incorrect to me. Anyway, my way is far simpler.

 

[whozum]

Do you mind showing me where I went wrong?

 

[Andrew Mason]

Each gear has a moment of inertia I. Gear 1 has the lowest I and Gear 5 has the highest. The engine provides a torque T that it applies to Gear X with moment I and the result is an angular acceleration alpha. The relationship between power and force would be proportional to the relationship between power and torque, but with that why would the moment of inertia affect the power produced?

The transmission is just the means of connecting the engine to the wheels. The load is the force applied by the wheels to the road. The moments of inertia of the gears have nothing really to do with it.

Torque is a function of power, not the other way around. (and inherently angular acceleration)

Im pretty sure hte power produced by an engine isn't dependant on where that power is going..

The power of an engine is proportional to the amount of fuel burrned per unit time (or, in the case of an electric motor, current supplied). An engine with a heavy load at 2000 rpm is producing more torque and power than a racing engine at 4000 rpm with no load.You can see this based on fuel consumption.

AM

 

[Andrew Mason]

Electric motors do not have constant power either. They produce roughly constant torque from 0 RPM (max current) to about 1/2 maximum freewheel RPM (roughly max efficienty) where the inductance takes over and the torque declines to zero at the max freewheel RPM. So they have a parabolic power curve as well and it changes depending on the loading and available electrical power and these are only rough rule-of-thumb gross approximations.

Torque characteristics of electric motors can vary a great deal depending on the design and type of motor. A series motor, such as your car stater, produces maximum torque at 0 rpm and rapidly decreasing torque at higher speeds (due to reduced field current as the armature back emf increases). Shunt wound motors are designed to run at constant speed and handle variable loads with little speed reduction (i.e. torque increases as load increases but speed is maintained). Compound motors run at variable speeds and produce maximum torque at maximum load at slow speed but also strong torques at higher speeds.

AM

 

[Cliff_J]

Andrew - my point was simply that most electric motor designs do not exhibit a constant power output but instead vary with RPM. Maybe my idea of a gross approximation is larger than it needs to be.

whozum - the average power applied over the period of time would be sufficient.

krab - awesome simplification. Only issue I have is that you used the peak power output and not the average - add a variable ratio transmission (CVT) that holds the engine at the peak power output and then it'd be likely surprisingly close.

 

[krab]

Do you mind showing me where I went wrong?

Assuming a quadratic relationship between power and time, we'll say P = c*t^2 where c is some proportionality constant.

Power is related to engine rpm (also to load, but we are assuming that it is loaded fully). You cannot impose a time-dependence of power, since you don't even know yet how it will go. You are putting cart before horse.

\frac{d^2x}{dt^2} = a(t) = c*rpm(t)

Acceleration is not proportional to rpm, in any gear, rpm is proportional to speed and the prop.constant is given by gear ratio.

m\frac{d^2x}{dt}\frac{dx}{dt} = k\int{r(t)}{dt}

Time integral of rpm? This is so wrong, I don't know what to say...

a(t) = \frac{k}{m}\frac{r(t)}{v(t)}

This finally is something correct (based on the constant-torque model), but somehow you missed the key simplifying concept that in a given gear, r(t)/v(t) is a constant, given by gear ratio. Use that fact and you get my formula.

 

[whozum]

The one thing I can't understand is how RPM is based on speed. What speed, speed of the car? Isnt the speed of the car the RESULT of running at an RPM?
I was leaving gears out on purpose, assuming the engine power was directly transferred to the wheels with no reductions. That would only require me to find the power produced through an engine cycle.

 

[krab]

Why is that hard to understand? Speed is proportional to rpm. So if 6200rpm is 60mph, 3100rpm is 30mph, 620rpm is 6mph, ... What's not to understand?

 

[whozum]

One RPM spin equals x tire spins. Ok that makes sense. Then you used the mechanics equations to find acceleration and such?

My goal was to solve this using power analysis.

 

[krab]

Right. So written in terms of Power, which is force times speed, the general equation is
m{dv\over dt}={P\over v}-kv^2
where the k is the coefficient of aerodynamic drag. This is a really fun equation to solve. For a drag race, a realistic model is to replace P/v with a constant up to redline in first gear, since you obviously cannot apply full power to the wheels at a standstill (actually, a lot of the energy is dissipated in the clutch at the start), and anyway you cannot exceed the traction limit of the drive wheels (about 1 g). Thereafter, it is a good approximation to use P=constant. Don't worry about moment of inertia of the spinning things, as this inertia effect is small compared with the main one which is getting the car as a whole moving. Also, don't worry about rolling resistance which would be a term like constant times v, since it is much less than the aerodynamics drag (at least at speeds that are fun :-)

 

[whozum]

Thats so cool. Thanks alot. I'll post if I think of anything else.

 

[plusaf]

back to the rat motor graph...

100% torque transfer meant that the torque applied by the engine is the same torque onto the wheels, aka there was no gear reduction.

Power is a function of torque because torque is a function of RPM? I am getting lost here. I just want a simple relationship between power and RPM because from power i can derive acceleration. Were involving all these middle steps now.

the graph showed that the engines could produce fairly constant torque across a broad range of engine rpm's. power output increases "sort of linearly" with increasing rpm, as you can see the curves that go pretty nicely from lower left to upper right. think of the higher-rpms of the engine drawing more fuel in faster for conversion to energy, so more rpms means more fuel being consumed means more power output.

no matter how fast the engine is turning, the torque comes from the ignition of the fuel-air mixture in the cylinders, which expands, pushing on the top of the piston, which pushes on the connecting rod, which pushes on its connection to the crankshaft, which is off-center from the axis of the crankshaft, hence torqe is applied to the crankshaft in this way. once the fuel-air mixture is ignited, unless it's pressurized or otherwise messed around with, once the throttle is wide open, you've got all of the air and fuel into the top of the cylinder you're going to get, and when it ignites, you've got all of the torque from each detonation that you're going to get.

playing with intake runner lengths, plenum chambers, and all that stuff just improve the efficiency of getting all of the fuel-air charge into the cylinders.

and the amount that goes into the cylinders is governed by throttle position. as computers try to keep the air-fuel mixture stochiometric across everything from idle to WOT. the max hp and torque curves do not have anything to do with how wide the throttle is being opened. it's already wide open when the measurement is being made.

 

[plusaf]

one more view, of course...

One RPM spin equals x tire spins. Ok that makes sense. Then you used the mechanics equations to find acceleration and such?

My goal was to solve this using power analysis.

why involve power?
to make a car accelerate, you have to get the tires, where they contact the pavement, to push backwards against the earth. then the car moves forward.

the tires have a rolling diameter, (not the wheel diameter, either, guys!), so to create Force to shove back on the Earth at the end of the lever arm known as the Radius of the Tire, it takes Torque!

Torque can be found on the graph of the rat motor (referenced below). since it's pretty flat, let's assume for now that it is a perfectly flat, constant torque curve, versus rpm.

next, you've got to take the car from zero to some max speed, which means the tires will have to go from Zero rpm up to some really nice high speed.

on the other hand, as you'll notice from your own experience with stick-shift cars as well as some of those nasty torque curves, that most AVERAGE engines we drive today really don't like to be very far from a 1500-3000 rpm range, or maybe 1500-4500 for some of the modestly zippier ones. I'm not talking Acura NS-X's here, with titanium connecting rods, gang...), so your engine wants to run in a 2:1 or 3:1 rpm range, while you want your car to cover the, say, 20-to-80 mph range very smoothly.

2:1 is not equal to 3:1 or 4:1.

something must be done.

the answer: gears. gear ratios. transmissions, automatic or manual.

the torque converter in your automatic or the clutch in your stick shift let's things stay loosy-goosy slippery so that you CAN get the car off the line from ZERO speed without forcing the engine to also run a zero rpm, which we know they don't like to do...

once the car is rolling in the first gear, you can mash the accelerator, and the torque is multiplied by the transmission and rear (front, nowadays)-axle ratio (call it Final Ratio) to the axle driving the tires.

the transmissions and clutches and such simply allow the engine to be kept in its happy 1500-4000 rpm range while the gear ratios deliver the torque to the tires to push the car forward.

yes, at high speeds, it does take power to push the air out of the way, so most cars ARE horsepower-limited on their high speed end.

at the low end, it's how much torque you can apply to the driving wheels/tires before they break loose from the pavement (recall that sliding coefficient of friction is lower than static...? here's where physics "meets the road" in real life...)

and that's a little more about "how cars really work."

see how much trouble horsepower can get you into?


+af

 

[krab]

You seem to think power and torque are unrelated measures of performance. They're not. Power is torque times rotation rate, or force times speed. So if you use Torque, you are forced to take the engine torque, divide by transmission gear ratio, divide by rearend gear ratio, divide by wheel radius, to finally get: Force to the road. With power, you simply take the engine power, divide by road speed and you have Force to the road. You see how much trouble it is dealing with torque instead of power? This is because energy is conserved: force, or torque is not.

 

[whozum]

I was going with power analysis initially, somehow I got lost in making my equations and ended up with some funky stuff. When you told me r(t)/v(t) = constant, then that simplifies things alot, but it also simplifies my equation to a = \frac{k}{m}. Dimensional analysis fails with k constant.

 

[plusaf]

torque is not conserved?

You seem to think power and torque are unrelated measures of performance. They're not. Power is torque times rotation rate, or force times speed. So if you use Torque, you are forced to take the engine torque, divide by transmission gear ratio, divide by rearend gear ratio, divide by wheel radius, to finally get: Force to the road. With power, you simply take the engine power, divide by road speed and you have Force to the road. You see how much trouble it is dealing with torque instead of power? This is because energy is conserved: force, or torque is not.

really?
why not, other than powertrain frictional losses... and why doesn't that equally non-conserve power?

 

[whozum]

Can anyone address the question two posts above, also?

 

[krab]

My car has 82 ft-lbs torque at 7000 rpm. Let's trace that through:
At the crank:

torque= 82 ft-lbs, rotation speed= 7000 rpm, power= 109 hp

Gear ratio reduction in first gear = 3.166, so

After the transmission:

torque= 260 ft-lbs, rotation speed = 2211 rpm, power= 109 hp

Gear reduction through differential to wheel axle = 4.313, so

At axle:

torque= 1120 ft-lbs, rotation speed = 512 rpm, power= 109 hp

Notice the torque is not conserved? It is 82, then 260, then 1120. Notice the power is the same throughout? Of course there are frictional losses, so a few % of the power is lost. Maybe there is only 100hp at the rear wheel. But then there is a similar 8% reduction in torque at the rear wheel to 1027 ft-lbs.

 

[shyboy]

That's right, but some crude equations could be helpful
Let's assume that the energy is conserving, so power on the gears is conserving.
Let's imagine gears as ideal cyliders. Then power at any given moment

P=F\times v=F\times \omega \times r=T \times \omega

here F is a force and T is a torque. Now, if we increase the rotational frequency, we lose torque and vise versa.

 

[whozum]

Ok, can we fit this thread into the data on this one: https://www.physicsforums.com/showthread.php?t=71727 ?

 

[whozum]

I have more detailed numbers if you wish
Times at
60'
330'
1/8mi
1000'
1/4mi
final speed, speed at 1/8mi

 

[krab]

Does 3560 lbs include driver?

I find for avg. of 100hp: 1/4mile time=17.66s, trap speed=78.5mph, top speed=107.6mph, 0-60time=9.9s, 60ft time=3.1s, 1/8mile time=11.4s, speed=64.4mph.

 

[whozum]

Sorry, the actual car weight is 2500 lbs, and I weigh 160lbs, plus an ambient 10-15 lbs for other stuff. These calculations, can you elaborate/

 

[krab]

Sorry, the actual car weight is 2500 lbs, and I weigh 160lbs, plus an ambient 10-15 lbs for other stuff. These calculations, can you elaborate/

That changes things a lot. I'm sorry to have to tell you you are only getting an average of about 80hp to the ground. The formula I use is in a previous post in this thread. Do you know your top speed? If you do, I can make the calculations more accurate. Right now I'm just guessing on air resistance.

 

[whozum]

My trapspeed was 77mph, the highest I've ever gotten it to was 115.

I know my car sucks, but can you show the method used to get to that answer?

 

[krab]

Here is the code, written for a Matlab-like program. To understand it, you need to know that variables like x and v (in feet and feet/sec) are vectors. The formula for v comes from a straight-forward integration of the differential equation in post #28. Then the time vector t is found from a numerical integration of dx/v.

! b is the air resistance coefficient; it is about 0.013
! for cars, about 0.0055 for motorcycles
b=?4
! ?1 is power in hp. p is power in ft.-lbs./sec.
p=?1*550
vinf=(p/b)^(1/3)
! ?2 is weight in pounds of vehicle (don't forget the fat driver)
m=?2/32
! ?3 is initial acceleration g's. This may be limited by traction or
! in any other case is given by first gear ratio.
v1=p/?2/?3
x1=v1^2/64/?3
t1=v1/32/?3
x=[(x1+.5):1319.5]
v=vinf*(1-(1-(v1/vinf)^3)*exp(-3*b*(x-x1)/m))^(1/3)
iv=1/v
ttt=t1+integral(x,iv)
generate tf 0 ,, t1 100
xf=16*?3*tf^2
display `1/4 mile:'
=ttt[#] !seconds
=v[#]*30/44 !mph
display `top speed:'
=vinf*30/44 !mph
vout[1]=88
display `0-60:'
=interp(v,ttt,vout)
display `60 ft time:'
=sqrt(120/?3/32)

 

[whozum]

I meant from a physics perspective, rather than a computational. Sorry, that just looks like a bunch of garbage to me, because I don't know MATLAB syntax.