# Homework Help: Car Engine problem.

1. Mar 22, 2005

### 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 dont have maple to graph the function to see if it makes sense, but a cubic function isnt really what I was expecting. So with a linear relationship between power and time we get:
a(t) = ct^2/(2md) which still doesnt sound right.

Something tells me I should be integrating with respect to Power, im 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.

Last edited: Mar 22, 2005
2. Mar 22, 2005

### whozum

bump? Someones gotta have some input.

3. Mar 23, 2005

### Andrew Mason

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

4. Mar 23, 2005

### 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 doesnt have max power as it accelerates, thats 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 isnt 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, thats the goal of the engine, to produce power per RPMs. You dont 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 $$P = k\int{rpm(t)} {dt}$$

5. Mar 23, 2005

### Shockwave

6. Mar 23, 2005

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

7. Mar 23, 2005

### Andrew Mason

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

8. Mar 23, 2005

### 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 isnt dependant on where that power is going..

9. Mar 23, 2005

### 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 [Broken]

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 [Broken]

Hope this is useful!

Last edited by a moderator: May 1, 2017
10. Mar 23, 2005

### whozum

Thats pretty cool. Thanks alot 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. Im 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?

11. Mar 23, 2005

### krab

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.

12. Mar 23, 2005

### 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? Im 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.

13. Mar 23, 2005

### 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.

14. Mar 23, 2005

### whozum

Thats all I need.

15. Mar 23, 2005

### 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.

16. Mar 23, 2005

### 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.

Last edited: Mar 23, 2005
17. Mar 23, 2005

### 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.

18. Mar 23, 2005

### whozum

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

19. Mar 23, 2005

### krab

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

20. Mar 23, 2005

### whozum

Do you mind showing me where I went wrong?