# Interpretation of an ideal engine torque,power curves

1. Jun 18, 2013

### marellasunny

Attached you will find the torque vs engine rpm and power vs engine rpm curves for an 'ideal engine' and also for a 'normal SI engine'.

1.(for the 1st ideal engine curve) Is it 'ideal' that the torque curve should decrease as the engine rpm increases? Why?
Does this mean that 'ideally' in a road, the traction force should decrease with increasing engine rpms ? I don't get the intuitive meaning as to why an 'ideal engine' would have curves like this.

Say, the vehicle is on the same gear and we increase the engine rpm by pressing the accelerator.
Then, in real life, the torque and traction force increase until a point where there is not more power for acceleration OR slip starts. But, in the ideal engine case, the vehicle would have reduced torque/traction force as you press the pedal more? How does this help?

By manipulating equations, I arrive at some understanding but its still vague:
I try to justify the reason why the torque curve is shaped like this: I try to arrive at a connect. between torque,traction force and engine rpm.

A.Speed equation : $\omega_e=\frac{n_in_d*v_x}{R_x}$

B.Traction equation:$T_e=\frac{R_wF_x}{\eta* n_in_d}$

Then equating the values for $n_in_d$, we get the relation between torque and traction force,

$T_e=\frac{v_xF_x}{\eta\omega_e}$

where, $\eta$ is the driveline efficiency, $\omega_e$ is the engine rpm, $F_x$ is the driving force/traction force.

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2. Jun 19, 2013

### jack action

Actually, the curve says that it is ideal that the torque increases as the engine rpm decreases. In other words, the torque at the maximum rpm is the same for the 'normal SI engine' and for the 'ideal engine'. This gives a lot more traction potential (hence greater acceleration or drawbar pull) to the vehicle at a lower regime.

What counts is the power available, not the torque. The theory behind an 'ideal engine', is that if an engine can produce a given amount of power, it should be able to do it at any wheel rpm. An electric motor can do that on its own; Internal combustion engines can do that with the help of a transmission with multiple gear ratios.

3. Jun 20, 2013

### marellasunny

That green line on traction limit cleared-up a lot of doubts. My question was actually regarding the term 'unavailable power', why is this power unavailable?

My interpretation:Taking for example the change from the 1st to the 2nd gear, the unavailable power arises because the 2nd gear doesn't produce the same max. wheel torque $T_w$ as the 1st gear in the same velocity range, and hence as a consequence of lower g.r of the 2nd gear the engine works at higher rpms =>thus keeping power almost constant,explaining the function of the gearbox you stated above.(am I giving a correct explanation y/n?)

Q.Why does this 'unavailable' power arise if as normally proposed,the whole function of the gearbox is to keep the engine running near the max.power range? What's stopping the 2nd gear from working near its max. wheel torque potential at the requisite speeds? Are they designed purposely this way?

I would also go on to guess,This problem of 'unavailable power' doesn't arise in EV's because they are gearless(i guess in most cases) and the torque is a function of the current.

Some math equations to explain how the 'inverted parabola' envelope that you see above in jack action's post is obtained:

$$(1)T=an^3v^2+bn^2v+cn$$ where n-gear ratio as a parameter,
the variation of the parameter generates a series of curves that you see in pink in the above post
$$(2)\frac{\partial T}{\partial n}=3an^2v^2+2bnv+c=0$$

solving (1) and (2) ,we get an equation in-terms of torque and road velocity v
$$T=C/v$$

Last edited: Jun 20, 2013
4. Jun 20, 2013

### jack action

On the above image, the lines for the 1st, 2nd and 3rd gears represent the wheel torque, which is directly proportional to the engine torque. (On the graph, the torque is converted to the vehicle acceleration potential, but it is proportional)

The engine torque has a peak value. The engine power also have a peak value. If you take that maximum power (which is conserve through the gearbox, hence the wheel power is the same as the engine power, no matter the gear ratio), you can calculate the equivalent maximum wheel torque (which is decreasing as the vehicle speed increases). That maximum torque achievable for any given speed is the black line (max power limit).

You can see on the gear ratio lines where the engine maximum power is reached, as these lines touch the max power limit black line (around 125 km/h, 180 km/h & 275 km/h).

In between those maximum power «spots», the engine rpm is either higher than the max power rpm (say 1st gear) or lower (2nd gear). Hence, between 125 km/h and 180 km/h, the vehicle will not be able to use the maximum power of the engine, thus the «unavailable power» statement.

The only solution would be to insert another gear ratio between the 1st and 2nd gears, making it possible to reach the maximum power of the engine at, say, 150 km/h. With an infinite amount of gear ratios (ex.: CVT), there wouldn't be any «unavailable power».

5. Jun 20, 2013

### rcgldr

The "advantage" of an ideal engine with constant power at all rpm is that torque output to the driven wheels versus speed of the vehicle would be the same regardless of the gearing as long as the engine was within it's operating rpm range.

I'm not sure how such an engine could be made. For a street car, the engine has a relatively flat torque curve (compared to race car engines, which tend to have "peaky" torque curves with the peak at high rpm for maximum power).

As an alternative to an "ideal" engine, a continously variable transmission would allow an engine to run at any rpm independent of the speed of the vehicle, allowing the engine to always run at peak power rpm for maximum acceleration (once beyond the traction limit), or optimal rpm for best fuel mileage. A regular transmission with a lot of gears can approximate a continously variable transmission (except for the shift times).

In the case of an ideal dc electric motor, peak torque occurs at 0 rpm and decreases linearly to 0 torque at maximum rpm. Maximum power occurs at 1/2 of maximum rpm.

Scroll down to the bottom half of this web page:

http://lancet.mit.edu/motors/motors3.html

Last edited: Jun 20, 2013
6. Jun 22, 2013

### marellasunny

Oh!now I get what the sentence 'gearbox keeps the engine running near the max.power range' means- this means(for instance imagine your'e on the 1st gear torque curve and the pedal is pressed gradually), as soon as the 1st gear torque curve touches the max.power curve(in black), there HAS to be progression to the 2nd gear and then we are on the torque curve again till we touch the max.power curve. So,we can ALMOST say(taking into account 'unavailable power') that 'the gearbox keeps the engine running near max. power range the whole time'.

Question 2.

So, am I right in also saying that the gearbox is also responsible for the shift in the torque curves to right? The advantage of the torque curves displaced to the right is easily shown in the above diagram. As the torque curves are displaced to the right, there is increase in the magnitude of the peak power.
i.e for instance maybe to the left is the engine torque curve and to the right is the wheel torque curve displaced because of the gear ratios(?) or someother device?. As a result of the shift of the torque curves to the right, there is also a increase in the magnitude of the 'peak power'.

Last edited: Jun 22, 2013
7. Jun 22, 2013

### jack action

No! No! No!

Power is always constant! You get what you can produce, that's it.

Power is the product of torque and rpm (or force and velocity).

A gearbox will dictate if the power will be outputted in the form of torque or rpm. If you increase the rpm, you must decrease the torque. If you increase the torque, you must decrease the rpm. No matter what, the product of the output torque times the output rpm will always be the same, i.e. the input power.

If the power always stays the same, as the car goes faster (increased velocity = increasing wheel rpm), it means that the force propelling the car has to decrease (= decreasing wheel torque). That is why you are reaching a maximum top speed: As speed goes up, the always decreasing tire force will equal the always increasing aerodynamic drag and at that point there is no more force left to accelerate the vehicle, hence the vehicle has to go at constant speed.

Torque is not constant, power is.

So in your graph, the power curves would be the same and the torque curves would decrease in magnitude as you go to the right. As a result of the shift of the power curves to the right, there is also a decrease in the magnitude of the 'peak torque'.