Need help understanding engine acceleration and relation with torque

1. Aug 21, 2014

rohn52

Hi ALL,

On this forum, I found some really sensible answers to a few of my queries... posting one thats unanswered..

I want to understand how an engine or any power plant for that matter behaves under load...
So i'm using a simple scenario to frame my questions... the values while made up reflect how an engine performs ...

A typical engine accelerates from idling to ~6000 rpm in about 4 secs (a = ~1300rpm/s), the driver disengages the clutch to shift gears and the rpm drops to 4500.
The driver engages the clutch and now the engine takes 3 secs to go from 4500rpm back to 6000rpm (a = 500 rpm/s).
For going into the next gear rpm drops to 4.5K again and takes 6 secs this time around to reach back to 6K (a = 250rpm/s).

The driver here is not throttling the engine at all except while changing gears...assume WOT operation throughout.

If the car is in neutral and the driver floors the pedal he/she will probably get an engine acceleration of over 3000rpm/s.

1) What is impacting the engine acceleration?

I have more questions... but i'll wait for an answer to this one first..

Assume no aero drag and rolling resistance and no losses in the transmission.

Thanks!

2. Aug 21, 2014

jack action

It is all related to F=ma (or T=Iα in rotation), i.e. Inertia.

When you are in neutral, all you have to accelerate are the engine components (crankshaft, piston, valvetrain, etc.). All of these components have a certain inertia.

But when the clutch is engaged, you are now adding the inertia of the transmission gears, driveshaft, axles, tires, etc. which leads to a bigger inertia, hence a smaller acceleration for the same force applied on the engine's pistons.

But, more importantly, in the end that torque is used to accelerate the car, which has a huge mass (inertia) compared to the rest of the mechanical components. That mass is added to the total inertia of the powertrain components. Actually, the mass of the vehicle represents between 75% & 95% of the total inertia of the car (It depends on the gear ratio of the transmission).

So that is why an engine that is into gear revs up at a slower rate than one that is in neutral.

As the car goes faster and faster, the drag force and rolling resistance (Sorry, you can't ignore them!) increase. Those forces have to be subtracted from the available force coming from the powertrain (i.e. the engine). The net force left over is the one that will be «transformed» into acceleration of the car and - because it is connected to the car wheels - the acceleration of the powertrain as well.

So that is why when you are in first gear (i.e. slow speed) you rev up from 4500 rpm to 6000 rpm a lot faster than when you are in 5th gear (i.e. high speed).

3. Aug 21, 2014

Simon Bridge

Simply put - giving power to an engine makes it accelerate until the power lost to friction (which depends on rpm) is equal to the power provided. At which point the acceleration stops, the engine goes at a constant rpm.
All engines have some sort of loss mechanism even with zero friction.

When the driver declutches, he also takes his foot off the gas. Thus there is nothing replenishing the power lost through friction. Thus the rpm's drop off quickly. If he didn;t, then the engine would suddenly over-rev, which you may have experienced if you've learned to drive stick.

The different gears deliver energy from the engine to the road at different rates. That is, they represent different power loads. The higher the load, the slower the acceleration.

In an ICE you don't deliver power to the engine continuously. It comes in short bursts when the fuel ignites in the cylinders. If you draw energy too fast, there won't be enough power to get to the next shot of fuel, and the engine stalls... so you have to start out with a low gear and work your way up.

4. Aug 21, 2014

rohn52

Makes a lot more sense now...

essentially torque required by the vehicle impacts engine acceleration.

At any given instant ...

(Torque produced by the engine X Gear ratio) - Losses = Torque required by the vehicle

Given a constant load - engine speed and engine torque will always have the same relation...is this right?

The car performance curves we see .. engine speed vs measured vehicle torque are at a constant load. So they are not really useful in understanding how a car will perform on the road when it experiences varying loads.

But these are useful for comparing it with other performance curves...am i right in saying this?

I realize rolling and drag are very important and should be considered.

What would really help understand this is a graph for vehicle speed vs required torque/force with all factors considered like mass of the car and equations for drag and rolling.

Can someone point me to some resource/spreadsheet?

@jack and @simon

Thanks for clearing this out for me.. really well explained.

5. Aug 21, 2014

jack action

Not only typical engine Torque vs RPM graphs are at constant load, they are at maximum load. It is referred to as WOT (Wide Open Throttle).

This acceleration simulator gives the maximum acceleration of a vehicle vs its speed. The equation for the acceleration is equation 1b on this page.

Note also that power requirement is more important than the torque needed. Torque can be adjusted at will with the proper gear ratio, but the power always stays constant, no matter what.

6. Aug 21, 2014

Simon Bridge

Like Jack says, the torque is not a useful metric for the kind of thing you are considering. You need to get used to thinking in terms of energy and power. Aso be aware that real machines can get quite complicated. The accel simulator looks pretty nifty.

You can match the calculator to vehicle full specs online to see how good it is.
i.e. http://www.mg-cars.org.uk/MGF/mgfspec.html
... use the "help with" links. Seems to be only a qualitative fit to this car.

Does handle the effects gears have (it's in a help with section: you can enter the gear ratios.)

7. Aug 22, 2014

xxChrisxx

I'm quite glad this has come up I did some posts on this for another forum that I have stored. I'll dig them out and post them.

It basically breaks down that calculator, neglecting any losses. It was made to show that you can use both torque and power to do an accurate calculation of performance. Both are easy and both need a speed and tacho reference to be accurate.

You can make an approximate model with maximum power that most of the time is perfectly fine. However it is a big assumption to make.

8. Aug 22, 2014

Kozy

The more load on the engine, the slower it will accelerate. Each time you change up a gear, the engine load increases so the rate of acceleration slows.

I've got a better version of that acceleration calculator on my website, instead of a simple horsepower input you can shape a torque curve and input gear ratios to see how it impacts on acceleration, and compare two different set ups at the same time. http://blackartdynamics.com/Transmission/ThrustIndexMX5.php

The default inputs are to compare two MX5/Miata's, one using the 5 speed gearbox and one using the 6 speed, to see how the gearing affects performance.

9. Aug 22, 2014

rohn52

@Kozy..."The more load on the engine, the slower it will accelerate. Each time you change up a gear, the engine load increases so the rate of acceleration slows."

Then i am forced to dive into engine dynamics a bit...at WOT (assuming max amount of usable charge possible) will an ignition event always yield the same amount of power? depending on the load it can either convert 90% of this power into speed and just 10% into torque or vice-e-versa. And this power is equal to the max rated power of the car minus losses?

have i understood this correctly?

Thanks for the links...i'll be sure to check them out..

and Chris looking forward to your posts.

10. Aug 22, 2014

jack action

Let's get one thing straight before we go on: You should stop using the word load. It is not a scientifically defined quantity and, depending on who you're talking with or what you're talking about, it can refer to almost anything: weight, force, work, power and more.

For our case, use power or torque for clarity instead of load.

To answer your first question, an ignition event in an engine will always create the same force, which in turn will create the same work (because the piston is always displaced the same distance). But the power produced depends on how fast the piston will travel the distance (i.e. the time it takes). A force (or torque) is one thing and velocity (or rpm) is an other; both are independent of each other. Power is the combination of those two quantities.1

The engine is connected to a device that will require some power. If the power produced by the engine is equal to the power required, the engine's rpm stays constant. If it's more, whatever force from the ignition event that wasn't use on the output, will now contribute to accelerate the engine and all the mechanical components connected to it. If it's less, whatever «leftover» force coming from the output that is not balance by the engine's power, will now contribute to decelerate the engine and all the mechanical components connected to it.

--------------
1 Note that the velocity combines the distance and time factor defined earlier for work and power.

Mathematically, defining work (W) and power (P) based on force (F), distance (d), time (t) and velocity (v):

W = F * d
P = W / t = (F * d) / t = F * (d / t) = F * v

11. Aug 27, 2014

rohn52

Got it! Thanks!

12. Aug 30, 2014

Damo ET

An alternative way to look at it (not suggesting that the previous are inaccurate or misleading) would be to completely separate the components which add to the lessening acceleration. I think it is wise to look the relationship between the gearing of the car/engine combo and its effect on acceleration exclusively, without the added complexity of drag and inertial losses.

For this example you only need engine torque gear ratios, and vehicle weight.
The gears in the transmission act as a torque multiplication device, the higher the ratio, the higher the multiplication of engine torque is available a the wheels, and vice versa. But also, the higher the ratio, the lower the final maximum speed can be, and vice versa also.
A vehicle which needs 5000Nm to accelerate at 10kmh/s in one gear may have as a result of different ratios 'only' have 2500Nm in the next, which would result in only a 5kmh/s increase in speed.

The only important factor (in this simple example) is the different transmission ratios, as all the other multiplication factors for engine torque remain constant.
A typical transmission will have something like:

1st 4:1
2nd 3:1
3rd 2:1
4th 1:1
diff 4:1

So 100Nm at the engine would be equivalent to:

100Nm x 4 (diff ratio) / .4 (trye radius) = 1000Nm at the wheels before factoring in the gear ratios. Using the x 10 multiplication of the tyre and diff combo outside of the gear multiplication gives:

1st = 1000Nm x 4 (gear ratio) = 4000Nm at the wheels in first
2nd = 1000Nm x 3 (gear ratio) = 3000Nm at the wheels in second
3rd = 1000Nm x 2 (gear ratio) = 2000Nm at the wheels in third
4th = 1000Nm x 1 (gear ratio) = 1000Nm at the wheels in fourth

So clearly the acceleration (change in engine rpm) is going to drop off with each decrease in ratio.
If you know the weight of your vehicle, you can calculate the acceleration which should be experienced.

As xxChrisxx suggested above, you can work out how much torque (and by calculation, power) a given vehicle generates by measuring time during acceleration and knowing weight only. You can calculate acceleration (change in velocity over time) convert that to acceleration in 'G' which will give torque at the wheels (which when measuring 'power' is the only important figure) from there if needed, you can calculate (or look up on the internet) the transmission ratios and diff ratio, check your tyre radius and that is all that is needed to calculate engine torque and by inference, power.

Damo