help needed understanding dynamic strain energy losses

[Mave]

hartech said:

I can imagine a static vehicle (say with the hand brake on or secured by straps) and a very small diameter drive shaft (say quarter of an inch or 6mm) so when the engine applies a force (torque) in 1st gear - if it is high enough as you let the clutch out it would snap the drive shaft. But letting the clutch out to apply the torque could also stall the engine or allow it to rev free after snapping the drive shaft – either way the torsional stiffness of the shaft will have been exceeded to snap it. This means to me it must have been under strain energy that exceeded the yield point?.

It's worth being careful with the nomenclature to track what's going on.

The driveshaft snaps because you've exceeded it's stress capability, not its torsional stiffness. (The torsional stiffness of the driveshaft is a propertyset by the design, (until it starts failing!), and it's not something which can be exceeded, just like you wouldn't say that the mass of the driveshaft can be exceeded).

In your example, the torque (torsional stress) has exceeded the ultimate capability of the driveshaft. The strain energy is the area under the stress / strain at the point when the driveshaft failed.

Anything before the yield point is elastic, that strain energy will be recovered by the components unloading after the failure.

Anything after the yield point is plastic and is lost to permanent deformation of the components.

hartech said:

However if we go back first to the 1st gear full throttle release – if the car was light or heavy it would take different diameters of drive shaft to reach a limit where it just didn’t snap because the resistance to initial motion would be higher in the heavy car and the rate of torque applied against that increased resistance would be higher in the drive shaft.

It shouldn't matter, if you have applied full torque in both cases. The only difference is that the heavy car accelerates more slowly - the driveshaft sees the same torque.

hartech said:

In other words the torsional stress in the drive shaft I think must be the sum of the input torque minus the proportion used to accelerate the car (but then I could be totally wrong).

Other than drivetrain losses, all the torque goes into accelerating the car. It doesn't have anywhere else to go

hartech said:

And I think this is why – the drive shafts snapped in the consul when we tried towing a trailer with 2 bikes on it but was OK without the trailer?.

Acceleration = torque/resistance (without splitting hairs over the various types of resistance including the moments of inertia and weight etc).

I suspect that normal engine torque wasn't enough to give you the acceleration you wanted, so you used engine revs and hence energy to give you a burst of extra torque which then overloaded the driveshaft.

hartech said:

Similarly when the car will not go any faster the resistance to motion is again maximum but not this time as a result of the rate of acceleration (because there is none). This time the higher gear has reduced the torque by 3 or 4 times less than in first gear – so the input torque must be much lower and the wind and rolling resistance has increased until the torque can no longer exceed the sum of the resistances and so this is a different calculation entirely (momentum ?) but there would also be a drive shaft diameter that would snap and another that would not under these torsional and linear forces.

So I want to work out where and what the maximum engine loading is – at what revs and in what gear and what circumstances – accelerating hard in first gear or flat out in top gear but going from the static car to the accelerating one and on to reaching the top speed limit there are numerous dynamic changes.

Firstly the torque applied will change (first increase then decrease in first gear as the revs increase following the torque curve) and at the same time the inertia resistance will reduce as the car gets moving so two dynamics occur simultaneously increasing then decreasing torque against decreasing resistance to motion - and I want to know the conditions of the maximum load against the resistance it is driving against and also if it is higher flat out in top gear or accelerating against the mass in first gear – and what the scale of difference is.

Initially the car is stationary with all its weight (imagine full of concrete blocks) and the load is high as soon as the drive is engaged but the revs are lower than peak torque so at what point does the engine torque load the engine highest?

How does this compare with forces at work flat out in top gear?

This could be determined by trial and error testing by fitting lots of different diameter drive shafts (going from too small and snapping to eventually not snapping both accelerating in 1st and flat out in top) and after carrying out lots of test runs I would then know the exact conditions when the limits were reached for each type of test and from that and the sizes and specifications of the drive shaft material - I can work out the torsional resistance being applied to the engine and therefore the engine loading – but I have neither the time nor money (nor patience).

I just wondered if there was a way to calculate it - but don't understand what are the right formulas to use in each of the two different situations using strain energy, potential energy, Kenetic energy, momentum or Newtons laws etc?

Perhaps I am asking too much and it isn’t possible – if so sorry - but if anyone can help – please do so – it would be really appreciated.

Thanks again,

Baz

The situation when you have most torque on the driveshaft is the same time when you are putting most load through the tyres. You're most likely to break traction dumping the clutch in 1st gear - that's also going to be the time you're most likely to break the driveshaft. The driveshaft torque will be lower at high speed because its been geared down by the gearbox.

[AW111]

Assume the bolt is parallel to the floor, and the wrench is horizontal.

Imagine you're not using your muscles to push the torque wrench down, but are sitting on the end of it.
Your weight is applying the torque, and you can sit there all day without using any energy.

[Mave]

hartech said:

I also can still not get my head around the accepted rule that it makes no difference to the energy used - whether you apply a force to something for a second or an hour. I don't see why it has to continue moving to receive the force. Once a shaft had deflected if you were applying the deflection with a torque wrench - would you not have used up more energy holding it at that torque for an hour than for a second. I agree you would only get the same force back when you stop applying the torque. Is it not possible that storing energy in the structure of material is an energy losing situation over time (must read up on hysteresis perhaps that will explain it).

Baz

It doesn't have to continue to move to receive force - a brick sitting on the ground continually exerts a force on the ground.

You don't have to use energy to apply a force, just as a brick doesn't use, or transfer, energy when it's sitting on the floor.

Our arm uses chemical energy to create force, but that's all about a particular way to make a force, rather than how that force reacts with the torque wrench, driveshaft etc.

[GreenV8S]

hartech said:

I think this is where the issue exists. The way I see it - if acceleration can break a driveshaft in 1st gear under initially trying to accelerate the mass of the car from a standstill then the energy applied to the drive shaft is high. imagine how much torque you would have to apply by a torque spanner to break a driveshaft in half if one end was in a vice and the other had a long torque wrench twisting it to breaking point.

That force would have to be applied starting at nothing and then right up to the point of fracture. The fact that the deflection is small is not IMHO relevant because that is in proportion to Youngs modulus for the material - it doesn't mean the force applied sufficient to break a drive shaft is small just because the deflection is small.

As the car starts to accelerate the revs rise and with it the torque increases to maximum torque at what - about 2/3rds maximum engine revs - so during that period the forces applying that torque to twist the drive shaft (however small the deflection is) are not available to accelerate the car - only what is left after the rise in torque progressively twists the shaft more from the previous point it was at while the torque is rising.

But then again as the car starts to move it gathers momentum and starts to accelerate faster with less torque anyway - so the way I see it there is indeed a reduction of the torque available to accelerate the car when the revs are rising from standstill and then because the torque curve doesn't drop off much from peak torque to peak revs - you don't get much of it back because you have declutched to change gear by then and the strain energy entrapped within the drive system must be lost within the components as they slow down.

I also can still not get my head around the accepted rule that it makes no difference to the energy used - whether you apply a force to something for a second or an hour. I don't see why it has to continue moving to receive the force. Once a shaft had deflected if you were applying the deflection with a torque wrench - would you not have used up more energy holding it at that torque for an hour than for a second. I agree you would only get the same force back when you stop applying the torque. Is it not possible that storing energy in the structure of material is an energy losing situation over time (must read up on hysteresis perhaps that will explain it).

Sorry if those of you that understand this better find my issues with it irritating - I am trying to understand it and you will all be helping and I am grateful for that - so thanks again.

Baz

You're confusing torque and energy.

You leaning on the end of a spanner doesn't do any work unless the spanner is moving relative to you.

The prop shaft has torque applied to both ends but doesn't absorb any energy unless the two ends are moving relative to each other. They aren't moving relative to each unless unless the prop shaft has failed.

[hartech]

Thanks Mave, Green V8S, AW111 etc for your patience - you are helping me a lot - perhaps being over 50 years from what was then a much less well constructed technical education system has proved a bit too much for my memory - but your explanations have helped.

Really sorry but there are still some points I don't follow. I am not arguing or trying to be proven right - I fully accept my limitations and you guys understanding it better- these questions just arise from my trying to make sense of things and if you are getting fed up with it and want to stop - I fully understand and thank you again for trying to help me.

Torsional stress comes from a load (or force) / area so if the area is the same and a shaft breaks the load or force must have increased to twist the shaft to that point and while it is doing so the energy that provided the force in the first place must have been used up and cannot also be available at the other end until the twist has finished and it still hasn't yet snapped?

I also don't understand how strain energy is "recovered" after the shaft has snapped - where is it recovered after the connection back to the engine providing the torque or force has broken?

I also cannot follow how the driveshaft sees the same force regardless of the amount of resistance. You cannot apply a torque unless there is resistance and the more the resistance the more torque you can apply. You cannot apply torque to a nut unless it reached its stop point and then you can only increase it if the resistance to turning was increasing as well?

If you were sat where the engine was and using a torque spanner to turn the main prop shaft - it would take more torque to get it moving from a standstill and as the car started moving it would take less torque but more speed to keep up with it? So if the torque was constant the resistance would decrease initially as the car moved (allowing more to be used to move the car) until the resistance increased again with speed until the torque could no longer overcome the resistance. The same would apply if you were trying to push the car from standstill. Initially needing a high force to initiate movement, then getting easier but then more difficult again as the speed increases?

I have already got enough benefit out of this to assist with my project - so thanks all for that - just a little bit more help with these answers may well finally allow me to see the light!

Baz

[Fast and Spurious]

I think the extra weight of the bikes and trailer caused the driveshafts to break because...
Nose weight (?) on tow hook is extra weight pushing down through the back tyres. Apply full power in first, without the extra weight the tyres would lose traction, wheelspin, but with extra weight (and grip), driveshafts give up instead.
????

[Mave]

Hawkshaw said:

hartech said:

I also don't understand how strain energy is "recovered" after the shaft has snapped - where is it recovered after the connection back to the engine providing the torque or force has broken?

Baz

Surely the applied torque is stored in the shaft as potential (or strain) energy, and that is what snaps the shaft when the yield point is reached?
It isn't "recovered" as such, but is conserved as other forms of energy.

Force and energy are different and separate things, although they may interact.
The force (or torque) isn't stored, it is transferred to whatever you apply the force to.
If you have an infinitely stiff system, you don't need to transfer any energy to apply a force.
In a real system with flexibility, you need energy to stretch (or compress) it, that energy is stored in the system as potential energy until you remove the load.

Imagine a spring hanging vertically with a mass hanging off it. The force at the top anchor of the spring is the same as the force at the mass.

When you initially apply the mass, the gravitational potential energy of the mass is converted into elastic potential energy to stretch the spring. It then sits there, stretched, with no more energy being transferred.

If you then cut the spring, the two halves of the spring would recover to their original length. The elastic potential energy then becomes kinetic energy and finally noise and heat as the spring twangs back to its original length.

Edited by Mave on Saturday 3rd July 13:24

[hartech]

Yes and thanks I get that Mave - but my issue (which I am sure I just cannot get my head around) is that while you are applying an increasing force on an increasing extending or compressing the spring (which I agree is then stored in that spring) if the other end of the spring experiences the same forces in magnitude and time as the forces you applied to it then great - you have agreed with Newton - but then if you (as you say) release the spring you get back that potential energy and the total is now more than you applied - so if that is impossible - you must have lost some forces from the other end while you were compressing it for the total equation to work.

In terms of a car transmission (and yes including everything - crank, flywheel, gearbox, drive shaft and tyres) while you are applying the increasing forces (say torque) they must be being stored in the components as potential or strain energy and therefore less than the input forces that applied were applied must be available to the restrained ned (the drive shafts and/or tyres).

I accept that you could argue that you get that stored potential or strain energy back when you reduce the driving force (torque) but if that it sight then you must have equally lost it from the end of the transmission as it was being applied.

But this is my diilema that a pound weight sat on a spring both manages to apply a pound force to the other end and store the spring rate of energy in the spring - it seems to double the applied force - i.e. it could squash what the spring is sitting on by a pound weight and also release back the force in the spring as you lifted the pound weight off and that is double the applied force for nothing?

If I were you I think I would give up on me - it is probably some weird issue my practical engineering brain finds it impossible to deal with - like some of the issue of relativity - probably I don't have enough brain power to cope with it lol.

Thanks again guys - very decent of you to try and help.

Baz

[Mave]

hartech said:

Yes and thanks I get that Mave - but my issue (which I am sure I just cannot get my head around) is that while you are applying an increasing force on an increasing extending or compressing the spring (which I agree is then stored in that spring) if the other end of the spring experiences the same forces in magnitude and time as the forces you applied to it then great - you have agreed with Newton - but then if you (as you say) release the spring you get back that potential energy and the total is now more than you applied - so if that is impossible - you must have lost some forces from the other end while you were compressing it for the total equation to work.

Why do you think you get back more than you applied when you release the spring? I think you're confusing force and energy.

If the spring was infinitely stiff then you could apply force at one end and the other end would see the same force. You wouldn't need to put any energy into the system to transfer the force.

When the spring isn't infinitely stiff you need to put in energy as well as force to compress the spring. Both ends of the spring see the same force, but the energy is held in the spring. If you release the spring, you get the same amount of energy back as it expands.

[Mave]

hartech said:

But then again as the car starts to move it gathers momentum and starts to accelerate faster with less torque anyway - so the way I see it there is indeed a reduction of the torque available to accelerate the car when the revs are rising from standstill and then because the torque curve doesn't drop off much from peak torque to peak revs - you don't get much of it back because you have declutched to change gear by then and the strain energy entrapped within the drive system must be lost within the components as they slow down.

Baz

It doesn't accelerate faster with less torque - if you ignore for a moment frictional losses and aero drag, the relationship between torque and acceleration is fixed. By definition, force = mass x acceleration - so for a constant force you get a constant acceleration.

In reality, as you accelerate the drag force incresses, so if the engine torque were constant, the force available to accelerate the car would drop, and the rate of acceleration would drop.

Any strain in the drivetrain disappears when you change gear because at that point you aren't putting any torque through the drivetrain. When you initially accelerated, the first very small proportion of a rotation after you released the clutch used energy to wind up the drivetrain. When you press the clutch, that energy is converted back into kinetic energy in the drivetrain and the drivetrain straightens out.

[GreenV8S]

It might help you visualise this to think in terms of linear forces instead of torsional ones. The underlying physics are equivalent in terms of strain energy and so on.

If you're pulling a car via a spring, when you start pulling on the spring it stretches in proportion to the amount of force you applied. If you imagine plotting force versus deflection on a chart, the amount of energy you put into the spring is the area under the curve. The energy you put into the spring remains there as long as it remains stretched. It doesn't matter whether it's being held stretched by you pulling on it, or the end being tied to a tree, or anything else. If the amount of force varies and the spring extends or contracts then the strain energy will vary. If you're bouncing on the end it may be going up and down continually as the force and deflection change. You can think of it as moving up and down that force / deflection curve. Overall you're getting out as much energy as you're putting in. The strain energy at any instant is the area under ths curve for the position you're at in that instant.

From the point of view of the spring, it doesn't matter whether you and the car are stationary or moving. The strain energy in it is purely determined by the force and deflection in the spring itself. If the car starts moving and you keep a constant pull on the spring, you're now putting kinetic energy into the car (and heating up tyres and bearings due to rolling resistance and so on) but the strain energy in the spring isn't changing.

If the force on the spring is removed, all the strain energy that was in the spring is returned as mechanical energy.

If the spring snaps, all the strain energy that was in the spring is released as noise and heat and so on.

[Mave]

hartech said:

But this is my diilema that a pound weight sat on a spring both manages to apply a pound force to the other end and store the spring rate of energy in the spring - it seems to double the applied force - i.e. it could squash what the spring is sitting on by a pound weight and also release back the force in the spring as you lifted the pound weight off and that is double the applied force for nothing?

Why do you think this is double the force? The weight applies a pound downwards due to gravity.
The spring pushes back upwards with an equal and opposite reaction. The spring then pushes down onto the table with that same pound force, and the table pushes up with an equal and opposite reaction. You don't need to add these forces together because they are the same continuous force - just like if you pull something with a rope, the tension force exists all along the length of the rope, it doesn't only exist at the ends of the rope.

[hartech]

Hi Mave and Green V8S, I don’t know how old you guys are but when I was at Kingston Uni (college back then with HND and a college diploma) there was little connectivity between topics or classes and I think we all left a bit confused about the relationships between potential energy, strain energy, power, force and energy. Nothing ever pulled any of it together.

I am still in touch with many of my fellow graduates – few of us managed to use what we had been taught in our careers.

I found experiences while developing racing engines demanded that I tried to work things out from 1st principles and then began to understand a bit more about what we were taught – and it has helped me. Strangely – in so doing – I have often found the general masses themselves didn’t understand things properly either.

I worked out that it was torque that accelerated a vehicle (I know pure Newton) but back then everyone was chasing revs and power and I used that knowledge to design and make engines that beat the works Japanese engines of the period by fitting my torque curves and making gear ratios so the rev drops in my gearboxes could handle it all the time rather than run out of power band (as the opposition often did when changing up). no one back then realised that if they raise the maximum revs (to get more power) they also increased the drop in revs between gears changes (i.e. increased the power band their engines needed to work between) and since the number of gears was limited by the regulations - they often tuned for higher revs and BHP but were slower in a straight line (often forcing them to slip the clutch after changing up to get back to the start of their power band) – so I did find that understanding basic physics was helpful and gave me an edge.

Funnily enough I was thinking about a person pulling a cart with a spring last night and agree it is a good analogy.

As I see it – if the cart is light and the spring weak and the guy pulls it without a lot of strength – if the cart is tied to something eventually he will reach a pint where he has extended the spring so far he cannot pull it any more. But if the cart is not tied the spring will slowly extend until there is enough force to over come the static resistance of the cart which starts accelerating. As it does so because the cart starts moving the resistance to initial motion reduces and the spring loses some of its length as the tension in it needed to get the car moving helps pull it towards the guy. No problem with any of that (not really sure if the spring has been subjected to strain energy, potential energy or they are both the same thing).

If a stronger guy tries the same thing he will initially pull the spring further away from the cart but eventually (if he ends up pulling at the same speed as the first guy – no longer accelerating) the spring will be the same length.

If the weight of the cart is increased – not only will the acceleration reduce (force/mass etc) but with either guy I think the spring will initially extend further overcoming the static resistance of the heavier cart but excluding aerodynamics and bearing loads etc (which would be relatively small) I think they will still end up at the same spring length at the same speed (more or less just weight factors having a small influence).

In both cases if at some point the guy lets go – I agree the spring returns to its original length but that has no impact on the cart which slows to a standstill.

If this is more or less right – what it proves to me is that even if the force applied to the spring (be it stored potential energy or strain energy in conditions where it is fixed to a mass being accelerated) is the same - the amount of strain energy varies with the resistance of the cart. In other words if there are two identical carts – one heavier than the other – two identical guys both pulling with the same force – I think the springs will initially stretch to a different length as the mass is accelerated because although the pull can only extend the spring the same amount when it is static the resistance is different so the amount of input force to the spring minus the changing resistance at the other fixed end will make the sum of the two different. I guess you guys will disagree – might just be a mental block of mine.

The spring analogy is also good because a spring is a torsion device in a more convenient space than a long straight piece of steel.
I think our drive shafts originally broke when we applied the same input torque but the resistance had increased from standstill and this left the twisting input more resisted by the mass of the trailer and therefore more of it retained in the shafts that exceeded their torsional strength designed limit – not because we used more throttle to overcome the slower acceleration that resulted.

In other words I think a change in mass alters some of the physics involved.

When a car engine tries to accelerate a mass - this time the weight might be the same but it is the torque that alters as the speed increases (as if the two guys had applied increasing and decreasing pull strengths while the cart accelerated depending on the speed it has reached) and I think this dynamic change alters the strain energy within the transmission (under these varying conditions) and therefore alters the dynamics of the outcome.

This exercise has been because I am supporting a patent application relating to engine load and LSPI problems and am trying to write in a more intelligent way and get my notations right. I already know that the load on the engine increases with the mass you are trying to accelerate with an identical torque (or throttle position) because my on board test sensors tell me so and I was trying to put into a few words why.

Thanks – you have all helped my report sound as if I might even know what I am doing. It is not a question of working out what is physically happening to the combustion temperatures (I already have established those differences) – it is about trying to explain why and what we have invented to minimise it.

Thanks for your patience with an old guy trying to work in a more modern and better educated World who is struggling to remember the right terminology (and what it all meant and the differences were) from an education system less well defined back then and too long ago to recall what little he understood of it all back then anyway!

Baz

[AW111]

Regarding gearing : my hobby is historic forest rallying, and I've seen more than one highly tuned car that "falls off the cam" far too readily.

---

Back to drive shaft failures ----

The big difference extra mass makes is the time you are in a particular rev range.

Starting from standstill:

You are adding some kinetic energy when you slip the clutch : the higher the revs and the faster you drop the clutch, the more force gets applied to the driveshaft(s). In this case resistance at the wheels is not only tyre friction, but also the inertia of the wheels & tyres, as you're trying to accelerate them / spin them up.
I think we all have or know of a drive line failure from "watch this" type clutch dumping.

Once the drive is fully engaged, the max torque you can apply is the engine torque at the current revs, multiplied by the current gear ratio.
The reaction to that torque is the tyre grip, which is proportional to the weight on the tyres, including any weight transfer.
Any torque applied above the tyre grip gets dissipated in wheelspin.

So the max torque you can apply is in first gear (or reverse in some cars ).

But : the heavier the load, the slower you accelerate for a given throttle opening, so you're spending longer at or near peak torque than when you're unloaded. This may make a difference.
You will also stay longer within a resonant frequency band, if there are any. Resonant modes include whirl mode - I've seen a dyno destroy itself due to shaft whirl.

[Mave]

hartech said:

Funnily enough I was thinking about a person pulling a cart with a spring last night and agree it is a good analogy.

As I see it – if the cart is light and the spring weak and the guy pulls it without a lot of strength – if the cart is tied to something eventually he will reach a pint where he has extended the spring so far he cannot pull it any more. But if the cart is not tied the spring will slowly extend until there is enough force to over come the static resistance of the cart which starts accelerating. As it does so because the cart starts moving the resistance to initial motion reduces and the spring loses some of its length as the tension in it needed to get the car moving helps pull it towards the guy. No problem with any of that (not really sure if the spring has been subjected to strain energy, potential energy or they are both the same thing).

If a stronger guy tries the same thing he will initially pull the spring further away from the cart but eventually (if he ends up pulling at the same speed as the first guy – no longer accelerating) the spring will be the same length.

Maybe.... but as I said before, you need to be really clear about what you are keeping constant in your thought experiment. If the stronger guy is pulling against a fixed spring, he will pull it further. If he is pulling against the free spring, and follows the same speed profile as the first guy, the spring length will be the same. If he is using all his power to accelerate as quickly as possible then the spring will be longer during acceleration.

hartech said:

If this is more or less right – what it proves to me is that even if the force applied to the spring (be it stored potential energy or strain energy in conditions where it is fixed to a mass being accelerated) is the same - the amount of strain energy varies with the resistance of the cart. In other words if there are two identical carts – one heavier than the other – two identical guys both pulling with the same force – I think the springs will initially stretch to a different length as the mass is accelerated because although the pull can only extend the spring the same amount when it is static the resistance is different so the amount of input force to the spring minus the changing resistance at the other fixed end will make the sum of the two different. I guess you guys will disagree – might just be a mental block of mine.

I don't agree with this. You don't get any resistance until you try to pull the cart. The force provided by the guy, and the force in the spring, and the force applied to the cart are the same. Take an extreme example of the though experiment - if the guy isn't pulling the cart at all, you wouldn't expect the spring to extend just because it could potentially provide a resistance if he tried to pull it. If the 2 guys pull the 2 carts with the same force, the spring extensions will the same, and the heavier cart will accelerate more slowly.

hartech said:

The spring analogy is also good because a spring is a torsion device in a more convenient space than a long straight piece of steel.
I think our drive shafts originally broke when we applied the same input torque but the resistance had increased from standstill and this left the twisting input more resisted by the mass of the trailer and therefore more of it retained in the shafts that exceeded their torsional strength designed limit – not because we used more throttle to overcome the slower acceleration that resulted.

I don't think that's the case. If you applied the same torque, then you applied the same torque. You may have applied the same throttle opening, but that's not the same as applying the same torque. If the resistance has increased then the torque you are applying to overcome it must also have increased (for every force there's an equal and opposite reaction)

hartech said:

In other words I think a change in mass alters some of the physics involved.

When a car engine tries to accelerate a mass - this time the weight might be the same but it is the torque that alters as the speed increases (as if the two guys had applied increasing and decreasing pull strengths while the cart accelerated depending on the speed it has reached) and I think this dynamic change alters the strain energy within the transmission (under these varying conditions) and therefore alters the dynamics of the outcome.

If you apply a different force, or torque, then I agree you will have more potential energy stored in the drivetrain due to it's extra "stretch"; but I think that this is pretty insignificant compared to the energy stored in the flywheel etc. I think that the reason you broke the drivehsfat is probably that either 1) you revved the engine so it was making more steady torque and slipped the clutch, or 2) you revved the engine so it was making more torque and you dumped the clutch and temporarily increased the torque by converting the flywheel inertia into torque

hartech said:

This exercise has been because I am supporting a patent application relating to engine load and LSPI problems and am trying to write in a more intelligent way and get my notations right. I already know that the load on the engine increases with the mass you are trying to accelerate with an identical torque (or throttle position) because my on board test sensors tell me so and I was trying to put into a few words why.

It shouldn't! How are you measuring torque? Are you measuring it directly, or are you deriving it from engine parameters?

hartech said:

Thanks – you have all helped my report sound as if I might even know what I am doing. It is not a question of working out what is physically happening to the combustion temperatures (I already have established those differences) – it is about trying to explain why and what we have invented to minimise it.

Thanks for your patience with an old guy trying to work in a more modern and better educated World who is struggling to remember the right terminology (and what it all meant and the differences were) from an education system less well defined back then and too long ago to recall what little he understood of it all back then anyway!

Baz

No problem, its interesting discussing this stuff.

[GreenV8S]

hartech said:

if there are two identical carts – one heavier than the other – two identical guys both pulling with the same force – I think the springs will initially stretch to a different length as the mass is accelerated because although the pull can only extend the spring the same amount when it is static the resistance is different so the amount of input force to the spring minus the changing resistance at the other fixed end will make the sum of the two different.

We're assuming the mass of the spring itself is not significant - all the force applied to one end of the spring is applied to the other end. Given that assumption, the forces on the two ends of the spring will be equal and opposite, and the extention of the spring is determined purely by that force. It makes no difference to the spring where that force came from or whether the spring itself is moving.

An effect that is far more important IMO in this context is the rotational inertia of the engine and drive train. That can consume a huge amount of torque, energy and power. When the engine is running at constant speed, the whole power output is delivered as mechanical power at the wheel. But if the engine and drive train are accelerating, some of the power output is converted to rotational kinetic energy. Imagine the extreme case where the wheels are off the ground and the engine is accelerating from idle at WOT - the actual delivered power is zero and the whole engine power is used just spinning up engine and drive train components.

You do get that energy back when the engine and transmission spin down, but that's usually lost to interbal friction and not used to propel the vehicle.

[hartech]

Thanks again everyone for patience and being polite - it has helped a lot and I will not take up any more of your time - that is not until you find out about our patent which I would not be surprised if it didn't interest you all enough to question it.

Until the next time

Baz

[hartech]

To help me and make sure I write up a technical report with the right terminology - It actually would help me if one of you would volunteer to write the correct words to use with the following description.

The "force" from the engine that drives the transmission I call "torque". Is it right to call it a "force" or is it correctly called something else when it is rotational?

The resistance to motion of a car from standstill - am I right to call this "Inertia"?

Is it then right to use the formula acceleration = torque/inertia ?

If the torque is high and the inertia also high so that on high torque applications some of the torque twists the transmission shaft with strain energy how is that accounted for in the above equation?

What are the components of the "Inertia" (if that is indeed the right description) when there is rotational mass and linear mass movement.

How do I work out the resulting acceleration when the applied torque varies with engine speed?

If you can help I will be happy to send you a small reward (my book worth £20 about designing and making race winning bikes in the 70's and 80's that won major races) although you might not be interested and if not I will send you something because I will appreciate the help.

I would like to tell you all about the reason behind it but when that comes out you will be glad you assisted me.

Just finally about this loss of energy I keep returning to in strain energy under acceleration - can I try and describe it this way?

if there was a heavy vehicle with a long small diameter shaft sticking out of one wheel and I turned it a long way away at the other end with a torque wrench that shows higher torque as more is applied - initially the vehicle would not move but the shaft would twist to overcome the static resistance - there is torque on it but no acceleration to start with. When there is enough torque to start the vehicle moving slowly - if I then apply more torque (like the revs in an engine rising and the torque it delivers rising with it), even though the vehicle is moving it must apply a bit more twist to that shaft before the result is felt by the vehicle as more acceleration?

If I suddenly let go of the toque wrench - although the twist in it unwinds it neither helps move the vehicle nor gives me back anything. The energy is lost untwisting the shaft so there must have been energy applied in it from the torque delivered which while it is twisting the shaft cannot also equally be used to accelerate the vehicle - surely only the balance left after twisting the shaft more against the inertia is available to accelerate the car?

If you have had enough - fair enough - don't feel obliged to respond!

Thanks again,


Baz

[Mave]

hartech said:

To help me and make sure I write up a technical report with the right terminology - It actually would help me if one of you would volunteer to write the correct words to use with the following description.

OK, I'll give it a go...

hartech said:

The "force" from the engine that drives the transmission I call "torque". Is it right to call it a "force" or is it correctly called something else when it is rotational?

Torque is fine for rotational force

hartech said:

The resistance to motion of a car from standstill - am I right to call this "Inertia"?

The resistance to change in velocity is inertia - could be from standstill, could be when already in motion

hartech said:

Is it then right to use the formula acceleration = torque/inertia ?

Yes

hartech said:

If the torque is high and the inertia also high so that on high torque applications some of the torque twists the transmission shaft with strain energy how is that accounted for in the above equation?

The torque only becomes high when the resistance becomes high. Take the spring example - when the spring is relaxed, you don't need a force to pull it (assuming the spring has no mass). As you pull it and it stretches, the resistance gradually increases. You can't apply a force to something which isn't providing the same resistance back. Equal and opposite reaction etc.

So what happens in your example? Depends on the inertia and stiffness of the drivetrain. Couple of thought experiments -

1) zero drivetrain inertia, infinite drivetrain stiffness. The torque you apply is instantly felt throughout the drivetrain, and there is no strain involved
2) zero drivetrain inertia, flexible drivetrain

Lets say you are trying to apply a particular level of torque. That level of torque equates to a certain amount for drivetrain deflection, or twist. The relationship between torque and twist is fixed (to a first approximation), whether or not it's rotating.
As soon as you apply the torque, the drivetrain instantly twists to that level of deflection, and the torque is instantly felt throughout the drivetrain.
In creating twist, you have transferred energy into the drivetrain. The relationship between deflection and energy is fixed (to a first approximation) for a given drivetrain, and is related to its stiffness (it's the area under the force / deflection curve)

3) some drivetrain inertia, flexible drivetrain.
Lets say you are trying to apply a particular level of torque. That level of torque equates to a certain amount for drivetrain deflection, or twist. The relationship between torque and twist is fixed (to a first approximation), whether or not it's rotating.
When you initially apply the torque, there is no resistance. All of the torque is used to accelerate the drivetrain, and twist it into its deflected state. The moment the drivetrain starts twisting (ie straight away!) some of the torque is transferred through the drivetrain (because its stiffness has increased), the rest is used to carry on accelerating the drivetrain until its gets to its final deflected shape. At this point all the torque is felt throughout the drivetrain and, the same as above, in creating twist, you have transferred energy into the drivetrain.

Note, when I'm talking about accelerating the drivetrain, this is the details of the split second, initial part of the process when you're taking "slack" out of the drivetrain - this isn't what happens when you've taken slack out and are applying steady state torque.

hartech said:

What are the components of the "Inertia" (if that is indeed the right description) when there is rotational mass and linear mass movement.

I don't know enough about your system to know what inertia components you have. Generally linear is easy, rotational is harder.
Linear inertia is mass (in kg!)
Rotational inertia is calculated by adding up the sum of all the individual masses x distance from rotational axis ^2 (ie kgm2)
If you google "rotational inertia formula" you'll find equations for cylinders, rods, discs etc, so if you can approximately break your system down into those features, you can calculate them individually and then just add them together.

hartech said:

How do I work out the resulting acceleration when the applied torque varies with engine speed?

If you know the inertia, then for a given torque, you know the acceleration.
If you are trying to work out the overall acceleration over a period of time, then you need to do a bit of integration.
If you have the relationship between torque and speed as an equation, then you can integrate the equation.
If you have it as a graph, and an approximate answer is sufficient, then I would stick it into excel as a time marching solution - ie start at zero speed, look up torque on the graph, calculate an acceleration, update the speed, look up new torque, calculate acceleration, update speed.....

hartech said:

If you can help I will be happy to send you a small reward (my book worth £20 about designing and making race winning bikes in the 70's and 80's that won major races) although you might not be interested and if not I will send you something because I will appreciate the help.

I would like to tell you all about the reason behind it but when that comes out you will be glad you assisted me.

just glad to try to help someone with a passion

hartech said:

Just finally about this loss of energy I keep returning to in strain energy under acceleration - can I try and describe it this way?

if there was a heavy vehicle with a long small diameter shaft sticking out of one wheel and I turned it a long way away at the other end with a torque wrench that shows higher torque as more is applied - initially the vehicle would not move but the shaft would twist to overcome the static resistance - there is torque on it but no acceleration to start with.

Agreed, this is what I'd call stiction, it's hard to analyse in the same way as normal dynamics

hartech said:

When there is enough torque to start the vehicle moving slowly - if I then apply more torque (like the revs in an engine rising and the torque it delivers rising with it), even though the vehicle is moving it must apply a bit more twist to that shaft before the result is felt by the vehicle as more acceleration?

Yes, you might need to use more energy (not force!) to twist the shaft a bit more until it is stiff enough to transfer the torque you are trying to apply - this is the same as the initial "taking up of slack" I was describing earlier

hartech said:

If I suddenly let go of the toque wrench - although the twist in it unwinds it neither helps move the vehicle nor gives me back anything. The energy is lost untwisting the shaft so there must have been energy applied in it from the torque delivered which while it is twisting the shaft cannot also equally be used to accelerate the vehicle

You applied torque and used energy to twist the shaft. The torque accelerates the vehicle, the energy twists the shaft. You need to twist the shaft before you can apply the torque.

hartech said:

- surely only the balance left after twisting the shaft more against the inertia is available to accelerate the car?

The way to look at this is you can't apply the torque until you have put in enough energy to twist the shaft enough to resist it.
It's like the spring example - a spring operates along a fixed force / extension relationship (Hooke's law), you can't apply a force to a spring until you have stretched it enough to react that force.

hartech said:

If you have had enough - fair enough - don't feel obliged to respond!

Thanks again,

Baz

No problem at all

[hartech]

Fantastic answer Fave - that has really helped clarify things - and set me up to complete my explanations.

Thanks again,

Baz