help needed understanding dynamic strain energy losses
Discussion
I am trying to do some calculations about the loads on an engine during acceleration against a mass from different starting points and acceleration rates and am struggling to find anything helpful.
You see I have long held the view that relatively powerful engines driving in 1st gear loose a lot of their torque through strain energy losses caused by dynamic elasticity within the yield point of the transmission – but have never been able to find any physics to cover it. Most reports state that strain energy is negligible in transmissions – which I can understand under constant speed but not under acceleration - since acceleration is proportional to torque/resistance (or inertia or whatever you want to call the sum of all the resistances).
I am sure modern education systems have covered this subject (which we hardly touched on in my time) but going back to then (50 years ago) I remember when we were towing a trailer with two racing bikes on it with a Ford Consul Classic and if we booted it too much in first gear the drive shafts would snap in half (we carried a spare just in case).
I occurred to me this can only be because the torque input trying to accelerate the mass of the car against an initial static resistance, (trailer and bikes) exceeded the torsional yield point and therefore the strain energy exceeded it as well. Once moving - the resistance to acceleration was less so as long as you accelerated from a standstill moderately – the drive shafts survived and you could continue on your journey OK but to snap a driveshaft with a torque wrench would require huge torque and torque is force * radius – so a huge force as well.
Looked at slightly differently - I mean of you apply a torque to the head of a bolt for 1 minute or alternatively for I hour – you must have used more energy for the hour – yet at the end of it everything returns to normal in both cases – so where did that extra energy used disappear to?
Similarly - if a horse tried to pull a barge tied to a bollard for 1 minute or 1 hour – more horse force was used over the hour yet the outcome is no different.
I accept that in both examples above there is no movement (once the strain extension has been reached - so there is no rate of doing work measured over a distance or revs etc) but strain energy is a form of potential energy so there is still power used and a force must be being applied and as I understand Newton an object stays at rest unless acted on by a net external force?
The first law of thermodynamics is also that energy cannot be created or destroyed but there seems something odd about energy used to strain something over different periods of time but the outcome being the same?
The internal combustion of an engine is different depending on the load the engine is driving against (and requires different ignition timing to avoid knock).
I am trying to work out what those differences are in relation to different rates of acceleration but so far it has proved to be beyond me.
OK I can work out the torque (and therefore the BHP) at any given rev point while Accelerating our rolling road dyno – but that is at a fixed resistance rate and usually at full throttle (WOT).
But if the mass being accelerated was altered or the power generated – the internal strain energy losses would be different.
If anyone can help I would be very grateful indeed especially if they can explain it in a tangible way (and not via mathematics I might struggle with).
Although not of direct interest to me right now – I once built a twin cylinder two stroke racing engine with identical everything except one fired two cylinders together and the other alternately (at 180 deg). They were as different a chalk and cheese – one had a very narrow power band from about 8,500 to 10,500 rpm while the other had progressive power from 2000 rpm gradually increasing and revved on to 14,000 rpm. I have had it explained to me that this is all to do with flywheel stored and released energy but I cannot help thinking that there must have been some strain energy differences involved?
So I get the gut feeling that there is more to strain energy losses than many give it credit for and current research would benefit if I understood it better.
I guess this might be of no interest to anyone else so a direct reply to me might be more acceptable to other readers?
I might very well be getting my sciences muddled up – but any advice would be welcome.
Baz
You see I have long held the view that relatively powerful engines driving in 1st gear loose a lot of their torque through strain energy losses caused by dynamic elasticity within the yield point of the transmission – but have never been able to find any physics to cover it. Most reports state that strain energy is negligible in transmissions – which I can understand under constant speed but not under acceleration - since acceleration is proportional to torque/resistance (or inertia or whatever you want to call the sum of all the resistances).
I am sure modern education systems have covered this subject (which we hardly touched on in my time) but going back to then (50 years ago) I remember when we were towing a trailer with two racing bikes on it with a Ford Consul Classic and if we booted it too much in first gear the drive shafts would snap in half (we carried a spare just in case).
I occurred to me this can only be because the torque input trying to accelerate the mass of the car against an initial static resistance, (trailer and bikes) exceeded the torsional yield point and therefore the strain energy exceeded it as well. Once moving - the resistance to acceleration was less so as long as you accelerated from a standstill moderately – the drive shafts survived and you could continue on your journey OK but to snap a driveshaft with a torque wrench would require huge torque and torque is force * radius – so a huge force as well.
Looked at slightly differently - I mean of you apply a torque to the head of a bolt for 1 minute or alternatively for I hour – you must have used more energy for the hour – yet at the end of it everything returns to normal in both cases – so where did that extra energy used disappear to?
Similarly - if a horse tried to pull a barge tied to a bollard for 1 minute or 1 hour – more horse force was used over the hour yet the outcome is no different.
I accept that in both examples above there is no movement (once the strain extension has been reached - so there is no rate of doing work measured over a distance or revs etc) but strain energy is a form of potential energy so there is still power used and a force must be being applied and as I understand Newton an object stays at rest unless acted on by a net external force?
The first law of thermodynamics is also that energy cannot be created or destroyed but there seems something odd about energy used to strain something over different periods of time but the outcome being the same?
The internal combustion of an engine is different depending on the load the engine is driving against (and requires different ignition timing to avoid knock).
I am trying to work out what those differences are in relation to different rates of acceleration but so far it has proved to be beyond me.
OK I can work out the torque (and therefore the BHP) at any given rev point while Accelerating our rolling road dyno – but that is at a fixed resistance rate and usually at full throttle (WOT).
But if the mass being accelerated was altered or the power generated – the internal strain energy losses would be different.
If anyone can help I would be very grateful indeed especially if they can explain it in a tangible way (and not via mathematics I might struggle with).
Although not of direct interest to me right now – I once built a twin cylinder two stroke racing engine with identical everything except one fired two cylinders together and the other alternately (at 180 deg). They were as different a chalk and cheese – one had a very narrow power band from about 8,500 to 10,500 rpm while the other had progressive power from 2000 rpm gradually increasing and revved on to 14,000 rpm. I have had it explained to me that this is all to do with flywheel stored and released energy but I cannot help thinking that there must have been some strain energy differences involved?
So I get the gut feeling that there is more to strain energy losses than many give it credit for and current research would benefit if I understood it better.
I guess this might be of no interest to anyone else so a direct reply to me might be more acceptable to other readers?
I might very well be getting my sciences muddled up – but any advice would be welcome.
Baz
Edited by hartech on Tuesday 29th June 13:54
Thanks very much for your contributions guys - perhaps I should have called it dynamic elastic strain energy losses - so according to your answers - why then did the consul drive shaft always snap when you were accelerating away if you gave it too much throttle??
Baz
Baz
Edited by hartech on Wednesday 30th June 18:26
Thanks Max - those drive shafts would fail from new - no time for metal fatigue - I believe that the issue is highest drive shaft torque (which is applied in 1st gear) set against inertia of the weight causing a maximum resistance initially. Once the car is moving the resistance reduces and as you change up the gears so to does the torque applied to the drive shafts - but I am seeking what type of physics this is to find ways to calculate the dynamic effect of torque against resistance or load. I can't describe it better because I don't know what it is or what to call it. I do believe dynamic elastic strain energy is something a bit odd - and was hoping it wasn't and there was an easy answer we didn't cover when I was studying in the mid '60's lol
Baz
Baz
That's also really interesting AER but I experienced it years ago under steady acceleration (no stored kenetic energy) and that is also what I am interested in now.
I really appreciate the advice you have all given – it is for a huge project that is working in a previously unexplored areas of engine performance and is straining my ability to the limit.
I threw in some analogies to see if anyone was up for it and found you are – so now I would like to get to the relevant issues directly.
It is all about trying to calculate the load on the engine under full throttle acceleration in low gears and flat out in top gear – the two main areas that apply the most engine loading. But it involves several dynamics together and I am trying to understand them and which are relevant and how to calculate them.
Although I finished my education in 1968 and I wouldn’t claim to have been the most academic student anyway lol.
I find the analysis between static loading and dynamic loading difficult to quantify.
I think work can only be done if a distance is covered (linear, radial or both together).
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?.
If we swap the clutch for a torque converter that gets round it stalling but it could still snap the small drive shaft but we could then fit bigger and bigger drive shafts until it didn’t and just got hotter instead.
I think if it didn’t snap (or up until it did) – the drive shaft would experience strain energy until it exceeded its torsional stress limit when it would snap?
Then again - if the hand brake or strapping was removed and the car was being driven at slow but constant speed in a higher gear (where the torque applied is lower anyway as the inverse proportion of the new gear ratio) then I agree there would be little strain energy in the transmission system.
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. 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). 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).
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
I really appreciate the advice you have all given – it is for a huge project that is working in a previously unexplored areas of engine performance and is straining my ability to the limit.
I threw in some analogies to see if anyone was up for it and found you are – so now I would like to get to the relevant issues directly.
It is all about trying to calculate the load on the engine under full throttle acceleration in low gears and flat out in top gear – the two main areas that apply the most engine loading. But it involves several dynamics together and I am trying to understand them and which are relevant and how to calculate them.
Although I finished my education in 1968 and I wouldn’t claim to have been the most academic student anyway lol.
I find the analysis between static loading and dynamic loading difficult to quantify.
I think work can only be done if a distance is covered (linear, radial or both together).
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?.
If we swap the clutch for a torque converter that gets round it stalling but it could still snap the small drive shaft but we could then fit bigger and bigger drive shafts until it didn’t and just got hotter instead.
I think if it didn’t snap (or up until it did) – the drive shaft would experience strain energy until it exceeded its torsional stress limit when it would snap?
Then again - if the hand brake or strapping was removed and the car was being driven at slow but constant speed in a higher gear (where the torque applied is lower anyway as the inverse proportion of the new gear ratio) then I agree there would be little strain energy in the transmission system.
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. 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). 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).
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
Thanks Voram I also agree and think the maximum load will be accelerating in 1st gear rather than flat out in top (because the input toque is so much lower in top gear) - but even in 1st - the tyres will not slip if they are grippy enough and the car will simply accelerate faster and in so doing the drive train will be under strain which you would also apply if you were the engine and turning the gearbox over yourself with a torque wrench - so some of the torque must be being lost to apply the strain in the drive train and it must be being lost over time and therefore it is a force over a period of time which is work done twisting the molecules?
If progressively smaller drive shafts would eventually break under those accelerating forces - then that to me proves there is wasted energy there and the forces capable of breaking smaller drive shafts must be the same as those losses just before they broke - therefore some energy must be being lost to the drive system in strain energy and since the input torque varies with revs and the output resistance varies with the rate of accelerating the mass - there must be an equation that would plot the outcome and that is what I am seeking.
Baz
If progressively smaller drive shafts would eventually break under those accelerating forces - then that to me proves there is wasted energy there and the forces capable of breaking smaller drive shafts must be the same as those losses just before they broke - therefore some energy must be being lost to the drive system in strain energy and since the input torque varies with revs and the output resistance varies with the rate of accelerating the mass - there must be an equation that would plot the outcome and that is what I am seeking.
Baz
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
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
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
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
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
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
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
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
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
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
Gassing Station | Engines & Drivetrain | Top of Page | What's New | My Stuff