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
hartech said:
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.
It (the system) doesn’t return to normal because the person doing the pushing or the horse doing the pulling has used energy. That’s where the engergy “disappears”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.
Push/pull for longer… and you use more energy.
Edited by thebraketester on Monday 28th June 17:06
Edited by thebraketester on Monday 28th June 17:06
hartech said:
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.
That's because it is not a thing. The angular deflection within the transmission under load is so small that the energy it absorbs is negligible.You idea seems to be based on a misunderstanding of the relationship between torque and power. Torque without deflection doesn't absorb any energy. Think of you leaning on the end of a spanner that won't come undone. Once you've taken up the slack in the various connections, you are not putting any energy in; you can replace yourself with a bit of string and the spanner won't know the difference. It is only when you start getting significant deflection that energy is absorbed; if you're removing a fastener that doesn't want to come off, it can be quite hot by the time you've wound it off.
The main reason a big engine loses a lot of power in the lower gears is nothing to do with the transmission. It may lose a little bit of power due to deflection within the tyre, but most of it will be lost accelerating the moving parts within the engine itself.
I thought one of the big efficiency losses in ICE cars is the pistons keep stopping and changing direction. In comparison the transmission must be pretty benign - with all parts heading in the same direction.
You put energy in to stretch a rubber band but most that energy comes out again when it springs back. So you don't lose much energy bending things which spring back afterwards. For example I'm thinking of wind-up in a long prop shaft when torque is applied. It will unwind again when torque is reduced. Sure, during the "winding up" phase some energy won't be making it's way to the rear wheels but that stops as soon as the deformation stops increasing.
To find more information it might be worth looking at dragsters. Although, as someone has already mentioned, the tyres will be doing a lot more bending about than anything else and the torque converter will also be a significant player.
You put energy in to stretch a rubber band but most that energy comes out again when it springs back. So you don't lose much energy bending things which spring back afterwards. For example I'm thinking of wind-up in a long prop shaft when torque is applied. It will unwind again when torque is reduced. Sure, during the "winding up" phase some energy won't be making it's way to the rear wheels but that stops as soon as the deformation stops increasing.
To find more information it might be worth looking at dragsters. Although, as someone has already mentioned, the tyres will be doing a lot more bending about than anything else and the torque converter will also be a significant player.
hartech said:
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 think you are mistaken in saying that time is a factor in these examples.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.
You apply a certain torque to a bolt, and the bolt resists it. If the resistance is equal to the torque, and nothing else changes, the bolt will not turn, ever. The torque will be stored as potential energy which is trying to twist the bolt, plus a proportion will be converted into heat energy. You will not put any more energy in unless you increase the torque. When you release the torque the potential energy will become kinetic energy, and again heat energy, and the bolt will un-twist. No energy lost overall, and time doesn't come into it.
If I understand your original point correctly - maybe I don't - there would seem to be a small loss of energy between the engine and the gearbox because the input shaft is being twisted, and this will be dissipated as heat, but it will be negligible.
hartech said:
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?
So, a general comment about your thought experiments - to pick your way through what's going on, it's useful to break them down into simple chunks rather than one complex scenario where you can't see what's happening.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?
Instead of applying the force in your example using a horse, which is a complex chemical reaction, consider applying the force by using a weight pulling the barge via a pulley.
If you leave it pulling for 1 minute, or 1 hour, has a different amount of force been applied? Has a different amount of strain been created? Has energy been used by the weight sitting there dangling over the pulley?
Mave said:
So, a general comment about your thought experiments - to pick your way through what's going on, it's useful to break them down into simple chunks rather than one complex scenario where you can't see what's happening.
Instead of applying the force in your example using a horse, which is a complex chemical reaction, consider applying the force by using a weight pulling the barge via a pulley.
If you leave it pulling for 1 minute, or 1 hour, has a different amount of force been applied? Has a different amount of strain been created? Has energy been used by the weight sitting there dangling over the pulley?
This puts it very well. The horse may be producing more energy over time to maintain a constant pull on the tow rope, but the tow rope doesn't see that, it only sees the constant pull.Instead of applying the force in your example using a horse, which is a complex chemical reaction, consider applying the force by using a weight pulling the barge via a pulley.
If you leave it pulling for 1 minute, or 1 hour, has a different amount of force been applied? Has a different amount of strain been created? Has energy been used by the weight sitting there dangling over the pulley?
.
Hawkshaw said:
This puts it very well. The horse may be producing more energy over time to maintain a constant pull on the tow rope, but the tow rope doesn't see that, it only sees the constant pull.
.
Yep, the horse confuses things, as it's a complex (and wasteful) way to convert energy into force..
If you swapped the horse for a helicopter, you could equally convince yourself that when it's hovering, a helicopter is converting energy (from the fuel) into force (to overcome gravity) and that therefore force somehow is consuming energy. But then the helicopter lands, the force reacted (to overcome gravity) is unchanged, and yet no energy is expended as it sits there...
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
What you seem to be talking about is hysteric losses in a metal spring?
Consider an actual metal wire spring.
You apply a force, it deflects. You release the force, the spring returns to it's original position, however the energy returned is very very slightly smaller than the energy expended, because the hysteresis in the material.
see here:
GOOGLE-hysteresis-of-steel
In any practical transmission or engine where the strain is always going to have to be maintained far below the plastic region of the material in order to avoid fatigue, this is such a small loss as to be almost un-measureable.
Consider an actual metal wire spring.
You apply a force, it deflects. You release the force, the spring returns to it's original position, however the energy returned is very very slightly smaller than the energy expended, because the hysteresis in the material.
see here:
GOOGLE-hysteresis-of-steel
In any practical transmission or engine where the strain is always going to have to be maintained far below the plastic region of the material in order to avoid fatigue, this is such a small loss as to be almost un-measureable.
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
hartech said:
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
Imagine you are not applying the torque with your muscles, but by hanging a weight on the end of the spanner.Once the slack has been taken up, the weight doesn't move, so no energy is being used. You can leave the weight there for minutes or days with no movement or energy loss.
There is a tiny amount of energy lost in flexing and unflexing the spanner/bolt, as Max_torque says, but that's a one-time loss : energy is stored when the load is applied and everything bends/distorts, and when the load is removed you don't get all of it back.
Most of the energy and torque that breaks driveline components comes from the rotational kinetic energy stored in the engine before you dump the clutch. You can store a lot of energy by revving the nuts off the engine and releasing it all at once into the drivetrain. If you do the calcs on a typical 2.0L four cylinder engine it's more than 20kJ at a typical 6k redline.
The torque you can put into the driveline by this rapid deceleration can be many times the maximum torque the engine can generate in steady-state conditions.
The torque you can put into the driveline by this rapid deceleration can be many times the maximum torque the engine can generate in steady-state conditions.
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
hartech said:
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
The reality is the opposite. 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
The amount of energy absorbed by flexing within the transmission is tiny because the cumulative amount of deflection is tiny and there is no hysteresis.
The amount of energy absorbed within the tyre is relatively high even if the tyre isn't sliding much, because there is more hysteresism and because the stresses within the tyre are released by the trailing edge of the contact patch sliding against the road (the energy stored within the tyre is not recovered).
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
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