Why can't diesel make more power than gasoline?
08-01-2007, 09:33 AM
#46
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OK i'll try to explain from A to Z since you seem to be confusing a couple of things
- thermal energy in combustion is converted into torque at the crank( mechncial energy)
-this torque causes motion (rotation) at the crank and thus power at the crank
-it's not a matter of formulas and that you can get this from that ..
-the relevant acceleration here is an angular acceleration at the crank NOT OF THE CAR. and yes you can still apply the rotational version of "Newton's law" here.
-then you can use a scaling ratio (e.g 0.9 for 10% losses) to transmit this torque to wheel torque and the same for crank to wheel power. so torque is not meaningless
-acceleration of the car is a dependent on the torque applied at the wheels
which you could also represent as power/velocity but these are just dependent variables
-torque at the wheels is what overcomes road friction and makes the car move
-another misconception is that peak values of power/torque is what makes a car go faster..NO it's just an indication but the real measure is torque curves
-to be exact for everything identical it's how much torque is transmitted at every point in time during an acceleration of the wheels:
at time ti acceleration(ti)=torque(ti)/inertia so to find out exactly what your speed is over a certain time you integrate the above equation over that time so the area under the torque curve as a function of time and not rpm is the real measure...you can get vehicle translational speed from the angular speed of the wheels above if there is no slip..otherwise it's more complicated
-yes available HP or torque at the crank will dictate how much torque or power you will have at the wheels but the power at the crank is generated through torque at the crank
-please review my explanation because you are confusing vehicle speed with rpm
-"comparing two totally different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed." how is this possible?
- "acceleration is ...inversely proportional to speed" that's not true as a fact for any moving object..it's only observed at higher speeds due to drag and that's an extra "negative" force/torque acting on the system at higher speeds that causes the acceleration to reduce because acceleration= force/mass and the net positive force is less due to drag..if we ignore drag then the statement would be wrong!
for instance with no drag and a constant torque (over some time) then acceleration is constant and velocity is just the integral i.e. it's increasing
-so according to the "derived" formula you are using at 0 speed we have infinite acceleration? the reason for this strange outcome is that the "real" newtonian law uses torques/forces and at nonzero net torque or force we can then have motion and then we can define speed and power..only then we can plug in
torque=power/speed instead in newton's law if we wish
I was trying not to get too technical or lengthy before but I hope this clarifies things for you
- thermal energy in combustion is converted into torque at the crank( mechncial energy)
-this torque causes motion (rotation) at the crank and thus power at the crank
-it's not a matter of formulas and that you can get this from that ..
-the relevant acceleration here is an angular acceleration at the crank NOT OF THE CAR. and yes you can still apply the rotational version of "Newton's law" here.
-then you can use a scaling ratio (e.g 0.9 for 10% losses) to transmit this torque to wheel torque and the same for crank to wheel power. so torque is not meaningless
-acceleration of the car is a dependent on the torque applied at the wheels
which you could also represent as power/velocity but these are just dependent variables
-torque at the wheels is what overcomes road friction and makes the car move
-another misconception is that peak values of power/torque is what makes a car go faster..NO it's just an indication but the real measure is torque curves
-to be exact for everything identical it's how much torque is transmitted at every point in time during an acceleration of the wheels:
at time ti acceleration(ti)=torque(ti)/inertia so to find out exactly what your speed is over a certain time you integrate the above equation over that time so the area under the torque curve as a function of time and not rpm is the real measure...you can get vehicle translational speed from the angular speed of the wheels above if there is no slip..otherwise it's more complicated
-yes available HP or torque at the crank will dictate how much torque or power you will have at the wheels but the power at the crank is generated through torque at the crank
-please review my explanation because you are confusing vehicle speed with rpm
-"comparing two totally different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed." how is this possible?
- "acceleration is ...inversely proportional to speed" that's not true as a fact for any moving object..it's only observed at higher speeds due to drag and that's an extra "negative" force/torque acting on the system at higher speeds that causes the acceleration to reduce because acceleration= force/mass and the net positive force is less due to drag..if we ignore drag then the statement would be wrong!
for instance with no drag and a constant torque (over some time) then acceleration is constant and velocity is just the integral i.e. it's increasing
-so according to the "derived" formula you are using at 0 speed we have infinite acceleration? the reason for this strange outcome is that the "real" newtonian law uses torques/forces and at nonzero net torque or force we can then have motion and then we can define speed and power..only then we can plug in
torque=power/speed instead in newton's law if we wish
I was trying not to get too technical or lengthy before but I hope this clarifies things for you
Originally Posted by mark kibort
It all starts with power. power gives you the torque at an RPM. of course, if you wish to ignor my chicken and egg comment, then yes, you can calculate power from torque and rpm. I think we are fine there.
HOWEVER, you can have the same power curves, and im talking shape curve, i was never talking identical curve values with respect with RPM, because what point would that make?
So, you can have identical shaped HP curves, however, one can have a max rpm of say 12,000rpm and the other 6000rpm. both could have the same peak HP at 90% of the max rpm and max peak torque at 75% of their max rpm, yet one would have HALF the torque of the other at any point. Now, understanding this is a breakthrough from the common misconceptions of HP and torque. Hp is made from torque and RPM. torque is one factor of HP of equal weight with RPM. yes, force will determine HP if all you know is the force or torque and the rpm or speed. however, this misses the point of the discussion. an available HP will dictate the force at the rear tires, and unless you are looking at torques or forces acting on the car, at the rear whees, torque is a meaningless value.
two cars with identical shapped HP curves will produce the same rear wheel torque and forces at any speed, hence the newtonian identiy:
acceleration= power/(mass x velocity)
Lets say that its impossible to have the same "shaped" HP curve. Fine, thats what close ratio gear boxes allow you to do. as long as you keep the average hp, or area under the HP curves the same (comparing two totaly different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed.
the CV joints will be experiencing the same forces. Now, up a few stages in the gear box, suddenly, the high torque engine requires greater strength gears, clutches, cranks, etc. you get the idea.
so, your point of not being able to have the same shaped HP curves without having the same torque curves is wrong. again, it would make no sense to talk about two cars with the same HP curve, as they would have the same torque curve, and that is far from what we are discussing here. (ie Diesel vs gas engine comparisons)
so, find one even tiny error in what i have posted since you are "amused" with people arguing with "wrong information" Its all very factual, and pretty basic stuff.
bottomline: acceleration = power/(mass x velocity) this means acceleration is proportioal to power and inversely proportional to speed. can you find any error in this.
also, as far as power, do you not agree that it is HP-seconds that will determine the fastest rates of acceleration. If so, and you wanted to equate this to torque (or force) you need a lot more information, as after all,
Force x distance is work
and Power is a rate of doing work. what we are really talking about with acceleration of our cars, is the difference of the rate of doing work.
to prove my points, one example that you can try and argue or explain.
two cars, as mentioned above. both 700hp, but one with 300ft-lbs of peak torque and the other with 600ft-lbs of peak torque. assuming the same car, weight, proportional gear boxes and same shaped HP curve.
coming off a 40mph turn approaching a 1/2mile straight , which one would have an advange and why?
MK
HOWEVER, you can have the same power curves, and im talking shape curve, i was never talking identical curve values with respect with RPM, because what point would that make?
So, you can have identical shaped HP curves, however, one can have a max rpm of say 12,000rpm and the other 6000rpm. both could have the same peak HP at 90% of the max rpm and max peak torque at 75% of their max rpm, yet one would have HALF the torque of the other at any point. Now, understanding this is a breakthrough from the common misconceptions of HP and torque. Hp is made from torque and RPM. torque is one factor of HP of equal weight with RPM. yes, force will determine HP if all you know is the force or torque and the rpm or speed. however, this misses the point of the discussion. an available HP will dictate the force at the rear tires, and unless you are looking at torques or forces acting on the car, at the rear whees, torque is a meaningless value.
two cars with identical shapped HP curves will produce the same rear wheel torque and forces at any speed, hence the newtonian identiy:
acceleration= power/(mass x velocity)
Lets say that its impossible to have the same "shaped" HP curve. Fine, thats what close ratio gear boxes allow you to do. as long as you keep the average hp, or area under the HP curves the same (comparing two totaly different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed.
the CV joints will be experiencing the same forces. Now, up a few stages in the gear box, suddenly, the high torque engine requires greater strength gears, clutches, cranks, etc. you get the idea.
so, your point of not being able to have the same shaped HP curves without having the same torque curves is wrong. again, it would make no sense to talk about two cars with the same HP curve, as they would have the same torque curve, and that is far from what we are discussing here. (ie Diesel vs gas engine comparisons)
so, find one even tiny error in what i have posted since you are "amused" with people arguing with "wrong information" Its all very factual, and pretty basic stuff.
bottomline: acceleration = power/(mass x velocity) this means acceleration is proportioal to power and inversely proportional to speed. can you find any error in this.
also, as far as power, do you not agree that it is HP-seconds that will determine the fastest rates of acceleration. If so, and you wanted to equate this to torque (or force) you need a lot more information, as after all,
Force x distance is work
and Power is a rate of doing work. what we are really talking about with acceleration of our cars, is the difference of the rate of doing work.
to prove my points, one example that you can try and argue or explain.
two cars, as mentioned above. both 700hp, but one with 300ft-lbs of peak torque and the other with 600ft-lbs of peak torque. assuming the same car, weight, proportional gear boxes and same shaped HP curve.
coming off a 40mph turn approaching a 1/2mile straight , which one would have an advange and why?
MK
08-01-2007, 09:39 AM
#47
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mark i agree i would rather have more power than torque up to some reasonable difference in a race because the power/torque would come at higher revs and would be easier to put down..but that's a different point
Originally Posted by mark kibort
krC2S, I think what "Insite" was trying to make as a point, was that power determines torque at the rear wheels at any speed. Sure, you can always go multiplying to your hearts content, to figure out what torque you have at the rear wheels through any gear box ratio at any speed. however, if you now are looking at accelerative forces at differnt speeds, it becomes much easier to look at the power comparisons. Engine torques also have to be coupled with rpm (ie HP) to get a final drive torque at the rear wheels at any vehicle speed through the gear box. more steps for a comparison. however looking at a HP curve, i can tell you which vehicle will win a race over a speed range by only knowing gear spacing. Its just easier and you need less information because more is included in HP. remember, its made up of torque and rpm.
mk
mk
08-01-2007, 09:56 AM
#48
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Originally Posted by murphyslaw1978
Gasoline has about 150,100 BTUs of energy per gallon. Diesel has 166,600 BTUs of energy per gallon. Regular diesel has a much higher octane rating than regular gasoline as found at the pump.
08-01-2007, 11:08 AM
#49
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Originally Posted by kens911
...a high octane number means the fuel burns slower...
...if you measured octane on deisel it would be very low...
...if you measured octane on deisel it would be very low...
Diesel would have a comparatively higher octane rating than gas, because it is much harder to ignite. This is why they need the heat generated from 18:1, 20:1, and even 22:1 compression ratios to ignite the fuel.
For these reasons, I would rather have someone drop a match into the pool of diesel that I was standing it, than have someone do the same while I was standing in a pool of gas.
08-01-2007, 11:20 AM
#50
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Originally Posted by mark kibort
another misnomer. even the R10 doesnt have a "flat" torque curve in its usuable range. its 820 ft-lbs falls dramatically as it approaches 5000rpm as it has to otherwise it would have a heck of a lot more power than 650!!
the 911 turbo, has a flat HP curve, not a flat torque curve (in the useable racing range) all torque curves fall, its just a matter of by how much.
mk
the 911 turbo, has a flat HP curve, not a flat torque curve (in the useable racing range) all torque curves fall, its just a matter of by how much.
mk
Graph # 2 is from typical Audi & VW diesel engines. Note that they have a fairly flat torque curve from 1000-4500RPM.
My argument is "why can't the diesel engine produce a flat torque curve from 2500-6500RPM?"
08-01-2007, 12:59 PM
#51
Rennlist Member
Thanks
Now, the entire point of this is maxizing performance. (ie acceleration)
this is done on the far right side of the HP and torque curve. (to redline)
If this is done, you are always spending time on a falling torque curve, yet some cars have a wider HP curve and this does two things.
1. gives you a higher area under the hp curve applied over time
2. allows for wider gears
I posted the 911 turbo curves too, and yes, there is a flat part down in the lower RPMs and you are absolutely right with the "driveability" differences. But, what actually happens is that improves HP levels at the lower rpms, and this makes a car feel faster staying in a gear, even at the lower rpms.
MK
Now, the entire point of this is maxizing performance. (ie acceleration)
this is done on the far right side of the HP and torque curve. (to redline)
If this is done, you are always spending time on a falling torque curve, yet some cars have a wider HP curve and this does two things.
1. gives you a higher area under the hp curve applied over time
2. allows for wider gears
I posted the 911 turbo curves too, and yes, there is a flat part down in the lower RPMs and you are absolutely right with the "driveability" differences. But, what actually happens is that improves HP levels at the lower rpms, and this makes a car feel faster staying in a gear, even at the lower rpms.
MK
Originally Posted by murphyslaw1978
Correct, the R10 doesn't have a flat torque curve, but the 911 Turbo does (in the majority of the street RPM range). I have attached a graph as found on Porsche.com. I believe the reason for this curve is because Porsche is controlling the amount of torque (lessening it) so as to preserve the reliability of the engine, but the point I'm making is that it can be done.
Graph # 2 is from typical Audi & VW diesel engines. Note that they have a fairly flat torque curve from 1000-4500RPM.
My argument is "why can't the diesel engine produce a flat torque curve from 2500-6500RPM?"
Graph # 2 is from typical Audi & VW diesel engines. Note that they have a fairly flat torque curve from 1000-4500RPM.
My argument is "why can't the diesel engine produce a flat torque curve from 2500-6500RPM?"
08-01-2007, 01:10 PM
#52
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Thread Starter
Originally Posted by mark kibort
Thanks
Now, the entire point of this is maxizing performance. (ie acceleration)
this is done on the far right side of the HP and torque curve. (to redline)
If this is done, you are always spending time on a falling torque curve, yet some cars have a wider HP curve and this does two things.
1. gives you a higher area under the hp curve applied over time
2. allows for wider gears
I posted the 911 turbo curves too, and yes, there is a flat part down in the lower RPMs and you are absolutely right with the "driveability" differences. But, what actually happens is that improves HP levels at the lower rpms, and this makes a car feel faster staying in a gear, even at the lower rpms.
MK
Now, the entire point of this is maxizing performance. (ie acceleration)
this is done on the far right side of the HP and torque curve. (to redline)
If this is done, you are always spending time on a falling torque curve, yet some cars have a wider HP curve and this does two things.
1. gives you a higher area under the hp curve applied over time
2. allows for wider gears
I posted the 911 turbo curves too, and yes, there is a flat part down in the lower RPMs and you are absolutely right with the "driveability" differences. But, what actually happens is that improves HP levels at the lower rpms, and this makes a car feel faster staying in a gear, even at the lower rpms.
MK
08-01-2007, 01:35 PM
#53
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Originally Posted by murphyslaw1978
Agreed. And to your point, now the only question still remaining is "how to extend the torque curve of a diesel engine farther to the right."
on another note, it's interesting how ME's with different styles of training think differently about the same problems. krC2S seems to think as a staticist while myself and Mark seem to think as dynamicists. i'm sure we can all think in either term; it's just interesting how brains work.
08-01-2007, 02:18 PM
#54
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krC2S,
This is going to take longer than i thought.
First of all, you need to be more specific. there is no "confusion" on my part, and if there is , i surely would like to know about it. second, I think you are missing the main point here by several of you responses below. You start to show some knowledge, and then you kill it with a couple of the comments below. Now, basically you are right about a few things. The overall message here, you seem to be missing. let me try and insert my comments below and you can feel free to disagree. BUT, state why. Ill try to make this brief.
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute. If you dont agree, you should possibly look at why, or ask some questions. You do make a note of this below and i will try to address it as simply as possible.
Here we go:
You are not getting too technical. what you are doing is missing the point. it all comes back to your single question above. you wanted to know about two different engines that make the same HP but have different peak torque values, maybe as far as 2x from each other, making the same forces at the rear tires at any vehicle speed. you ask how is this possible. I tell you and i hope you understand. This is tough doing a seminar over the web. over a beer with a white board or napkin, this would take 10 mins!!!
This is going to take longer than i thought.
First of all, you need to be more specific. there is no "confusion" on my part, and if there is , i surely would like to know about it. second, I think you are missing the main point here by several of you responses below. You start to show some knowledge, and then you kill it with a couple of the comments below. Now, basically you are right about a few things. The overall message here, you seem to be missing. let me try and insert my comments below and you can feel free to disagree. BUT, state why. Ill try to make this brief.
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute. If you dont agree, you should possibly look at why, or ask some questions. You do make a note of this below and i will try to address it as simply as possible.
Here we go:
Originally Posted by krC2S
OK i'll try to explain from A to Z since you seem to be confusing a couple of things
- thermal energy in combustion is converted into torque at the crank( mechncial energy)
-this torque causes motion (rotation) at the crank and thus power at the crank
-it's not a matter of formulas and that you can get this from that ..
>>>>>HUhh? yes, energy is converted from heat to the expansion of the gasses in the form of a net force on a piston. Think of this as the power applied. energy, killowatts-hours/min/seconds, all determine the rate of doing work. Work, as we know is a force x a distance. how fast we do this work is power. with the equation, we can determine whether the piston will accelerate at a fast or slow rate depending on what it is attached to . (long stroke rod, etc)
the force that comes out of this application of power will be determined by the mechanical factors. you can think of torque or force as starting the event, or causing the motion, Becasue it is. But, it is one component of the larger picture. The amount of force will be determined by the power potential and the mechanical configurations.
Here is an anaology. dropping a rock on to one side of a teeter-totter with a projectile on the other side. at 10ft, the rock has potential energy. (power potential ) . The hight and weight of this rock determines the forces we are going to generate on the projectile. (our cars) We can drop it on a long levered teeter-totter, or a short levered one. the long levered one will create more force but over a shorter distance. The short lever will create lower forces, but over a greater distance. the projectile will be fired at the same speed. the power will be the same.
-the relevant acceleration here is an angular acceleration at the crank NOT OF THE CAR. and yes you can still apply the rotational version of "Newton's law" here.
>>>>>>>>> The entire discussion is whether a high torque engine CREATES greater CAR acceleration over cars with lower torque engines but rated at the same HP. keep the comparisons and analysis on track. you are making it confusing. I have never debated that the equations couldnt be applied to the engine components.
-then you can use a scaling ratio (e.g 0.9 for 10% losses) to transmit this torque to wheel torque and the same for crank to wheel power. so torque is not meaningless
>>>>>>>>>>>>>You are getting on thin ice here. Remember i mentioned to be clear of the differences of torque at the engine, and torque at the rear wheels. WE are not talking about engine torque as measured at the rear wheels, we are talking about the multiplied torques at the rear wheels THROUGH the gear ratios. (i.e. 500ft-lbs of peak engine torque through a 5:1 3rd gear will be 2500ft-lbs of torque at the rear wheels.) at this point we could care less about gear box efficiencies and other losses (thats another discussion as well). that 2500ft-lbs of torque could be developed by two 500hp engines, one with 100ft-lbs of engine torque and the the other with 500ft-lbs of engine torque. does this make it more clear?
-acceleration of the car is a dependent on the torque applied at the wheels
which you could also represent as power/velocity but these are just dependent variables
>>>>>>>yes, this is true. In fact, when looking at drag values and how it effects acceleration at speed, its easier to deal with Force x Speed values.
this has never bee questioned. example, at 100mph, if i have a drag of 20lbs on a wing, and 24" diameter tires, the force at the axles is 20ft/lbs. knowing im in 3rd gear with 5:1 ratios, i can say this has the effect of reducing my engine torque value (as measured at the rear wheels )by 4ft-lbs.
torque at the wheels is what overcomes road friction and makes the car move
->>>>>>>yes. and this is determined by power. plain and simple.
-another misconception is that peak values of power/torque is what makes a car go faster..NO it's just an indication but the real measure is torque curves
>>>>>>>>>ahhh, now you are getting into the confusion area for most. read this carefully. rate of acceleration in any gear will be at max torque. However, the greatest rate of acceleration at ANY vehicle speed, will be at max HP, or as near max hp as possible. by maximizing HP (using close ratio gears for example) you maximize the torque produced at the rear tires as multiplied through the gear box at any vehicle speed. PLEASE let me know if you dont agree with this, as this is key to the understanding of the topic we are discussing.
-to be exact for everything identical it's how much torque is transmitted at every point in time during an acceleration of the wheels:
>>>>>>>>>>Yes, this torque at the rear wheel is at the rear tires, as multiplied through the gear box and HP determines this. however, yes, you can look at torque curves and multiply it out too.
at time ti acceleration(ti)=torque(ti)/inertia so to find out exactly what your speed is over a certain time you integrate the above equation over that time so the area under the torque curve as a function of time and not rpm is the real measure...you can get vehicle translational speed from the angular speed of the wheels above if there is no slip..otherwise it's more complicated
>>>>>>>>>yes, not in debate here. its not just rpm vs torque or rpm vs hp, it is the time spent at those rpms that determines the ft-lb seconds applied or HP-seconds applied. eitherway, this is right.
-yes available HP or torque at the crank will dictate how much torque or power you will have at the wheels but the power at the crank is generated through torque at the crank
>>>>>>>>>>back to the chicken and egg analogy. power or torque at the crank??? its power at the crank, and torque and rpms are a result of that power. If i back off the accelerator, what happens? less power is produced, and this creates less torque at any rpm. Hey, we are talking semantics here. I already gave the potential energy analogy above. torque is an essential component of HP.
-please review my explanation because you are confusing vehicle speed with rpm
>>>>>>>>i have never done this. in fact, i have often put in ( ) torque and then force or RPM or Speed. Ive even clarified by using the term, "vehicle speed" to avoid confusion.
-"comparing two totally different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed." how is this possible?
>>>>>>>>>I'M glad you asked. its really QUITE simple. hp determines torque at the rear wheels. just do a simple test. take 500hp. (both engines, but in the same car with same spaced gear boxes). one engine has 250ft-lbs of peak torque, the other 500ft-lbs of peak torque do all the math you have quoted and tell me the torque at the rear wheels at ANY vehicle speed.
you will find that both will have the same torque at the rear tires ,as mulltiplied through the gear box.
- "acceleration is ...inversely proportional to speed" that's not true as a fact for any moving object..it's only observed at higher speeds due to drag and that's an extra "negative" force/torque acting on the system at higher speeds that causes the acceleration to reduce because acceleration= force/mass and the net positive force is less due to drag..if we ignore drag then the statement would be wrong!
>>>>>>>>Man, you are stubbon! Its an Identity, it cannot be questioned, but it can be questioned to be understood! You are also leaving out a main part of the equation. acceleration = power/(mass x velocity) if you are at 0 speed, then there is no power being applied, so there is no acceleration. then you start going into "drag", "only at higher speeds, "extra negative force"
You are way off base there. We dont care about any of that as it doesnt matter, due to all of our discussion being focused around NET Force. If you are at 100mph and you have a 10hp vehicle and you stop accelerating, its do to the fact that you have no more net force avalable, and the power needed to overcome the f *s is matched. you dont need to get this deep . the concepts are REAL basic here. you are making it complicated.
for instance with no drag and a constant torque (over some time) then acceleration is constant and velocity is just the integral i.e. it's increasing
>>>>>>>>>yes, acceleration is the 1st derivative of speed. so?
do you have a constant torque with an engine in a car? (trick question) a flat torque curve, does this give you constant torque? well, because i have said that with constant HP that acceleration is inversely proportional to speed, you can see that acceleration goes down with speed. you are talking about a rocket. it will have an acceleration that goes up with speed, because thrust is constant (force) . with an engine, when racing it, you are in the decending torque range of the curve anyway when maximizing acceleration.
If you want, we can go into the jet vs the car race. constant force, constant acceleration vs constant power, inversely proportional accelertion to speed, decending net forces.
-so according to the "derived" formula you are using at 0 speed we have infinite acceleration? the reason for this strange outcome is that the "real" newtonian law uses torques/forces and at nonzero net torque or force we can then have motion and then we can define speed and power..only then we can plug in torque=power/speed instead in newton's law if we wish
>>>>>no, see above where i address the "0" speed thought. at 0 speed you would have the greatest potential acccleration when you applied the power and the force was generated. The force would be depedant on power. if you can find a way to apply all the power to the wheels at 0 speed, then yes you would have infinte acceleration
you are forgeting about the net force. what ever force can be applied to the wheels, will determine the acceleration. power will dicate the amount of force that can be applied. And by the way, this also is true for A=F/m. acceleration is proportioal to net force. so, you question and comment have to apply to this as well If you find a way to apply the full power at 0 speed, or apply a force at 0 speed, let me know. (remember, before you comment, its NET FORCE!)
I was trying not to get too technical or lengthy before but I hope this clarifies things for you
- thermal energy in combustion is converted into torque at the crank( mechncial energy)
-this torque causes motion (rotation) at the crank and thus power at the crank
-it's not a matter of formulas and that you can get this from that ..
>>>>>HUhh? yes, energy is converted from heat to the expansion of the gasses in the form of a net force on a piston. Think of this as the power applied. energy, killowatts-hours/min/seconds, all determine the rate of doing work. Work, as we know is a force x a distance. how fast we do this work is power. with the equation, we can determine whether the piston will accelerate at a fast or slow rate depending on what it is attached to . (long stroke rod, etc)
the force that comes out of this application of power will be determined by the mechanical factors. you can think of torque or force as starting the event, or causing the motion, Becasue it is. But, it is one component of the larger picture. The amount of force will be determined by the power potential and the mechanical configurations.
Here is an anaology. dropping a rock on to one side of a teeter-totter with a projectile on the other side. at 10ft, the rock has potential energy. (power potential ) . The hight and weight of this rock determines the forces we are going to generate on the projectile. (our cars) We can drop it on a long levered teeter-totter, or a short levered one. the long levered one will create more force but over a shorter distance. The short lever will create lower forces, but over a greater distance. the projectile will be fired at the same speed. the power will be the same.
-the relevant acceleration here is an angular acceleration at the crank NOT OF THE CAR. and yes you can still apply the rotational version of "Newton's law" here.
>>>>>>>>> The entire discussion is whether a high torque engine CREATES greater CAR acceleration over cars with lower torque engines but rated at the same HP. keep the comparisons and analysis on track. you are making it confusing. I have never debated that the equations couldnt be applied to the engine components.
-then you can use a scaling ratio (e.g 0.9 for 10% losses) to transmit this torque to wheel torque and the same for crank to wheel power. so torque is not meaningless
>>>>>>>>>>>>>You are getting on thin ice here. Remember i mentioned to be clear of the differences of torque at the engine, and torque at the rear wheels. WE are not talking about engine torque as measured at the rear wheels, we are talking about the multiplied torques at the rear wheels THROUGH the gear ratios. (i.e. 500ft-lbs of peak engine torque through a 5:1 3rd gear will be 2500ft-lbs of torque at the rear wheels.) at this point we could care less about gear box efficiencies and other losses (thats another discussion as well). that 2500ft-lbs of torque could be developed by two 500hp engines, one with 100ft-lbs of engine torque and the the other with 500ft-lbs of engine torque. does this make it more clear?
-acceleration of the car is a dependent on the torque applied at the wheels
which you could also represent as power/velocity but these are just dependent variables
>>>>>>>yes, this is true. In fact, when looking at drag values and how it effects acceleration at speed, its easier to deal with Force x Speed values.
this has never bee questioned. example, at 100mph, if i have a drag of 20lbs on a wing, and 24" diameter tires, the force at the axles is 20ft/lbs. knowing im in 3rd gear with 5:1 ratios, i can say this has the effect of reducing my engine torque value (as measured at the rear wheels )by 4ft-lbs.
torque at the wheels is what overcomes road friction and makes the car move
->>>>>>>yes. and this is determined by power. plain and simple.
-another misconception is that peak values of power/torque is what makes a car go faster..NO it's just an indication but the real measure is torque curves
>>>>>>>>>ahhh, now you are getting into the confusion area for most. read this carefully. rate of acceleration in any gear will be at max torque. However, the greatest rate of acceleration at ANY vehicle speed, will be at max HP, or as near max hp as possible. by maximizing HP (using close ratio gears for example) you maximize the torque produced at the rear tires as multiplied through the gear box at any vehicle speed. PLEASE let me know if you dont agree with this, as this is key to the understanding of the topic we are discussing.
-to be exact for everything identical it's how much torque is transmitted at every point in time during an acceleration of the wheels:
>>>>>>>>>>Yes, this torque at the rear wheel is at the rear tires, as multiplied through the gear box and HP determines this. however, yes, you can look at torque curves and multiply it out too.
at time ti acceleration(ti)=torque(ti)/inertia so to find out exactly what your speed is over a certain time you integrate the above equation over that time so the area under the torque curve as a function of time and not rpm is the real measure...you can get vehicle translational speed from the angular speed of the wheels above if there is no slip..otherwise it's more complicated
>>>>>>>>>yes, not in debate here. its not just rpm vs torque or rpm vs hp, it is the time spent at those rpms that determines the ft-lb seconds applied or HP-seconds applied. eitherway, this is right.
-yes available HP or torque at the crank will dictate how much torque or power you will have at the wheels but the power at the crank is generated through torque at the crank
>>>>>>>>>>back to the chicken and egg analogy. power or torque at the crank??? its power at the crank, and torque and rpms are a result of that power. If i back off the accelerator, what happens? less power is produced, and this creates less torque at any rpm. Hey, we are talking semantics here. I already gave the potential energy analogy above. torque is an essential component of HP.
-please review my explanation because you are confusing vehicle speed with rpm
>>>>>>>>i have never done this. in fact, i have often put in ( ) torque and then force or RPM or Speed. Ive even clarified by using the term, "vehicle speed" to avoid confusion.
-"comparing two totally different engines, with different peak torque values, but equivilant HPs), you will have two cars producing the same forces at the rear wheels at any vehicle speed." how is this possible?
>>>>>>>>>I'M glad you asked. its really QUITE simple. hp determines torque at the rear wheels. just do a simple test. take 500hp. (both engines, but in the same car with same spaced gear boxes). one engine has 250ft-lbs of peak torque, the other 500ft-lbs of peak torque do all the math you have quoted and tell me the torque at the rear wheels at ANY vehicle speed.
you will find that both will have the same torque at the rear tires ,as mulltiplied through the gear box.
- "acceleration is ...inversely proportional to speed" that's not true as a fact for any moving object..it's only observed at higher speeds due to drag and that's an extra "negative" force/torque acting on the system at higher speeds that causes the acceleration to reduce because acceleration= force/mass and the net positive force is less due to drag..if we ignore drag then the statement would be wrong!
>>>>>>>>Man, you are stubbon! Its an Identity, it cannot be questioned, but it can be questioned to be understood! You are also leaving out a main part of the equation. acceleration = power/(mass x velocity) if you are at 0 speed, then there is no power being applied, so there is no acceleration. then you start going into "drag", "only at higher speeds, "extra negative force"
You are way off base there. We dont care about any of that as it doesnt matter, due to all of our discussion being focused around NET Force. If you are at 100mph and you have a 10hp vehicle and you stop accelerating, its do to the fact that you have no more net force avalable, and the power needed to overcome the f *s is matched. you dont need to get this deep . the concepts are REAL basic here. you are making it complicated.
for instance with no drag and a constant torque (over some time) then acceleration is constant and velocity is just the integral i.e. it's increasing
>>>>>>>>>yes, acceleration is the 1st derivative of speed. so?
do you have a constant torque with an engine in a car? (trick question) a flat torque curve, does this give you constant torque? well, because i have said that with constant HP that acceleration is inversely proportional to speed, you can see that acceleration goes down with speed. you are talking about a rocket. it will have an acceleration that goes up with speed, because thrust is constant (force) . with an engine, when racing it, you are in the decending torque range of the curve anyway when maximizing acceleration.
If you want, we can go into the jet vs the car race. constant force, constant acceleration vs constant power, inversely proportional accelertion to speed, decending net forces.
-so according to the "derived" formula you are using at 0 speed we have infinite acceleration? the reason for this strange outcome is that the "real" newtonian law uses torques/forces and at nonzero net torque or force we can then have motion and then we can define speed and power..only then we can plug in torque=power/speed instead in newton's law if we wish
>>>>>no, see above where i address the "0" speed thought. at 0 speed you would have the greatest potential acccleration when you applied the power and the force was generated. The force would be depedant on power. if you can find a way to apply all the power to the wheels at 0 speed, then yes you would have infinte acceleration
you are forgeting about the net force. what ever force can be applied to the wheels, will determine the acceleration. power will dicate the amount of force that can be applied. And by the way, this also is true for A=F/m. acceleration is proportioal to net force. so, you question and comment have to apply to this as well If you find a way to apply the full power at 0 speed, or apply a force at 0 speed, let me know. (remember, before you comment, its NET FORCE!)I was trying not to get too technical or lengthy before but I hope this clarifies things for you
Last edited by mark kibort; 08-01-2007 at 03:45 PM.
08-01-2007, 02:44 PM
#55
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Originally Posted by mark kibort
krC2S,
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute.
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute.
so when you first apply power (you suggested that power is what accelerate an object/vehichle not force/torque) to the wheels while the vehicle is stationary (velocity is zero) the acceleration is infinite
i would love to experience that.
08-01-2007, 02:57 PM
#56
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Originally Posted by murphyslaw1978
now the only question still remaining is "how to extend the torque curve of a diesel engine farther to the right."
in a diesel engine the fuel is injected in the cylinder just before the piston reaches top-dead-center position. this gives a very short time for the fuel to be delivered by the injector and mixed with the air to ensure proper combustion and burning of most of the fuel-air mixture. as engine speed increases the time avilable for fuel injection and mixing is shorter and gets closer to the time response of the injectors. at very high engine speeds you would end up not being able to deliver the right amount of fuel and have it properly mixed before self ignition occurs. hence you would not be able to keep the making enough torque at higher speeds.
in comparison a spark-ignition engine has a lot more time to inject the fuel to the air entering the intake valve before the compression cycle starts.
also, the frictional losses are higher in a diesel engine compare to a spark-ignition engine due to the higher cyclinder pressure. these losses increase at higher speeds.
08-01-2007, 03:25 PM
#57
Rennlist Member
As soon as you find away to apply a power, that includes a net force at the rear wheels at 0 velocity, you might experience it.
This is why both the A=F/m and A=power/(mass x velocity) can be used. It is the force at the rear wheels that will accelerate the mass (car). power will dictate this. Ive never suggested that power, not force , is what accelerates the mass. Ive said, very clearly, that power determines the force that will accelerate the car.
at 0 speed, you dump the 500hp to the wheels, do you ever have 500hp or the 10,000ft-lbs of torque at the wheels ? if you could, you would have infinite acceleration. but, then again, if you are accelerating, you are not stationary.
You are arguing regarding a newtonian identity. if you want to question it to understand it better, thats ok, ask away.
so, to better understand it. think of two cars both with infinitely variable gear boxes. lets continue to use the 12,000rpm 250ft-lbs but 500hp engine vs the 6000rpm, 500ft-lb 500hp engine. Both rev the engines up to max HP rpm and dump the clutch. assuming NO wheel spin, both will have the exact same torque at the rear wheels, meaning that both will accelerate at the same rate.
If you truely have an infintely variable gear box ( kind of impossible), the moment the clutch is dropped, you would have trememdous acceleration, however, due to Newtonian laws, it would fall porportionate with velocity.
you could also look at falling rock. At a height, it has a potential energy, drop it, and it starts to gather momentum and store kinetic energy. the rate of change of kinetic energy is power! as it accelerates it is gaining power to the extent of the potential energy that it held at a certain hight above the ground.
It has a constant force (ie gravity) and it has constant acceleration. 32f/s/s.
acceleration = power/(mass x velocity)
it means acceleration is proportional to power and inversely proportional to velocity. but, you have to look at the power applied at the wheels. it means grip cant be a variable. This the part that you and KrC2S are missing
Mk
This is why both the A=F/m and A=power/(mass x velocity) can be used. It is the force at the rear wheels that will accelerate the mass (car). power will dictate this. Ive never suggested that power, not force , is what accelerates the mass. Ive said, very clearly, that power determines the force that will accelerate the car.
at 0 speed, you dump the 500hp to the wheels, do you ever have 500hp or the 10,000ft-lbs of torque at the wheels ? if you could, you would have infinite acceleration. but, then again, if you are accelerating, you are not stationary.
You are arguing regarding a newtonian identity. if you want to question it to understand it better, thats ok, ask away.
so, to better understand it. think of two cars both with infinitely variable gear boxes. lets continue to use the 12,000rpm 250ft-lbs but 500hp engine vs the 6000rpm, 500ft-lb 500hp engine. Both rev the engines up to max HP rpm and dump the clutch. assuming NO wheel spin, both will have the exact same torque at the rear wheels, meaning that both will accelerate at the same rate.
If you truely have an infintely variable gear box ( kind of impossible), the moment the clutch is dropped, you would have trememdous acceleration, however, due to Newtonian laws, it would fall porportionate with velocity.
you could also look at falling rock. At a height, it has a potential energy, drop it, and it starts to gather momentum and store kinetic energy. the rate of change of kinetic energy is power! as it accelerates it is gaining power to the extent of the potential energy that it held at a certain hight above the ground.
It has a constant force (ie gravity) and it has constant acceleration. 32f/s/s.
acceleration = power/(mass x velocity)
it means acceleration is proportional to power and inversely proportional to velocity. but, you have to look at the power applied at the wheels. it means grip cant be a variable. This the part that you and KrC2S are missing
Mk
Originally Posted by doc2s
so when you first apply power (you suggested that power is what accelerate an object/vehichle not force/torque) to the wheels while the vehicle is stationary (velocity is zero) the acceleration is infinite i would love to experience that.
i would love to experience that.Last edited by mark kibort; 08-01-2007 at 03:41 PM.
08-01-2007, 03:34 PM
#58
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Thread Starter
Originally Posted by insite
it just occured to me that top fuel dragsters use compression ignition after about the 60ft mark. they actually burn up the spark plugs by then and the engine diesels all the way to the 1/4 mi trap. you actually have to cut fuel to shut it down. to extend the torque curve of a diesel farther to the right, we'd need a way to create enough heat or enough pressure with a short stroke engine to compression ignite the fuel. and that's if diesel burns fast enough to support the higher RPM's (it could; i simply don't know).
08-01-2007, 03:47 PM
#59
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Thread Starter
Originally Posted by doc2s
in a diesel engine the fuel is injected in the cylinder just before the piston reaches top-dead-center position. this gives a very short time for the fuel to be delivered by the injector and mixed with the air to ensure proper combustion and burning of most of the fuel-air mixture. as engine speed increases the time avilable for fuel injection and mixing is shorter and gets closer to the time response of the injectors. at very high engine speeds you would end up not being able to deliver the right amount of fuel and have it properly mixed before self ignition occurs. hence you would not be able to keep the making enough torque at higher speeds.
in comparison a spark-ignition engine has a lot more time to inject the fuel to the air entering the intake valve before the compression cycle starts.
also, the frictional losses are higher in a diesel engine compare to a spark-ignition engine due to the higher cyclinder pressure. these losses increase at higher speeds.
in comparison a spark-ignition engine has a lot more time to inject the fuel to the air entering the intake valve before the compression cycle starts.
also, the frictional losses are higher in a diesel engine compare to a spark-ignition engine due to the higher cyclinder pressure. these losses increase at higher speeds.
Ahh, yea, now we're getting somewhere. So now we're thinking in terms of limitations by the injection system. The interesting thing is that we're starting to see direct injection in performance engines and HCCI is coming around the corner. You can bet that engineers are already looking at ways to make injection faster/better/more precise. They may still be injecting fuel well before TDC, even in direct injection, and have plenty of time to do so. When you look at the limitation of HCCI, it's mostly due to the injection system, so they are trying to circumvent these limitations by adjusting things like:
Variable compression ratio
Variable induction temperature
Variable exhaust gas percentage
Variable valve actuation
If someone can invent an injection system that can inject fuel at just the right moment at 8000RPM, then we've got a serious solution on our hands, because we've been able to build our engine materials up enough to withstand the kind of power that sort of engine would generate.
08-01-2007, 04:09 PM
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i will only make two comments now
1- you are arguing newton law and physics and not how the torque curve of car will be or that of a rocket
so your claim that newton law as you re-write it suggests that acceleration is inversly proportional to velocity is just wrong!
if i am pushing a block of mass with constant force it has constant acceleration
but an increasing velocity..very simple
2-">I'M glad you asked. its really QUITE simple. hp determines torque at the rear wheels. just do a simple test. take 500hp. (both engines, but in the same car with same spaced gear boxes). one engine has 250ft-lbs of peak torque, the other 500ft-lbs of peak torque do all the math you have quoted and tell me the torque at the rear wheels at ANY vehicle speed.
you will find that both will have the same torque at the rear tires ,as mulltiplied through the gear box."
sorry that's wrong torque at the real wheels is just crank torque times an effective transmission ratio which you say is the same so with different crank torques we have different wheel torques
i just wanted to help you understand some of these things better but you clearly don't want to try
1- you are arguing newton law and physics and not how the torque curve of car will be or that of a rocket
so your claim that newton law as you re-write it suggests that acceleration is inversly proportional to velocity is just wrong!
if i am pushing a block of mass with constant force it has constant acceleration
but an increasing velocity..very simple
2-">I'M glad you asked. its really QUITE simple. hp determines torque at the rear wheels. just do a simple test. take 500hp. (both engines, but in the same car with same spaced gear boxes). one engine has 250ft-lbs of peak torque, the other 500ft-lbs of peak torque do all the math you have quoted and tell me the torque at the rear wheels at ANY vehicle speed.
you will find that both will have the same torque at the rear tires ,as mulltiplied through the gear box."
sorry that's wrong torque at the real wheels is just crank torque times an effective transmission ratio which you say is the same so with different crank torques we have different wheel torques
i just wanted to help you understand some of these things better but you clearly don't want to try
Originally Posted by mark kibort
krC2S,
This is going to take longer than i thought.
First of all, you need to be more specific. there is no "confusion" on my part, and if there is , i surely would like to know about it. second, I think you are missing the main point here by several of you responses below. You start to show some knowledge, and then you kill it with a couple of the comments below. Now, basically you are right about a few things. The overall message here, you seem to be missing. let me try and insert my comments below and you can feel free to disagree. BUT, state why. Ill try to make this brief.
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute. If you dont agree, you should possibly look at why, or ask some questions. You do make a note of this below and i will try to address it as simply as possible.
Here we go:
You are not getting too technical. what you are doing is missing the point. it all comes back to your single question above. you wanted to know about two different engines that make the same HP but have different peak torque values, maybe as far as 2x from each other, making the same forces at the rear tires at any vehicle speed. you ask how is this possible. I tell you and i hope you understand. This is tough doing a seminar over the web. over a beer with a white board or napkin, this would take 10 mins!!!
This is going to take longer than i thought.
First of all, you need to be more specific. there is no "confusion" on my part, and if there is , i surely would like to know about it. second, I think you are missing the main point here by several of you responses below. You start to show some knowledge, and then you kill it with a couple of the comments below. Now, basically you are right about a few things. The overall message here, you seem to be missing. let me try and insert my comments below and you can feel free to disagree. BUT, state why. Ill try to make this brief.
Let me start with Acceleration = Power/(mass x velocity)
This means for a given mass and at any vehicle speed (not rpm), acceleration will be proportional to power, and inversely proportional to vehicle speed.
This is a newtonian identy that is absolute. If you dont agree, you should possibly look at why, or ask some questions. You do make a note of this below and i will try to address it as simply as possible.
Here we go:
You are not getting too technical. what you are doing is missing the point. it all comes back to your single question above. you wanted to know about two different engines that make the same HP but have different peak torque values, maybe as far as 2x from each other, making the same forces at the rear tires at any vehicle speed. you ask how is this possible. I tell you and i hope you understand. This is tough doing a seminar over the web. over a beer with a white board or napkin, this would take 10 mins!!!