Race cars - Torque vs Hp - The Undiscovered Country (for many)

[zanick]

Im obsessed with the "shape of the HP curve" because it seems its the only thing that determines rear wheel torque (after the gear ratio reductions). I know you could get the same info through the torque curve and multiplying it through the gears manually for any speed, but it seems that HP incorporates more information and is easier to work with. Plus, when you compare identical HP engines but with different shaped hp curve, the lower torque, higher rpm, could have more torque at the rear wheels through the gear box at any vehicle speed.

What governs the shape of the HP curve is the engines ability to burn air and fuel at any rpm.

I think I answered the dyno question above, but it measures rate of change of kinetic energy, by looking at drum speed changes and knowing mass and size of the drum. (we are talking about wheel dynos here.)

mk

It acutally seems we're getting nowhere. You seem obsessed about the shape of the HP curve. What governs the shape of the hp curve.

What does a dyno measure?

 

[xxChrisxx]

ok i'll give you the answer. a dyno measures torque

 

[zanick]

techncally, it measures speed change of the drums. by knowign the size and weight of the drum, you end up with torque (thats what you were getting at, i know) But, its torque at the rear wheel, and of course, rate of change of kinetic energy. Beause you have to know speed, speed change, and drum inertia, by definition, you have Hp. You can't even attempt to figure out engine torque unless you someone tells you the rpm running. A spark signal, allows for the computer to then work backward and provide engine torque.

mk

We're going to have to deal with this one question at a time I'm having trouble following many mish mashed questions at one. Sorry to be a bit pedantic about this, but following your reasoning is very difficult.

You haven't clearly stated what the dyno measures. Does it measure the torque or horsepower that the engine puts out?

 

[xxChrisxx]

oh my god.

no no no.

You run the dyno up through the rev range. It MEASURES the torque output at each rpm, it then CALCULATES the horsepower reading.

Edit I apologise if that seemed a little bit brash, I am just getting a wee bit sleepy.

 

[zanick]

How is this possible if the dyno is not getting a rpm signal. without an rpm signal, it has no way of knowing what the engine rpm or torque is, correct? the output is HP and MPH.

No problem, I just want to get the right information.

Im sure the computer can caluate the HP just as easily as it can calculate torque. It has to calculate torque, because all it is doing is measuring the rate of speed change of the rollers. by knowign the speed, and the speed change, it has rear wheel torque for any speed change period. But, isn't it just as easy for it to out put HP first? rate of change of kinetic energy is power too, right??

mk

oh my god.

no no no.

You run the dyno up through the rev range. It MEASURES the torque output at each rpm, it then CALCULATES the horsepower reading.

Edit I apologise if that seemed a little bit brash, I am just getting a wee bit sleepy.

 

[xxChrisxx]

oh dear.

it can't output horsepower first. as that is the calcualted value.
the variable measured by the acutal apparatus is torque.

You are totally 100% incorrect in what you have said in the previous post.

http://auto.howstuffworks.com/horsepower1.htm
http://en.wikipedia.org/wiki/Dynamometer

I understand you are proabably more used to the 'rolling road' but this is giving you a skewed understanding of what is going on.

 

[zanick]

Isn't it all calculated? By the way, I am talking about a dynojet 248e, which is a chassi dyno that has giant rollers, known diameter and weight (like 2000lbs or something). Not water brake, electromagnetic brake, or engine dyno.

Help me understand. You have drums with a known size and mass. they are being accelerated. Its not lifting anything, and there is no drag working against the drivng force.
Worst case, you have acceleration by the difference and time between two known rotational speeds, gives you a rate of change of kinetic energy. Isnt that by definition, Power?
Sure you can find torque too, but knowing the rate of speed change at some rotatioal speed, (dv/dt) it gives torque on the drum, correct? Still need to converte to engine torque, but ONLY if you have the engine rpm. again, that has to be calculated as well. Hp in that case could be found as easy as drum torque, right?

Im 100% wrong? wow, I really don't understand. Help me understand.

See if you can be as detailed in your response as possible.

Of course we are getting a little off the track here. But that's ok.


Thanks,

Mk

oh dear.

it can't output horsepower first. as that is the calcualted value.
the variable measured by the acutal apparatus is torque.

You are totally 100% incorrect in what you have said in the previous post.

http://auto.howstuffworks.com/horsepower1.htm
http://en.wikipedia.org/wiki/Dynamometer

 

[xxChrisxx]

But this thinking in horsepower is giving you a false sense of what is going on. The force on the rollers (that contributes to their acceleration) can be defined as a wheel torque.

As Torque = Force * distance.

So in this case the inertia and mass of the drum is known. The change in the speed of the drum in the amount of time is the acceleration. from this the force is known. From the size of the wheel the torque at the wheel is known.

So you see the torque is really still the measured 'thing'.

After the run up, the clutch is dipped and the run down let's you know the losses through the transmission. This will give you the torque at the flywheel. The revs are found from a tachometer iirc.

EDIT: I'll apologive now, as I know I am making a pigs ear out of trying to explain this.

Well you weren't 100% wrong, sorry for saying that. What i meant was, your reading of what was going on was wrong. Yours knowledge of how the rolling road works is spot on.

 

[zanick]

Well, at least I have the effects on the road correct. :) thanks for that. that's most important, but just want to get the terminology right.

I think its coming into a philosophical debate now, which came first, the force or the power that created it. I think I understand, that even if the power creates the force, its the force that is causing the acceleration, even though the power of the act is known at the same time. (how is that for a compromise?).

So, how do we address the hp curve and the progressive time spent at different rpm as the engine accelerates over a given vehicle speed range when comparing different shaped HP curves of two engines, in the same car, with the same max HP?

Mark

But this thinking in horsepower is giving you a false sense of what is going on. The force on the rollers (that contributes to their acceleration) can be defined as a wheel torque.

As Torque = Force * distance.

So in this case the inertia and mass of the drum is known. The change in the speed of the drum in the amount of time is the acceleration. from this the force is known. From the size of the wheel the torque at the wheel is known.

So you see the torque is really still the measured 'thing'.

After the run up, the clutch is dipped and the run down let's you know the losses through the transmission. This will give you the torque at the flywheel. The revs are found from a tachometer iirc.

Well you weren't 100% wrong, sorry for saying that. What i meant was, your reading of what was going on was wrong. Yours knowledge of how the rolling road works is spot on.

 

[xxChrisxx]

Its the force (engine) that creates the power (engine) :P hehe.

Torque - Ability of engine to do work
Power - Rate at which said work can be done.

Well the horsepower curve is a function of the torque. I believe the angle you are coming from is that: two cars traveling a constant speed along the road (the one with less torque requires it to be in a higher rpm band) both floor it. The answer would very corrctly be that they would both probably accelerate at the same rate.

However, the other car would require different gearing to do this. This was why I was harping on about it earlier.

This didnt come across in your original question. As two identical engines but one with more torque to me means identical gearing and ataring form the same rpm. In which case the engine with the larger torque would win out. (This was why your arguements were quite confusing)

The practical upshot is that, high torque engines accelerate very well in the bottom end of the rev range. but higher revvers accelarate better in the upper rev range. In very slow corners (such as a hairpin) it's quite likely that a car with a narrow power band is likely to fall out of the optimal range. This is very true with sometihng like a formula 1 car where the power band is about 1500RPM.

This led me to my point about the slow corners thing. A huge V8 would out accelerate a smaller high revving engine initally, but then the high revver would have a relative increase in acceleration as it gets down the straight.

This then leads into useful power band and that Jeff said about lots of narrow gears are needed for the high revver as its peaky but a big lazy V8 would need less to get the same job done.

EDIT: I think we've just argued ourselves to a similar conclusion! Damn! You thinking was pretty much ok, but like you suspected its your terminoligy that was incorrect (and what got me all :S). I'd like to say I am sorry for being a bit curt with you earlier.

One thing you always have to keep in mind though and this is the most important thing. Its NOT the power that's accelerating you, its still the torque produced at the engine that is determining the rear wheel force and therefore acceleration. So don't think of it as 'RWHP determinign torque, its torque determining RWHP!

 

[Gnosis]

Well, at least I have the effects on the road correct. :) thanks for that. that's most important, but just want to get the terminology right.

I think its coming into a philosophical debate now, which came first, the force or the power that created it. I think I understand, that even if the power creates the force, its the force that is causing the acceleration, even though the power of the act is known at the same time. (how is that for a compromise?).

So, how do we address the hp curve and the progressive time spent at different rpm as the engine accelerates over a given vehicle speed range when comparing different shaped HP curves of two engines, in the same car, with the same max HP?

Mark

Zanick, the matter hasn’t been reduced to a philosophical debate in the least. Your supposition concerning horsepower and torque is quite simply backwards. Torque isn’t a factor of horsepower rather; horsepower is a derivative of torque. You must first produce torque if you are to “calculate” a “horsepower rating” for the engine.

Horsepower is an intangible quantity that cannot be measured directly, but more importantly, horsepower isn’t a “force”. Horsepower is simply a rating of how much work can be performed over time and distance. It is “calculated” per the torque measured since torque CAN be measured directly in ft lbs. For instance, when an engine is stated as having 200 HP, it is merely a rating of how much work can be performed at a given RPM per the engine’s available torque.

The dyno measures torque by applying a known braking force to the dyno’s rotating drum. When the vehicle’s drive wheels accelerate the dyno’s drum under this braking force up to a given RPM in a given time, it reveals the torque being applied by the vehicle’s drive wheels, as it would require a given torque to have accomplished this rate of drum acceleration in the time recorded.

Force produced in a Combustion Engine

Consider where the combustion engine actually produces its force. It is per the optimized A/F ratio ignited per each power-stroke, as the piston and its connecting rod are forced downward by the expanding hot gases in the cylinder with the other end of the connecting rod applying its force to the crankshaft in a manner that provides leverage to rotate the crankshaft and produce crankshaft torque. So, torque is the accelerative component, not horsepower.

You need to get horsepower out of your head and acknowledge the actual source of the vehicle’s acceleration; that being torque. Horsepower is only a rating of how much work can potentially be performed at a given RPM per the available torque.

Horsepower is a “rating” of potential work that can be accomplished, but it is NOT an actual force

For instance, my workouts had me moving approx 30,000 pounds of free weights a distance of 3 feet in an hour’s time. This can be given a horsepower rating HOWEVER, that horsepower rating does not imply an actual force used. It merely specifies a total of 30,000 pounds was moved 3 feet over a span of one hour (a given weight moved a given distance over time). In fact, some weight movement was several hundred pounds per rep while things like curls were more around the 95 to 115 pounds. Again, horsepower doesn’t specify a given force rather; it implies a given amount of work that can potentially be accomplished.

Hopefully this will provide the insight you need to resolve your confusion associated with horsepower. Unfortunately, those who tend to adore horsepower and have long confused its meaning have a most difficult time relinquishing their love for its mighty-sounding namesake, when it’s actually “torque” that should be held in highest accord.

I demonstrated the example below back in my mini-bike building days. It demonstrates the significance of an engine’s torque.

Firsthand Example:

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill. When I hit the hill, the engine was at its upper rev range therefore producing its 1.5 HP, so why could it achieve 35 MPH on flat ground, but not make it to the top of the steep hill?

Answer: The engine lacked sufficient torque to accomplish the climbing of the steep incline. The incline imposed additional loading on the crankshaft, so the engine’s crankshaft steadily lost RPM. As its crankshaft slowed, less A/F mixtures were ignited per second as a result, so less over all energy was provided per second, which further slowed my ascent. It must be realized that “IF” the engine’s torque were sufficient to climb the steep hill in the first place, the engine would have never slowed down. A 2.5 HP combustion engine remedied the issue, as it produced greater torque and even had enough extra torque to start out from a dead stop up the steep hill. From a dead stop (engine just above idle), it couldn’t be the extra HP yielding the more than adequate acceleration up the hill; it was a direct result of the extra torque provided by the 2.5 HP engine.

Point: A given task requires a given minimal torque to accomplish the task. If less torque is produced, then it makes no difference what the “horsepower rating” is, the task will not be achievable.

 

[xxChrisxx]

Thanks Gnosis, that's acutally a pretty damn good explination. You did far better in one post than I manages in lots!

People may say that 'all that matters is power'. This is to an extent true, but you've got to go back to how that 'more power' is produced. You either spin in more quickly, or generate more torque. So to say that torque is irrelevant is completely erroneous (I know you didnt say this, but I've seen it as an argument on quite a few forums). An engine that generates no torque generates no power.

The acceleration can be calculated both ways, by using the power method or torque method. I'm a massive proponent of the torque method because torque is the acutal measurable driving force.

 

[rcgldr]

integrating hp versus rpm curves

This make more sense if the calculated area is divided by the range of rpm, which calculates the average hp across a range of rpm. Even this information may not provide a good estimate of acceleration if one curve is very peaky compared to another. Torque versus rpm curves are easier to read because it eliminates the rpm multiplier effect which adds a linear slope to the power versus rpm curve.

torque ... power

A torque can be applied to a non moving object, such a wrench applying a torque to a bolt; in this case zero work is being done. Power defines a rate of work. In the case of an idealized CVT (continously variable transmission), then as previously mentioned, acceleration = power / (mass x speed). (power = force x speed = mass x acceleration x speed).

Note, torque = force times radial distance, not distance moved by a tire.

chassis dynos

Not all chassis dynos are inertial. Some use a variable load via a mechanical or electrical brake. The torque on the chassis drum equals the force from the driven tire contact patch times the radius of the drum. The angular speed of the chassis drum equals the driven tire contact patch speed divided by the radius of the drum. The dyno converts this information into contact patch force times contact patch speed, which equals contact patch power. The driven wheel torque can't be calculated from this unless the effective radius of the driven tires, or the engine rpm and overall gearing is included with the sampled data. The effective engine torque can be calculated given engine rpm data without requiring overall gearing data.

 

[zanick]

Ill start with you Chris, because your post is shorter. Maybe I didnt make myself clear in the beginning. when i mentioned same engine HP but different torque values. (one High rpm, and the other lower rpm) the gearing had to be appropriate for the same speeds in each of the cars respective gears. So, yes, not "same gear ratios" but proportionaly the same gear ratios. Does that make it more clear. If you come from the torque side of things, only looking at engine torque, you could be way off, as a lower torque engine, could make more power in certain sections of its HP curve and thus, yeild better acceleration to to MORE avaialbe torque at the rear wheels as it is seen through the gear box. correct?

You are also right, I've never said torque doesn't matter, I only said that the value is not an indication of better or worse acceleration based on a comparison with two equal peak HP engines.

So, going back to the original question, can you look at hp and say in the right venacular, that HP dictates the torque at the rear wheels at any vehicle speed? (remember, not engine torque, multiplied rear wheel torque and ultimately force to the ground).

Also, looking at area under the torque curves is a popular concept, but that can't be right, as you might have the most area under the curve shown by integrating the curve, from 1000 to 2000rpm. certainly, the mulplied force at the wheels wouldn't reflect optimum use of power at that engine rpm range. You would have to look at gear ratio factors as well. Also, area under the HP curve is better, because it would point to max accelerative potential at any vehicle speed, but it doesn't address the progressive time element that is seen as a car accelerates through the higher rpm. In otherwords, two HP curves could share the same HP peak, but one might have more area in the lower hp range, and the other in the higher hp range. the one with area in the higher HP range will have more accelerative potential. Thats why I was asking about HP-seconds or watt-seconds as a way to explain it or quantify accelerative potential.

Then, going back to Acceleration=power/momentium, doesn't this say that acceleration, and thus rear wheel Force will be proportional to power at any given vehicle speed?

Thoughts on this line of thinking or terminology.

thanks

Mark

Thanks Gnosis, that's acutally a pretty damn good explination. You did far better in one post than I manages in lots!

People may say that 'all that matters is power'. This is to an extent true, but you've got to go back to how that 'more power' is produced. You either spin in more quickly, or generate more torque. So to say that torque is irrelevant is completely erroneous (I know you didnt say this, but I've seen it as an argument on quite a few forums). An engine that generates no torque generates no power.

The acceleration can be calculated both ways, by using the power method or torque method. I'm a massive proponent of the torque method because torque is the acutal measurable driving force.

 

[zanick]

Chris,

But, in those two examples of two equal HP engines, one high rpm, lower torque, and the other high torque, lower rpm. If the lower torque engine had a broader HP curve, (and this is definitely possible as I have seen), the lower torque engine can produce more rear wheel torque at the same vehicle speed, because it is making more HP at that same speed. (again, gear ratios kept proportional to keep things even for the comparison. in otherwords, both cars have 1st gear to 50mph, 2nd to 80mph, 3rd to 120mph for example)

So, we know how engine torque creates rear wheel HP and rear wheel torqe as muliplied through the gear box, BUT, if there is more available hp at any vehicle speed with another engine, it will produce more rear wheel forces, correct?

I agree that the force is what accelerates the cars through the torque at the rear tires.
But, the engine torque is not an indication of what the total force will be when comparing two engines, unless you have the gear ratios or rpm handy. Hp values from an engine can determine what the rear wheel torque numbers will be at any vehicle speed, right?

mk

Its the force (engine) that creates the power (engine) :P hehe.

Torque - Ability of engine to do work
Power - Rate at which said work can be done.

One thing you always have to keep in mind though and this is the most important thing. Its NOT the power that's accelerating you, its still the torque produced at the engine that is determining the rear wheel force and therefore acceleration. So don't think of it as 'RWHP determinign torque, its torque determining RWHP!

 

[xxChrisxx]

Regarding your original question, yes it is more clear now. The answers I gave are still the same for everything else. But the fact that its in a higher RPM range and is well geared, both should accelerate the same as long as they ar eboth in their peak power band. The engine with more torque has a wider band, so will pull better from lower rpm. The revver will do all of its useful work at higher rpm.

You are still confused over torque and horsepower. You said I could be way off by looking at only that, this is not the case. As hp is calcualted directly from torque, to it really makes no difference if we use power or torque, the sums will be different but we'd ge tthe same answer.

Power = Torque * angular velocity.

I prefer using torque purely as that is the acutal real world driving mechinsm of the engine. The point you make that torque figures alone are pretty much useless is true, without knowing the engine speed they arent helpful. Using the torque value a trear wheel is a more useful figure from a design perspective, as the size of the wheel has no bearing on the torque, but it will affect RWHP reading and therefore acceleration. So say you calcualte your acceleration times with a 14 inch wheel on a dyno, if you are using RWHP you'd need to re run it as the figure would no longer be valid for the new wheel. By using the engine torque you eleiminate this issue. Although is technically not strictly correct you can say that RWHP determines the torque at the rear wheel, but again this makes it sound like power dictates torque when really its torque dictating power.

You are correct in that to use the engine torque, you'd need the gear ratio and rpm of the engine. This for practical purposes makes the RWHP reading from the dyno more convenient to use, however this also leads people into the false sense that the driving force is power and not torque.

By using Engine T * Gearing * %Losses = Wheel torque
and Using the RWHP

Its making more yes you are correct, its making more power so the forces a the rear wheel will be higher. This is only becuase the torque is being applied more rapidly.

Im going to have to dig out my books on integrating curves. I'll have a read up tonight.

So summerise: both ways will give you the same answer. Using the power and vehichle speed is more convenient to calcualte quickly. but using the engine torque is a more powerful tool for analysing engine performance.

 

[zanick]

Gnosis, Thanks for the response. The purpose of the questions were to look at engine hp curves (and torque curves) as they apply to vehicles. There are a lot of instances where broader HP curves from lower peak torque engines produce more rear wheel torque . I've been asking this board, how to describe this situation with proper, less confusing terminology. So far, we all know that with equal HP curve shapes, if the gearing is appropriate and proportional, two same cars will accelerate at exactly the same rate at any vehicle speed, regardless of the numerical "engine torque" value.

Below, I go into what causes the "F" in a=F/m. sure the force at the engine or the rear tires is what causes the acceleration, but isn't the power available, caused by the ignition of gasses, release of potential energy to kinetic energy, the cause for the force. If I have 746watt-seconds available, 1-hp second, can't i use 7,460watts for 1/10th of as second , creating 10hp, and lifting 5500lbs 1/10th of a foot, or 550lbs 1ft, in 1/10 of a second?

since work=energy, doesn't the the energy determine the quantity of work that can be done, and if the rate of work to be done is faster or slower that would be higher or lower HP.
Sure, the torque or force is what creates the acceleration. But to achieve the force, you need something to cause it. power, watts ,etc, come in many forms. Is gravity the force that causes the acceleration, or is the size of the planet that causes the gravity? :)
I consider the engine' abiltiy to process fuel (potential energy) as its power potential which generates torque . If it spins at 1,000,000rpm and produces 1ft/lb of torque, it isn't very torquie. but, the HP is huge. harness the power effectively with gears and the rear wheel torque will be determned by the engine power at any speed. I feel like I am talking in circles. so, maybe you can straighten me out. Thanks!


Let me insert my comments below with the >>>>>>>>>>

Zanick, the matter hasn’t been reduced to a philosophical debate in the least. Your supposition concerning horsepower and torque is quite simply backwards. Torque isn’t a factor of horsepower rather; horsepower is a derivative of torque. You must first produce torque if you are to “calculate” a “horsepower rating” for the engine.
>>>>>>>>>>>>>>>>>>while this is true, when looking at strictly engine torque (not the resultant force or torque at the driven tires, it might not be a direct indicaton of accelerative potential, unless you know engine speed. So, HP ratings and curves can be a better indication of acceleration potential. Yes, at the engine Hp is the rate that the force does work. Its all about rate of doing work, which is hp but its cause by the force at the rear tires, not the engine. (meaning the engines numerical torque value can be misleading as a comparitive factor)

Horsepower is an intangible quantity that cannot be measured directly, but more importantly, horsepower isn’t a “force”. Horsepower is simply a rating of how much work can be performed over time and distance. It is “calculated” per the torque measured since torque CAN be measured directly in ft lbs. For instance, when an engine is stated as having 200 HP, it is merely a rating of how much work can be performed at a given RPM per the engine’s available torque.

>>>>>> agreed, but isn't power also the rate of change of kinetic energy as well. so , the rate of change of speed of the drums can be calculated to show HP without knowing torque.
I know that the force is makeing the acceleration, but as far as what causes the force, isn't that caused by the energy source and its potential energy? (a 75amp/second 10v battery that produces 1hp-second or 750watt-seconds? Ill tie back into this below.)

The dyno measures torque by applying a known braking force to the dyno’s rotating drum. When the vehicle’s drive wheels accelerate the dyno’s drum under this braking force up to a given RPM in a given time, it reveals the torque being applied by the vehicle’s drive wheels, as it would require a given torque to have accomplished this rate of drum acceleration in the time recorded.
>>>>>>>>>>for a brake dyno this is true, but there are inertial dynos that are just drums weighing up to 3000lbs and are free spinning. they only measure the acceleration change of the drum and calculate everything from that.

Force produced in a Combustion Engine

Consider where the combustion engine actually produces its force. It is per the optimized A/F ratio ignited per each power-stroke, as the piston and its connecting rod are forced downward by the expanding hot gases in the cylinder with the other end of the connecting rod applying its force to the crankshaft in a manner that provides leverage to rotate the crankshaft and produce crankshaft torque. So, torque is the accelerative component, not horsepower.
>>>>>>>>>>>>>>but doesn't the rate of expanding gasse the cause for the force? without the energy, power/t available, you don't have any force? correct?

You need to get horsepower out of your head and acknowledge the actual source of the vehicle’s acceleration; that being torque. Horsepower is only a rating of how much work can potentially be performed at a given RPM per the available torque.
>>>>>>>>>>>Doesnt the equation Acceleration=power/momentium point to acceleration being determined by power, which creates the force? this is where i was suggesting it gets a little bit philosophical. :) I understand that the force creates the acceleration. a=F/m. I get that. But, if we are going to dump 1 oz of gas in an engine and get the energy out of it, we can get lots of rpm and a little torque or visa versa. in the end, the force we will generate at the driven wheels will be multiplied through a gear box so that we maximise power, not engine torque at the rear wheels and the resultant torque at the rear wheels will be proportional to the power. This torque can be made from low engine torque or high engine torque, based on the RPM range needed to create max power at any particular vehicle speed.



Horsepower is a “rating” of potential work that can be accomplished, but it is NOT an actual force

For instance, my workouts had me moving approx 30,000 pounds of free weights a distance of 3 feet in an hour’s time. This can be given a horsepower rating HOWEVER, that horsepower rating does not imply an actual force used. It merely specifies a total of 30,000 pounds was moved 3 feet over a span of one hour (a given weight moved a given distance over time).

>>>>>>>you did work during the hour. it can be rated in watt-seconds, KWh or Hp-seconds with are unit measures of work, not rates of doing work. rates of doing work wold be if you then lifted 550lbs in 1 second 1 foot. that would be using 1 hp. we could then calculate the acceleration caused by that force over that period of time.


In fact, some weight movement was several hundred pounds per rep while things like curls were more around the 95 to 115 pounds. Again, horsepower doesn’t specify a given force rather; it implies a given amount of work that can potentially be accomplished.
>>>>>>> but the rate of which the force does work, is HP. you bench press 550lbs ove 1 foot in 1 second, you needed 1hp to do it. If you use pulleys to do it slower and it takes 4 seconds to do this work, then the HP required was only 1/4 hp, but the force at the weight was still 550lbs being lifted over 1 ft.

Hopefully this will provide the insight you need to resolve your confusion associated with horsepower. Unfortunately, those who tend to adore horsepower and have long confused its meaning have a most difficult time relinquishing their love for its mighty-sounding namesake, when it’s actually “torque” that should be held in highest accord.
>>>>>>>>>>>>This is ironic as I've been fighting the torque folks for 10 years in the field of racing, and I am a very small minority. There are countless discussions on the topic, all feble arguments becuase in the end, torque is a part of HP. HP can determine torque at the rear wheels, even though torque at the engine multiplied through the gears does this too.
I understand that the force is what does work. or torque in the rotational plane. If we look at constant variable gear boxes, where will engines operate to maximize acceleration, peak engine torque or peak HP? right, peak engine HP! this will yeild the maximum amount of torque at the rear wheels at any vehicle speed, right?

I demonstrated the example below back in my mini-bike building days. It demonstrates the significance of an engine’s torque.

Firsthand Example:

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill. When I hit the hill, the engine was at its upper rev range therefore producing its 1.5 HP, so why could it achieve 35 MPH on flat ground, but not make it to the top of the steep hill?
>>>>>>>>>>>>>>>becuse the hp requried to climb that hill was not availble from the 1.5HP engine. HP is the rate of doing work. that amount of fuel and air was not enough to do the job of lifting you and the mini bike up the hill at that speed. Power=Forcex speed. so , you slowed, got out of the power band and could have come to a complete stop. however, with right gear, you could climb that hill at its maximum potential (P=Fv) which would be at max hp of 1.5hp. sure , you could gear it down even further and go slower producing more torque, but all that would do would be to accelerate to a slower speed faster and then top out at max rpm at a slower speed, thus now you back off the throttle and you continue up the hill at a slower speed, and lower HP setting and lower torque requirement.

Answer: The engine lacked sufficient torque to accomplish the climbing of the steep incline. The incline imposed additional loading on the crankshaft, so the engine’s crankshaft steadily lost RPM. As its crankshaft slowed, less A/F mixtures were ignited per second as a result, so less over all energy was provided per second, which further slowed my ascent. It must be realized that “IF” the engine’s torque were sufficient to climb the steep hill in the first place, the engine would have never slowed down. A 2.5 HP combustion engine remedied the issue,
<<<<<<<<<<<<yes, more HP required to accelerate or create the greater rate of work. I get this.
as it produced greater torque and even had enough extra torque to start out from a dead stop up the steep hill. From a dead stop (engine just above idle), it couldn’t be the extra HP yielding the more than adequate acceleration up the hill; it was a direct result of the extra torque provided by the 2.5 HP engine.
>>>>>>> the hp at just above idle would then have more Hp available comparitively than the other 1.5hp engine. yes, I get it, more rear wheel torque was developed then.

Point: A given task requires a given minimal torque to accomplish the task. If less torque is produced, then it makes no difference what the “horsepower rating” is, the task will not be achievable.

>>>>>>>>>I don't quite follow you. Now, If the the hp rating is high enough, the minimal torque can be acheieved. ? I don't understand the point.

 

[xxChrisxx]

Your responces to Gnosis' questions show that you've not changed your thought process about this one bit.

"I know that the force is makeing the acceleration, but as far as what causes the force, isn't that caused by the energy source and its potential energy? (a 75amp/second 10v battery that produces 1hp-second or 750watt-seconds? Ill tie back into this below.)"

the force at the contact patch is caused by the torque on the wheels NOT THE POWER. Power is a way of visualising what the torque is doing. This is why it is more of an abstract concept albeit one that people are more familiar with. Caused by the torque deliviered by the engine, from burning the fuel. so talking about heat into the engine can also be used, but the trype of work that the engine does is torque. and this is the crucial thing you must understand.

everything else is an application of that torque generated.

 

[zanick]

I totally agree, that rear wheel torque is equal to the engine torque though the ratios at any vehicle speed. If folks talked in terms of that, the world would be a much better place, but instead, folks talk in terms of engine torque as a comparitive value. Is not, an as you seem to agree, it depends on engine rpm and then gear ratios to be of any value.
area under the engine torque curve will be meaningless without gear reduction ratios mixed with rpm. area under the rear wheel torque curve, through the gear box is correct, but who has this data? using area under the HP curve or Watt-seconds, would be a much easier way to optimize vehicles acceleration over a operational speed range, right?

I have a dyno run (https://www.physicsforums.com/showpost.php?p=2128716&postcount=1) that shows a graphical representation of two engines, one with more torque than the the other, even though their HP's are equivilant. clearly, the lower torque engine has more accelerative potential. this is not common, but possible. I have actual dynos of cars I've raced that have a similar message, that sometimes, the lower torque engine can have a broader HP curve and produce more rear wheel torque at any vehicle speed. I just want to be able to explain why, correctly! thanks!

Mk

Regarding your original question, yes it is more clear now. The answers I gave are still the same for everything else. But the fact that its in a higher RPM range and is well geared, both should accelerate the same as long as they ar eboth in their peak power band. The engine with more torque has a wider band, so will pull better from lower rpm. The revver will do all of its useful work at higher rpm.

You are still confused over torque and horsepower. You said I could be way off by looking at only that, this is not the case. As hp is calcualted directly from torque, to it really makes no difference if we use power or torque, the sums will be different but we'd ge tthe same answer.

Power = Torque * angular velocity.

I prefer using torque purely as that is the acutal real world driving mechinsm of the engine. The point you make that torque figures alone are pretty much useless is true, without knowing the engine speed they arent helpful. Using the torque value a trear wheel is a more useful figure from a design perspective, as the size of the wheel has no bearing on the torque, but it will affect RWHP reading and therefore acceleration. So say you calcualte your acceleration times with a 14 inch wheel on a dyno, if you are using RWHP you'd need to re run it as the figure would no longer be valid for the new wheel. By using the engine torque you eleiminate this issue. Although is technically not strictly correct you can say that RWHP determines the torque at the rear wheel, but again this makes it sound like power dictates torque when really its torque dictating power.

You are correct in that to use the engine torque, you'd need the gear ratio and rpm of the engine. This for practical purposes makes the RWHP reading from the dyno more convenient to use, however this also leads people into the false sense that the driving force is power and not torque.

By using Engine T * Gearing * %Losses = Wheel torque
and Using the RWHP

Its making more yes you are correct, its making more power so the forces a the rear wheel will be higher. This is only becuase the torque is being applied more rapidly.

Im going to have to dig out my books on integrating curves. I'll have a read up tonight.

So summerise: both ways will give you the same answer. Using the power and vehichle speed is more convenient to calcualte quickly. but using the engine torque is a more powerful tool for analysing engine performance.

 

[xxChrisxx]

and by the way, if you were comparing two different engines. It'll be more correct to compare bmeps as they are dimensionless.

 

[zanick]

I get it, its getting to be less and less of a chicken and egg thing. force is what causes the acceleration. got it. Hp is a measure of of the rate the force does the work. HP is a vision into what the potential force is though right? or is it just a distilled version of what you need to really look at, an indicator when you don't have the time or the information to see the actual torque created at the rear tires.

Lets say you have a dyno, (inertia). you don't have sparksignal, you only have the car, and the gas pedal. what is the output? you get rear wheel torque, and Power, vs MPH as you floor it and hit the rev limiter, but you never know what the engine profile is of torque, correct? It could be 5000 or 10,000rpm redline.

I guess that is my point. for years I was muliplying gear ratios with the torque curves for every gear doing manual integration. Years later, i just started optimizing HP and accomplished the same thing.

Thanks for the comments. you brought me back to reality. awesome!

Mk

Your responces to Gnosis' questions show that you've not changed your thought process about this one bit.

"I know that the force is makeing the acceleration, but as far as what causes the force, isn't that caused by the energy source and its potential energy? (a 75amp/second 10v battery that produces 1hp-second or 750watt-seconds? Ill tie back into this below.)"

the force at the contact patch is caused by the torque on the wheels NOT THE POWER. Power is a way of visualising what the torque is doing. This is why it is more of an abstract concept albeit one that people are more familiar with. Caused by the torque deliviered by the engine, from burning the fuel. so talking about heat into the engine can also be used, but the trype of work that the engine does is torque. and this is the crucial thing you must understand.

everything else is an application of that torque generated.

 

[xxChrisxx]

If you only had rolling road data without rpm signal, you'd be forced to use the power method. This is perfectly valid.

Now we've got that sorted we can look at the practical side of this. Let's talk about a track with a slow hairpin onto a midlength straight. Both cars have 4 gears, one prodeuces 50 Nm more torque than the other. You could gear the cars differently to give the same low down acceleration from the corner. The car with the lower torque would need to be geared shorter to allow it to be running in the higher rpm range.

As you have a shorter gear ratio in first and the same ratio of ratios (if you know what i mean) so that the acceleration performance ws the same the higher torque car would have less top speed.

However! If you geared the cars for the same speed in each gear, the lower torque car would likely fall off its peak power band, and would accelerate slower away from the corner, but begin to catch the higher torque car down the straight.

This is not a problem if you had a lower torque engine with a 6 speed box, as you could gear for lower accelration and still maintain top speed.

you are making the classic mistake that lots of people have made for race engines. They tune their horsepower and gears so it looks perfectly on paper (fom the dyno), but this won't necessarily transfer well to the tack.

 

[zanick]

Thanks for the dyno validation, now the good stuff

I think you might be confused when you start looking at same ratios to achieve same vehicle speed in each gear. If the HP curves were the same, the acceleration performance would be the same at ANY speed on any track at ANY time. (even though one engine had 60NM less torque than the other) Of course, that lower torque engine has proportioally deeper gearing per gear and proportionally higher rpm ranges per gear. This is the classic example that there is NO difference, not in top speed, not in accelerative force anywhere.
Do we agree there?

Now, your point below that is valid that most high torque lower rpm engines have a boader HP curve, so the peaky HP curve, high rpm, lower torque engine usually has less grunt out of corners when the rpm is less than even the gear to gear rpm drop percentage. (e.g. Most are 27% rpm drops per gear, but you can be at near 55% of max rpm on some corners)

now, the closer ratio gear boxes can bridge this gap, almost entirely. in fact, they can even be an advange with being able to use more available peak HP.


As far as race cars on the track, I have 10 years of racing experience that points to not making the mistakes of some of my competitors. what i have done is maximized my gearing and HP design to fit more tracks without changes. (i.e. broader hp curve). As long as you know the rpm range, and the time spent at those rpm ranges, you can determine the best gearing to optimze available HP. (or rear wheel torque throguh the gears if you want to go through the additional math). by optimizing time spent at or near max HP, you get the greatest acceleration on the track. There are few trade offs , such as do you want to have to shift in an area of the track that might upset the car or take time for a shift, calling for an immediate downshift too soon afterward. those are considerations that can be tuned with small gearing changes. (ie tire size, rear end changes that shift the entire gear ratio set up or down)

mk

If you only had rolling road data without rpm signal, you'd be forced to use the power method. This is perfectly valid.

Now we've got that sorted we can look at the practical side of this. Let's talk about a track with a slow hairpin onto a midlength straight. Both cars have 4 gears, one prodeuces 50 Nm more torque than the other. You could gear the cars differently to give the same low down acceleration from the corner. The car with the lower torque would need to be geared shorter to allow it to be running in the higher rpm range.

As you have a shorter gear ratio in first and the same ratio of ratios (if you know what i mean) so that the acceleration performance ws the same the higher torque car would have less top speed.

However! If you geared the cars for the same speed in each gear, the lower torque car would likely fall off its peak power band, and would accelerate slower away from the corner, but begin to catch the higher torque car down the straight.

This is not a problem if you had a lower torque engine with a 6 speed box, as you could gear for lower accelration and still maintain top speed.

you are making the classic mistake that lots of people have made for race engines. They tune their horsepower and gears so it looks perfectly on paper (fom the dyno), but this won't necessarily transfer well to the tack.

 

[zanick]

As a note , you guys have got me looking at the power scene a little differnet. It is a highlevel indicator of what torque you have available at the rear tires. (as multiplied though the gear box). It doesn't cause the motion, Force does. Its a snap shot of what you have. However, with cars we have only engines with HP ratings. (Kw ratings like with elecric motors) with that and torque values, we get a picture of the kind of force we can apply to the ground at any given vehicle speed. the engine torque is not a good indicator unless it has the rpm attached to it, and then we can determine exactly what torque /force will be able to be achieved at any vehicle speed. without it, HP can allow us to work backward and find that engine torque at the rpm and vehicle speed to determine Force! Hows that guys?

Now, in maximizing force at the rear wheels at any speed, how do you use the time factor to know if you are maximizing force on the track? Ft-lb-seconds? If that is the better, more clear terminology, Ill use that instead of watt-seconds or HP-seconds or something in that relm.

mk

 

[xxChrisxx]

That sounds exactly on the button for the physics terminoligy for what is going on. I didnt know you had that much racing experience, I've not got that much practical experience and most of what I know comes from engine dynos. So that's why I would always lean towards torque figures.

You can either integrate the torque at engine (gives an indiation of engine work) or horsepower curve to give an indication of work. In this case the horsepower curve would be the more useful for finding what you want to know. I'm really not sure what the units would be for that though, I am just assuming Power.S.

For the rear wheel, time based one. I just don't know I am afraid, i'll try looking it up but most of my books are purely engine based and are unlikely to have this information. I think a trip to the library is in order.

As a performance indicator are you thinking of inegrating for an acceptable power band. So say find the area between 80% max power?

I'm still a little confused as why what you mean when you say the horse power curves are the same. As that to me means the torque must be the same as the graphs overlay perfectly. Do you mean that when the x-axis is scaled, the curves have the same trend?Also you mention getting a broader HP curve, so you sacrifice peak HP for gains along more of the rev range. (in effect maximising the area under the power curve). This is essentially the same as altering the torque curve to sustain at higher rpm (I think). My friends is acutally better than me at this, i'll ask him about it tonight.

I'm really enjoying this discussion, its making me think quite hard and polish up on stuff I've not read in quite a while.

 

[zanick]

Thanks Chis, I think I am back on track now. (literally, going racing tomorrow too! )

Anyway, as far as the intgration of the power curve, yes, I think 75% would probably do it for sportscar and closer to 80% for factor race cars or spec open wheel type cars.

Sorry about the confusion of the HP curves. I mean they are the same shape, with the x-axis scaled. So, when both A car and B car are coming off a turn at 55% of max rpm, they both have the same HP created by :) engine torque thorugh the gears, making the same rear wheel force to the ground. (it might not even be same wheel torque, if the tires are different diameters. (all part of the gearing part of the equation, right?)

as far as the broader HP curve, in the example, the HP peaks were identical, but one curve just had more HP over the operational range. Generally, this is more true for a high torque engines, but occasionally, you see one that has a slightly broader HP curve. generally, it means the other high torque engine has much upside potential for mods.

Ive enjoyed the talk as well . Trust me, very few, but the motorsport top engineers get this stuff, even at this basic level. Its nice to get grounded here with the right thinking and terminology. I worked in the industrial controls arena for 12 years as well, so I have a great handle on the basic stuff. (torque, gear ratios and efficiency, basic motion control profiles,etc). Delt with all sorts of flavors of small electric motors and their associated electronics, including servos and stepmotors. So, I have the high level understanding, but the devil is in the details. :) . Obviously for racing, its an avantage to know how to optimize your equipment performance on a given race day!

Thanks,

mk

That sounds exactly on the button for the physics terminoligy for what is going on. .

As a performance indicator are you thinking of inegrating for an acceptable power band. So say find the area between 80% max power?

I'm still a little confused as why what you mean when you say the horse power curves are the same. As that to me means the torque must be the same as the graphs overlay perfectly. Do you mean that when the x-axis is scaled, the curves have the same trend?


Also you mention getting a broader HP curve, so you sacrifice peak HP for gains along more of the rev range. (in effect maximising the area under the power curve). This is essentially the same as altering the torque curve to sustain at higher rpm (I think). My friends is acutally better than me at this, i'll ask him about it tonight.

I'm really enjoying this discussion, its making me think quite hard and polish up on stuff I've not read in quite a while.

 

[Gnosis]

I get it, its getting to be less and less of a chicken and egg thing. force is what causes the acceleration. got it. Hp is a measure of of the rate the force does the work. HP is a vision into what the potential force is though right?

Mk

Just when I think you're maybe finally getting it, you go right back to that association you have with horsepower and force as per your last sentence above.

Horsepower isn't a "vision" or "insight" into the potential force. All force in the combustion engine is strictly produced by torque and torque alone. Torque IS measureable on a dyno. Horsepower is merely a rating of how much "WORK" (not force) can potentially be accomplished per a given RPM via the available torque of the engine. "Work" and "force" are two entirely different quantities just as horsepower and torque are entirely different.

Horsepower should only be thought of as a rating, as in a "horsepower rating", a work rating.

Try to grasp the concept of “horsepower rating” from this next example:

YOU could potentially twist a ratchet (thereby apply some measure of torque) all day long. This also means we could actually give you a horsepower rating of some kind rated per an entire day’s work. Let’s assume that the absolute torque limit that you can apply per your ratchet is 200 ft pounds and the conveyor belt you’ve been manually cranking with your ratchet has only required 20 ft lbs (piece of cake). This conveyor belt is carrying buckets of material to the top of a silo, then dumping the contents. Suddenly, someone accidentally overloads one of the buckets on the conveyor belt with 400 pounds of material. Doh!

At this point, you HAVEN’T lost your willingness to work neither have you lost your “horsepower rating” (that work which you are capable of accomplishing per day). You simply lack the strength to apply any more than 200 ft pounds of torque to lift the excessive 400 pound load. Since torque is the ONLY factor causing the conveyor belt’s movement and you lack the torque to lift the 400 pound load, all work has ceased. If you stepped over to an identical conveyor belt that didn’t have the excessive 400 pound load, you’d be able to continue at your rated horsepower rating for the rest of the day. Insufficient torque is the issue here, NOT insufficient horsepower.

That's why I stated it WASN’T lack of horsepower that prevented my 1.5 HP mini-bike from climbing that steep incline; it was the engine’s sheer lack of torque, as torque is the ONLY force being produced by a combustion engine and I cannot state this enough. Horsepower is a work rating, torque actually causes the rotation and acceleration of the crankshaft.

 

[zanick]

I understand your push for clarifying that it is the force that does the work. But, it is the HP "rating" as you call it, that can indicate the potential force at any speed. (if you don't know the force, as power=fv. In your analogy, clearly if 1.5HP doesn't give the force you need to climb the hill, more power will give more force. (and I know you are going to say, if I give I more force, it will have a higher power rating as a result ;) ). But, I don't understand why HP isn't an indication of force at any vehicle speed. I am talking about HP and its indication of a force's ability to do work. This is why I was leaning toward watt-seconds (or HP-seconds) as a determinant factor and useful term on the subject of automobile comparisons. If you have a battery , its energy potential is rated in KW-Hours. (same thing as Hp-seconds, watts-seconds etc) That battery can give you the power to be able to lift 550lbs in 1 second over 1 foot. (if this 10v battery is 74amp/second, 1hp-second battery, 746watt-seconds) 746w-seconds, 74amp-seconds, 20ma-hour, etc) . This power rating tells me how much work I can do and how fast I can do it. I can lift 550lbs in 1 foot in 1 second or 1100lbs 1 foot in 2 seconds, or 225lbs, 1 foot in 1/2 second. Its a power limit of how fast you can accelerate a mass at a given velocity, or the rate of doing 'work' . Again, I understand that the force does the work, but you don't really need to know it to get an acceleration of a mass at a velocity, right? Rate of change of kinetic energy is one example I can think of where if I know this, the answer can be power, without the knowledge of force or torque. A chassis dyno has drums and can measure hp without using a torque value, even though a torque value is easily produced from the change of sped of the drums. Without an engine speed input, engine torque could be any value, so the output is engine HP and rear wheel MPH.

Now, we might have glazed over the main question here, but when folks look at HP ratings of vehicles, or torque, which is more meaningful as an indication of potential rear wheel forces at any speed? Certainly engine torque is only an indication, if we are talking about comparing two equal HP cars, with grossly different engine torque and RPM values. If you don't look at the gearing and vehicle speeds, HP will be able to be compared by only plotting both HP curves on one another. if they are same shape for an given range of vehicle speed, the cars will accelerated the same (assuming the same car and gear boxes adjusted for same range of speeds). In other words, both identical cars have gear boxes that allow for the same MPH speed of the cars in each gear. Average power, but more importantly, HP-seconds being a determinant factor on which car will accelerate faster. In other words, if two curves have the same area under the usuable rpm range, but one has more area under the top rpm are vs the lower rpm area in the usable range, the one with he area in the higher rpm range will have more "Hp-seconds" and will accelerate an equal car faster through a wide range of vehicle speed

The point I was trying to make earlier, was that even an integration of the HP curve doesn't exactly find the answer, as this is due to the varying time spent at the higher rev areas. Area under the engine torque curve doesn't work, as most engines make most their torque below an area where they would even be operating in. The most rear wheel torque (after the gear box reductions and torque multiplications) would be found at the maximum amount of HP available at any vehicle speed.

Getting back to your mini bike analogy, if you don't have over 1.5hp or some force at the rear wheels , you are not going to be able to climb that hill. the rate of change of kinetic energy or adding potential energy (mini bike climbing hill) can not be met by the power source and resultant forces from the rear wheel on the road. If we were using the battery, we could turn up the power setting and climb the hill. (using more amps) the capacity would be reduced, but the system would be using a faster rate of doing work, and use more power. !nstead of 1100watt-seconds of power, you could be using 2200watt-seconds (or near 3Hp) to keep that current rate of speed, while climbing the hill. 2 x the amp draw, 2x the power and 2x the torque to the drive wheels. Clearly, the force has been increased by increasing the current draw, which is by definition a higher rate of doing work (higher power). How is this logic not correct?

These are the questions many are looking to have answered in the right terms.

Thanks,

Mk

Just when I think you're maybe finally getting it, you go right back to that association you have with horsepower and force as per your last sentence above.

Horsepower isn't a "vision" or "insight" into the potential force. All force in the combustion engine is strictly produced by torque and torque alone. Torque IS measureable on a dyno. Horsepower is merely a rating of how much "WORK" (not force) can potentially be accomplished per a given RPM via the available torque of the engine. "Work" and "force" are two entirely different quantities just as horsepower and torque are entirely different.

Horsepower should only be thought of as a rating, as in a "horsepower rating", a work rating.

Try to grasp the concept of “horsepower rating” from this next example:

YOU could potentially twist a ratchet (thereby apply some measure of torque) all day long. This also means we could actually give you a horsepower rating of some kind rated per an entire day’s work. Let’s assume that the absolute torque limit that you can apply per your ratchet is 200 ft pounds and the conveyor belt you’ve been manually cranking with your ratchet has only required 20 ft lbs (piece of cake). This conveyor belt is carrying buckets of material to the top of a silo, then dumping the contents. Suddenly, someone accidentally overloads one of the buckets on the conveyor belt with 400 pounds of material. Doh!

At this point, you HAVEN’T lost your willingness to work neither have you lost your “horsepower rating” (that work which you are capable of accomplishing per day). You simply lack the strength to apply any more than 200 ft pounds of torque to lift the excessive 400 pound load. Since torque is the ONLY factor causing the conveyor belt’s movement and you lack the torque to lift the 400 pound load, all work has ceased. If you stepped over to an identical conveyor belt that didn’t have the excessive 400 pound load, you’d be able to continue at your rated horsepower rating for the rest of the day. Insufficient torque is the issue here, NOT insufficient horsepower.

That's why I stated it WASN’T lack of horsepower that prevented my 1.5 HP mini-bike from climbing that steep incline; it was the engine’s sheer lack of torque, as torque is the ONLY force being produced by a combustion engine and I cannot state this enough. Horsepower is a work rating, torque actually causes the rotation and acceleration of the crankshaft.

 

[rcgldr]

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill.

Assume mini-bike weighs 60 pounds. Weight of rider and mini-bike is 200 lbs. To climb a 20 degree hill, it will take sin(20) x 200 = 68.4 pounds of force to climb the hill (ignoring drag factors here, assuming rear wheel horsepower is 1.5).

If the rear wheel horsepower is 1.5 hp, then you need to solve:

1.5 = 68.4 x speed / 375
speed = 8.22 mph.

In order to climb a 20 degree hill, the mini bike would have to be geared down so it's makes 1.5 rear wheel horsepower at 8.22 mph.

If the hill was vertical, with the mini-bike attached to a chain going up the side of the hill, then it would lift 200 lbs at 2.81 mph

 

[xxChrisxx]

Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy won't tell you the force nor acceleration. K.E = .5mv^2
Just because they use the same terms doesn't make them mathematically compatble. The dyno doesn't acutally measue energy, as energy can't be directly measured.The thing you are describign for the dyno is rate of change of momentum of the drum. momentum = mv
Dy differentiating this you get the magic formula of F = ma.

 

[zanick]

Thanks Chris,

But if acceleration = power/(mass x velocity or a=P/p , the rate of change of momentium would be proportionally higher from 50-60mph as it would from 60 to 80mph and so would the force for the two speed ranges. However, the power would be much higher at the higher speed range. Thats why I seemed to remember power equaling rate of change of kinetic energy, not momentium. I agree that the rate of change of momentium would get you to "Force".

so I guess what I'm saying is that the rate of change of kinetic energy gets you power and thus acceleration at any given velocity, and from power you could get force. (working backward )right?
mk

Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy won't tell you the force nor acceleration. K.E = .5mv^2

The thing you want is rate of change of momentum. momentum = mv
Dy differentiating this you get the magic formula of F = ma.

 

[xxChrisxx]

This is where are going wrong. you are correct that Power is not the rate of change on momentum. But what the dyno actually measures, and what it outputs on a screen is not the same thing. From what it measures it does some mathematical operators. to give out a nice convenient power figure. To same you messing around calcualting it yourself.

The dyno:
1) measures the rate of change of anguar momentum of the drum in a given time.
2) this gives the acceleration of the drum.
3) From f=ma you can find the force at the wheel giving said acceleration.
4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

This is that's going on from what the dyno measures, what the computer then calcualtes and operates. Then produces the power figures.

This cannot be done the other way round as energy cannot be directly measured (see this link:http://en.wikipedia.org/wiki/Energy#Measurement) to get acceleration directly.

We know that as energy is a concept that is inferred, so is power. You can mathematically operate the equations once you konw evreything to give you what you want, but that doesn't mean you can do it in reality.Also i'll discuss the climbing a hill thing with you in a bit, as that's a tricky one to talk about over a forum. Its another time that figurees can be misleading. you look at the figures and see one thing, but reality tells you something different.

 

[rcgldr]

The dyno:
1) measures the rate of change of anguar momentum of the drum in a given time.
2) this gives the acceleration of the drum.
3) From f=ma you can find the force at the wheel giving said acceleration.
4) From this rear wheel force and the distance traveled you get the work done.
5) The change of work done in a given times tells you the power.

Not quite:

4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

 

[xxChrisxx]

Not quite:

4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

Both are actally correct :P, the last two steps manipulate the units in the same way. That just inculdes a time function at step 4 and not 5.

I prefer the way you said it though, I think its slightly more clear. So i'll steal your answer and edit my other post :D

 

[zanick]

Ok, we got a little side tracked on how power is measured on a dyno. I still think if you have the rate of change of the KE, can't you solve for acceleration and then work backward to find the force (and then easily convert to torque)? Going back to the linear world, if a car is accelerating at a certain rate, and we know two velocities, the mass of the car, doesn't average power=dw/dt ? It seems that we would have to calculate the change in work from the rate of change of KE. Kind of digging a hole here, I know. :)

Anyway, getting back to the original point and question, what would the terms used in comparing two same vehicles with equal HP "rated" engines, one being higher torque than the other, at different rpm ranges each? I am pointing at watt-seconds or HP-seconds to determine which one would yield the greatest rear wheel forces at any vehicle speed.
(gearing proportioal to each engines rpm range to achieve the same speed in any gear).
Integration of the rear wheel torque curves would obviously do this, but a succession of them would have to be used due to overlap. integrating the HP curves wouldn't do it either, as time spent in higher rpm ranges is different and would give incorrect weighted values for the results. I keep on falling back to HP-seconds or watt-secons as the answer, and I am really asking, in the physics world, what would be the best way to address this with the correct terminology. The car is an interesting subject, with varying force even if power was constant, but its not due to gearing and power varies as well. What I have found, is the easiest way to determine optimal acceleration is to look at the HP curves adjusted to the same vehicle speed operating ranges. Or, just look at gearing spacing, and use same percentage drop of each engine to determine the area that will be maximized and used under the HP curves. However, this doesn't adress the time factor, that's why it seems like HP-seconds is the right way to look at this to compare performance.
Same could be done for rear wheel torque figures, but would be a little more cumbersome to calculate.

It seems that comparing the shape of two HP curves, with the same proportioal engine rpm ranges is the easiest way to compare vehicle performance.*

Thoughts?

mk

*two engines, one with a 10,000rpm redline and the other 5,000rpm redline and shift points.
Both would have 25% rpm drops per shift per gear. Both engines have the same peak HP values, but would have different shaped curves or the same shaped curves, but either way, grossly different engine torque values.