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

[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.

 

[xxChrisxx]

You are refusing to budge from incorrect thinking and its very difficult to give a techincally correct explination until you do. You need to stop thinking about KE now, and start thinking about momentum if you want a technically correct explanation.

By simply using KE you are ignoring what the dyno is actually measuring. Therefore any further premise from that is based on false mathematics. You are approaching the problem backwards.

Just because you can maniplulate the equations the correct way on paper doesn't mean it is done like that in reality. Without a context energy means very little, and its the context that gives it importance.

What you are saying is correct for the car acceleration, but you are butching the way this is acutally calculated. Its also leading you to incorrect thinking. That more power is more force. This isn't correct and is also why you are getting the 1.5hp motor climing a hill messed up a bit.

More torque is more force, more power is force more quickly. (you cannot do this the other way around)
More power does not = more force.

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

your battery/electric motor example. You say doubling the power will allow you to lift something heavier, this is true. But as they are operating at the same rpm, the way it doubles power is by doubling the torque outrput.

I'm going to come up with a worked example to show this.

 

[zanick]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

I'm going to come up with a worked example to show this.

 

[xxChrisxx]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience,

 

[xxChrisxx]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience, it shas no physical basis. I am talking about the power part, not the force * speed.

By injecting nos you are increasing the engine TORQUE at a given rpm because you are increasing the force on the piston and by extension the power. NOT THE OTHER WAY ROUND.

To talk of gearing, why when you are coming to a hill do you downshift? If power is the key factor in rear wheel force it shoudlnt matter about the gear you are in as the engien is pumping out the same power. They downshift to take advantage of the TORQUE multiplication of the lower gear ratio. Seriesly do the calculations for the F1 and truck engine. I gaurantee you will make the mistake that power = force. IT DOESNT!

See how it works : http://auto.howstuffworks.com/question381.htm

^^ read it^^

You are really confusing yourself by flitting between the rear wheel horsepower and rear wheel torque. give all the equations you can.

you seem to clearly have problems thinking in terms of torque and power.

Seriously please go and buy a book on this. i'll have a look through my library for the best one.

 

[zanick]

ll concede that calcuating dyno force is really based on acceleration of the loads and power is calculated from that. again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, you truck and climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will relate in higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.

you also say: "More torque is more force, more power is force more quickly. (you cannot do this the other way around)
More power does not = more force. But with power=fv, it does indicate that more power would create more force. (or to keep on the force side of thinking, a higher power rating would indicate that more force would have to be produced at the same vehicle speed to climb that hill.


Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly."


With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

I'm going to come up with a worked example to show this.

 

[xxChrisxx]

You are pretty much wrong in everything you've just said regarding power.

Read the links on how stuff works, or go to your amazon and buy a book on this subject.

Until you realize your fundamental error in thinking you will never learn why you are going wrong. You are clinging to P=F*s, but if you'd acutrally read up on this you know that that relation is not what is heppening in reality. you will discorver its a mathemetical relation devised by people AFTER pissing about with torque and rpm and then doing lots of calculations to it for so long. You will also discover why its the torque that creates the force at the wheel NOT POWER.

Until then there is nothing I can do to increase your inderstanding of this.

 

[zanick]

I get what you are saying, especially with the NOS example. we get more torque, and subsequently more force at the rear wheels. I just have to change my direction of thinking. (coming from car land.)

But, when you talk of gearing below, you lose me. power is the key. you ask the question below. power is key and as you are crusing along at 50mph with your truck, you are using 100hp of your 800hp rating. Here comes the hill, you need much more force to climp the hill, you downshift to take advantage of the engines power available at the higher rpm and higher fuel requirements. At that prior power setting, you might have only been at 10% of the power available. now, you take advantage of higher gear reduction by downshifting, you now are in the higher rpm ranges and at the higher hp range at full throttle. You also are maximizing you rear wheel forces at the max HP range.

Now, you asked me to compare a high torque, low rpm truck engine and F1 engine at the same power, so here it is. Let's keep it simple. 1000hp 1000ft-lbs of peak torque for the truck and 1000hp and 250ft-lbs of peak torque for the F1 engine. The both go to climp a hill. The Big diesel engine is in the truck. At 50mph, the truck is at 4000rpm and producing 8000ft-lbs of torque at the rear wheels. (1000ft-lbs with 8:1 gearing) At the same 50mph, the truck with the F1 engine is running at 16,000rpm and 250ft-lbs of torque, but the rear wheel torque is the same 8000ft-lbs. (250ft-lbs with 32:1 gearing). vehicle speed is the same, HP is the same, engine torque is way different, yet, rear wheel forces are identical. Where is the mistake there?

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience, it shas no physical basis. I am talking about the power part, not the force * speed.

By injecting nos you are increasing the engine TORQUE at a given rpm because you are increasing the force on the piston and by extension the power. NOT THE OTHER WAY ROUND.

To talk of gearing, why when you are coming to a hill do you downshift? If power is the key factor in rear wheel force it shoudlnt matter about the gear you are in as the engien is pumping out the same power. They downshift to take advantage of the TORQUE multiplication of the lower gear ratio. Seriesly do the calculations for the F1 and truck engine. I gaurantee you will make the mistake that power = force. IT DOESNT!

See how it works : I am finding the link now

You are really confusing yourself by flitting between the rear wheel horsepower and rear wheel torque. give all the equations you can.

you seem to clearly have problems thinking in terms of torque and power.

Seriously please go and buy a book on this. i'll have a look through my library for the best one.

 

[zanick]

Im hearing you about that its really the force that determines the power, and power is just the power rating. But, you asked me to provide the example, and I did. Your truck powered by F1 or its normal engine, doesn't make your point, as i proved, both can yeild the same rear wheel torque at the same HP ratings, and grossly different engine torque values.
Power ratings is what you and I were comparing here and they seem to yield the same rear wheel torque at any vehicle speed. (if they are running at the same power level)

You mention i was wrong in "everything " i had a said regarding power. can you be more specific? you make the comparison of the truck powered by F1 engine vs its normal low reving high torque engine. that was actually incorrect. If power ratings are the same, the same rate of work can be done by either, and this means the same rear wheel torque can be produced!

In the electric motor and battery rating world, you get things rated in KW or KW-hours. unit measures of work. This indicates the potential rates of doing work. Is this wrong as well?

mk

You are pretty much wrong in everything you've just said regarding power.

Read the links on how stuff works, or go to your amazon and buy a book on this subject.

Until you realize your fundamental error in thinking you will never learn why you are going wrong. You are clinging to P=F*s, but if you'd acutrally read up on this you know that that relation is not what is heppening in reality. you will discorver its a mathemetical relation devised by people AFTER pissing about with torque and rpm and then doing lots of calculations to it for so long. You will also discover why its the torque that creates the force at the wheel NOT POWER.

Until then there is nothing I can do to increase your inderstanding of this.

 

[xxChrisxx]

You keep moving the goal posts by changing gearing. Gearing is a torque multiplyer.

He cannot change the gearing he has on his bike when he comes to a hill as he is geared for a certian road speed. For that gearing he doesn't have the RW torque available. When you said that he doesn't have the power available to do it, this is incorrect as the engine always produces the same power output, the phenomenon you are talking about is torque. The determination of whether you can climb the hill is torque, the determination of how fast you can climb is power. A rwhp value doesn't tell you if you can climb the hill or not, a rw force value would.

So to find if you can climb, you use:

RWTorque * wheel radius = Force

The RWHP tells you how fast you can climb it. from P = F * S

If you were theoretically detmining gearing before you set off to go this ans you had a RWHP curve, It would be correct to use P = F* S to determine the gearing for a set speed.

If you have set gearing available (ie you've already set off) The P = F* S does not tell you anything useful.
In this case you have to use RWT*wheel radius = F to determine if you have sufficient torque at the rear wheel to do the job. If you find that you don't you either have to go away and get the gearing that will give the correct RWT or you could slap bigger wheels on the bike. You can then find how fast you can climb the hill form P=T*angular velocity.

As jeff pointed out, he could climb that hill if he was geared for it. Then again with gears you can get any mortor to lift any amount. It'll just do it very slowly. So you see you can use the RWHP formula in some situations, but you can use the torque calculation in all situations.

What you were doing was the former, but applying it to the latter situation. So you were not calcualting anything useful. He stated he was geared for something already, and that it was his lack of torque that was causing the problem. Stating that he can do becuase P = F * S tells him he can is pointless as he can't chgane the gearing to get to the poitput that that equations states.

If you cannot climb a hill at the peak engine torque value, you cannot climb it at peak engine power. (as torque is lower at peak power). This transfers down through the transmission so that. If you can't climb a hill at peak RWT you cannot climb the hill at peak RWHP. It means you need to fiddle with the gearing to be able to do so. If you can't fiddle with the gearing and can't fit a larger wheel, you can't get up the hill. This is something that is easier to visualise using a torque calculation as that is the acutal physicla driving force.

I've got a question that I'm just finishing up that I'm going to post after your next reply. It should demonstrate that power is not the critical factor.

 

[zanick]

Not really, I think we are having the classis problem of rear wheel torque vs engine torque adding to the cross lines of the discussion. you tell me.

I think I see a potential problem. So, getting back to the mini bike. If he is running 1.5 hp and running some top speed say, 30mph, you are right. he has a certain amount of force being generated to sustain that speed. Stilll, power=force x speed. he is not at max torque of the engine, he is at max HP which as you say is higher than max torque of the engine. this is generating the maximum amount of force at that particular speed, 30mph. Now comes the hill. he can't choose a lower gear, because that would slow him down, so you are right. limied by max HP and rear wheel torque. if he adds NOS, it increases force by x% so he can climb the hill at that same speed.

You keep on saying that the bike cannont climb the hill if it doesn't have the torque, what you mean to say, is that if it was at max HP prior, and you wanted to keep the speed constant, yes that would be true. with more HP, you would have more force and then you would. you say its more force so then you can then determine the hp has increased as well. Thats fine. In the electrical /mechanical world, if i dropped in a higher output battery, to the elecrict motor powered bike, it could cause the motor to run with more torque and thus at a higher power level.

truck analogy with its stock engine vs F1 engine. If both engines are installed with appropriate gearing, both will be able to climb the hill at the same speed, using gears appropriate for the trade offs of rpm and speed and mulitplication of torque. One engine would do it at high rpm , low torque and the other would do it with high torque, low rpm. the net forces at the rear wheel would be the same because the rate of work would be the same.

Now , your last paragraph. If I have a transmission, and I can't climb a hill at a certain speed at max engine torque, there is a distinct possibility, and in most cases, that i could climb it at max HP, even though the engine torque would be less at that same vehicle speed, rear wheel multiplied torque would be higher at that same vehicle speed than it would be found at max engine torque. agree? I think you two part sentence is saying that if you have a fixed gear, and you can't climb the hill at max torque, certanly at max HP you wouldn't be able to climb it, because the speed would be much higher. and your second part of the sentence, is saying that if you can't climb the hill with max rear wheel torque, as multipled thorugh the gear box at that speed, certainly you wouldn't be able to climb it at max hp, because those two points are the same engine rpm. max HP would be the point at which you would have max torque at any given speed. just like if we had an infinitely variable gear box, the engine would operate at max hp and not max torque to achieve max torque levels at the rear wheels at any given vehicle speed.






mk

You keep moving the goal posts by changing gearing. Gearing is a torque multiplyer.

He cannot change the gearing he has on his bike when he comes to a hill as he is geared for a certian road speed. For that gearing he doesn't have the RW torque available. When you said that he doesn't have the power available to do it, this is incorrect as the engine always produces the same power output, the phenomenon you are talking about is torque. The determination of whether you can climb the hill is torque, the determination of how fast you can climb is power. A rwhp value doesn't tell you if you can climb the hill or not, a rw force value would.

So to find if you can climb, you use:

RWTorque * wheel radius = Force

The RWHP tells you how fast you can climb it. from P = F * S

If you were theoretically detmining gearing before you set off to go this ans you had a RWHP curve, It would be correct to use P = F* S to determine the gearing for a set speed.

If you have set gearing available (ie you've already set off) The P = F* S does not tell you anything useful.
In this case you have to use RWT*wheel radius = F to determine if you have sufficient torque at the rear wheel to do the job. If you find that you don't you either have to go away and get the gearing that will give the correct RWT or you could slap bigger wheels on the bike. You can then find how fast you can climb the hill form P=T*angular velocity.

As jeff pointed out, he could climb that hill if he was geared for it. Then again with gears you can get any mortor to lift any amount. It'll just do it very slowly. So you see you can use the RWHP formula in some situations, but you can use the torque calculation in all situations.

What you were doing was the former, but applying it to the latter situation. So you were not calcualting anything useful. He stated he was geared for something already, and that it was his lack of torque that was causing the problem. Stating that he can do becuase P = F * S tells him he can is pointless as he can't chgane the gearing to get to the poitput that that equations states.

This comes down to:

If you have a given gearing - use torque eq.
If you have a given speed (and can vary gearing) - can use torque or power eq.

If you cannot climb a hill at the peak engine torque value, you cannot climb it at peak engine power. (as torque is lower at peak power). This transfers down through the transmission so that. If you can't climb a hill at peak RWT you cannot climb the hill at peak RWHP. It means you need to fiddle with the gearing to be able to do so. If you can't fiddle with the gearing and can't fit a larger wheel, you can't get up the hill. This is something that is easier to visualise using a torque calculation as that is the acutal physicla driving force.

 

[xxChrisxx]

Will repost q in a second.

Ok Q is:

14.4Kg block on a 45% slope

The motor used is:

100 Watt at 200rpm
4.77Nm at 200rpm

167.2 Watt at 400 rpm
4Nm at 400 rpm.

The reduction gearing is 15:1.
The rolling radius is 1m

Can the motor move the block up the hill?

EDIT: you can assume no losses through the drivetrain, so RWHP = engine HP.

Now after ansering this I'll give you the second part and address the truck and F1 engine thing.

Also buy this book :https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20
and read the last chapter. It's not a very technical book, but is perfect for what we are talking about.

 

[zanick]

Well first off you have two motor speeds and thus driving speeds, fixed by a 15:1 gear box. If I can't change that, then I can't get speeds that utilize the available power of the motor.
the power range for the two speeds is 1:1.65, but the speed difference is 2x. so, I would need much more power to achieve 2x the speed of the 400rpm motor setting. If the slower motor setting works and uses all available power to lift the mass, even with optimal gearing, 1.67x greater power setting could only allow for a small increase in drive speed.

Thats at first glance. it looks like yes for the 100watt setting, but no for the 167watt setting due to the amount of power needed to raise the mass at 2x the speed.

Will repost q in a second.

Ok Q is:

14.4Kg block on a 45% slope

The motor used is:

100 Watt at 200rpm
4.77Nm at 200rpm

167.2 Watt at 400 rpm
4Nm at 400 rpm.

The reduction gearing is 15:1.
The rolling radius is 1m

Can the motor move the block up the hill?

EDIT: you can assume no losses through the drivetrain, so RWHP = engine HP.

Now after ansering this I'll give you the second part and address the truck and F1 engine thing.

 

[xxChrisxx]

First of all buy this book:
https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20

And read the last chapter (hell you can read all of it if you want). It'll tell you exatly wehre your thinking is a bit off.Regarding the question.

The acutal answer is neither will get it moving at that ratio.

Those are the peak figures, so the maximum power the motor can produce is 167.2 W. The maximum torque it can produce is 4.77 Nm.

The point of the above was to show that for a given gearing its the torque that creates the moving force. A minimum moving force requites a minimum torque at the wheels. The power figure has no bearing on this force. If it did the power figure would cause larger rear wheel forces and get it moving.

The next thing I was going to ask was try it with 21:1 (ive misplaces my notes on this i think that was the ratio). You'll find that the force at the wheel is then enough to move the block at the lower power figure but not the higher.

If you did rev it up (as more power apparently creates more force at the rear wheel) you should move faster. In fact it does the opposite, it you did rev it up you would no longer be providing enough force to overcome the resistance.

Do the acutal calucaltions and you'll see this is the case.Here is my working for the 1st part:

14.4*9.81*cos(45)= 100N
You need at least 100N to get the block moving up the slope.

It can be shown that the rear wheel forces for each settign are:

100W setting.

Engine_Torque*gearing = Wheel torque.
4.77*15= 71.55 Nm at the wheel.
As the wheel is 1 m Rolling radius.
T=F*d
Force at wheel is 71.55 N

This gives a net negative force of 28.45N

for the higher power setting the wheel force (using the same method above is)
Force =60N

Using the power method: we know the power of the motor and the force necessary.

P=100W
F=100N

using P = F * S

100/100 = 1 m/s So to get the thing moving we need a wheel speed of 1m/s or below.

Wheel speed at lower power = circumfrence of wheel * angular velocity
for 1 second.

S = (2*pi)*(200/(15*60)) = 1.39 m/s
S = (2*pi)*(400/(15*60)) = 2.79 m/s

The wheel speeds are too high to provide the adequate forces to start the block moving.

So more power doesn't = more rear wheel force. And the torque method is fer better at this (its more clear that it won't work) as it is a snapshot in time.

 

[rcgldr]

100 Watt at 200rpm, 4.77Nm at 200rpm
167.2 Watt at 400 rpm , 4Nm at 400 rpm.
The reduction gearing is 15:1.

It's not a fair comparason, setting the gearing for the same speed = 1.3963 m/s, with a radius of 1 meter:

100.00 watt at 200 rpm, 4.77465 Nm at 200 rpm, reduction gearing 15:1, force = 71.620 N
167.55 watt at 400 rpm, 4.00000 Nm at 400 rpm, reduction gearing 30:1, force = 120.00 N

As expected, 120.00 / 71.62 = 1.6755 = 167.55 / 100.00

 

[xxChrisxx]

You've totally missed the point of the arguement.

It was to show that the instantaneous force was dictated by torque only (not power). To do that they must have the same gearing. If you could change the gearing willy nilly, then the question becomes moot.

Thats why I told him to use 21:1 next. after that 25:1 and after that 30:1.
It shows that power output of the engine is the same, but the gearing changes to torque at the wheels.

I was then going to move on to what if you wanted them to move at the same rate. Now this is a power over time.

So with the 15:1 reduction at the higher power, you have a lack of torque to get it moving. But not when you use a different gearing.

I need him to break the thinking that its power dominating the size of the force (thats torque). and get him to understand that power essentially shows the frequency of the force. So although at the higher power, you have a smaller force. that force acts more often.

HOWEVER: The instantaneous force MUST be large engouh to move the block, otherwise the frequency is irrelevent. Think about trying to hammer in a nail with a feather. You can hit it as often as you want, but you can provide enogh linear force to get it to move. (thats a bit of an over simplified comparison, but it'll do for the purposes of explaining what I want)

 

[rcgldr]

It was to show that the instantaneous force was dictated by torque only (not power).

I understand that torque is the angular equlivalent to (linear) force.

To do that they must have the same gearing.

Why? If you have gearing, then the gearing should be matched to the expected demands of the motor. This is why cars have transmissions and differentials. This is why a P51D Mustang (WW2 airplane) gears it's prop down to 1437 rpm while it's engine spins at 3000 rpm. Since the title of this thread is race cars, I would assume that those cars would be geared appropriately.

In your example, you specify a torque, a gearing and a wheel radius, allowing the linear force at the wheel to be calculated, but without knowing that this torque occurs at non-zero rpm, it's not known if the motor can move the block.

Take the case of an electronic stepper motor. Dynamic torque (moving) is about 70% of holding torque (not moving), so for example.

reduction gearing 15:1, wheel radius 1 meter, 60N load (required force)

torque == 5.0Nm at rpm = 0, force = 75.0N
torque <= 3.5Nm at rpm > 0, force = 52.5N

The motor has enough torque to hold the load in place, but it can't turn against the load.

Knowing only the torque isn't sufficient to know if it will move unless it's also known that that torque can be generated when the motor is turning. Some knowledge of angular velocity (the fact that it is non-zero) in addition to torque (so therefore power) is required to know if the motor can perform any work.

 

[zanick]

you really missed the point of the entire discussion with that one. I had similar values at my quick glance, using .707 x the 14Nm of the load up the hill. Even still, you missed the part that if you didnt use gears to utilize the available power, of course the load can't be lifted up the hill. there is not enough power available to move those two loads. However, if you slowed things down a bit, the power could be used to move the mass up the hill.

geaering doesn't create HP , it gives you the ability to utilize it. It also multiplies torque, by sacraficing rpm. Because at 200rpm, and the resultant 2m/second requires more torque than the motor can provide. P=fv. however, at a slower speed, the hp can be utilized though deeper gearing and would move/lift the load. I you have more power at 400rpm on the motor, 100watts going to 167watts, then obviously, double the gearing again, for the same output speed , and you will have increased the final torque output proportionally.
You are using more electrical power, creating more mechanical power, and thus have more force generated at any given output speed.

A stepping motor, like Jeff brought up, has its greatest torque at the first signal to move. It has something called a torque vs angle of displacment. however if you want to get a load to move, you can easily gear it, and take advantage of its maximum power capabilities. If the load is above its holding torque, it can't move it Gear it down, and now it has more torque and is better utilizing the available power capacity.

go back to your original idea of the truck powered by its engine or a F1 engine. you said that at the same hp, the F1 engine doesn't do a good job, and that is just not true with half the engine torque and double the gearing with its double the engine speed, the same rear wheel torque is achieved. This is the point of the discussion as I see it.

HP (or HP rating) will determine the torque at the wheels at any given speed, and is proved by the combination of the identities showing:

Acceleration =power/mass x velocity.

why don't you summarize what the last chapter in the book says. I've been racing awhile and always have plotted out torque curves in each gear. I now use HP curves with geared % rpm drops. I can compare other powerplants as well, without looking at engine torque curves and get accurate assesement of which engine will perform best over a given speed range. basically, he who uses more of the available HP, if two cars have the same hp, will win the race (all other things being equal). even if they are different in HP , a lower hp can win if it can be used in such a way where it has more average HP over the operational speed range. Conversely, and more commonly, the broader HP engines can yeild more available torque at the rear wheels, even if its engine torque is lower compared to a high torque same HP engine.

The point of this really was, to confer with the physics community to talk about what terms would be appropriate to describe the charactistics of engines in race cars . Is it HP-seconds that we want to optimize? Becauase as I see it, the amount of time spent at the highest engine hp levels available, will yield the max amount of rear wheel torque, and acceleration at any given vehicle speed.



mk

First of all buy this book:
https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20

And read the last chapter (hell you can read all of it if you want). It'll tell you exatly wehre your thinking is a bit off.


Regarding the question.

The acutal answer is neither will get it moving at that ratio.

Those are the peak figures, so the maximum power the motor can produce is 167.2 W. The maximum torque it can produce is 4.77 Nm.

The point of the above was to show that for a given gearing its the torque that creates the moving force. A minimum moving force requites a minimum torque at the wheels. The power figure has no bearing on this force. If it did the power figure would cause larger rear wheel forces and get it moving.

The next thing I was going to ask was try it with 21:1 (ive misplaces my notes on this i think that was the ratio). You'll find that the force at the wheel is then enough to move the block at the lower power figure but not the higher.

If you did rev it up (as more power apparently creates more force at the rear wheel) you should move faster. In fact it does the opposite, it you did rev it up you would no longer be providing enough force to overcome the resistance.

Do the acutal calucaltions and you'll see this is the case.


Here is my working for the 1st part:

14.4*9.81*cos(45)= 100N
You need at least 100N to get the block moving up the slope.

It can be shown that the rear wheel forces for each settign are:

100W setting.

Engine_Torque*gearing = Wheel torque.
4.77*15= 71.55 Nm at the wheel.
As the wheel is 1 m Rolling radius.
T=F*d
Force at wheel is 71.55 N

This gives a net negative force of 28.45N

for the higher power setting the wheel force (using the same method above is)
Force =60N

Using the power method: we know the power of the motor and the force necessary.

P=100W
F=100N

using P = F * S

100/100 = 1 m/s So to get the thing moving we need a wheel speed of 1m/s or below.

Wheel speed at lower power = circumfrence of wheel * angular velocity
for 1 second.

S = (2*pi)*(200/(15*60)) = 1.39 m/s
S = (2*pi)*(400/(15*60)) = 2.79 m/s

The wheel speeds are too high to provide the adequate forces to start the block moving.

So more power doesn't = more rear wheel force. And the torque method is fer better at this (its more clear that it won't work) as it is a snapshot in time.

 

[xxChrisxx]

Look if you care in the slightest about being correct about the physics of this.

Buy/borrow/steal and read the following books.

John Heywood - Internal Combustion Engine Fundamentals
Richard Stone - ICE
A.G Bell - Four-Stroke Performance Tuning
Paul Van Valkenburgh- Race Car Engineering & Mechanics

1. is probably least relevant but is pretty much the engine bible.
2. stone is good for practical workings.
3. Good non technical book reguarding practical tuning
4. a very good all round car setup book. especially the section on gearing and performance.

I have read all of these at some point over the last 3 years. They are all thorough and will show you where you've gone wrong.

Read Bell first. then any of the others. I am in no way inclined to type out whole sections from the book.

 

[xxChrisxx]

I understand that torque is the angular equlivalent to (linear) force.

Why? If you have gearing, then the gearing should be matched to the expected demands of the motor. This is why cars have transmissions and differentials. This is why a P51D Mustang (WW2 airplane) gears it's prop down to 1437 rpm while it's engine spins at 3000 rpm. Since the title of this thread is race cars, I would assume that those cars would be geared appropriately.

In your example, you specify a torque, a gearing and a wheel radius, allowing the linear force at the wheel to be calculated, but without knowing that this torque occurs at non-zero rpm, it's not known if the motor can move the block.

Take the case of an electronic stepper motor. Dynamic torque (moving) is about 70% of holding torque (not moving), so for example.

reduction gearing 15:1, wheel radius 1 meter, 60N load (required force)

torque == 5.0Nm at rpm = 0, force = 75.0N
torque <= 3.5Nm at rpm > 0, force = 52.5N

The motor has enough torque to hold the load in place, but it can't turn against the load.

Knowing only the torque isn't sufficient to know if it will move unless it's also known that that torque can be generated when the motor is turning. Some knowledge of angular velocity (the fact that it is non-zero) in addition to torque (so therefore power) is required to know if the motor can perform any work.

jesus.

im going to answer this point 1 last time.

this was a demonstation about a key physical concept to the op. for the purposes of demonstating that physical concept, the gears had to be kept the same.

Gears are torque multipliers. The point was to show that the force at the wheel is given by torque only. not power. if it was given my the power figure, an increase in power should give an increase in wheel force. it doesnt.

in my exapmle i specify a torque AT AN RPM. however this point is moot as:

torque defines the force at the wheel. So a huge torque at 0 rpm would give a huge force that had the POTENTIAL! (READ THIS 50 TIMES) POTENTIAL! to move the block.

if you applied this 0 times per second. which is where power comes in. you would not move the block. as the potential is not being used.

likewise if you had a force that ws not large enough DID NOT HAVE THE POTENITAL due to incorrect gearing. you could have all the power available in the gear but you wouldn't shift he block.
THIS IS THE POINT YOU NEED TO KNOW BOTH! POWER AND TORQUE TO KNOW IF IT'LL MOVE. YOU CAN KNOW POWER and RPM or torque abd RPM. or anything else, but in the end you'll always be boiling down the equations for a torque and power.

 

[xxChrisxx]

thats me done for this now. Read the books I posted.

 

[zanick]

As far as being wrong, you haven't pointed one single point out to that effect. I would certainly like to hear that. What we have been arguing about is the chicken and egg aspects of force and power. I still think its a valid debate. I don't need to win it, as long as we can discuss the way we can apply power and torque curves to race car performance.

The biggest problem that I see with your argument as I see it, is that you keep on bringing the comparisons to the same gear for two equal power sources. Two problems happen here. This gives the power source with a higher rpm, lower torque characteristics a situation where much less than its full HP, or comparitive HP can be realized and we are comparing output speeds that are grossly different. Since, power is the RATE of doing work, and Work is Force x distance. If you have just a force, applied at ANY vehicle speed, it can be created by a high rpm, low torque power source, or a low rpm high torque source. Power dictates a force at any vehicle speed. No net force is applied, until there is movement. if there is movement, there is work, and the faster the rate of doing work the greater the power.


Your analogy if you don't have the torque and you just apply the torque at a faster rate, you still don't move the load is flawed. You are looking strictly at the power source and not what is available at the drive wheels. as you apply the same torque at a higher drive wheel speed, the power requirements to move the load go way up! You have to keep the vehicle speed the same in any comparison , or the comparison or test is flawed. Gearing is used to keep a vehicle as close to its max HP range as possible. in the end, the most amount of hp-seconds used , wins the acceleration race over any speed range.

You still never answered the truck vs the F1 engine powered truck. You made a bold statement that the F1 powered engine, at the same HP counldnt move the load up the hill. IN fact, it can. This is very easy to prove by using the F1 engine at its max HP at higher rpm and with deeper gearing.

Power is a force's capacity to do work. my 6 year old can generate 600ft-lbs of torque. Its not going to do much work, or accelerate anything very fast, because his rate of doing work is very small. However, a 600ft-lb torque engine with 6000rpm available, will be able to accelerate a 3000lb car to 100mph in near 10 seconds!

If you are comparing two same HP engines, or motors, and you don't keep the relative vehicle speeds the same, you are not allowing one of the engines to be at or near its useable HP range.

I do care about being correct. I have asked many times to discuss the venacular of the proper use of watt-seconds, HP-seconds, in looking at the time spent at the higher rpm of the HP curve and the total rear wheel forces created in those engine speed ranges.
certainly, when ploting to vehicle speed, you could use gear ratios, engine torque curves and find optimal shift and thus acceleration values. This will exactly parallel engine HP curves.(as measured at the rear wheels off a dyno, as well as engine torque values as calculated back to the engine). "Parallel" meaning the effects, not the rear wheel torque curves .

If I have this wrong, please show me one example where proportiona gearing is available, where this is not true with two equal HP cars with one having much more engine torque than the other. Here is a couple of engine HP curves of the exact, real life example. which powerplant would you want if you were going racing on a road course?


Again, not trying to get into a wrestling match here. Just in search of truth. Trust me, I know what I am saying works out on paper, works at the track and follows alll the basic equations. But, I've come here for the correct terminology and ways of explanation.

If I am wrong, I CERTAINLY want to know it.

Again, here are two different engines, both at 290rwhp, but grossly different torque values. I contend that the lower torque engine of the two is going to be better on the race track at any speed or point on the race track. use what ever gears you want that suits one over the other, as long as the shift points are at the same MPH in vehicle speed.
See if you agree or not.
mk

Look if you care in the slightest about being correct about the physics of this.

Buy/borrow/steal and read the following books.

John Heywood - Internal Combustion Engine Fundamentals
Richard Stone - ICE
A.G Bell - Four-Stroke Performance Tuning
Paul Van Valkenburgh- Race Car Engineering & Mechanics

1. is probably least relevant but is pretty much the engine bible.
2. stone is good for practical workings.
3. Good non technical book reguarding practical tuning
4. a very good all round car setup book. especially the section on gearing and performance.

I have read all of these at some point over the last 3 years. They are all thorough and will show you where you've gone wrong.

Read Bell first. then any of the others. I am in no way inclined to type out whole sections from the book.

 

[zanick]

I think I might understand out disconnect from what you wrote below. You seem to be saying, if we can play with gearing the point is moot. same HP, will create the same acceleration at any vehicle speed. And you say, if you don't have the abilty to change gears to optimal values to equal two vehicle's speeds for comparison, you need to know power AND torque. however, i contend if you know the power it is enough. You are talking about a power rating that you may or not be able to use due to lack of gear selections. If this is the case, then of course, without the lack of gear selections, you won't be able to utilize the POWER potentially available. Again, if you can realize the power, you will be able to move the load with a predicted force and will know the rate at which it can be accelerated at any velocity. It is the power rating that will determine the rate of acceleration at any speed. If you are not able to utilize the power rating at that speed, then you are operating at a lower power setting. all this proves is that you didnt gear the car properly for the test.

mk

likewise if you had a force that ws not large enough DID NOT HAVE THE POTENITAL due to incorrect gearing. you could have all the power available in the gear but you wouldn't shift he block.



THIS IS THE POINT YOU NEED TO KNOW BOTH! POWER AND TORQUE TO KNOW IF IT'LL MOVE. YOU CAN KNOW POWER and RPM or torque abd RPM. or anything else, but in the end you'll always be boiling down the equations for a torque and power.

 

[xxChrisxx]

I am going to make one final statement (its not this one its coming next) addressing your last post and then that's it.

I have no inclination to try to help you to understand the flaws in your thinking if you just keep repeating what you have said before. I have reccomended some books that, IF you read them, you will see where you went wrong.

 

[xxChrisxx]

I invite you to respond to the below, if you wish to dispute anything said you must give evidence to back up your claim. The evidence must be listed so that it can be found by all. By evidence I mean something remotely scientific, not an observation you made and are asserting as true.

As far as being wrong, you haven't pointed one single point out to that effect. I would certainly like to hear that. What we have been arguing about is the chicken and egg aspects of force and power. I still think its a valid debate. I don't need to win it, as long as we can discuss the way we can apply power and torque curves to race car performance.

WHERE YOU ARE INCORRECT:

YOU STATE: chicken and the egg debate about force and power.
REALITY: It is NOT a chicken and egg thing. The physical process is you apply a force and something moves a given distance. You cannot apply power at a speed and create a force.

The physical process is Newtons first law of motion.

"A body persists its state of rest or of uniform motion unless acted upon by an external unbalanced force." Newton 1st Law

The above do NOT work in reverse, something simply cannot start moving spontaneously and a force is created as a result. Because of this:

P = F * V is true, a force acting at a speed will give a power.
P/V = F Cannot occur in nature, as it breaks Newtons 1st law. What you are doing with the equation in this form is finding the NECESSERY force to create a power at a gien speed. It does not mean, if you apply a power at that speed you will get that force.

Energy, and by extention power cannot be directly measured. It is always calculated from more fonamental things.

There is no absolute measure of energy, because energy is defined as the work that one system does (or can do) on another. Thus, only of the transition of a system from one state into another can be defined and thus measured.

"Methods

The methods for the measurement of energy often deploy methods for the measurement of still more fundamental concepts of science, namely mass, distance, radiation, temperature, time, electric charge and electric current." wikipedia. but you can get this from pretty much any physics book.Conclusion to this: Power cannot be directly measured, it requires more fundamental principles to quantify and give context. Applying a power at a speed cannot occur as it violates Newtons 1st law of motion. The fundamental factor in this must therefore be force distance and time. Of which the only one that has potential to do work is force.

Force therefore is first, and power is a method by which we quatify what that force has done.

The biggest problem that I see with your argument as I see it, is that you keep on bringing the comparisons to the same gear for two equal power sources. Two problems happen here. This gives the power source with a higher rpm, lower torque characteristics a situation where much less than its full HP, or comparitive HP can be realized and we are comparing output speeds that are grossly different. Since, power is the RATE of doing work, and Work is Force x distance. If you have just a force, applied at ANY vehicle speed, it can be created by a high rpm, low torque power source, or a low rpm high torque source. Power dictates a force at any vehicle speed. No net force is applied, until there is movement. if there is movement, there is work, and the faster the rate of doing work the greater the power.

This (the power speed force bit) is incorrect for as it violates Newtions first law as stated above.

By the same reason that force and speed can bve used to find power:

P = T * angular velocity torque and angualr velocity can be used to find power
P/angualt velocity = T you can't do it this way. The find the necessary torque. not the actual.You say that the faster the rate of doing work by the engine the greater its power output. This is indeed correct. Now for where you are going wrong with it.

"The engines power output is set. It is defined by:

P=(thermal efficieny * volumetric efficieny * displaced volume * number of revolutions * fuel heating value * density of the air * (fuel air ratio)/2" Equation from Heywood

This defines the amount of work a given engine can do. The POWER OUTPUT IS SET IT DOES NOT CHANGE DEPENDING ON THE GEARING. So if you geared it 1:1 you'd have the same power at the rear wheel as if you used a 100:1 gearning. It would be going this work more slowly, but it would be doing more work.

You are NOT doing more work at the rear wheels by using a higher gear ratio.

"W = F * d" wiki or books.

You are using a higher force but moving

 

[rcgldr]

I think we all understand that torque corresponds to force, and power is the same for both angular and linear environments.

If you only know the torque, gearing, and radius of tire, then you know the force, but you won't know the speed at which the force can be applied.

If you only know the peak power, then you don't know the torque versus rpm curve and wouldn't know how to set the gearing.

If you know torque or power versus rpm, then you can calculate power or torque versus rpm from that information, and you would know how to optimize the gearing for a specific load range, and the rate of acceleration versus speed for a given load.

 

[xxChrisxx]

Yup that's bang on the money Jeff.


OP doesn't understand that its torque not power deermining the magnitude of the force. as he's stated that power determines the acutal force at the rear wheels when it doesnt.

 

[zanick]

Exactly, but with one comment. you would need to know the HP vs RPM (not just peak HP) as you say in you last sentence, which has been the point of the discussion. by knowing the HP curve, you can optimize max acceleration for any range of speeds by using gearing, without knowing the engine torque values. you could also calculate the applied forces as well at any vehicle speed.

mk

I think we all understand that torque corresponds to force, and power is the same for both angular and linear environments.

If you only know the torque, gearing, and radius of tire, then you know the force, but you won't know the speed at which the force can be applied.

If you only know the peak power, then you don't know the torque versus rpm curve and wouldn't know how to set the gearing.

If you know torque or power versus rpm, then you can calculate power or torque versus rpm from that information, and you would know how to optimize the gearing for a specific load range, and the rate of acceleration versus speed for a given load.

 

[zanick]

I fully understand that torque does the work (or force does the work) . But, HP ratings (rate of doing the work) can determine rear wheel torque, or forces at any vehicle speed. in otherwords, go back to the electric motor and the power source. If the ratings are such that we can create x-KW-hours, with that unit measure of work, we can determine how fast we can do that work and for how long. just as if we had a power sourse being a spinning flywheel that has stored KE. That can be applied quickly or slowly. The force will be a result of how that energy is released. In the end, the change of KE is the work done. the rate of change is the Power.
For any given power, or rate of change of KE, we can determine the force acting on a moving body at any velocity.
This discussion has clearly changed my way of thinking of power as more of a rating, but none the less, if you know the rating, and it is being used, you can determine the force at any speed.

Ill take a look at a couple of the books you have listed, but I think if anything, it will change my way of explaining what we see in reality. Did you have a chance to look at the HP curves I supplied. tell me which one would be better on a race track at any speed. Only given is say the cars have 4 gears, all with 25% rpm drops per shift. Keep MPH shift points the same, to level the playing field and let me know what you think. In the end , the lower torque engine, will have more rear wheel torque. why, because it has more average HP and more HP-seconds available to create the forces at the rear tires.

mk

Yup that's bang on the money Jeff.


OP doesn't understand that its torque not power deermining the magnitude of the force. as he's stated that power determines the acutal force at the rear wheels when it doesnt.

 

[zanick]

Chris, I am going to do my best to answer your challenge. Again, I am not questioning that the force is what creates movement. I am saying that power can create the force. doesn't the explosion create the force. (force doesn't create the explosion) Energy can't be created nor destroyed, right? Energy is converted to mechanical energy (watt-seconds), J, etc. or do I have that backwards?

See my inserts to you comments and see If I am on the right track. with the:
>>>>>>>>>>>>>>>>>>>>>>>>

I invite you to respond to the below, if you wish to dispute anything said you must give evidence to back up your claim. The evidence must be listed so that it can be found by all. By evidence I mean something remotely scientific, not an observation you made and are asserting as true.




WHERE YOU ARE INCORRECT:

YOU STATE: chicken and the egg debate about force and power.
REALITY: It is NOT a chicken and egg thing. The physical process is you apply a force and something moves a given distance. You cannot apply power at a speed and create a force.

The physical process is Newtons first law of motion.

"A body persists its state of rest or of uniform motion unless acted upon by an external unbalanced force." Newton 1st Law
>>>>>>>>>Agreed, but if I have a spinning flywheel with a known KE, if I release that through a clutch, isn't a force created. I stopped the flywheel and use the rate of change of KE to produce a rate of doing work, (i.e. Power) Didnt I create the force by converting he KE to another form. Force did the work but didn the KE produce the force?


The above do NOT work in reverse, something simply cannot start moving spontaneously and a force is created as a result. Because of this:

P = F * V is true, a force acting at a speed will give a power.
P/V = F Cannot occur in nature, as it breaks Newtons 1st law. What you are doing with the equation in this form is finding the NECESSERY force to create a power at a gien speed. It does not mean, if you apply a power at that speed you will get that force.

Energy, and by extention power cannot be directly measured. It is always calculated from more fonamental things.

>>>>>>>>>I agree. Even in its best case, HP would be calculated form a rate of change of KE. you need to know speed change vs time and mass.

There is no absolute measure of energy, because energy is defined as the work that one system does (or can do) on another. Thus, only of the transition of a system from one state into another can be defined and thus measured.

"Methods

The methods for the measurement of energy often deploy methods for the measurement of still more fundamental concepts of science, namely mass, distance, radiation, temperature, time, electric charge and electric current." wikipedia. but you can get this from pretty much any physics book.


Conclusion to this: Power cannot be directly measured, it requires more fundamental principles to quantify and give context. Applying a power at a speed cannot occur as it violates Newtons 1st law of motion. The fundamental factor in this must therefore be force distance and time. Of which the only one that has potential to do work is force.

Force therefore is first, and power is a method by which we quatify what that force has done.
>>>>>>>>>>>>>>Doesnt KW-hours, watt-seconds, HP-seconds quantify how much work can be done? If I have a battery or a gas tank filled with electrons or fuel, I have potental energy here. is the capacity for how much work can be done. I can do it fast or slow, in the end its work (fs). How fast I do that work, or rate of doing work is power. If push a button and release electrons in motor at a known rate, and it is converted, i know, at any given speed, what the force will be. The force does the work, but the power is the indicator of what the force will be at any particular speed.




This (the power speed force bit) is incorrect for as it violates Newtions first law as stated above.

By the same reason that force and speed can bve used to find power:

P = T * angular velocity torque and angualr velocity can be used to find power
P/angualt velocity = T you can't do it this way. The find the necessary torque. not the actual.


You say that the faster the rate of doing work by the engine the greater its power output. This is indeed correct. Now for where you are going wrong with it.

"The engines power output is set. It is defined by:

P=(thermal efficieny * volumetric efficieny * displaced volume * number of revolutions * fuel heating value * density of the air * (fuel air ratio)/2" Equation from Heywood

This defines the amount of work a given engine can do. The POWER OUTPUT IS SET IT DOES NOT CHANGE DEPENDING ON THE GEARING. So if you geared it 1:1 you'd have the same power at the rear wheel as if you used a 100:1 gearning. It would be going this work more slowly, but it would be doing more work.

>>>>>>>>>>>>>>>>>>>You just made my point right there. sure, if I was at max power and we had 1:1 gearing, the force would be a heck of a lot different greater, at the driven wheels at 100:1 gearing. I think you made a critical mistake next. The rate of doing work would be IDENTICAL the amount of work per unit of time would be identical. Since we are talking the same " car " here. if power is the same in both cases, acceleration and thus rear wheel forces will be proportional to power and inversly proportioal to speed. IN this case, fundamemtally, you changed the conditions by comparing power at two different speeds. (100:1 vs 1:1). power is constant, rate of work is the same, acceleration and force are different based on the car's speed.
To answer your comment directly, if power didnt change with gearing, then the HP curve would be flat, and torque would be very high at start and very low at the higher speeds . If this was the case, rate of work still would be equal and forces would go down with speed proportional to speed.


You are NOT doing more work at the rear wheels by using a higher gear ratio.

"W = F * d" wiki or books.

You are using a higher force but moving

>>>>>>>>>as said above, you are doing the same amount of work, if you are using ANY ratio, if the power is the same. high rpm, low torque vs low rpm, high torque. If they trade off, the amount of work stays the same. Getting back to lifting a mass, 550lbs lifted in 1 second, one foot, is 1hp. 1000lbs lifted 1 foot in 2 seconds is 1 hp. 275lbs lifted 1 foot in .5 seconds is still 1 Hp. (746watt-seconds, 746J, 1hp-second, etc).

I can only think about this in a much broader analogy. what meteor objects to fall to earth, the Earth's mass (power) or gravity (force)? The fact that the Earth has such a large mass creates force that cause objects to be attracted and fall to earth. the more objects that fall to earth, the more the mass the Earth gets and the the greater the force of gravity . Its "power" goes up and creates a greater force. This is where I am either very confused, or think think the discussion is a kind of "chicken and egg" one. :)

 

[xxChrisxx]

Please buy some books and read them.

No you didnt create a force from the energy. It was the momentum stored that caued the force, due to conservation. Energy stores are not good examples of what you are trying to say as they require forces in.

You can express the energy that its storing, but that energy doesn't create the force.

Read any statics book. Or the flywheel section in Shigley - Mechanical engineering design.YOU DONT USE KE OR RATE OF CHANGE OF IT TO CALCULATE FORCES! YOU USE MOMENTUMS! USING KE IS NOT REPRAT NOT WHAT HAPPENS IN REAL LIFE!
Tell me how do we measure power or energy directly. That means not using any more fundamental principle? I want to see proof of this.I'll respond to your odd thinking about gravity at a later time.