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

by zanick
Tags: cars, country, race, torque, undiscovered
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xxChrisxx
#55
Mar27-09, 01:13 PM
P: 2,048
and by the way, if you were comparing two different engines. It'll be more correct to compare bmeps as they are dimensionless.
zanick
#56
Mar27-09, 01:14 PM
P: 49
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 pontential force is though right? or is it just a distilled version of what you need to really look at, an indicator when you dont have the time or the information to see the actual torque created at the rear tires.

Lets say you have a dyno, (inertia). you dont 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

Quote Quote by xxChrisxx View Post
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, isnt 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
#57
Mar27-09, 01:25 PM
P: 2,048
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. Lets 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 wont necessarily transfer well to the tack.
zanick
#58
Mar27-09, 03:15 PM
P: 49
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

Quote Quote by xxChrisxx View Post
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. Lets 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 wont necessarily transfer well to the tack.
zanick
#59
Mar27-09, 03:19 PM
P: 49
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 doesnt 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
#60
Mar27-09, 03:43 PM
P: 2,048
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 thats 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, im just assuming Power.S.

For the rear wheel, time based one. I just dont know im 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
#61
Mar27-09, 06:01 PM
P: 49
Thanks Chis, I think Im 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



Quote Quote by xxChrisxx View Post
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
#62
Mar27-09, 08:16 PM
P: 143
Quote Quote by zanick View Post
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 pontential 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
#63
Apr2-09, 06:26 PM
P: 49
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. Im 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 with out 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




Quote Quote by Gnosis View Post
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
#64
Apr2-09, 07:07 PM
HW Helper
P: 7,135
Quote Quote by Gnosis View Post
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
#65
Apr2-09, 07:10 PM
P: 2,048
Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy wont tell you the force nor acceleration. K.E = .5mv^2
Just because they use the same terms doesnt make them mathematically compatble. The dyno doesnt acutally measue energy, as energy cant 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
#66
Apr2-09, 07:30 PM
P: 49
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

Quote Quote by xxChrisxx View Post
Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy wont 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
#67
Apr2-09, 07:38 PM
P: 2,048
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 thats 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 doesnt mean you can do it in reality.


Also i'll discuss the climbing a hill thing with you in a bit, as thats 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
#68
Apr3-09, 03:55 AM
HW Helper
P: 7,135
Quote Quote by xxChrisxx View Post
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 travelled 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
#69
Apr3-09, 08:59 AM
P: 2,048
Quote Quote by Jeff Reid View Post
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
#70
Apr3-09, 09:53 AM
P: 49
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, cant 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, doesnt 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? Im 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 wouldnt 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 Im 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 doesnt adress the time factor, thats 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
#71
Apr3-09, 10:11 AM
P: 2,048
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 explaination.

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 doesnt 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 isnt 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 wont 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 wont 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
#72
Apr3-09, 11:17 AM
P: 49
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 explainations.

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

Quote Quote by xxChrisxx View Post
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 wont 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 wont have the pulling power to climb with several tons attached.

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


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