I will actually model the HP curve with a higher order polynomial, and not use the torque curve at all in the problem when calculating motion. I use RPM steps, and calculate all the changes in energy. Then by dividing the energy change by the average power over the RPM interval, I get the elapsed time directly. All the distances fall out easily too.
Driving it TOO easy....
- Thread starter LouDawg
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JHZR2
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I get:
HP=((ft*lbf)/min)/(5252 ((ft*lbf)/(HP*min)))
so the units on 5252 are ft*lbf/(HP*min), so when the function is divided by this, the HP goes up to be the new unit, the minutes go up to divide out the RPM, and then youre left with HP.
However, that doesnt make 5252 any less of a scalar or conversion factor... its not a function in any way... its no different than saying that there are 2.54cm in 1 in or 3.875L in 1 Gal. Its just more complex.
That said, lets go back... Think of a watt in terms of an engine power output spec, or as power for your home.
1W = 1 J/s
0.746 kW = 0.746 kJ/s = 1 HP
1HP = 550 ft*lbf/s = 0.746 kJ/s
Ttherefore:
550 ft*lbf = 0.746 kJ
So,
550 ft*lbf = 0.746 kJ = work = energy = 0.746 kg*m^2 / s^2 = 0.746 N*m
And a Newton-meter is a definition of torque...
So HP is just torque in terms of time.
So one is the analog of the other.
JMH
HP=((ft*lbf)/min)/(5252 ((ft*lbf)/(HP*min)))
so the units on 5252 are ft*lbf/(HP*min), so when the function is divided by this, the HP goes up to be the new unit, the minutes go up to divide out the RPM, and then youre left with HP.
However, that doesnt make 5252 any less of a scalar or conversion factor... its not a function in any way... its no different than saying that there are 2.54cm in 1 in or 3.875L in 1 Gal. Its just more complex.
That said, lets go back... Think of a watt in terms of an engine power output spec, or as power for your home.
1W = 1 J/s
0.746 kW = 0.746 kJ/s = 1 HP
1HP = 550 ft*lbf/s = 0.746 kJ/s
Ttherefore:
550 ft*lbf = 0.746 kJ
So,
550 ft*lbf = 0.746 kJ = work = energy = 0.746 kg*m^2 / s^2 = 0.746 N*m
And a Newton-meter is a definition of torque...
So HP is just torque in terms of time.
So one is the analog of the other.
JMH
JHZR2
Staff member
This is fun... and not even really OT!!!
Post up your results so we can see what youve done and what you get!!!
JMH
I'm afraid that you're incorrect.
Well, let's break this down. TORQUE is a force, and POWER is a measure of work done (in the physics sense) which is defined as "force through a distance".
Acceleration of a simple body follows Newton's law F=ma, or to solve for acceleration a=F/m. All units are metric, so force in Newtons, mass in kg, and acceleration in m/s^2.
How is a car accelerated forward? It can only be accelerated along the ground by a force parallel to the ground. Where does this come from? The wheels. Wheels translate a rotational force from the drivetrain into a linear force at the ground. This linear force alone is responsible for accelerating the vehicle forward and is expressed in weight-like quantities (Newtons, in the most basic case without doing conversions). At this point it's more like dealing with thrust, which sometimes is measured in pounds. If you want to accelerate a 1000kg car at 1g (ie. Porsche Turbo), 1g=9.8m/s^2 so
F = 1000kg * 9.8m/s^2 = 9800Newtons
A Newton is defined as the force which, when applied to one kilogram mass, causes an acceleration of 1 meter/sec2. So a kilogram weight under gravity creates a force of 9.8N. So we simplify back to 1000kg (equivalent) or 2200lbs of linear force are required for this acceleration example. I know mass and weight aren't the same thing, but that's not the topic here.
As I said, this linear force is derived from the drivetrain which provides a rotational force. Rotational forces are defined as a force acting at a distance from the center of rotation. This leads to the units ft-lb (one pound of force applied to a one foot radius from center along a tangent of the circle) to measure twisting force. In metric, Newton-meters is used (you've all seen this on your metric torque wrenches).
So, if you need 9800 Newtons (2200lbs) to accelerate your vehicle at 1g, and you have 24" tires (outer diameter, not rim size) you will require a torque at the wheel of 2200 ft-lbs. A 24" diameter wheel has a radius of 12".
The transmission and final gearing provide torque multiplication factors from the engine's base torque output. By gearing down you increase torque, hence 2200 ft-lbs at the wheels can be possible in, say, first gear on some very nice cars.
So, through this crappy example I've illustrated that FORCE is required to cause the acceleration of a rolling body. Horsepower, being a measurement of POWER, is force through a distance. In this case, the "distance" is rotational so RPM is used. Actually, to get specific, RPM is already a complex unit (revolutions per minute) that also involve time, so it gets more complex. The '5252' from the equation is mostly a scalar to translate from metric into imperial units, radians into revolutions etc...
How do you propose to take F=ma and turn it into F*revs/minute=ma? The units are wrong for the equations, and mathematics won't allow you to simply add distance (RPM) to one side of the equation of carry on as if nothing happened. The very fact that horsepower is DERIVED from torque should tell you that torque is the actual motivating force involved.
Horsepower ratings are only indicating that your engine can continue to provide a certain torque while spinning at a given speed. Generating 100 ft-lbs of twist at 10,000RPM requires a lot more energy than at 10RPM, hence a lot more "work" is done, hence it is still a measure of engine output. It is not, however, the basic source of acceleration.
The road acceleration of the vehicle depends SOLELY on the linear force applied at the road, which is the twisting force from the drivetrain, minus wind resistance etc... The mass being accelerated doesn't care how fast your engine is spinning, or how fast it is already going, as long as the twist is there.
So - find the *torque* peak of your power plant and use the transmission to convert between wheel speed and engine speed so you can keep the engine near peak "regardless" of your road speed.
To clarify my 'integrate under the torque curve', it is most correct to integrate under the output torque curve of the transmission, accounting for the multiplication factor of each gear to derive optimum shift points. I've never actually bothered, I drive by feel and sound.
Craig.
The problem solutions are massive, so I will try to give a simple example after I read Craig's post.
Also, you are calculating the magnitude at which HP and torque in ft.-lbs are equal. I am just saying that they are unequal because one is torque and one is power.
power(HP) = torque(ft-lbs)*rot. speed(RPM)/5252(ft-lbs*RPM/HP)
Me, too.
When your shift points are correct for WOT acceleration, the acceleration of the car just before and after the shift will be the same. This coincides with the power at the wheels being the same before and after the shift. Use the whole powerband to get this.
Once the car is in motion, the maximum force at the wheels at any given car speed will be realized by running in a gear that gets best POWER from the engine. Do a numerical example, and you will see. 100% of racers cant' be wrong (nor me).
The rawest solution to the forward acceleration of the car is:
a=P/(mV)
So for any car speed V, you must maximize power to maximize acceleration. You don't need to know anything about the engine torque, gearing or tire diameter. You just need to relate engine power to car velocity, and drive in the powerband.
So, build a high and broad powerband and select gears to use it.
erm ,numerical error, my first example with 700 rpm gear gap didn't use 700 rpm in the calculations below. Hopefully the point is still clear.
Holy Mackerel! Remind me to always call you guys sir!
And this is probably the basics, right?
Guess I'm going to have to buy some Dyno-time, fellas! I have a feeling these factors are of such minor import on this import, pardon the pun, that the factory to my knowledge hasn't put this data out anywhere I can find it. The gear ratios are in the spec, but torque/hp curves as defined by a dyno aren't to be found in my documents. If I know you three characters, or four, including Craig, I put could put the gear ratios for 1-2-3-4-5 and tire size out there and I be willing to bet between you, you'd come pretty close to calculating torque and hp. The magic of mathematics, eh?
Pretty fascinating stuff though.
[ May 12, 2006, 03:40 AM: Message edited by: toocrazy2yoo ]
(in a little, weak voice)...ummm, I sometimes run my rpms up close to redline. Maybe once/twice per tankful of gas. I think it's a good idea. Oh, and 3+4=7. I hope.
Back to the advanced math lesson....
Back to the advanced math lesson....
Ssssssooooooooo...back to my original question....should I drive it harder or not?
Just kidding...you guys are WAY to technical for me!
Just kidding...you guys are WAY to technical for me!
Wow, who knew that a simple formula for HP would start all this?
Just remember than any equations & formulas you use to calculate accleration, speed, etc are, at best, *approximations* of what will(or "should") happen in the real world.
Something I've noticed for years now, on automotive websites & in the classroom, is that people usually pick a few variables they're interested in & leave out the rest. Remember: It's the power applied to the ground that gets the job done(what kind of traction do you have? do your calculations have a place for this?), air resistance we have with us always(temperature/density/wind speed/direction/etc), ambient temperature & humidity affect engine power, & the list goes on.
So have fun with it, but remember to include *all* the variables you can think of. And someday, if you get good enough at it- well, there's always rocket science. Now, *that's* a hot rod!
Just remember than any equations & formulas you use to calculate accleration, speed, etc are, at best, *approximations* of what will(or "should") happen in the real world.
Something I've noticed for years now, on automotive websites & in the classroom, is that people usually pick a few variables they're interested in & leave out the rest. Remember: It's the power applied to the ground that gets the job done(what kind of traction do you have? do your calculations have a place for this?), air resistance we have with us always(temperature/density/wind speed/direction/etc), ambient temperature & humidity affect engine power, & the list goes on.
So have fun with it, but remember to include *all* the variables you can think of. And someday, if you get good enough at it- well, there's always rocket science. Now, *that's* a hot rod!
Wow... This reminds me of math class. I believe I remember reading somewhere that a BMW 1980s manual suggested that people run in high rpm to clean the engine after a long drive at low rpms...
1980...SF oil, sludge, Fram, man, those were the days, eh?
JHZR2
Staff member
It does... my 91 e30 manual says to keep it over 3000 for a few minutes aftr traffic to let the engine 'breathe'.
JMH
Reckon I may as well jump in and throw a monkey wrench in the works.
Torque is a twisting force. Nothing more. A piston type steam engine has the ability to develop 100% torque at zero RPM. Also certain electric motors can do this (for a while). When this occurs, THERE IS NO HORSEPOWER being developed! (no work accomplished)
Horsepower defines Work accomplished in a period of Time. James Watt developed this oft mis-understood method of rating an engine's power in an attempt to relate the capabilities of his new steam engine to folks who had no clue about mechanics, but they did know what a horse could do. After some testing he detirmined that an average work horse could do 550 lbs/ft per second of work somewhat continuously.
Obviously, internal combustion engines cannot normally produce torque at zero RPM.
A hydraulic cylinder can do work, which can be expressed in lbs/ft per unit of time (horsepower), but develops zero torque.
Torque and horsepower can be related, but they are NOT the same thing.
Joe
Torque is a twisting force. Nothing more. A piston type steam engine has the ability to develop 100% torque at zero RPM. Also certain electric motors can do this (for a while). When this occurs, THERE IS NO HORSEPOWER being developed! (no work accomplished)
Horsepower defines Work accomplished in a period of Time. James Watt developed this oft mis-understood method of rating an engine's power in an attempt to relate the capabilities of his new steam engine to folks who had no clue about mechanics, but they did know what a horse could do. After some testing he detirmined that an average work horse could do 550 lbs/ft per second of work somewhat continuously.
Obviously, internal combustion engines cannot normally produce torque at zero RPM.
A hydraulic cylinder can do work, which can be expressed in lbs/ft per unit of time (horsepower), but develops zero torque.
Torque and horsepower can be related, but they are NOT the same thing.
Joe
Ok, Lazy. not to seem stupid (ok, ok, I'm an idiot, but at least I know what I don't know, which makes me smarter than when I was a younger man!), how does a piston steam engine develop torque at zero RPM?
JHZR2
Staff member
there would be zero HP at zero RPM because torque (as a function of RPM in, say an IC engine, or not in some other work-performing device) times RPM would be zero if RPM is zero...
And similarly, at 0.05 RPm or whatnot, HP would be EXTRMELY low, while that steam engne could be doing a LOT of work.
Such comparissons arent particularly relevant in an IC engine discussion, IMO.
JMH
And similarly, at 0.05 RPm or whatnot, HP would be EXTRMELY low, while that steam engne could be doing a LOT of work.
Such comparissons arent particularly relevant in an IC engine discussion, IMO.
JMH
Not terribly relevant to a discussion about gasoline engines perhaps, but it is important to understand the difference between torque and horsepower.
To wit: Torque can be exerted without any work being accomplished,
And: Horsepower can be developed in a linear path with no torque being exerted.
Using a steam locomotive for example, if it were coupled to a load it could not move, and assuming the wheels had sufficient traction so they could not spin, the engineer could apply full steam pressure to the pistons, yet it would not move. Torque (twisting force) is being applied, but no work is done, therefore no HP is developed.
If you try to loosen a bolt that is stuck too tight to budge, you are exerting torque, but no work is done, so no HP is developed (if you keep pulling you will still get tired).
When an aircraft carrier launches an F-14, the steam catapult develops LOTS of horsepower (work done in a given time frame), but again, no torque is developed.
And to further confuse us, most engine dynomometers measure Torque, and then Horsepower is calculated.
Clear as mud, Eh?
Joe
To wit: Torque can be exerted without any work being accomplished,
And: Horsepower can be developed in a linear path with no torque being exerted.
Using a steam locomotive for example, if it were coupled to a load it could not move, and assuming the wheels had sufficient traction so they could not spin, the engineer could apply full steam pressure to the pistons, yet it would not move. Torque (twisting force) is being applied, but no work is done, therefore no HP is developed.
If you try to loosen a bolt that is stuck too tight to budge, you are exerting torque, but no work is done, so no HP is developed (if you keep pulling you will still get tired).
When an aircraft carrier launches an F-14, the steam catapult develops LOTS of horsepower (work done in a given time frame), but again, no torque is developed.
And to further confuse us, most engine dynomometers measure Torque, and then Horsepower is calculated.
Clear as mud, Eh?
Joe
JHZR2
Staff member
excellent interesting points (the F-14 program was officially ended at NAVAIR the other day, BTW) but I guessone ought to say that the measurement and meaning is VERY application specific, eh?
Now, when you say about the catapult... which is just the last point and thus the easiest... that work is done in a given time frame... That is correct... when we are working on designing the power systems for the rail guns and EM catapults for the DDX, CVN21, etc... we think in terms of megawatts, which are a.k.a. horsepower...
OK, fine, then work, joules, are done in time, seconds... Joule per time is a watt... or power...
But a joule is equivalent to a ft*lbf by some conversion factor... see my post previous, so they still are in fact saying the same thing...
a watt=J/s=ft*lbf/s=HP...
JMH
Now, when you say about the catapult... which is just the last point and thus the easiest... that work is done in a given time frame... That is correct... when we are working on designing the power systems for the rail guns and EM catapults for the DDX, CVN21, etc... we think in terms of megawatts, which are a.k.a. horsepower...
OK, fine, then work, joules, are done in time, seconds... Joule per time is a watt... or power...
But a joule is equivalent to a ft*lbf by some conversion factor... see my post previous, so they still are in fact saying the same thing...
a watt=J/s=ft*lbf/s=HP...
JMH
Yup. Watts, Horsepower, etc. are measurements of work done in time frame.
Torque, pressure,etc. are measurements of force.
In a car engine we don't usually get one without the other, so they do get confused.
I didn't know about the F-14 program being ended. When I was in the Marines we were still operating Phantoms.
Joe
Torque, pressure,etc. are measurements of force.
In a car engine we don't usually get one without the other, so they do get confused.
I didn't know about the F-14 program being ended. When I was in the Marines we were still operating Phantoms.
Joe
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