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Old 01-02-09, 04:17 PM   #61 (permalink)
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Quote:
Originally Posted by Gothche View Post
This is not theory it's the fact.
I drive three different types of cars pretty often.
My own 1994 Acura Legend LS Coupe 6-spd, my company car and a 1995 Rover 620ti 5-spd(Turbo) and they all work in completely different ways.
I have told you about the company car with the huge amount of torque but only 140HP(slow car with a really hard pull at the first meters), the Rover 620ti with an early torque peak and late HP peak(very fast all the way, but especially in the early rpm's), The Legend slow in the low rpms but very fast at the high rpms). The Legend is a late peaker.
Of all these cars the fastest from 0-60mph is the Legend, but the Rover is way quicker anywhere else in the lower speeds and from stops. That is very simple, it's the relatively high amount of torque at low rpm's and a high amount of HP at the late rpm's. The only problem with the Rover compared to the Legend type2 is that the Rover will loose a bit of momentum in between 2100rpm's and 6000rpm's because the torque will drop a bit before max HP is reached. The Legend will start slow and then pull harder and harder all the way til past 6000rpm's, then when you shift gear(go up one gear) on a manual at about 6000rpm, it will only drop 600-800rpm's and still be in the high end of the powerband, so the power will be more consistantly all the way except when you start from a stop.
That's a good description. I think a summary of some different torque curves with some qualitative descriptions of what they mean behind the wheel makes the most sense. HP follows from that, but I think the torque curve description is much more intuitive.

Similarly my Camaro is a low RPM torque producer (not as low as I'd presume for a diesel, but peak torque is at [email protected] RPM) that redlines at 5000. I try to avoid direct comparisons in how it drives compared to the legend, since one's an auto and one's a manual, but those descriptions describe my experience pretty well.

I'm not trying to beat everyone over the head with theory, but this is an area in which it applies pretty directly and I think there's some value in adding it to the discussion
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Old 01-02-09, 05:10 PM   #62 (permalink)
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Yes I agree....
I just wanted to get more clear in what I was trying to say in my first response here.
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Old 01-02-09, 06:59 PM   #63 (permalink)
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we understand the differences in Torque and horspower, as what Gotche was just explaining...I explained it in the first post...

More torque is what gets you moving...it keeps diverting into a conversation about units then into what is Low-end torqur versus High-end torque...

It is going around in circles like dasfiktauf said, It's cool though...makes no difference to me...

If you think about it (going back to units) if you stopped an engines flywheel from 3 feet away from its crankshaft...If the engine produced 210lbs/ft, you'd have to apply 70lbs of force to stop the engine (negating the engines inertia)


that says that the LEVER is acting on the engines CRANKSHAFT at a rate of 70ft.lbs. if that breaker bar was 1 foot long you'd need 210ft.lbs to STOP the engine...

working the other way, the engine is putting out 210lbs/ft...

I can't make it any simpler than that...i really can't...

When you talking about Levers it's ft.lbs when you're talking about a power output, it's lbs/ft.


EDIT:

and since you wanted to keep trying to bring in Mechanical advantage (I.E. Gearing) what is a gear?

A gear is a series of levers. How does a gear work? the smaller gear runs at a much higher RPM than the multiplying gear...the reason this works is because the smaller lever has a HIGHER torque rating than does the larger lever FROM is fulcrum...I said from it's fulcrum...

So say you have 200 lbs/ft...it's coming into a gear 12" in diameter and pushing a gear that's 24" in diameter.

the power output of that motor being 200lbs/ft is now 400lbs/ft at the surface of the 6" radius gear. the 24" diameter gear (radius 12") now has 400lbs of force being applied to it...the LEVER action of the LARGER gear then acts in Ft.Lbs on its fulcrum to the tune of 1ft by force applied...

so the 200lbs/ft output motor now is rotating a gear who's FULCRUM OUTPUT is now 400lbs/ft. (Mechanical Advantage/Torque Multiplication)

I believe my math is correct, if not I'll jut have to go back and fix it but I'm almost sure that's right.

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Old 01-02-09, 07:19 PM   #64 (permalink)
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Hmmm.... why do you write all that when your point is that the power output is lbs-ft? that is exactly what I write.
And it doesn't matter if you call it lbs-ft or ft-lbs.
The torque number is the force the engine pulls with put in weight(I know it's a rotating force, but nevermind it's the same thing), no news in that.
What I'm trying to say is that because the type1 is rated with 210lbs-ft and the type2 is rated with 206lbs-ft doesn't make the type1 quicker because you need the point at where it reaches the max torque and the HP rating and where it is produced is very important to that calculation as well. Also the type1 has a smaller powerband of only 1000 instead of 1200 of the type2. Those two engine types has been discused a lot, so they are perfect examples of how torque and HP works. Sure there are other factors like transmission and all that, but when only talking about engines I think we can agree that you can't have one without the other if you want to have a quick/fast car. Cause even though you would like torque and HP too be the same thing, it is not.
Cars with turbo equipped will reach maximum torque at much lower rpm's than normal fuel injection cars, but they won't have anymore of anything, they will just reach the torque peak much earlier and the HP about the same rpm range, giving them a wider powerband. That will cost in the high end due to a small loss opf high end torque, but it will still be strong up there.

Quote:
Originally Posted by B+O View Post
we understand the differences in Torque and horspower, as what Gotche was just explaining...I explained it in the first post...

More torque is what gets you moving...it keeps diverting into a conversation about units then into what is Low-end torqur versus High-end torque...

It is going around in circles like dasfiktauf said, It's cool though...makes no difference to me...

If you think about it (going back to units) if you stopped an engines flywheel from 3 feet away from its crankshaft...If the engine produced 210lbs/ft, you'd have to apply 70lbs of force to stop the engine (negating the engines inertia)


that says that the LEVER is acting on the engines CRANKSHAFT at a rate of 70ft.lbs. if that breaker bar was 1 foot long you'd need 210ft.lbs to STOP the engine...

working the other way, the engine is putting out 210lbs/ft...

I can't make it any simpler than that...i really can't...

When you talking about Levers it's ft.lbs when you're talking about a power output, it's lbs/ft.


EDIT:

and since you wanted to keep trying to bring in Mechanical advantage (I.E. Gearing) what is a gear?

A gear is a series of levers. How does a gear work? the smaller gear runs at a much higher RPM than the multiplying gear...the reason this works is because the smaller lever has a HIGHER torque rating than does the larger lever FROM is fulcrum...I said from it's fulcrum...

So say you have 200 lbs/ft...it's coming into a gear 12" in diameter and pushing a gear that's 24" in diameter.

the power output of that motor being 200lbs/ft is now 400lbs/ft at the surface of the 6" radius gear. the 24" diameter gear (radius 12") now has 400lbs of force being applied to it...the LEVER action of the LARGER gear then acts in Ft.Lbs on its fulcrum to the tune of 1ft by force applied...

so the 200lbs/ft output motor now is rotating a gear who's FULCRUM OUTPUT is now 400lbs/ft. (Mechanical Advantage/Torque Multiplication)

I believe my math is correct, if not I'll jut have to go back and fix it but I'm almost sure that's right.
Where do you wanna go with this? It doesn't matter how you put it, the torque number will still be the same.
If you wanna calculate the time a boy with only one leg is at getting to his school 15 miles from home by using a smart formula instead of just observing and using experience...... well, be my guest, but it doesn't really concern this matter.
You see my point.
The matter is the torque vs HP or the torque+HP=power output.

But don't put it out there as a simple calculation, since it's never that simple.
What you are trying to say is that if a car has a total of 500(HP+torque) with more torque and less HP and a car with 500(HP+torque) with more HP and less torque they will perform the same? I don't see your point.
Help me understand what you are trying to say except this: I'm a math-prodegy and I know all the formulas.

You can find those formulas everywhere, what's the point when that's all you are looking at.

Last edited by Gothche; 01-02-09 at 08:00 PM.
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Old 01-02-09, 07:28 PM   #65 (permalink)
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^

??

I'm talking about the Units used to understand torque.

I'm not talking about what car, has what horsepower/torque, where, and why. I'm pretty sure i've harped about torque long enough in my first thread...

Anyway, lbs/ft is not the same as lbs-ft is interpreted...

lbs-ft is interpreted as "Force times a distance acting on a fulcrum"...lbs/ft is "Force acting FROM a fulcrum to a standardized distance." that distance would be 1 foot...

So in effect, no Lbs-ft and Lbs/ft are not the same.


Quote:
Originally Posted by Gotche
What I'm trying to say is that because the type1 is rated with 210lbs-ft and the type2 is rated with 206lbs-ft doesn't make the type1 quicker because you need the point at where it reaches the max torque and the HP rating and where it is produced is very important to that calculation as well.
Quote:
Originally Posted by B+O
The type II has less force over all, but can produce more work because the force it produces at the higher rpm range is greater than the force the Type I produces, therefore you end up with more work done. a side by side race between a Type II and Type I shows that it's the ability of the Type II to continue producing work after they accelerate from a stop. this will make it (the Type II) pull away.
I said this already...that exact example is in the first post.
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Old 01-02-09, 07:37 PM   #66 (permalink)
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Quote:
Originally Posted by B+O View Post
^

??

I'm talking about the Units used to understand torque.

I'm not talking about what car, has what horsepower/torque, where, and why. I'm pretty sure i've harped about torque long enough in my first thread...

Anyway, lbs/ft is not the same as lbs-ft is interpreted...

lbs-ft is interpreted as "Force times a distance acting on a fulcrum"...lbs/ft is "Force acting FROM a fulcrum to a standardized distance." that distance would be 1 foot...

So in effect, no Lbs-ft and Lbs/ft are not the same.






I said this already...that exact example is in the first post.

Ok yea, forgot that you are much into where it is produced. But nevermind, I think we agree at the end of the day. Just two different ways of explaining.
And what I mean about the lbs-ft and the lbs/ft is that it doesn't really matter what you call it and how many calculations you can make, when at the end of the day you already have the numbers.
These ratings are in lbs-ft, so lets keep it at that.

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Old 01-02-09, 07:41 PM   #67 (permalink)
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To be smart and give you different ratings I could give you all these numbers when answering a question about HP and torque, but at the end of the day would it make any difference?

Acura Legend LS Coupe (type II)

Specs:
Year Introduced: 1993
Kerb Weight: 1,605 kg
Engine type: V 6
Valves: 24 valve
Cylinders: 6 cylinder
Aspiration: Multi-point fuel injection system
Displacement: 3,206cc (3.2 liters)
Compression Ratio: 9.6:1
Fuel: Petrol
Drive: FWD
Transmission: 6 speed manual
Engine location: Front Mounted

Power:
KW KiloWatt: 172 @ 6200 rpm

SAE net HP: 230 @ 6200 rpm

DIN HP(EU): 233 @ 6200 rpm

SAE Gross HP: 267 @ 6200 rpm

Torque:
LBS.-FT. : 206 @ 5000 rpm

NM: 279 @ 5000 rpm


Performance:
0-60mph: 6.7s(6.3s without tcs)

Top Speed: 155mph / 250kph


What you guys use is, SAE net HP and lbs-ft so why would I need to give you NM and DIN HP?
That is what you are doing by putting out numbers to confuse people and make them get lost, when all they need to know is how big the numbers are and where they are reached.

Quote:
Originally Posted by B+O View Post
^

??

I'm talking about the Units used to understand torque.

I'm not talking about what car, has what horsepower/torque, where, and why. I'm pretty sure i've harped about torque long enough in my first thread...

Anyway, lbs/ft is not the same as lbs-ft is interpreted...

lbs-ft is interpreted as "Force times a distance acting on a fulcrum"...lbs/ft is "Force acting FROM a fulcrum to a standardized distance." that distance would be 1 foot...

So in effect, no Lbs-ft and Lbs/ft are not the same.






I said this already...that exact example is in the first post.
But yes I can see you have already made some very good points.
When I'm writing something here, it's not to win a discussion or hang someone up on a mistake he made. I simply told my view on this matter, nothing else.
It seems as if you think I'm trying to have an arguement with you when that is not the case at all.

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Old 01-03-09, 12:13 AM   #68 (permalink)
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Quote:
Originally Posted by B+O View Post
we understand the differences in Torque and horspower, as what Gotche was just explaining...I explained it in the first post...

More torque is what gets you moving...it keeps diverting into a conversation about units then into what is Low-end torqur versus High-end torque...
Yeah, like I said, your sticky is good

I don't think low-end torque vs. high-end torque is a diversion, high-end torque is what gives you a good HP rating. The units talk will go away once you realize you're wrong

Quote:
If you think about it (going back to units) if you stopped an engines flywheel from 3 feet away from its crankshaft...If the engine produced 210lbs/ft, you'd have to apply 70lbs of force to stop the engine (negating the engines inertia)
Your numbers are right, but it backs up my units instead of yours.

3 feet * 70 lbs = 210 ft. lbs. That's it. That's the units. How can you argue that? You take a distance and multiply it by a force. You figured out 70 lbs. by doing 210 ft.lbs. / 3 ft. = 70 lbs. If it was 210 lbs./ft. then 210 lbs./ft. divided by 3 ft. would give you 70 lbs./ft.^2 (pounds per square foot)

If you moved out to four feet you'd only have to use 52.5 lbs of force, because 4 feet * 52.5 lbs = 210 ft. lbs. Completely consistent with real world physics, and the units of distance*force for torque.

Quote:
that says that the LEVER is acting on the engines CRANKSHAFT at a rate of 70ft.lbs. if that breaker bar was 1 foot long you'd need 210ft.lbs to STOP the engine...

working the other way, the engine is putting out 210lbs/ft...
Torque is not a "rate", and the resulting moment at the crankshaft by a 70lb. force on a 3 ft. lever is 210 ft.lbs in order to counter the torque of the crankshaft. At no point is there ever 70 ft.lb. of anything applied to anything, only a 70 lb. force at a distance of 3 ft. causing a 210 ft.lb. torque.

If the breaker bar was 1 ft. long you'd need 210 lbs. to make that 210 ft.lbs. countering the torque of the engine. You apply a force to the lever, which creates a moment or torque at the fulcrum, you don't directly apply a torque to the lever.

Quote:
I can't make it any simpler than that...i really can't...

When you talking about Levers it's ft.lbs when you're talking about a power output, it's lbs/ft.
distance * force has units of distance * force, it doesn't matter if its a torque creating a force at the end of a lever, or a force at the end of a lever creating a torque because it's the same system. There's no such thing as torque acting to or from a fulcrum, either there's a force producing torque at the fulcrum, or a torque at the fulcrum producing a force on a lever or gear. In both cases force is in lbs., distance is in ft., and torque is in ft.lbs.

Units are not some token name applied to the end of numbers, they have physical meaning that is borne out by relations between the relevant components of the system, the same relations used when the numbers are calculated. You can't multiply a distance by a force and get units of force/distance, and to can't do torque / distance = force if the units of torque are force/distance. I can't make it any more simple than that.

Quote:
and since you wanted to keep trying to bring in Mechanical advantage (I.E. Gearing) what is a gear?
I keep bringing it up because if you're lacking torque you can increase it by gearing, just like in the example you give here. It also fits in perfectly with the F1 example. I already agreed that it doesn't necessarily warrant a place in your sticky, but are you of the opinion that a direct torque multiplier between the crankshaft and the axles (while keeping the same HP rating) doesn't warrant a mention in a thread about HP vs. Torque?

Of course if you weren't implying that my discussion of gearing was not relevant to the topic at hand, then you can just ignore me rambling

Quote:
A gear is a series of levers. How does a gear work? the smaller gear runs at a much higher RPM than the multiplying gear...the reason this works is because the smaller lever has a HIGHER torque rating than does the larger lever FROM is fulcrum...I said from it's fulcrum...

So say you have 200 lbs/ft...it's coming into a gear 12" in diameter and pushing a gear that's 24" in diameter.

the power output of that motor being 200lbs/ft is now 400lbs/ft at the surface of the 6" radius gear. the 24" diameter gear (radius 12") now has 400lbs of force being applied to it...the LEVER action of the LARGER gear then acts in Ft.Lbs on its fulcrum to the tune of 1ft by force applied...

so the 200lbs/ft output motor now is rotating a gear who's FULCRUM OUTPUT is now 400lbs/ft. (Mechanical Advantage/Torque Multiplication)

I believe my math is correct, if not I'll jut have to go back and fix it but I'm almost sure that's right.
You have it exactly right, except the units and notation (using "power" instead of "torque" and applying torque at the edge of the gear instead of force)

You have 200 ft.lbs. of torque driving a 12" diameter gear. The force at the edge of this gear is 200 ft.lbs. / 0.5ft = 400lbs (notice that ft.lbs. divided by ft. gives the correct units for force). (In case it's not obvious to anyone reading through this, I use 0.5ft because the radius of the gear is the "lever length"). That 400 lbs of force applied to the 24" diameter gear gives a torque of 400lbs * 1ft. = 400 ft.lbs. of torque.

Also not that it's 400 lbs at the edge of the gear, not 400 ft.lbs. You don't apply a torque at the end of the lever, you apply a force and that force creates a torque. Think about a breaker bar again, you push or pull on the end of it, not try to twist it. Similarly if you get smacked with a swinging bar, it puts a force against you, it doesn't give you a twist.

Your units aren't consistent for trying to apply 400 ft.lbs. there anyway, but that might be where this particular train is getting derailed.

You have it all exactly right, but the units you give aren't consistent with the systems. You haven't responded to my challenge to solve any of these problems with units intact, which is an easy check when you're doing these kinds of things. Sit down with a piece of paper, it all works out

I'm still interested in references for lbs./ft. Just to make sure I wasn't going insane I busted out "Engineering Mechanics: Statics and Dynamics" 9th edition by R.C. Hibbeler.
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Old 01-03-09, 07:01 AM   #69 (permalink)
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Wow, I gotta say you explain it very detailed and as far as I understand, it all seems to add up.
Nice ways of making the torque concept clear.
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Old 01-03-09, 08:00 PM   #70 (permalink)
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Geeze Grammar Police, wtf. If you understand what I'm saying then come off it.

since you can see that i'm telling you exactly what's good, but you're looking at it coming from one direction and I'm looking at it coming from another direction.

Anyway, I'm looking at like this...

In a stoich system, every action has an equal and opposite reaction...

So with infinite traction (a function of coefficient of friction and gravity) You're going to be applying equal for--wait...


As i went up to highlight>Copy>Paste, I noticed you just said EVERYTHING i just said...

Quote:
Originally Posted by sap
force on a 3 ft. lever is 210 ft.lbs in order to counter the torque of the crankshaft. At no point is there ever 70 ft.lb. of anything applied to anything, only a 70 lb. force at a distance of 3 ft. causing a 210 ft.lb. torque.

If the breaker bar was 1 ft. long you'd need 210 lbs. to make that 210 ft.lbs. countering the torque of the engine. You apply a force to the lever, which creates a moment or torque at the fulcrum, you don't directly apply a torque to the lever.
In my haste to be done arguing over something so piddling as "Potato PoTAHto" I mentioned 70ft.lbs versus just 70 lbf on a lever acting as a multiplier of 3 (based on ft.lbs) to create the 210ft.lbs torque load.

what the heck are we arguing about??? we're saying the same thing... COUNTER ACTING the Crankshaft who's rotational force is a quotient of the force applied to it by a lever at an equal distance, is the same as the Crankshaft overpowering a lever to its Stall torque (which would be 0 after it stalled) at 1 foot if that torque applied to it is 210ft.lbs or more...

Does that make sense?

Here's a few math bits to help you see what I'm talking about...

Engine Torque:
210lbs/ft stall torque
A lever's action
210ft.lbs appl. Torque.

Ft.Lbs v lbs/ft

at 210lbs/ft from a fulcrum's applied force

At 1 foot 210ft.lbs is required to stop this right? thereby 210lbf is required to stop the engines rotation. You said this yourself.

at two feet, that pound force (lbf) is 105lbs. engines applied force reduces as you move farther from the point of force application...that would be the center of the crankshaft...

That's the end of my bit on this...It's getting tiresome as I'm repeating myself and we're essentially saying the same thing.

I bid you adieu

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Old 01-03-09, 09:37 PM   #71 (permalink)
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Quote:
Originally Posted by B+O View Post
i figure this should go into the Lounge since the subject matter is amorphic (non make/model specifc)

i kinda look around and see that the two concepts are misunderstood by a great deal of car fanatics. on this forum as well as on the internet and outside of the internet.

what is torque:

Torque, as by its definition, is a twisting force. As by its unit of measurement, it's pretty dern straight forward. Nm (Newtons per meter), Kgm (Kilograms per Meter), Lbs.ft (Pounds per foot, or as we car people changed it to "Foot-Pounds") and any other mixture of unit measurement and unit force.

As this concept relates to automotives, the more torque you have the more massive your acceleration will be. This is a law, so there's no need to find a way out of it. Why? look at one of the fundamental equations the automotive is built upon....F=ma, Force is equal to Mass multiplied by acceleration. flipping this equation around you'll have acceleration is equal to Force divided mass or F/m=a. the higher the amount of force you have, the greater your acceleration will be. nothing more nothing less. that is given all things equal.

The only thing a driver feels when he "mashes" (as we say in the south) the accelerator is torque. there's no mysterious "horsepower" trunk monkey gremlin. it's all torque.

There's a saying that most people have heard, some haven't but it rings true if only people would listen. "Horsepower sells cars, Torque wins races" When you gather all things together this saying is more true than many people give credit to it.

with that said...you maybe thinking:

what is Horsepower:

Horsepower is something that people really dont understand too much. Without going into too much about kinetic physics and stuff, it's pretty much energy capable of being done in an amount of time, just like a Watt (or similarly a Joule with direction). Ever wonder why there's so many ways to calculate the rated power of an engine? you've got horsepower (hp) Pferdestarke (PS), Kilowatt(kW) and some others im missing. The one that should really make you question hp is kW.

What is a Kilowatt? well from the top, a Kilowatt is the amount of energy that can be produced over a certain amount of time with a direction, and this is equal to "work" or action taken on an object. If no motion is made, no work is done. i believe the time is 1 second. energy is a function of Force and distance in relation to time. To get more work done in a set time frame, you need to have more speed, or cover more ground. however you cover more ground is irrelevant at the current time. just know that to cover more ground in the same amount of time. you've had to expend more energy. in order to move 738 lbs 1 foot in one second, you basically need act 1 kilowatt of energy on that object.

now horsepower says that you have the ability to move five-hundred fifty pounds (550lbs) 1 foot, in one second.

Now putting it in terms of engines and power, imagine a crankshaft. you can see it spinning? take the stroke of our C32A engine at 84mm...multiply that by pi, then by rpm, divided by 1000, and then divide that by 60 and that will give you a number i dont know what to do with (j/k) but right now im just showing you what we did

soooo
84 x pi=264
264x (any rpm i'll do 1000)=264,000
divided by 1000=264
divided by 60=4

so...what we have here is four...four whats? 4 meters per second. how? 84mm is the stroke over a 180 angle, or the diameter. multiplying it by pi gives you the circumference, multiplying that by RPM gives you total distance traveled in 1 minute. dividing that number by 1000 turns that measurement from millimeters to meters, and then dividing that by 60 breaks that meters per minute into meters per second.

Now with this you can then say hmmm what if i double the rpm keeping the force the same. remember in this case we're dealing with rotational force i.e. torque. and since i've left torque/force out of this equation we'll call it a constant

blah blah bull crap up to rpm
264 (multiplied by 2000)=528,000
divided by 1000=528
divided by 60=9

Now, 9...9 meters per second if you cancel out the seconds from the watts, you've got distance covered is equal to 9 meters

9 meters covered is greater than 4 meters covered. therefore it's going to produce more energy, therefore more "work" is done.

energy in a direction over a set amount of time is equal to 1 watt. 745 watts equal 1 hp.

this is how engine dynos work. it gives a certain amount of resistance, the force needed to overcome that resistance is measured, if the force stays the same as the distance per second increases, the work done increases or...HP increases.

now when you look at graphs...



if you examine these, you'll find that as rpm, or distance covered in a set amount of time (think about the circumfrencial stroke of the crank shaft as a linear motion as opposed to eccentric or circular) get's greater. as does the work produced as well.

i tried to find the most linear graphs i could. linear torque curve i mean. With the first one it's pretty lumpy but it clearly shows how as the RPM goes up and the torque remains somewhat constant the work produced continues to go up.

in the second one you can see that the graph is still going up even up to its revlimit. that is...the HP is still climbing as well as the torque. But if you look closer, you'll see that HP is climbing at an exponential rate compared to torque. if the engine were to keep a constant torque (force) from 5000 through to 10,000 rpm the hp would be double that at 5000rpm, simply because the linear distance of the crankshaft doubled and inherently producing more work...



another one. this is a really good one. it's CLEARLY evident that the relationship of force to a distance over a set time is what is working here.

notice how the torque curve is flat, and the rpm just climbs and climbs...keeping the torque curve flat like that through 20,000rpm would produce. about 480hp. think on that now.....you'd say "FAWK! 480hp from a 2.0 4Cyl. naturally aspirated"....now then look at the torque....

125lbft of torque at 20,000rpm will give you 480hp. this is how F1 engines produce such ridiculous amounts of "WORK DONE" that can continue to accelerate long after a car of the same amount of power would have had to shift. there for more work is done and less time is wasted. I.E. turns into faster times.

application specific, the Type II versus Type I... The type II has less force over all, but can produce more work because the force it produces at the higher rpm range is greater than the force the Type I produces, therefore you end up with more work done. a side by side race between a Type II and Type I shows that it's the ability of the Type II to continue producing work after they accelerate from a stop that will make it (the Type II) pull away. the Type I has the Hole shot (because it has greater force behind it) but the type II produces more work as the distance traveled increases in order to catch up.

lets break a formula engine down to it's numbers...i believe it was a BMW engine that produced 950hp but that's all that was advertised. why? to make people think it was an amazing thing. what's really at work here is the power done. if we were to take it's numbers and find out the torque it produces at its rated peak HP rpm we'd stumble onto this...

at 19,000rpm (which is the advertised peak HP rpm) the BMW made 950hp...throw it through the math of 950x5252(a constant used in this calculation)/19000 we get

262.6lbft

do what? that's exactly right. 262lbft of torque...not much oomph is it? not compared to its hp, but in a light chassis like an F1 you get supremely fast acceleration because such a light chassis would accelerate like that with that amount of force.

now can see how an F1 really doesn't play on anything other than rpm. As a matter of fact....

since the car is light...you just played with the first equation too. F=ma...or F/m=a.

Reduce the M and you get more A with the same amount of F.

next time someone tells you This car has this much "horsepower"...ask about the torque.


food for thought. 100hp/liter is easily capable on almost any engine with just a good set of cams and light weight rotating assembly. however, how many cars can produce 100lbft/liter?...

im still trying to find one.

Stay up everybody, and thanks for giving me your time.

-Ivan
..

Wow... This very good very educational I can uese this data, like it but need to take a brake reading almost had a blown head gasket by reading it
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Old 01-04-09, 01:23 AM   #72 (permalink)
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Originally Posted by B+O View Post
Geeze Grammar Police, wtf. If you understand what I'm saying then come off it.
I'm not the grammar police, I'm the unit police And just to clarify, that's the *only* point I'm arguing here. If it wasn't otherwise good it wouldn't matter a bit to me if you were using the wrong units.

FWIW I'm trying to correct things even though I *think* I know what you mean because I'm not sure what part you're misunderstanding. You obviously know how to do the calculations since we always get the same numbers, but the units are not lbs./ft. Your own examples show this if you actually do the math with the units to make sure it it's dimensionally consistent, so I'm not going to assume that I know what you mean where you're not being accurate.

Also, since I haven't used the term dimensionally consistent before and I use it quite a bit here, I will note that it just means the physical units on both sides of any given equation work out to be the same. Make no mistake, these problems must be dimensionally consistent since there are no quantities involved which can be assigned arbitrarily.

Quote:
since you can see that i'm telling you exactly what's good, but you're looking at it coming from one direction and I'm looking at it coming from another direction.
I don't know what different direction I'm looking at this from? If you talking about that FROM/TO the fulcrum BS you'd better find me a convincing argument that you're not making that up, because it doesn't match reality.

As before, if I'm misunderstanding then disregard

Quote:
Anyway, I'm looking at like this...

In a stoich system, every action has an equal and opposite reaction...

So with infinite traction (a function of coefficient of friction and gravity) You're going to be applying equal for--wait...


As i went up to highlight>Copy>Paste, I noticed you just said EVERYTHING i just said...
That's why I try to make it clear that I'm only disputing your units. Your numbers are right and the sticky is good

Quote:
In my haste to be done arguing over something so piddling as "Potato PoTAHto" I mentioned 70ft.lbs versus just 70 lbf on a lever acting as a multiplier of 3 (based on ft.lbs) to create the 210ft.lbs torque load.

what the heck are we arguing about??? we're saying the same thing... COUNTER ACTING the Crankshaft who's rotational force is a quotient of the force applied to it by a lever at an equal distance, is the same as the Crankshaft overpowering a lever to its Stall torque (which would be 0 after it stalled) at 1 foot if that torque applied to it is 210ft.lbs or more...
What we're arguing about is what units should be on the number for torque produced by an engine. I say it should be ft.lbs. and you say it should be lbs./ft.. It's nothing worth letting our blood boil.

I'll also note, since I've been pretty excessively pedantic anyway, that everywhere I talk about pounds I'm referring to pounds force. I would have moved to metric units immediately if we had to deal with mass

But I digress, on to the meat ...

Quote:
Does that make sense?
No it doesn't. You're right about the systems being the same, but the crankshaft's torque (torque is not force, it's rotational tendency caused by force applied elsewhere or the ability of such a rotational tendency to cause a force elsewhere) is a product of force applied to the lever (not applied to the fulcrum by the lever, the lever in this case being the portion of the crankshaft between the rod bearing(s) and the crank centerline/fulcrum) and the distance from the fulcrum at which that force is applied. Conversely, the force provided by a crankshaft at some distance away from the fulcrum (the crankshaft produces zero force at the center, which I think you may have noted before) is indeed a quotient, but it's the quotient of the torque provided by the crank to the distance along the lever the force will get applied.

Real world example: Take a lever and put a force meter on it at some distance away from the fulcrum, then apply a torque at/from the fulcrum (lets go back to 210 and leave the units off of it). I can tell you that at a distance of 2 ft the meter will tell you there is 105 lbs. of force, at a distance of 3 ft it will tell you there is 70 lbs of force, at a distance of 4 ft there will be 52.5 lbs etc., decreasing as you go further away from the fulcrum. You always seem to get the right numbers so I doubt you'll disagree with these, but if you do disagree then flag me down because we have a disconnect at a very fundamental level (not that any of our conversation has strayed out of the "fundamental" realm).

It's quite clear from the above example that the force that can be generated by the torque at the fulcrum is the ratio of torque to distance. 210 / 2 ft. = 105 lbs., 210 / 3 ft. = 70 lbs., 210 / 4 ft. = 52.5 lbs.. My gut feeling is that this is the exact relation you're thinking of to justify your lbs./ft. number, since this is the only relation in which there is an inverse proportionality to distance (ie. some units over feet, which is a necessity if talking about units of lbs/ft), but the quantity which is inversely proportional to distance is force, not torque. In order for the system to generate a force in lbs. at some distance along the lever (and this is what fundamentally defines torque) the only units that work for 210 are ft. lbs. You can not show me a dimensionally consistent equation that properly relates distance in ft., force in lbs. and torque in lbs./ft. because such an equation simply doesn't exist (ignoring the trivial case of adding a non-existent constant with the sole purpose of artificially fixing the units).

For dimensional consistency with torque measured in lbs./ft. you would have to relate torque, distance, and force as force / distance = torque. The problem with that is that for a given torque you would get increasing force at increasing distance, which is exactly the opposite of the real world (as I know you know, because you always get the numbers right).

Since we've beaten the breaker bar example to death, and you seem to agree that a force on the breaker bar produces a torque in ft.lbs. at the fulcrum, I hope the preceding serves to illustrate that the same units are correct when you are talking about the fulcrum being driven at some torque.

Quote:
Here's a few math bits to help you see what I'm talking about...

Engine Torque:
210lbs/ft stall torque
A lever's action
210ft.lbs appl. Torque.
This isn't math, this is a list of numbers with units (some wrong) . I've been trying to talk math this whole time and you won't write an equation. I don't know who you're used to talking math with, but this doesn't pass where I'm from

I suspect the simple fact of the matter is that you can't actually show any math because the units of lbs/ft are wrong.

Quote:
Ft.Lbs v lbs/ft
I will tell you directly by inspection that the difference between these units is that one describes an ability provide increasing force as a distance is reduced, and that the other describes an ability to produce increasing force as distance is increased. Which of those describes torque (in whatever "direction" you please). If you think that both of those can apply it will prove to me that you're confusing where torque and force are applied. If you think that only one can apply then you position that torque can be measured in both units is untenable.

Can't you see just by looking at "210lbs/ft stall torque" (which you listed above) that the physical implication is that to stall the engine you need to increase the applied force by 210 lbs. for every foot of distance between the center of the crank and the point at which you're applying the force? By that logic you could more easily stall the engine by grabbing the crank than by grabbing a 1-mile long breaker bar

Quote:
at 210ft.lbs.[1] from a fulcrum's applied torque[2]

At the fulcrum[3] 210ft.lbs is required to stop this right? thereby, applied at a distance of 1 ft,[3] 210lbf is required to stop the engines rotation. You said this yourself.

at two feet, that pound force (lbf) is 105lbs.
You're kind of right. The points that are wrong I've corrected in red and elaborated below:

1) Obviously lbs/ft is incorrect. Normally I'd refer to this as my contention rather than fact, even though it is backed up by countless sources and my own calculations (while I've yet to be satisfied in my quest for the same regarding your contention), but maybe if I'm more direct about how certain and correct I am you'll feel my contention worthy of a proper response.

2) The driven quantity from the fulcrum is torque, not force. If the fulcrum was exerting a force it would be countered by the main bearings and you'd get no rotation or torque at all, and a torque generates no force at the fulcrum. You may think it's nitpicking, but you can't just interchange "force", "torque", (and "power" as the case may be) and still think you're making anything that even resembles a strong case. Proper use of terms is essential to determining the units of a parameter in question

3) 210 ft.lbs. is required at the fulcrum. At 1 foot 210 lbs is required to stop the engine's rotation. You do not apply a torque at the end of a lever, nor do you apply a force to the fulcrum (well, not in terms of generating torque). Your force is applied to the lever which creates a torque at the fulcrum. At no point is there a torque anywhere but at the fulcrum. I don't think you're just being lazy because you've done this pretty consistently. A force can only create a torque if the line of action of the force does not pass through the fulcrum. It is physically impossible to apply a torque to a point on a lever by applying a force to that point on the lever (and every push or pull is a force). Example: if the connecting rod was applying a torque to the crankshaft (which it is incapable of doing, in any significant quantity in the plane of interest, because of the rod bearing) it would cause a force at the fulcrum of the crank that would be countered by the main bearings. There is plenty of force transferred through the connecting rods to the crank and thus to the main bearings, but it's not a result of the rods torquing the end of the lever.

... using unity values makes for really crappy examples BTW. You get the right answer whether you multiply or divide (which is convenient if your units are wrong I suppose, since 1 ft where the dimensionally consistent equations for ft.lbs. and lbs/ft cross). If you were trying to show how to add vectors would would choose one of the vectors to be the zero vector so it didn't matter if you added or subtracted.

Futhermore, you don't counter a quantity with another quantity in different units, you counter a torque with an equal and opposite torque and the both have units of force * distance. It's like countering forces, you need equal and opposite forces to do it. If you have 1 N in one direction you need 1N in the other direction. If we start pretending that we have to change units to counter something (which is what you're contending, even if you don't realize it) we would need to counter 1 N with 1 kg / (m*s^2)

Quote:
engines applied force reduces as you move farther from the point of force application...that would be the center of the crankshaft...
This part is completely, unambiguously, correct. It also backs up the units of ft.lbs. perfectly (as I've discussed above). Your own examples demand torque be measured in ft.lbs. to be dimensionally consistent.

Quote:
That's the end of my bit on this...It's getting tiresome as I'm repeating myself and we're essentially saying the same thing.

I bid you adieu
Heh, I assure you that you're not the only one who feels like you're repeating yourself, but I thrive on explaining things to people and I've got no shortage of material to bring up if you decide you'd like to continue.

Note: I hit the submit button instead of the preview button when I was reviewing things, so I apologize for mistakes I missed or things I edit.
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Old 01-04-09, 02:55 AM   #73 (permalink)
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Heh, I assure you that you're not the only one who feels like you're repeating yourself, but I thrive on explaining things to people and I've got no shortage of material to bring up if you decide you'd like to continue.
Ah, I forgot an example I wanted to include that I think causes problems for measuring torque in different units.

Consider concentric gears (both with the same fulcrum) sharing an ideal shaft. You drive the first gear to create a torque on the shaft then you use that torque to drive something on the second gear. If I understand your contention correctly, B+O, you are of the opinion that the torque on the shaft of the first gear will be in ft.lbs., but that torque on the shaft driving the second gear will be in lbs/ft. How the hell does that work? We're talking about the same parameter in the same system. If you put 210 ft.lbs. on this shaft you can't claim that this exact same thing is now 210 lbs/ft on the same shaft

Even their relative sizes are unimportant in this example, but I'll go through the calculations so lets consider gear G1 to have a 2ft. radius and gear G2 to have a 3 ft. radius. Just to break from the usual 210 ft.lbs., lets consider a 120 lb force driving G1. The torque on the shaft is

120 lbs * 2ft = 240 ft.lbs.

This same torque is now available for G2 to exert force on something. The force G2 can exert at the teeth is

240 ft.lbs. / 3 ft. = 80 lbs.

Notice that in two lines I've got both "directions" from lever (we already agreed that gears are levers) to fulcrum and back. It also works in reverse, you can apply 80 lbs to G2 and get 240 ft.lbs. of torque on the shaft and 120 lbs at the edge of G1. Hell, you could drive the shaft somewhere else at 240 ft.lbs. and use those calculations to determine either the 120 lbs. available on G1 or the 80 lbs. available on G2, it's all the same two lines. All of this is done with only ft.lbs., is completely dimensionally consistent, and matches perfectly with the physical reality of getting reduced force with increased distance.

FWIW this example works out exactly the same for applying a force on a freely pivoting lever and getting a resulting force elsewhere on the same lever.

Barring correcting details or clarification of this example, if necessary, it seems to me like a ridiculously clear, complete characterization of torque in both "directions". If this example doesn't clarify why torque can't have different units based on whether it's driven or driving then I can't imagine what would be more clear.
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Old 05-31-09, 08:03 PM   #74 (permalink)
 
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Looks like we got a showdown on sesame street here...
I better not participate in this...
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Old 05-31-09, 08:19 PM   #75 (permalink)
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Looks like we got a showdown on sesame street here...
I better not participate in this...
Then why bump it .

I've said my bit, and I think Ivan and I are man enough that we can agree to disagree on this . I'd hardly call it a showdown either, I probably wouldn't have had this discussion with someone I didn't get along with well .
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