I know that Torque shows how fast your car will go off the line, and HP shows how fast your car is capable of going. So a lower torque and higher power (most imports) means that the car will take off slower but is capable of going faster than a car with higher torque and lower HP?
I get that Torque is the overall force applied to the crank from the pistons transmitted to the flywheel. And HP is the product of an equasion involving torque which means how much power you engine is able to produce in a given point in time involving the rpm and torque. I just don't get why something with higher torque doesn't automatically have more HP than something with lower torque... you'd think that with more torque, it'd be able to go faster at higher speeds since there's more force at the flywheel, it would just require a transmition that produces higher speeds at lower rpms. If anyone got any of that, can anyone explain please? ( asked my father who has been working on HUGE block such as tractors dumptrucks etc all the way down to little imports, his whole life and he still coudn't answer this question.)
I already saw that on the howstuffworks.com a while ago, but that didn't really help much. I completely understand the concept of torque and Horsepower, but I don't see why a car with lots of torque and little horsepower (muscle car is a good example) will accelerate faster but has a much lower top speed, where as a car with lots of horsepower but very little torque (imports) have bad acceleration but excelent top speed. It just doesn't make any sense. I have always been taught to keep the rpms at the peak horse power to get the best performance, so how can a car with lower horsepower and a lot more torque perform better if you still keep the throttle at the horsepower's peak rpm? If everyone understands the relationship, there's gotta be something I'm missing since I've been asking this question for a year now and nobody's been able to answer that question.
One way to undestand is that with most muscle cars where they get their torque is from the stroke. Well with an engine with more top end the stroke is usually shorter like in Indy and F1 cars. So if you have a car with more torque due to more stroke think how much travel that is for the pistons at high speeds. But with a shorter stoke engine teh pistons don't need to travel as much so it can perform faster. Remember HP is how fast you hit the wall torque is how far you move it. Well if you keep moving hte wall then you arent going ot go as fast.
Horsepower that you read on the sticker of a car is ONLY the liberation rate of energy from fuel...nothing more. This rating can be used to find top speed of a vehicle by doing a force equlibirum equation suming the losses such as rotating enrtia(wheels, rotors, flywheel, ect.) and drag due to air resistance with the energy liberation from the motor. Once the drag + rotational losses+ ect.(all measured at a certain mph) = energy liberation from the motor( or horsepower). By adjusting the common varible, mph, on left side and keeping the motors output constant at the peak power output you will find the top speed of the vehicle.
Torque is a measured valve, typically at the flywheel and is what actually moves the car. Remember the olds addage: horsepower sells cars, torque wins races. This is pretty much true if your running 1/4 mile or small circle track where its a torque dominated feature. Others like the salt flats.....straight line of as long as you need to prove your point.....its all about horsepower.
Obvioulsy you can see in a car, you need a good bit of both which is why car manufacturers give both numbers as a reference.
Hope this is clear enough to follow.
This of it like this.
engine "a" puts down 100 ftlbs at 1 rotation / second
engine "b" puts down 50 ftlbs at 4 rotations / second
engine "b" will make twice as much power. it applies half the torque at 4 times the speed.
What screws people up is powerbands. A engine that's made for pulling a trailer makes most of its power at low rpms. this gives it a low peak hp due to lack of breathing at higher rpms.
this might be a better way of putting it
(rpm) 1000 2000 3000 4000 5000 6000 7000
engine "a" (tq) 50 75 100 80 65 50 30
engine "b" (tq) 25 30 40 50 70 75 80
engine "b" will make much more power
But engine "a" makes an abundance of torque at lower rpms.
this would make "a" better from a start, but as soon as "b: is above 4000 rpm it will make more power.
end result : torque and rpm is what dictates power. power accelerates your car.
Don't listen to anyone saying large stroke = large torque/lower power...etc. typically engines with a large stroke have a large displacement....therefore not requiring a high rpm to make lots of power.
your VE vs. rpm dictates what your torque curve will look like, not stroke.
What the hell is energy libertation?!? do you mean brake specifc fuel consumption?
-> BSFC will always be close to where peak torque is (where peak VE is). due to a: the most air the cylinder will see in that stroke, and b: due to "a", it will have a higher dynamic compression ratio, also increasing torque at that rpm.
2000 z24
1985 z28
http://www.cardomain.com/ride/825536
^No i mean energy liberation. From basic law of enrgy consevration. Fuel is a form of stored energy....buring the fuel liberates this energy into a useable form of power.....horespower. How the engine translates that energy into shaft work is totally dependant on the motors design......ie: torque
What you said is correct, however not answering his question.
Edited 1 time(s). Last edited Monday, March 19, 2007 6:33 AM
Ok, I kind of get it. I guess at higher rpms (eg 2000 Vs 4000) the engine has to take in two times the intake to produce the same amount of torque, but then again at that higher rpm it's producing less torque at a greater multiple meaning that it is actually putting out more power per minute (in refference to rounds per minute)
EG: 120fp@3000 rpm to 80fp@6000 rpm, when the engine is reved to 6000 rpm, it's actually putting out 40more foot pounds of torque per second than if it was reved to 3000 rpm? I think I get it now, only the term torque is the total force that the pistons are applying to the crank, so adding in the rpm would convert that to horsepower. So horsepower is how much power the engine is acutally putting out, so to get the torque to peak at a higher rpm would make the engine produce more power at that rpm. So I think I get it now. Kind of complicated and I still dont know how even I can word it, but I have finally grasped the concept. thanks guy, you've been a big help.
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120fp@3000 rpm to 80fp@6000 rpm, when the engine is reved to 6000 rpm, it's actually putting out 40more foot pounds of torque per second than if it was reved to 3000 rpm?
I think you're getting it. Power is just a product of rpm and torque. If you make a lot of torque at low rpms, and little torque at high rpms you'll have a engine will low peak hp, but high hp in the low end. (making it feel more powerful at low rpms until if falls on it's face at higher rpms).
Power is what accelerates your car. having lots of torque at low rpms makes it accelerate faster at low rpms (more hp - it can apply more work).
Quote:
I guess at higher rpms (eg 2000 Vs 4000) the engine has to take in two times the intake to produce the same amount of torque, but then again at that higher rpm it's producing less torque at a greater multiple meaning that it is actually putting out more power per minute (in refference to rounds per minute)
not really. don't confuse rpms into the equation when talking about torque. when thinking about torque, think about one rotation.
I'll try to explain it a little better by adding in volumetric efficiency (ve).
this isn't quite exact, but it will give you a better idea.
rpm 1000 2000 3000 4000 5000 6000
ve 60 75 85 90 75 60
tq 100 120 130 140 120 100
ve is a measure of how much air is drawn into the cylinder, compared to how much volume the cylinder displaces (in
.
at 4000 rpm, the engine is drawing the most air into the engine, per stroke. where your peak ve is, is where your peak torque is.
in reference to your quote, the engine does not require more air to make the same torque, it requires more air to make more power (due to rpms).
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so to get the torque to peak at a higher rpm would make the engine produce more power at that rpm."
Exactly! this is why changing anything that affects airflow into the engine, also changes your ve. a large cam will lower ve at low rpms, yet improve it at higher rpms, thus making more peak power at the sacrifice of low rpm power.
Joshua i think you're talking about BSFC. it's a measure of how efficiently a engine is at converting a volume of gasoline, into power. The lower your BSFC is, the more efficiently you are making power.
I hope i didn't come off hasty, I've just never heard of fuel liberation when talking about internal combustion engines.
2000 z24
1985 z28
http://www.cardomain.com/ride/825536
Robert.....your trying really hard to confuse yourself........rpm is dependent on time, torque isn't....you can't make assumptions like you did.....its like saying there 12volts in an amp......or 5 bolts in a nut.
Look....you have two "power curves" for a motor for a reason. This might be a little off topic here but it might help.
The horespower curve shows a RATE of energy liberation vs. RPM - basically amount of fuel burned vs. engine speed. This curve is only a energy balance of the motor, nothing else.
The second curve is a physical torque curve, this curve shows engine RPM vs. shaft work or rotational work. This curve is only a physical measurement of the power put out by the motor.
The differences:
HP curve can be calculated by fuel put in and exhaust sniffer to see how much burned.....non-physical(sort of)
TQ curve is only calculated by a physical measurement, measured at the crank shaft.
There can be inferences from these two curves when they are laid atop of eachother and matched by RPM......showing you high efficiency points of the motor giving the best fule input to power output and so on...this is how car manufacturers design there gearing in transmissions...obviously want low fuel with good torque and get a gearing ratio to take advantage of that spot of the rpm band for highway driving, and so on. Trucks abviously are interested in staying in a high torque area of the motor, not too much worries about fuel consuption.....depending on the truck obviously.
I hope this is making some since.....there not at all directly related, not even close....once you get that clear you'll notice your original questions are so dependant on the actual design of the motor that its almost impossible to answer them without making huge assumptions.
I can design a motor that can consume 1000hp of fuel and not make more than 5 ft/lbs of torque....its possible. The inverse is also possible, I can make a motor get 1000ft/lbs of torque with only 5hp. - it wont be small, but i can do it.
Edited 2 time(s). Last edited Monday, March 19, 2007 6:23 PM
Ok, now you're going to screw him up.
Do not read the above, if you do, don't take it seriously.
Horsepower is directly, and only related to torque and rotational speed.
HP = torque * rpm / 5252. Now you can see why if you have less torque, but more rpm, you will make more power.
It doesn't matter how much, or little fuel is burnt. Look at any dyno sheet, hp and torque always cross at 5252 rpm. They are DIRECTLY related.
Brake specific fuel consumtion is how much fuel is burnt relative to power made. This does not in no way calculate hp, it just show you the rate of fuel consumtion to make 1 hp.
^ your energy balace thing is way out to lunch. also the highest efficiency point will always be at, or relatively near, your peak torque. end of story.
2000 z24
1985 z28
http://www.cardomain.com/ride/825536
Quote:
I can design a motor that can consume 1000hp of fuel and not make more than 5 ft/lbs of torque....its possible. The inverse is also possible, I can make a motor get 1000ft/lbs of torque with only 5hp. - it wont be small, but i can do it.
lol. Take a grad level mecanical engineering course on combustion engines and you wouldn't be making a statement like that. That's like me saying I can make a time machine with a bag of pop cans, a 10 speed bike, and a bottle of whiskey.
2000 z24
1985 z28
http://www.cardomain.com/ride/825536
Sorry your right, I am confusing his question. The equation you listed above is used for dyno and performance curves in the production market.
However, when your in the process of engine optimization you use the energy balance to find bad spots in inefficiencies of the engines and its mated components like oil pump, ac compressor, water pump, and losses from timing chain harmonics ect. I worked in an engine lab for several years and we have fuel temperature gauges and flow meter pre and post injectors to measure HP into the engine, then exhaust sniffers to see how much of the possible HP was utilized. Then took this number and matched it to the torque output using the equation you listed....sorry thats where I missed the boat. Sorry about that. However, I'm surprised a grad level mechanical engineer hasn't heard of this technique. I didn't stay with school and go into grad level, I still want to....maybe someday.
What you do is take the HP in vs. HP out....the difference between the two gives you losses. You also, take the same data and generate an actual HP(fuel input in mass multiplied by the heating value of the fuel gives you the amount of power that can be liberated from the fuel) vs. RPM. Then you can "dyno" the accessories independently to find the inefficient spots and normalize them to the engine RPM and drop them onto the same plot as the HP vs. engine RPM. When the HP requirements curve of the oil pump increases you'll see the HP output of the motor decrease....and so on. You can do this for every component and use this data to change components to use less power at a critical rpm where you want the most power to go to the wheel and not the accessories. Basically you can optimize an engines output using an energy balance.
Skip: I've seen a motor design get 103ft/lbs of torque and use 476hp at 4800rpm, using energy balance analysis of course. Using your equation, it should have only used ~94 hp worth of fuel or 520 ft/lbs or torque.....it didn't. All the rest of the fuel went to losses which is where of course, car manufacturers dont want to show off there inefficiencies so they use the calculation HP = torque * rpm / 5252. Its not wrong, but its also not right, you could be using much more power from fuel than what the HP curve ever shows in a production model engine.
Yes the numbers I gave above have no legitimate backing and I dont have time or energy to try and prove them if there even possible, they are what I'd call gross exaggerations since I've never seen anything like that, however it doesn't mean its not possible. This first would be possible by using a horribly miss-desiged fulcrum/lever crank shaft, giving horrible torque. The second would need a small B&G 5hp motor, a good clutch, smooth bearings and a huge flywheel, but thats not exactly part of the motor so........
Sorry Robert, I hope I didn't confuse you, as far as dealerships and dyno shops and all that other crap goes, skip is right. HP is a calc. value from the amount of torque generated.
Lol, yea at first that confused the crap out of me. But I got it now basically the horsepower says how many times a certain amout of force (torque) is applied to the crank per a certain amount of time (rpm) using the torque output at that rpm. Basically that's the written form of the equasion used to get hp out of torque and rpm and the constant 5252. Which why in the heck is it 5252. Is that just something that someone made up, or did they figure it out from crossing multiple lines in algebraic equasions involving a horse as a constant? So, next question: If I had a manual car and kept the rpms to the rpm which has the peak torque before shifting it into first, will the car acclerate faster than if I held it at the peak horsepower rpm? In theory keeping it at the peak power should give better accleration since power translates to the amount of work that can be done within a given amount of time, so more kinetic energy is produced at the rpm with higher hp and thus more power can be sent to the wheels and so the car can go faster. if that's true, then HP is the true performance rating of a car and the torque is what you try to increase along the rpm range of the engine. In which more torque in a higher rpm would mean more horsepower at that rpm since it's a multiple of the torque and rpm. So if I'm right about my final question that I got the concept. Thanks
5252 is just a lot of unit conversions rolled together.....a constant. Honestly, dont worry about it, just take it for what it is.
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I'm surprised a grad level mechanical engineer hasn't heard of this technique
I'm not a grad level mech e, just took a combustion engines course.
Also, that technique was taught in grade 12 chem. I fondly remember making a device to "catch" the heat output of a candle. by calculating the temp change of the water (in which the candle heated) you could see how much energy you could "catch" and compare that to the actual amount of energy the candle possesses (per gram). this would give you the efficiency of your device.
you're talking about the difference between gross and net power ratings. I believe they changed that back in the early 70's (ie. a 1970 z28 lt1 came with 370 gross hp, but would get walked all over by a 93 lt1 with 285 net hp).
I understand what you're saying regarding testing accessories, you don't want use components on an engine that will hamper the performance. ie. like using a water pump that draws an excessively large amount of power to run at high rpms on a high rpm engine.
Look up brake specific fuel consumtion. All you do is meter the flow rate of fuel entering the engine, and relate that to the corresponding output (for any means, you would only measure the physical output. ie torque and rpm to calculate power).
This gives you your bsfc.
what you're talking about -> If you want to see how much energy the engine is using/wasting, you would only have to multiply the volume flow rate of fuel used by the energy contained in the fuel per unit volume.
I completely agree with you regarding testing accessories.
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If I had a manual car and kept the rpms to the rpm which has the peak torque before shifting it into first, will the car acclerate faster than if I held it at the peak horsepower rpm? In theory keeping it at the peak power should give better accleration since power translates to the amount of work that can be done within a given amount of time, so more kinetic energy is produced at the rpm with higher hp and thus more power can be sent to the wheels and so the car can go faster.
yes...well sort of. It sounds like you get it. to make a car faster, you have to make more power. if you can increase torque, you will increase power. on a naturally aspirated engine, increasing rpm (in addition to ve at higher rpms) is typically how you make the most power.
2000 z24
1985 z28
http://www.cardomain.com/ride/825536
Wow...
Leave it to a few engineering students to really drag out an answer.
j / k
Question 1:
Quote:
I just don't get why something with higher torque doesn't automatically have more HP than something with lower torque...
It does, plain and simple. Pick an rpm, any rpm, and calculate hp from two different torque values using the standard definition hp = tq * rpm / 5252. Higher torque at a specific rpm will always produce higher hp at that rpm. Remember this relationship. It's the key.
Interpretation 1:
Why does it seem like an engine which produces large torque values at low rpm is usually unable to produce high hp?
This is partly due to the formula above. Internal combustion engines as we're used to produce a maximum peak torque within a relatively narrow range. If the peak torque occurs far enough below 5252 rpm then the peak power number will not be greater than the peak torque. Diesel engines are often like this since it's the nature of a traditional diesel to work at low rpm. On the other hand, some engines find their peak torque at a fairly high rpm. Peak torque occuring above 5252 rpm will generate peak hp numbers higher than peak torque. Most street engines have a torque curve which is broad enough such that peak torque occurs below 5252 rpm but there is still sufficient torque above 5252 rpm to generate a higher peak hp number than the peak tq value.
Interpretation 2:
Of two different engines, why doesn't the engine with a higher torque peak at low rpm also produce higher torque at a higher rpm?
Now this hits at the core of Engineering. This question is, "Why is the torque curve shaped the way it is?" What Joshua was referring to, measuring fuel consumption and determining energy losses, is part of the search to find answers to this question. Is all the energy in the fuel being liberated, or is some escaping in the form of raw hydrocarbons or combustion byproducts which are the result of incomplete reactions? Of the energy being liberated, what percentage is being wasted as heat and what percentage is being lost to friction? Is there any way we can adjust how and when components draw energy away from the flywheel? Skip was mentioning another aspect by referring to volumetric efficiency. Since fuel combustion depends on oxygen intake, it's important to ask "Why is cylinder filling, (air density) greater at one rpm than another? How can we improve it across the entire rpm range?" Everyone trying to get more power from their engines is, directly or indirectly, working on some aspect of this problem.
Question 2:
Quote:
So, next question: If I had a manual car and kept the rpms to the rpm which has the peak torque before shifting it into first, will the car acclerate faster than if I held it at the peak horsepower rpm?
No. As said above, peak efficiency occurs at peak torque. Keeping the engine at peak torque results in maximum fuel efficiency. This technique, shifting a vehicle at peak torque rather than peak power, is becoming extremely popular among truck drivers as it maximizes their mileage per tank of fuel. With a vehicle that might see fuel consumption as high as 1 gallon per mile, everything helps.
Question 3:
Quote:
In theory keeping it at the peak power should give better accleration since power translates to the amount of work that can be done within a given amount of time,
Absolutely! When shifting you can sometimes gain a little more acceleration by shifting at a slightly higher rpm than the power peak thereby using increased energy stored in the rotating assembly to help "bump" the vehicle closer to the rpm peak.
As an aside, torque can be calculated given enough information. One commonly used method involves measuring acceleration of a drum with known mass then calculating power required to produce the acceleration. With the addition of engine rpm as a variable it's also possible to derive the torque needed to produce the power which produced the measure acceleration. This is the method used with Dynojet chassis dynos.
Question 4:
Quote:
Basically that's the written form of the equasion used to get hp out of torque and rpm and the constant 5252. Which why in the heck is it 5252. Is that just something that someone made up, or did they figure it out from crossing multiple lines in algebraic equasions involving a horse as a constant?
I hope you've had some physics.
James Watt determined one horsepower is equal to 33,000 lbs * foot / minute in order to sell steam engines to replace horses. The horses he was replacing were tied to a wheel 24 ft dia, and Mr. Watt came up with values for RPM of the horse and force applied to the wheel to create a standard unit of measure. Torque is measured in pounds at a distance of one foot from the center of a rotating part applying the force. The resulting value is foot lbs, but it's not the same as the lbs * foot units used in the hp standard above. The difference is what the "Foot" unit describes. In torque, it describes the radius at which the applied force is measured. In horsepower it describes the linear distance the force is applied through. Since torque is applied in a circle with 1 foot radius, the linear distance through which an engine's force is applied is equal to the circumference of a wheel with a 1 foot radius. Circumference is pi * d or 2 * pi * r. So power = force times distance over time, or
ft/lbs * 2 * pi * r / minute
To convert this value to hp we divide the formula above by 33000 lbs ft / min using some algebraic cancellation and come up with
torque * rpm / 33000 / (2 * pi)
33000 / (2 * pi) = about 5252.1131.
I may have breezed through this last part since I'm getting a bit tired. Look for "hp 33000" in google and the first few articles cover it.
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