Horsepower Vs. Torque ???
#76
Originally posted by bob13bt
Black Dragon,
It really is POUND-FEET, not FOOT-POUNDS.
I was confused for years, and then took a Physics class. So go argue with my professor.
Jim Downing is a rich ************. He is very smart and probably more technical than any of us. He doesn't have to give exact hard numbers to be reputable, it was probably some walk-by comment or quick interview, not a ******* spot on Larry King! It was a generalized statement. Not one to be poked and probed for validity! Get over yourself!!
Black Dragon,
It really is POUND-FEET, not FOOT-POUNDS.
I was confused for years, and then took a Physics class. So go argue with my professor.
Jim Downing is a rich ************. He is very smart and probably more technical than any of us. He doesn't have to give exact hard numbers to be reputable, it was probably some walk-by comment or quick interview, not a ******* spot on Larry King! It was a generalized statement. Not one to be poked and probed for validity! Get over yourself!!
''By convention, in the US/British 'inch/pound' measurement system, ft-lb is the correct unit for the expression for WORK, the movement of an object from one position to another. Work is measured in terms of straight line distance and force. If a 10 lb weight is lifted off the ground 2 feet, the work done on the weight is 2 ft x 10 lbs, or 20 ft-lbs.
A turning or twisting effort not in a straight line produces TORQUE, which in the US/British system, is defined as lb-ft. If an engine develops 200 lb-ft of torque, 200 lbs of force on a 1 foot lever is required to stop its motion. One lb-ft produced during one revolution is equal to 6.28 ft-lb of work or mechanical energy(2śr= 2x3.146 x 1lb-ft=6.2832 ft-lb).''
(I can't find a pi symbol, that will have to do )
#77
I went through seven years of Mechanical Engineering education, and I can tell you that at least at my alma mater, lb-ft and ft-lb were used more or less interchangeably. In textbooks, in discussions, in problem solving, and on tests... ft-lbs was commonly used by many people in reference to torque, and I never had a professor question or criticize that.
After all, it's a scalar multiplication, so order isn't important. It's more a matter of convention, or more importantly personal preference. I've had people harp on me for calling torque 'ft-lbs' in a forum before, and all I could do was laugh. They actually tried to explain how the order could make a difference in the numerical value.
About torque vs. power... you really only need one thing to define an engine's performance: torque vs. rpm curve. Horsepower is a calculated value, torque is a measured value. A dyno, be it chassis or engine, measures torque and rpm, and then calculates power. For that reason, I have always had the opinion that the "raw" data is torque and rpm.
However, power is a much more convinient way of expressing that information, because it embodies both the rpm and torque measurements at a specific point on the curve. You can look at one number instead of two. Of course, then you start looking at power vs. rpm curves, and calculating the "power area under the curve", in which case you might as well calculate "torque area under the curve" because it is telling you exactly the same thing without redundant rpm information embedded in the equation.
For those that believe torque is everything, you better think again. Formula One engines don't produce that much torque, but they produce astonishing amounts of power. You want to drag race your big block Chevy against an F1 car? I know, I know, not fair because of the weight and such. Just making a point... low or moderate torque engines with high horsepower can accelerate fast, provided the gearing is suitable.
That brings us into the final part of the equation. Formula One gearboxes are hardly simple contraptions. You can in theory take an engine with a high peak horsepower at a high rpm and make it accelerate hard from stop to top speed, but it would require lots of gears to do so. You can also take an engine with lots of low down torque, and a low redline, and do just the same... and the solution is the same: lots of gears.
Having a high reving engine produce more low rpm torque, or raising the redline on a low end torque motor is all about one thing: reducing gearing complications. It makes the vehicle much more driveable, even if you don't increase the peak horsepower number one red cent.
Of course, you guys already knew this. That is why "power under the curve" numbers and average power in a gear are important. My perspective however is that that information is available in raw form as "torque under the curve."
High boost four cylinders that have high peak HP in a narrow band are difficult to gear and drive. In fact, a lower hp motor could be faster all around if it has a broader torque curve. V8's are good from driveability because the good torque range starts low... rotaries are good for driveability because the good torque ranges continues up high. Both achieve the same result, and for more or less the same reason (large ranges of useable torque, just in different places). High strung four bangers just suck, and so do long stroke motors running out of steam at 4000rpm.
After all, it's a scalar multiplication, so order isn't important. It's more a matter of convention, or more importantly personal preference. I've had people harp on me for calling torque 'ft-lbs' in a forum before, and all I could do was laugh. They actually tried to explain how the order could make a difference in the numerical value.
About torque vs. power... you really only need one thing to define an engine's performance: torque vs. rpm curve. Horsepower is a calculated value, torque is a measured value. A dyno, be it chassis or engine, measures torque and rpm, and then calculates power. For that reason, I have always had the opinion that the "raw" data is torque and rpm.
However, power is a much more convinient way of expressing that information, because it embodies both the rpm and torque measurements at a specific point on the curve. You can look at one number instead of two. Of course, then you start looking at power vs. rpm curves, and calculating the "power area under the curve", in which case you might as well calculate "torque area under the curve" because it is telling you exactly the same thing without redundant rpm information embedded in the equation.
For those that believe torque is everything, you better think again. Formula One engines don't produce that much torque, but they produce astonishing amounts of power. You want to drag race your big block Chevy against an F1 car? I know, I know, not fair because of the weight and such. Just making a point... low or moderate torque engines with high horsepower can accelerate fast, provided the gearing is suitable.
That brings us into the final part of the equation. Formula One gearboxes are hardly simple contraptions. You can in theory take an engine with a high peak horsepower at a high rpm and make it accelerate hard from stop to top speed, but it would require lots of gears to do so. You can also take an engine with lots of low down torque, and a low redline, and do just the same... and the solution is the same: lots of gears.
Having a high reving engine produce more low rpm torque, or raising the redline on a low end torque motor is all about one thing: reducing gearing complications. It makes the vehicle much more driveable, even if you don't increase the peak horsepower number one red cent.
Of course, you guys already knew this. That is why "power under the curve" numbers and average power in a gear are important. My perspective however is that that information is available in raw form as "torque under the curve."
High boost four cylinders that have high peak HP in a narrow band are difficult to gear and drive. In fact, a lower hp motor could be faster all around if it has a broader torque curve. V8's are good from driveability because the good torque range starts low... rotaries are good for driveability because the good torque ranges continues up high. Both achieve the same result, and for more or less the same reason (large ranges of useable torque, just in different places). High strung four bangers just suck, and so do long stroke motors running out of steam at 4000rpm.
#80
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Joined: Mar 2002
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From: Monterey Bay Area (Capitola) California
Well said
Well said Bigus Dickus. It's great to see that someone on the forum has a mechanical engineering background and someone to say on this topic. Of course, I agree with everything you say because not only is it right inline with my argument (or atleast, the spirit of my argument since attacks to my theory led me off base a few times) but it was also well put. Thanks for your $.02
And yes, I was very pissed the day I last posted, things have only gotten worse for that texas car and I'm just trying to deal with it on top of a lot of other ****. But anyway, sorry all and I'll see you guys later
And yes, I was very pissed the day I last posted, things have only gotten worse for that texas car and I'm just trying to deal with it on top of a lot of other ****. But anyway, sorry all and I'll see you guys later
#81
Rotories have lot's of torque ( even if its not important )
espessially for their size ( 1.3L! the displacement is measured basically the same way as a piston engine. When the rotor has sucked up as much air/fuel as it can (BDC in piston engine), the displacement is measured (about 650cc) when it is fully compressed (TDC in piston engine) the combustion chamber is measured and then subtracted from the total. This happens three times in 1 rev of the rotor but with the same combustion chamber. Because the e-shaft spinns 3 times faster this equals 1 powerstroke for every 1 rev, compared to 1 PS for every 2 revs for the piston motor. New RX-7 fan calls it a 2.6 L just because it has twice as many power strokes!LOL HA HA HA . By that definition, a
2 stroke KX250 dirtbike would really be a 500 and a 500 would be 1000!) so anyways, the 87 TII has 182HP and 183 ft.lbs. of torque the 12A has 100 HP and 105 ftlbstq
the 89 TII has 200 Hp and about the same tq. the 91 2.0L cosmo has Around 275hp and 290 ftlbstq...
Where the hell did this torque rumor about the rotory start anyway? the torque curve (very important) also starts low and goes high in most rotories. Most 4 bangers have low torque #'s like the type R- 195hp and only 130 ftlbstq. Still gets to 60 in 6.2 secs though.
espessially for their size ( 1.3L! the displacement is measured basically the same way as a piston engine. When the rotor has sucked up as much air/fuel as it can (BDC in piston engine), the displacement is measured (about 650cc) when it is fully compressed (TDC in piston engine) the combustion chamber is measured and then subtracted from the total. This happens three times in 1 rev of the rotor but with the same combustion chamber. Because the e-shaft spinns 3 times faster this equals 1 powerstroke for every 1 rev, compared to 1 PS for every 2 revs for the piston motor. New RX-7 fan calls it a 2.6 L just because it has twice as many power strokes!LOL HA HA HA . By that definition, a
2 stroke KX250 dirtbike would really be a 500 and a 500 would be 1000!) so anyways, the 87 TII has 182HP and 183 ft.lbs. of torque the 12A has 100 HP and 105 ftlbstq
the 89 TII has 200 Hp and about the same tq. the 91 2.0L cosmo has Around 275hp and 290 ftlbstq...
Where the hell did this torque rumor about the rotory start anyway? the torque curve (very important) also starts low and goes high in most rotories. Most 4 bangers have low torque #'s like the type R- 195hp and only 130 ftlbstq. Still gets to 60 in 6.2 secs though.
#84
#85
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Joined: Sep 2001
Posts: 163
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From: Portland, OR
Here's a hypothetical race.
Two cars drag race, both have CVT's.
Both cars have a flat torque curve from 4000-8000 rpm, say 150 ft lbs.
Car 1 launches at 4000 rpm and stays there for the whole race, due to it's continously variable transmission.
Car 2 launches at 8000 rpm and also stays there for the whole race.
Who will win? If torque is king then the two cars will tie. If horsepower rules, then car 2 will kick *** since it is putting out twice the power!
Two cars drag race, both have CVT's.
Both cars have a flat torque curve from 4000-8000 rpm, say 150 ft lbs.
Car 1 launches at 4000 rpm and stays there for the whole race, due to it's continously variable transmission.
Car 2 launches at 8000 rpm and also stays there for the whole race.
Who will win? If torque is king then the two cars will tie. If horsepower rules, then car 2 will kick *** since it is putting out twice the power!
#86
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Joined: Mar 2002
Posts: 168
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From: Monterey Bay Area (Capitola) California
#87
heres my 2 cents on a post brought back from the dead.
its all about the power you make and the gear. having problems launching a a car that makes power way up high with out smoking your tires down the drag strip, its called an adjustable slipper clutch, i believe mcleod makes it. it lets your engine leave under power say around 3-5 rpm and will slip the clutch for you just enough to keep the tires from spinning. i know a hemi in the stock tire drags thats running on 6.5 inch redline tires. its a 4200 pound car thats 60' as low as 1.79, lets say the launch is the best part as the car looks like a top fueler when running down the track as in it leaves tire marks behind it across the finish line cause well the 60' is really the only place where it doesnt spin the tires
i dont see how a longer stroke helps low end power, bigger displacement might, but compare 2 engines of the same displacement, they both have about 5 liters of displacement, engine A is a 302 it has a 4 inch bore and a 3 in stroke, engine B is a 305 that has a 3.75 inch bore, and a 3.48 stroke. the 302 will make as much low end power as a 305 because the force acting on the piston is greater because of its greater area, the 305 will make the low end power because of its longer stroke and better leverage pushing on the crank, the 305 will not make high end power due to its bad rod to stroke ratio compared to the 302.
displacement of an engines can be measured different ways. for a 4 stroke, it can be measured by the calculating the radius of the bore squared X pi X the length of the stroke X the number of cylinders.
the displacement of a 2 stroke can be measured the radius of the bore squared X pi X the length of the stroke X the number of cylinders.
in my "OPINION" a rotary is closer to a 2 stroke 2 cylinder than to a 4 stroke 4 cylinder because it fires the same 2 "cylinders" on every revolution i believe the displacement of one of these "cylinders" to be 654 CCs there for the displacement of 2 of these is 1308 CCs therefore its a 1.3 liter or 79.81 CI, i think this is how mazda arrived at 1.3 liter displacement, if it was like a 4 stroke you "could consider" it a 2.6 becuase for a 4 stroke to displace all of its displacement it needs to make 2 revolutions. but theres no way of reckoning it a 3.9 liter
power is work done over time
torque is force applied on a lever
i heard a wind mill puts out around 10,000 ft lbs of torque at about 5 rpm well that calculates down about 9 Horse power
imagine useing a 13b turbo out of an fd, its making about 255 hp, gear that down to 5 rpm and your making nearly 270,000 ft lbs of torque probably enough force to spin the mill house around the engine
i think it was stated before HP is torque * (RPM/5252), you cant measure hp but you can measure torque, even though you cant compare torque numbers on a chassis dyno you can compare the hp it makes.
its all about the power you make and the gear. having problems launching a a car that makes power way up high with out smoking your tires down the drag strip, its called an adjustable slipper clutch, i believe mcleod makes it. it lets your engine leave under power say around 3-5 rpm and will slip the clutch for you just enough to keep the tires from spinning. i know a hemi in the stock tire drags thats running on 6.5 inch redline tires. its a 4200 pound car thats 60' as low as 1.79, lets say the launch is the best part as the car looks like a top fueler when running down the track as in it leaves tire marks behind it across the finish line cause well the 60' is really the only place where it doesnt spin the tires
i dont see how a longer stroke helps low end power, bigger displacement might, but compare 2 engines of the same displacement, they both have about 5 liters of displacement, engine A is a 302 it has a 4 inch bore and a 3 in stroke, engine B is a 305 that has a 3.75 inch bore, and a 3.48 stroke. the 302 will make as much low end power as a 305 because the force acting on the piston is greater because of its greater area, the 305 will make the low end power because of its longer stroke and better leverage pushing on the crank, the 305 will not make high end power due to its bad rod to stroke ratio compared to the 302.
displacement of an engines can be measured different ways. for a 4 stroke, it can be measured by the calculating the radius of the bore squared X pi X the length of the stroke X the number of cylinders.
the displacement of a 2 stroke can be measured the radius of the bore squared X pi X the length of the stroke X the number of cylinders.
in my "OPINION" a rotary is closer to a 2 stroke 2 cylinder than to a 4 stroke 4 cylinder because it fires the same 2 "cylinders" on every revolution i believe the displacement of one of these "cylinders" to be 654 CCs there for the displacement of 2 of these is 1308 CCs therefore its a 1.3 liter or 79.81 CI, i think this is how mazda arrived at 1.3 liter displacement, if it was like a 4 stroke you "could consider" it a 2.6 becuase for a 4 stroke to displace all of its displacement it needs to make 2 revolutions. but theres no way of reckoning it a 3.9 liter
power is work done over time
torque is force applied on a lever
i heard a wind mill puts out around 10,000 ft lbs of torque at about 5 rpm well that calculates down about 9 Horse power
imagine useing a 13b turbo out of an fd, its making about 255 hp, gear that down to 5 rpm and your making nearly 270,000 ft lbs of torque probably enough force to spin the mill house around the engine
i think it was stated before HP is torque * (RPM/5252), you cant measure hp but you can measure torque, even though you cant compare torque numbers on a chassis dyno you can compare the hp it makes.
#89
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Joined: Sep 2001
Posts: 163
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From: Portland, OR
It's an interesting document, and alot of thought went into it, but the conclusions are wrong. It's area under the power curve, not the torque curve, that needs to be maximized for optimal acceleration. Power is the rate of torque. If you can put it out twice as fast, you have twice the power. Imagine torque as a fist hitting a punching bag, or better yet, a kid kicking the back of your car. Power is how many kids kick your car per minute. 8000 kicks per minute will give you considerably more acceleration than 4000 kicks per minute. You see, torque is an instantaneous, timeless measurement, like the work done by a kid kicking your car. It has nothing to do with RATE. Only when you know the rate can you say anything about power or acceleration. How would you like a motor that can do 200 ft lbs of work 10,000 times a minute? Would you rather have a motor than can do 300 ft lbs of work 5000 times a minute? I think not!
Torque is work.
Power is the rate of torque.
Torque is work.
Power is the rate of torque.
#90
um no your mistaken on a couple parts, the area under the power curve is redundent, the area under the torque curve is easier to understand why because its a linear graph all the way across the rev band a hp curve has an extra slope across it because rpm is in the equation for hp. you cant imagine torque as a fist hitting a punching bag because that is a force that hits the bag. and the kid kicking the car he is applying a force to the back of your car. when kid kicks your car he has applied an impulse to the car, impulse is force over time, well if he hits you car with an impact force of 10 newtons the duration of that was .5 seconds he just hit your car with an impluse of 5 newton seconds. well say he hits your car twice in a second both times with 10 newtons of force but he only holds his duration over .25 second. thats 2.5 newton seconds each time so he still has only applied 5 newton seconds of an impulse on the back of your car.
you cant measure torque. you can measure a force and calculate torque through how long of lever its working on.
torque is not work, it is has the potential ability to work.
so you cant have a motor do 200 ft lbs of work 10,000 times a minute cause torque doesnt do anywork
work is the amount of force applied to an object and the distance to which it moves, power is work over a time period. a watt is power, it is one joule of work over one second, a joule is a force of 1 newton over 1 meter
i hope your not so misinformed on your neuroscience studies
you cant measure torque. you can measure a force and calculate torque through how long of lever its working on.
torque is not work, it is has the potential ability to work.
so you cant have a motor do 200 ft lbs of work 10,000 times a minute cause torque doesnt do anywork
work is the amount of force applied to an object and the distance to which it moves, power is work over a time period. a watt is power, it is one joule of work over one second, a joule is a force of 1 newton over 1 meter
i hope your not so misinformed on your neuroscience studies
Last edited by sbertolone; 10-26-02 at 03:39 AM.
#91
this is a minor point, but somebody said earlier that you don't buy a truck based on the torque, you buy it based on the horsepower.
Sure, if you don't feel like towing anything, and if you don't want your truck to be able to do anything trucklike, you go ahead and buy your horsepower-y truck. I'll get the truck with the highest torque rating I can find, and I'll tow your **** around when it breaks.
Sure, if you don't feel like towing anything, and if you don't want your truck to be able to do anything trucklike, you go ahead and buy your horsepower-y truck. I'll get the truck with the highest torque rating I can find, and I'll tow your **** around when it breaks.
#92
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Joined: Sep 2001
Posts: 163
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From: Portland, OR
sbertolone,
IT IS area under the POWER curve that is important for acceleration, otherwise we'd all be shifting at 5000 rpm instead of 8000 rpm. The power curve is "weighted" for rpm, and this information is important for maximizing acceleration (power is the rate of work, work per unit of time, ft*lbs/sec, or if you like N*m/s). Sure you can integrate a torque curve to find the area under it, but that doesn't give you the average power or help you optimize acceleration.
When a kid kicks your car, the mass of his/her moving leg stops. This produces a force (Force = mass * acceleration). For every action there is an equal and opposite reaction (the mass of your car accelerates in response). Thus a force has been applied to your car and the car has moved. This is the definition of work (Work = Force x Distance). So, my anology is like this, and it's just an anology: combustion generates a brief force, like a kid kicking your car. If the car is kicked once a second, you will slowly accelerate. If the car is kicked 100 times a second, acceleration will be considerably more rapid, because the power, or rate of work is higher. Bigger piston, more torque, bigger kid, more work done per kick. Rate (rpm) is not redundant, you know nothing about acceleration without knowing the rate at which work is being done.
A Joule, or Newton * meter, or kg*m^2*s^-2 , or foot pound, is the unit of both work and energy. Do you know the difference? (sorry, just feeding the flames)
More on torque:
You CAN measure torque, that's exactly what is measured at a dynometer. You can't measure twisting force without a mechanical moment.
Torque is a physical vector quantity characteristic for an object in rotational motion around a certain given axis. By definition torque is the cross product between the vector of position (r) of the point where a force is applied (from an origin considered to be on the axis of rotation) and the applied force (F) (torque= r x F). According to the definition of the cross (or vector) product, the magnitude of the torque is equal to the magnitude of r times the magnitude of F multiplied by sine of the angle (theta) between r and F (torque = r F sin theta). In this expression for magnitude of the torque, the product (r sin theta) is equal to the magnitude of the distance (d) between the axis of rotation and the "line of action" of the force (F) (r sin theta = d). This distance (d) is called "arm of the force." Now, the torque can be expressed as torque = Fd (torque is equal to force multiplied by the arm of the force.) Intuitively, torque can be interpreted as "force with a twist" since it results in rotation of the object on which the force is applied.
IT IS area under the POWER curve that is important for acceleration, otherwise we'd all be shifting at 5000 rpm instead of 8000 rpm. The power curve is "weighted" for rpm, and this information is important for maximizing acceleration (power is the rate of work, work per unit of time, ft*lbs/sec, or if you like N*m/s). Sure you can integrate a torque curve to find the area under it, but that doesn't give you the average power or help you optimize acceleration.
When a kid kicks your car, the mass of his/her moving leg stops. This produces a force (Force = mass * acceleration). For every action there is an equal and opposite reaction (the mass of your car accelerates in response). Thus a force has been applied to your car and the car has moved. This is the definition of work (Work = Force x Distance). So, my anology is like this, and it's just an anology: combustion generates a brief force, like a kid kicking your car. If the car is kicked once a second, you will slowly accelerate. If the car is kicked 100 times a second, acceleration will be considerably more rapid, because the power, or rate of work is higher. Bigger piston, more torque, bigger kid, more work done per kick. Rate (rpm) is not redundant, you know nothing about acceleration without knowing the rate at which work is being done.
A Joule, or Newton * meter, or kg*m^2*s^-2 , or foot pound, is the unit of both work and energy. Do you know the difference? (sorry, just feeding the flames)
More on torque:
You CAN measure torque, that's exactly what is measured at a dynometer. You can't measure twisting force without a mechanical moment.
Torque is a physical vector quantity characteristic for an object in rotational motion around a certain given axis. By definition torque is the cross product between the vector of position (r) of the point where a force is applied (from an origin considered to be on the axis of rotation) and the applied force (F) (torque= r x F). According to the definition of the cross (or vector) product, the magnitude of the torque is equal to the magnitude of r times the magnitude of F multiplied by sine of the angle (theta) between r and F (torque = r F sin theta). In this expression for magnitude of the torque, the product (r sin theta) is equal to the magnitude of the distance (d) between the axis of rotation and the "line of action" of the force (F) (r sin theta = d). This distance (d) is called "arm of the force." Now, the torque can be expressed as torque = Fd (torque is equal to force multiplied by the arm of the force.) Intuitively, torque can be interpreted as "force with a twist" since it results in rotation of the object on which the force is applied.
#93
tourque is the engine's ability to provide a twisting force, but it doesnt necessarily result in forward motion.
Horsepower is the engine's ability to turn torque into motion.
(taken from the Road&Track Automotive Dictionary)
Horsepower is the engine's ability to turn torque into motion.
(taken from the Road&Track Automotive Dictionary)
#94
Originally posted by mguire
sbertolone,
IT IS area under the POWER curve that is important for acceleration, otherwise we'd all be shifting at 5000 rpm instead of 8000 rpm. The power curve is "weighted" for rpm, and this information is important for maximizing acceleration (power is the rate of work, work per unit of time, ft*lbs/sec, or if you like N*m/s). Sure you can integrate a torque curve to find the area under it, but that doesn't give you the average power or help you optimize acceleration.
sbertolone,
IT IS area under the POWER curve that is important for acceleration, otherwise we'd all be shifting at 5000 rpm instead of 8000 rpm. The power curve is "weighted" for rpm, and this information is important for maximizing acceleration (power is the rate of work, work per unit of time, ft*lbs/sec, or if you like N*m/s). Sure you can integrate a torque curve to find the area under it, but that doesn't give you the average power or help you optimize acceleration.
Originally posted by mguire
When a kid kicks your car, the mass of his/her moving leg stops. This produces a force (Force = mass * acceleration). For every action there is an equal and opposite reaction (the mass of your car accelerates in response). Thus a force has been applied to your car and the car has moved. This is the definition of work (Work = Force x Distance). So, my anology is like this, and it's just an anology: combustion generates a brief force, like a kid kicking your car. If the car is kicked once a second, you will slowly accelerate. If the car is kicked 100 times a second, acceleration will be considerably more rapid, because the power, or rate of work is higher. Bigger piston, more torque, bigger kid, more work done per kick. Rate (rpm) is not redundant, you know nothing about acceleration without knowing the rate at which work is being done.[/B]
When a kid kicks your car, the mass of his/her moving leg stops. This produces a force (Force = mass * acceleration). For every action there is an equal and opposite reaction (the mass of your car accelerates in response). Thus a force has been applied to your car and the car has moved. This is the definition of work (Work = Force x Distance). So, my anology is like this, and it's just an anology: combustion generates a brief force, like a kid kicking your car. If the car is kicked once a second, you will slowly accelerate. If the car is kicked 100 times a second, acceleration will be considerably more rapid, because the power, or rate of work is higher. Bigger piston, more torque, bigger kid, more work done per kick. Rate (rpm) is not redundant, you know nothing about acceleration without knowing the rate at which work is being done.[/B]
Originally posted by mguire
A Joule, or Newton * meter, or kg*m^2*s^-2 , or foot pound, is the unit of both work and energy. Do you know the difference? (sorry, just feeding the flames)[/B]
A Joule, or Newton * meter, or kg*m^2*s^-2 , or foot pound, is the unit of both work and energy. Do you know the difference? (sorry, just feeding the flames)[/B]
Originally posted by mguire
More on torque:
You CAN measure torque, that's exactly what is measured at a dynometer. You can't measure twisting force without a mechanical moment. [/B]
More on torque:
You CAN measure torque, that's exactly what is measured at a dynometer. You can't measure twisting force without a mechanical moment. [/B]
Originally posted by mguire
Torque is a physical vector quantity characteristic for an object in rotational motion around a certain given axis. By definition torque is the cross product between the vector of position (r) of the point where a force is applied (from an origin considered to be on the axis of rotation) and the applied force (F) (torque= r x F). According to the definition of the cross (or vector) product, the magnitude of the torque is equal to the magnitude of r times the magnitude of F multiplied by sine of the angle (theta) between r and F (torque = r F sin theta). In this expression for magnitude of the torque, the product (r sin theta) is equal to the magnitude of the distance (d) between the axis of rotation and the "line of action" of the force (F) (r sin theta = d). This distance (d) is called "arm of the force." Now, the torque can be expressed as torque = Fd (torque is equal to force multiplied by the arm of the force.) Intuitively, torque can be interpreted as "force with a twist" since it results in rotation of the object on which the force is applied. [/B]
Torque is a physical vector quantity characteristic for an object in rotational motion around a certain given axis. By definition torque is the cross product between the vector of position (r) of the point where a force is applied (from an origin considered to be on the axis of rotation) and the applied force (F) (torque= r x F). According to the definition of the cross (or vector) product, the magnitude of the torque is equal to the magnitude of r times the magnitude of F multiplied by sine of the angle (theta) between r and F (torque = r F sin theta). In this expression for magnitude of the torque, the product (r sin theta) is equal to the magnitude of the distance (d) between the axis of rotation and the "line of action" of the force (F) (r sin theta = d). This distance (d) is called "arm of the force." Now, the torque can be expressed as torque = Fd (torque is equal to force multiplied by the arm of the force.) Intuitively, torque can be interpreted as "force with a twist" since it results in rotation of the object on which the force is applied. [/B]
#97
I dont know so how does it being 4 stroke make it displace more then if its 2 stroke? Its still the same engine with the same dimensions and all. I dont see how the displacement can change that dramatically.....:S
#98
the displacement of a reciprocating engine is calculated by pi * radius of piston ^ 2 * stroke * # of cylinders. for a 4 stroke to displace its volume it needs to make 2 rotations of the crank. for a 2 stroke to displace its volume it needs only to make 1 revolution of the crank. with a rotary making one revolution of the e-shaft it has displaced 1.3 liters worth of volume. with the rotary making 2 revolutions it has displaced 2.6 liters. its a matter of interpretation
#99
Ok I get how you guys would get that.
I guess you could say that a 13b isnt a 4 stroke becouse it doesnt have to revolve 2 times like a 4 stroke. Even though it does have some of the same operations (intake compression power and exaust) I dont personally see how it can be interpereted as a 4 stroke becouse if you think about it it doesnt "stroke" it kinda just wurlies around in there. I guess it is all just how you interpretat it.
I guess you could say that a 13b isnt a 4 stroke becouse it doesnt have to revolve 2 times like a 4 stroke. Even though it does have some of the same operations (intake compression power and exaust) I dont personally see how it can be interpereted as a 4 stroke becouse if you think about it it doesnt "stroke" it kinda just wurlies around in there. I guess it is all just how you interpretat it.