I know this is one of those questions that I can search online but I think I learn better when someone tellsm me stuff:P I wanted to know what is they difference between horse power and torque. What if there is more torque than horse power..if possible..and other way?
horsepower is used for top speed, torque is acceleration. so for 0-60 torque is better, and for lets say for a 1 mile staight track race H/P is better.
Diesel engines usually have more torque than horsepower.
Torque is a turning force. Imagine yourself turning a wrench.
Horsepower is a measurement of work in a given time. Specifically 33,000 foot pounds per minute is equal to one horsepower.
Actually torque isn't for accelleration like you make it sound. If that was
the case diesel engines would be owning all us gas guys, think about it.
Torque is turning force. The strength the motor is turning at, so it's relavent to how much you're going to be able to tow, or, how easy it is to maintain a speed, think of it that way.
Horsepower is relavent to accel/speed, the more horsies, the faster you're going to get from point A to point B. Think drag racing, you want a crazy amout of horse to get that beast to the other side of the track ASAP
It's actually a hard concept to explain, but once you understand it, it's really quite simple.
Oh ok I get it.. There are upgrades that can give you horse power like better exhaust or air intake ...etc... are there upgrades to get more torque??
Stroker kits are pretty much the only ones aimed at torque. But really, you improve one, the other goes up as well.
Torque is just a force in a twisting motion. Take for example a wrench or a
bicycle pedal. You power the wrench and in turn it twists and gives off
torque, you pedal the bicycle and you twist the wheels and go foreward.
Horsepower is just power. The more horsepower you haev the faster and more efficiently you'll go.
You can't have 1000 torque and only 100 HP. YOu have to have an even amount of both. :mrgreen:
Take a look at these two cars
05' Ford Mustang: 4.0 liter 210 HP@5250rpm and 240 lb-ft touque
05' Acura RSX-S: 2.0 liter 210 HP@7800rpm and 142 lb-ft torque
Both cars put out the same horsepower, but Mustang has almost 100 lb-ft more torque than RSX-S, so then is it better?
Not necesary. Larger torque means you get more HP at lower rpm. For every dday driving, larger torque might be better, But on track, higher redline gives extra reving and better performance. and I think RSX runs faster 0-60m time...I counldnt find the info.
damn, really helful, i oftenn wondered about that, so if one car has a higher hp at a lower rpm, then it redlines faster? which is worse for a quetermile then a same hp at a higher rpm?
I don't think larger torque makes a difference in a quater mile though.
But, yeah with RSX-s you juust have to rev higher to get the same HP then with mustang because it peaks at 7800rpm. But after all, both cars gaet the same HP at their redlines.
didnt we just discuss this last week in another thread?
Something else VERY important to remember:
Torque is a measured force. HP is CALCULATED from that. Torque is work, hp is work over time. Without time, torque doesn't do anything. Without force, hp doesn't exist.
Torque is a twisting force (actually force perependicular to a rotation). When related to hp and the rotation of the crankshaft of an engine, they are intrinsically related. BELOW 5252 rpm, the torque figure will ALWAYS be greater than the hp figure. Above 5252 rpm, the hp will ALWAYS be greater than the torque figure. AT 5252 rpm, both figures will allways be identical.
Here's a fun page to read about it: http://vettenet.org/torquehp.html
"Torque is the only thing that a driver feels."
""It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*."
If you were on a chassie dyno and did the torque testingonly, you can
calculate the horsepower by just knowing the torque right? So
speed,aceleration,power is all done by torque right?
One more question.. if you measure the wheel hp in a car with 1000hp and a car with 100hp wich one would lose the most hp through the driveshafts?
A RSX S makes 200hp at crank and 180 atwheels so its a 10percent lost. So if a car with 1000hp at the crank, would it lose 100 hp through the driveshafts?
So basicly im asking if the more hp you have at the crank, the more hp you lose to the driveshafts assuming everything is the same.
Well if someone had the same drivetrain and tranny in a 1000hp engine and 200hp engine then u'd never know which one would lose more power cause something would snap each and every time on that 1ghp beast. U couldn't test it. And to know which would lose the greater percentage it would generally be whichever had a longer distance to travel before it got to the wheels. you have to pass the power through the pistons, rods, crankshaft, flywheel, pullies, clutch, then the input shaft, layshaft, driveshaft, differential, axles, then finally tires. A rear wheel drive car loses more power than an FWD because the long driveshaft is heavier and requires more force to turn. AWD cars lose the most power because there are twice as many axles to spin and more differentials. Also whichever car has more things attached to the pullies will take up more power. And whether the clutch is in good condition or not or if its an automatic which generally loses 5% of power to the transmission to avoid the car dying when stopped. There are alot of factors which determine 'da business'.
This post in particular, and the entire thread in general with the
exception of ChrisV, is a perfect example of why this forum is unable to
retain members who are willing and able to make intelligent and worthwhile
contributions. There is SOME correct information and a BUNCH of absolute
"A rear wheel drive car loses more power than an automatic". WTF is that even supposed to mean? Did I miss something that says RWD cars can no longer be automatics? "or if its an automatic which generally loses 5% of power to the transmission to avoid the car dying when stopped"! Huh? What? Please turn off your computer now and do not post here... EVER AGAIN. It not only makes my head hurt reading your blathering but I'm sure it reduces everyone's IQ who has the misfortune to gaze upon your words.
Just as a side note, what exactly does it mean to be the "Self-appointed anchorman of CF"? Is it just in your nature to type moronic things? While we're at it, what is there to be proud of being a dependent of the USMC? Are you unable to to stand on your own merits? Are you really that inane or are you just pretending to be an idiot? You sir are proof that stupid people shouldn't breed.
I think this thread would benefit if chrisv the vwhobo and I were the only ones who were allowed to post after the initial question was asked. Everyone else seems to spew the bullshit the heard was true from their inbred grandpa dad.
and....what makes you more qualified than anybody else?
VMJyogi, I like how you think outside the box, but now it's not really necessary, try to answer the question only in the context it was asked.
I'll presume you mean a FWD car, but I'm sure you lose more power, but you get more traction, and arguable better handling from a RWD car. But the parasitic power loss is sometimes less on a RWD car also.
I going to presume you mean the 5% slippage thats present in the torque converter. Now, to my knowledge, I believe that a transmission fluid runs through the torque converter as the RPMs raise, correct, so you shouldn't have that 5% slippage when you're accelerating. Am I right?
A question to anybody who can answer it;
A honda civic puts out somehting like 100 ft-lbs of torque. is it possible for to take something like a 5 ft long wrench, attach it to the wheel of a civic, and stop the wheels from spinning using my own two hands and a wrench?
okay thanks for being cool about it.
Yes, i did mean FWD. You know how sometimes you think faster than you speak and end up running words together? Thats what i did typing.
Yes you are most likely right. I never knew exactly where or how the 5% slippage came to be but now i am well informed!
Yes, it is in my nature to type stupid things. Hey it's funny cause the movie was funny and i felt like doing it.
Well since i'm not rude. I'll simply state that DEP stands for Delayed Entry Program which means that i have enlisted for the Marine Corps and have not yet entered because i must first graduate high-school hence the term delayed entry.
Yes i really am that uninformed (not stupid). But it is my charisma not my knowledge of cars or ability to yell at others that charms women. So unfortunately i will breed if that is god's intentions.
I wish for the day when i could learn from someone as intelligent as VWHOBO, not get bashed by him...Thank You Godlaus for teaching me about the torque converter.
My info is right about 40% of the time. You're better off learning from somebody else. Don't take my word as God's.
With a 3.5:1 first gear ratio and a 4.30:1 final drive ratio, that 100 lb
ft of torque becomes 1500+ lb ft at the hubs. No, I doubt you could stop
OTOH, if you put the car in 5th gear and let out the clutch, the resistance of the car just sitting there is enough to stall the engine.
apparently, you underestimated Godlaus' power. I can easily hold up 300
pounds!!!!! :laughing: lol, jk, I can barely pick up my dog, and she's
only 3 months old.
But, if the wrench was attached to the crank, then I could stop the crank from spinning, correct?
Its funny but why are you so interested? I could actually see somebody sitting there with a vice-grip clamped to the crank saying 'i'm ready, start the engine'. I'd be worried that the guy would break his arm but it'd still be funny to watch. Now i'm wondering if you could do it too.
just curious, that and bragging rights that i could hold back a honda civic.
Ok, so I have a question.
What relation does the cars weight have with HP and Torque? Obviously less weight the fast you can go. But is there a point when weight dosnt matter?
Example, my Fiero weighs roughly 2300 pounds, the 89 GTA that the engine came out of weighs roughly 3,500 lbs. What effect does this weight difference have on the HP and Torque?
The torque converter always has fluid in it, that's how it works. There is
an impeller portion that is physically welded to the torque converter
housing (which is bolted to the flexplate, which looks like a thin
flywheel, and is bolted to the crank, this the impeller is directly driven
by the crank).
http://www.drivesubaru.com/Fall02/Fall02_TorqueConvert/ConverterParts2.jpg< br />
That impeller pushes the fluid against a turbine that is connected to the input shaft of the transmission itself. At idle, the fact that the timpeller and turbine are connected only by fluid allows the turbine to sit still while the engine is spinning. As the rpm's rise, the force of the spinning fluid forces the turbine to match speed with the impeller causing the car to move, like when you slip the clutch to take off from a stop in a manual. Above a certain rpm, teh turbine "stalls" in relation to the impeeller (that is, it moves the same speed). In modern transmissions, the parts physically lock together for zero slippage (like having the clutch engaged on a manual trans).
Inbetween is a thing called a stator.
"The stator works between the impeller and turbine to magnify the torque as the vehicle accelerates. At low speeds, the stator remains stationary against the flow of transmission fluid. It works to redirect the fluid flow from the turbine to boost impeller action and multiply the engine torque. As the speed increases, the stator starts to rotate in the same direction as the impeller and turbine. As it does, the torque multiplication stops."
But there's more going on in an automatic that causes slippage. Unlike manual tramsnissions, the gears inside an automatic a planetery gearsets, not hypoid (look it up). The changing of gears is accomplished by changing teh fluid pressure through a valve body, controlled by a combination of vacuum and rpm in older transmissions, or the computer and throttle position in newer ones. There are also clutch packs inside the transmission, and bands that tighten on teh differnt gearsets to lock them in place. If the bands slip, or teh valves are tuned too soft, you end up with smooth, but sloppy shifts that rob power due to excess slipping between gears and while IN gear. High performance bands and clutchpacks help, as does a shift kit (which changes spring pressures and fluid pressurs in different areas of the valve body, which reduces teh time the shift takes, and increases band pressure for reduced or eliminated slippage giving more performance).
So yes, in stock and luxury automatic transmissions, thre is more overall loss of power. But that can be fixed to the point where there is actually less overall loss of power through the transmission.
The weight doesn't change the engine's torque and horsepower. But the
weight DOES have an effect on how much work the engine can do. And there's
really no point where weight doesn't matter unless there is't any weight TO
The difference you see is simply how many lbs each hp has to move. If you had an engine with 200 hp, in the GTA each hp had to move 17.5 lbs. That same engine in the Fiero would only have to move 11.5 lbs. To put this in perspective, you could lift a 100 lb box of magazines. But it would be hard. If that box only weighed 10 lbs, it would be easy to move and you'd be less liekly to hurt yourself, even though you haven't changed how strong you are.
Got it.. so me lifting the 10 lbs boxes, in relation to someone who is built just like me, would have to lift more weight to accomplish the same task. Inheritably I would finish first with less body ach then the guy struggling with the 100lbs box.
Well, first you have to realize that the engine has to be running first in
order to stop it, and at idle, that crank is spining at around 1000 rpm.
Try putting a wrench ON it.. ;) But at idle, that engine might only be
putting out 20-30 lb ft of torque.
Basically what you're asking is if you put a clutch on the back of the engine and applied pressure by hand to a running engine with a long enough handle, could you stall it. And that answer is yes, that's how an engine dyno works, in general (only an angine dyno applies a brake to measure force, not to stall the engine).
Now, just to muddy up the waters, you can increase the torque:weight ratio of a car by gearing, without changing the car's weight OR the engine's output. But the only way to change the hp:weight ratio is by reducing weight or increasing the engine's power output.
Forget all the tripe. Fuel is rated in joules and the when burnt at a rate
of one joule per second releases a watt of power. How that power is applied
whether it be for cooking steaks or to make a wheel turn is a product of
Without power torque cannot exist, but power can exist without torque.
You cannot have more or less torque than power simply because they are different measurement, like oranges and lemons although they may share common elements. Just because someone decided to scale a graph to conveniently correlate with a 5252 derived constant does not mean one is greater than the other. The Torque (ft-lbs) = HP x 5252/rpm equation can be rearranged whichever way you like. If you use the same nonsense internet mechanics do by rearranging the the equation to prove hp is dependent on torque, you could equally argue power is dependent on rpm, which we all know is not true.
A convenient way for power to be measured is via a chassis dynamometer. One of the methods of measuring on the chassis dyno is tractive effort. From this you can apply gear ratios and wheel diameter data and find either a torque figure or a power figure or both at your leasure. Some people find it easier to find torque and then from that find power, but that is only stepping the maths.
For the relationship between acceleration and torque = simple school physics "torque and angular acceleration".
Just like in any math, you can rearrange the equation many ways. But that
isnt' what we are talking about here and you should know it.
Torque is tangential force * the distance from the fulcrum. Power is defined as work per unit of time.
Applying 1 lb of force 1 ft from the fulcrum for a complete revolution will lead to;
W = F*2*pi*r = 1 lb * 2*pi * 1 ft = 2*pi lb-ft = 6.283 lb-ft
If it takes one minute to complete this revolution, then the power is;
P = W / time = 6.283 lb-ft / min
1 hp is defined as 550 lb-ft / s = 33,000 lb-ft / min
Therefore, applying 1 lb-ft of torque in one minute (1 rpm) = [6.283 lb-ft / min] / [33,000 lb-ft / min] = 1 / 5252 of 1 hp.
From this you can then calculate the number of hp from any given torque and rpm:
hp = torque (lb-ft) * rpm / 5252
No one "scaled a graph conveniently" to make that happen.
Where does the equation HP=TORQUE * RPM / 5252 come from? We will use Watts observation of one horsepower as 150 pounds, 220 feet in one minute. First we need express 150 pounds of force as foot pounds torque.
Pretend the force of 150 pounds is "applied" tangentially to a one foot radius circle. This would be 150 foot pounds torque.
Next we need to express 220 feet in one minute as RPM.
The circumference of a one foot radius circle is 6.283186 feet. ft. (Pi x diameter 3.141593 * 2 feet)
The distance of 220 feet, divided by 6.283185 feet, gives us a RPM of 35.014.
We are then talking about 150 pounds of force (150 foot pounds torque), 35 RPM, and one horsepower.
Constant (X) = 150 ft.lbs. * 35.014 RPM / 1hp
35.014 * 150 / 1 = 5252.1
5252 is the constant.
So then hp = torque * RPM / 5252
"The word horsepower was introduced by James Watt, the inventor of the steam engine in about 1775. Watt learned that "a strong horse could lift 150 pounds a height of 220 feet in 1 minute." One horsepower is also commonly expressed as 550 pounds one foot in one second or 33,000 pounds one foot in one minute. These are just different ways of saying the same thing. Notice these definitions includes force (pounds), distance (feet), and time, (minute, second). A horse could hold weight in a static position but this would not be considered horsepower, it would be similar to what we call torque. Adding time and distance to a static force (or to torque) results in horsepower. RPM, revolutions (distance) per minute (time), is today's equivalent of time and distance. Back to horses, imagine a horse raising coal out of a coal mine. A horse exerting one horsepower could raise 550 pounds of coal one foot every second."
Of course, the problem, as someone else pointed out in another thread, is that "Watt was simply observing that the horse lifted the 150 lb. object by 220 feet in one minute. We can't change that. Had Watt simply observed a bigger (say, a Clydesdale) or smaller horse (like Ford used to measure the '99 Cobra!) the definition of HP would be different."
The reason we dont measure internal combustion engine horsepower directly is that engies make a rotary motion, not linear motion. And unless the engine is geared down, the speed at which they do work (time and distance or RPM) is too great for practical direct measurement of horsepower. It seems logical then that the solution was to directly measure torque (rotational force eventually expressed in pounds at one foot radius) and RPM (time and distance, i.e. distance in circumference at the one foot radius) and from these calculate horsepower. Torque and RPM are easily measured directly.
Therefore when discussing internal combustion engines, torque IS the measured force, and hp is ALWAYS the calculated amount.
I agree that measuring, for instance, tractive effort is convenient, but
newtons is not newton metres nor is it watts. There is no need to even
think about torque when deriving power from newtons over time.
I'll demonstrate the graphing in a diffrent way. Here is a nissan sss, from Newcastle Oz, dyno readout conveniently in british units. Where is the crossover at 5252?:
I guess you were trying to convey is the relationship of torque/hp by this:
hp = torque * RPM / 5252
@ 5252 rpm hp = torque
as rpm increases and power is static then torque decreases or if torque is to remain static the power input must increase?
One small detail I'd like to point out about that. Work and Torque aren't
exactly the same thing. Torque is the lb-ft. Work is the ft-lb. Movement
must occur for work to be done, torque doesn't have to have movement.
Hobo seems to have this confused in the "torquing my valve cover" (or whatever the name is) thread. Torque a bolt and you are doing work that's ft-lb. Ah, this is for another thread.
I'm just pointing out that torque isn't exactly work. It doesn't fit when used mathematically.
No, the only thing I'm confused about is why those of us with knowledge
make any attempt at helping individuals like you learn. It's pointless
because you're convinced that at age 17 you already know everything but we
still try. However, here's something else for you to try to comprehend.
Of course torque isn't work!!! But when you apply a torque to do some
work, then at any instant in that process, you are using power.
That's why the speed is in the equation ... the distance in the torque units is different to the distance in the work units. For example, if we forget that we're turning something with torque, and instead pretend we're moving something in a straight line:
Rather than torque we substitute force in Newtons [N] (sorry, we work in metric here in oz)
For work (energy) we have Nm (or joules [J])
For power we have Watts [W] (or J/sec)
For speed we have m/sec (instead of rpm (or radians/sec) in turning case)
For distance we have m
Force x distance = work [N*m = Nm = J]
Force x distance / time = Force x speed = Power [N*m/s = J/s = W]
Therefore, the power used at any instant = the force used to move an object at a certain speed x that speed.
Work done = the force used to move an object x the distance you've moved the object
Force = the force you are applying to the object to move it at the speed corresponding to the power being used.
These can all be converted to rotational motion using torque instead of force, and radians/sec (rpm*2pi/60) instead of m/sec and angle turned in radians (revolutions * 2pi) instead of distance in m.
They quite clearly are linked in the theory.
Your graph shows two different measurements of energy, and no measurement of torque. From Hobo's link: "Note that the SI units of torque is a Newton-metre, which is also a way of expressing a Joule (the unit for energy). However, torque is not energy."
Your graph thus doesn't display torque, merely two different units of energy. That's why the curves are very similar in shape and don't cross over.
OK let me put it yet another way:
The simplified imperial formula is hp = ftlbs-force * RPM / 5252
when the rpms hit 5252 the result is power = torque *1
Now lets take the metric equivalent kW = N-m *RPM/ 9551
when the rpms hit 9551 the result is power = torque *1
In both instances we are dealing with the same effects, but with different units of measurements. As you can easily conclude if we used the same convention for graphing that is used for imperial power/torque curves the metric torque curve would not intersect the power curve until 9551 rpm.
As most daily grind engines run out of puff well before 9551 rpm we would not expect to see any intersection on our graph if it's x axis ceased @ 6500 rpm.
In so far as the shape of the tractive effort curve, do you really think it's shape will change by multiplying it by a constant rolling tyre radius and dividing by a constant drivetrain ratio? I don't think so somehow, so the curves are indeed valid and the bottom one is in fact the torque profile with a different scale.